Re: [R] [EXT] Re: Initializing vector and matrices
Dear Steven,
I used "sample" just to generate a non-trivial example, you could insert
your code of generating the real xi at this point :-)
If you want to stick to for-loops for some reasons, something like this
could work
x<-NULL
for (i in 1:5){
xi<-1:5
if (is.null(x)) x<-xi else x<-x+xi
}
cheers
Am 29.02.2024 um 09:23 schrieb Steven Yen:
Hello Eik:
Thanks. I do not need to sample. Essentially, I have a do loop which
produces 24 vectors of length of some length (say k=300) and 24 matrices
of 300x300. Then, I simply need to take the averages of these 24
vectors and matrices:
x=(x1+x2+...+x24)/k
y=(y1+y2+...+y24)/k
I am just looking for ways to do this in a do loop, which requires
initialization (to 0's) of x and y. My struggle is not knowning length
of x until x1 is produced in the first of the loop. Thanks.
Steven
On 2/28/2024 6:22 PM, Eik Vettorazzi wrote:
Hi Steven,
It's not entirely clear what you actually want to achieve in the end.
As soon as you "know" x1, and assuming that the different "xi" do not
differ in length in the real application, you know the length of the
target vector.
Instead of the loop, you can use 'Reduce' without having to initialize
a starting vector.
# generate sample vectors, put them in a list
xi<-lapply(1:5, \(x)sample(5))
# look at xi
xi
# sum over xi
Reduce("+",xi)
this works also for matrices
# generate sample matrices, put them in a list
Xi<-lapply(1:3, \(x)matrix(sample(16), nrow=4))
# look at them
Xi
# sum over Xi
Reduce("+",Xi)
Hope that helps
Eik
Am 28.02.2024 um 09:56 schrieb Steven Yen:
Is there as way to initialize a vector (matrix) with an unknown
length (dimension)? NULL does not seem to work. The lines below work
with a vector of length 4 and a matrix of 4 x 4. What if I do not
know initially the length/dimension of the vector/matrix?
All I want is to add up (accumulate) the vector and matrix as I go
through the loop.
Or, are there other ways to accumulate such vectors and matrices?
> x<-rep(0,4) # this works but I like to leave the length open
> for (i in 1:3){
+ x1<-1:4
+ x<-x+x1
+ }
> x
[1] 3 6 9 12
> y = 0*matrix(1:16, nrow = 4, ncol = 4); # this works but I like to
leave the dimension open
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 0 0
[3,] 0 0 0 0
[4,] 0 0 0 0
> for (i in 1:3){
+ y1<-matrix(17:32, nrow = 4, ncol = 4)
+ y<-y+y1
+ }
> y
[,1] [,2] [,3] [,4]
[1,] 51 63 75 87
[2,] 54 66 78 90
[3,] 57 69 81 93
[4,] 60 72 84 96
>
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Christoph-Probst-Weg 1
4. Obergeschoss, Raum 04.1.021.1
20246 Hamburg
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Fax: +49 (0) 40 7410 - 57790
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_
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Prof. Dr. Blanche Schwappach-Pignataro, Matthias Waldmann (komm.)
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Re: [R] [EXT] Initializing vector and matrices
Hi Steven,
It's not entirely clear what you actually want to achieve in the end.
As soon as you "know" x1, and assuming that the different "xi" do not
differ in length in the real application, you know the length of the
target vector.
Instead of the loop, you can use 'Reduce' without having to initialize a
starting vector.
# generate sample vectors, put them in a list
xi<-lapply(1:5, \(x)sample(5))
# look at xi
xi
# sum over xi
Reduce("+",xi)
this works also for matrices
# generate sample matrices, put them in a list
Xi<-lapply(1:3, \(x)matrix(sample(16), nrow=4))
# look at them
Xi
# sum over Xi
Reduce("+",Xi)
Hope that helps
Eik
Am 28.02.2024 um 09:56 schrieb Steven Yen:
Is there as way to initialize a vector (matrix) with an unknown length
(dimension)? NULL does not seem to work. The lines below work with a
vector of length 4 and a matrix of 4 x 4. What if I do not know
initially the length/dimension of the vector/matrix?
All I want is to add up (accumulate) the vector and matrix as I go
through the loop.
Or, are there other ways to accumulate such vectors and matrices?
> x<-rep(0,4) # this works but I like to leave the length open
> for (i in 1:3){
+ x1<-1:4
+ x<-x+x1
+ }
> x
[1] 3 6 9 12
> y = 0*matrix(1:16, nrow = 4, ncol = 4); # this works but I like to
leave the dimension open
[,1] [,2] [,3] [,4]
[1,] 0 0 0 0
[2,] 0 0 0 0
[3,] 0 0 0 0
[4,] 0 0 0 0
> for (i in 1:3){
+ y1<-matrix(17:32, nrow = 4, ncol = 4)
+ y<-y+y1
+ }
> y
[,1] [,2] [,3] [,4]
[1,] 51 63 75 87
[2,] 54 66 78 90
[3,] 57 69 81 93
[4,] 60 72 84 96
>
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Universitätsklinikum Hamburg-Eppendorf
Institut für Medizinische Biometrie und Epidemiologie
Christoph-Probst-Weg 1
4. Obergeschoss, Raum 04.1.021.1
20246 Hamburg
Telefon: +49 (0) 40 7410 - 58243
Fax: +49 (0) 40 7410 - 57790
Web: www.uke.de/imbe
Webex: https://webteaching-uke.webex.com/meet/e.vettorazzi
--
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg | www.uke.de
Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vorsitzender), Joachim Prölß,
Prof. Dr. Blanche Schwappach-Pignataro, Matthias Waldmann (komm.)
_
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Re: [R] Decompose df1 into another df2 based on values in df1
A tidyverse-ish solution would be
library(dplyr)
library(tidyr)
library(tibble)
# max cols to split values into
seps<-max(stringr::str_count(unlist(d1),"[/|]"))+1
d1 %>% pivot_longer(S1:S5, names_to="S") %>%
mutate(value=na_if(value,"w")) %>% separate(value,"[/|]",
into=LETTERS[1:seps], fill="right") %>% pivot_longer(-S, names_to=NULL,
values_to="rownames") %>% filter(!is.na(rownames)) %>%
mutate(index=1L)%>%pivot_wider(names_from=S, values_from=index) %>%
mutate_all(replace_na,0L) %>% column_to_rownames(var = "rownames")
Best, Eik
Am 26.05.2021 um 23:16 schrieb Adrian Johnson:
Hello,
I am trying to convert a df (given below as d1) into df2 (given below as
res).
I tried using loops for each row. I cannot get it right. Moreover the df
is 25 x 500 in dimension and I cannot get it to work.
Could anyone help me here please.
Thanks.
Adrian.
d1 <-
structure(list(S1 = c("a1|a2", "b1|b3", "w"), S2 = c("w", "b1",
"c2"), S3 = c("a2", "b3|b4|b1", "c1|c4"), S4 = c("w", "b4", "c4"
), S5 = c("a2/a3", "w", "w")), class = "data.frame", row.names = c("A",
"B", "C"))
res <-
structure(list(S1 = c(1L, 1L, 0L, 1L, 0L, 1L, 0L, 0L, 0L, 0L),
S2 = c(0L, 0L, 0L, 1L, 0L, 0L, 0L, 0L, 1L, 0L), S3 = c(0L,
1L, 0L, 1L, 0L, 1L, 1L, 1L, 0L, 1L), S4 = c(0L, 0L, 0L, 0L,
0L, 0L, 1L, 0L, 0L, 1L), S5 = c(0L, 1L, 1L, 0L, 0L, 0L, 0L,
0L, 0L, 0L)), class = "data.frame", row.names = c("a1", "a2",
"a3", "b1", "b2", "b3", "b4", "c1", "c2", "c4"))
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Re: [R] geom_ribbon function in ggplot2 package
Hi, just add +scale_fill_discrete(name=NULL) Cheers Am 19.12.2018 um 07:05 schrieb John: Hi, When using the geom_ribbon function in gglot2 package, I got the text "fill" above the legend "A" and "B". How can I get rid of the text "fill" above the legend? Thanks! The code is as follows: df<-data.frame(x=c(1,2), y=c(1,2), z=c(3,5)) ggplot(df, aes(1:2))+geom_ribbon(aes(ymin=0,ymax=df[,"y"],fill="A"),alpha=0.5)+geom_ribbon(aes(ymin=0,ymax=df[,"z"],fill="B"),alpha=0.5) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistrasse 52 building W 34 20246 Hamburg Phone: +49 (0) 40 7410 - 58243 Fax: +49 (0) 40 7410 - 57790 Web: www.uke.de/imbe -- _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg | www.uke.de Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Marya Verdel _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] system solver in R
How about this:
library(rootSolve)
f1<-function(x)5/((1+x)^1) + 5/((1+x)^2) + 5/((1+x)^3) + 105/((1+x)^4) -105
uniroot.all( f1,c(-1e6,1e6))
[1] -1.9881665 0.0363435
Cheers
Am 20.11.2018 um 13:09 schrieb Engin Yılmaz:
Dea(R)
I try to solve one equation but this program did not give me real roots
for example
yacas("Solve( 5/((1+x)^1) + 5/((1+x)^2) + 5/((1+x)^3) + 105/((1+x)^4) -105
==0, x)")
gave me following results
How can I find real roots?
expression(list(x == complex_cartesian((1/42 - ((1/63 -
((root(7339451281/3087580356,
2) - 4535/71442)^(1/3) - (4535/71442 + root(7339451281/3087580356,
2))^(1/3)))/21 - -2/21)/(4 * root(((root(7339451281/3087580356,
2) - 4535/71442)^(1/3) - (4535/71442 + root(7339451281/3087580356,
2))^(1/3) - 1/63)^2/4 + 1, 2)))/2 - 1, root(4 *
(((root(7339451281/3087580356,
2) - 4535/71442)^(1/3) - (4535/71442 + root(7339451281/3087580356,
2))^(1/3) - 1/63)/2 + root(((root(7339451281/3087580356,
2) - 4535/71442)^(1/3) - (4535/71442 + root(7339451281/3087580356,
2))^(1/3) - 1/63)^2/4 + 1, 2)) - (((1/63 -
((root(7339451281/3087580356,
2) - 4535/71442)^(1/3) - (4535/71442 + root(7339451281/3087580356,
2))^(1/3)))/21 - -2/21)/(4 * root(((root(7339451281/3087580356,
2) - 4535/71442)^(1/3) - (4535/71442 + root(7339451281/3087580356,
2))^(1/3) - 1/63)^2/4 + 1, 2)) - 1/42)^2, 2)/2),...more
Engin Yılmaz , 20 Kas 2018 Sal, 12:53 tarihinde şunu
yazdı:
Thanks a lot!
Berend Hasselman , 20 Kas 2018 Sal, 12:02 tarihinde şunu
yazdı:
R package Ryacas may be what you want.
Berend
On 20 Nov 2018, at 09:42, Engin Yılmaz wrote:
Dea(R)
Do you know any system solver in R ?
For example, in matlab, is very easy
syms a b c x eqn = a*x^2 + b*x + c == 0; sol = solve(eqn)
How can I find this type code in R (or directly solver)?
*Since(R)ely*
Engin YILMAZ
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*Saygılarımla*
Engin YILMAZ
--
Eik Vettorazzi
Universitätsklinikum Hamburg-Eppendorf
Institut für Medizinische Biometrie und Epidemiologie
Martinistraße 52
Gebäude W 34
20246 Hamburg
Telefon: +49 (0) 40 7410 - 58243
Fax: +49 (0) 40 7410 - 57790
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--
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Gerichtsstand: Hamburg | www.uke.de
Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Joachim Prölß, Marya Verdel
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Re: [R] Reporting binomial logistic regression from R results
Dear Jedi, please use the source carefully. A and C are not statistically different at the 5% level, which can be inferred from glm output. Your last two wald.tests don't test what you want to, since your model contains an intercept term. You specified contrasts which tests A vs B-A, ie A- (B-A)==0 <-> 2*A-B==0 which is not intended I think. Have a look at ?contr.treatment and re-read your source doc to get an idea what dummy coding and indicatr variables are about. Cheers Am 12.11.2018 um 02:07 schrieb Frodo Jedi: Dear list members, I need some help in understanding whether I am doing correctly a binomial logistic regression and whether I am interpreting the results in the correct way. Also I would need an advice regarding the reporting of the results from the R functions. I want to report the results of a binomial logistic regression where I want to assess difference between the 3 levels of a factor (called System) on the dependent variable (called Response) taking two values, 0 and 1. My goal is to understand if the effect of the 3 systems (A,B,C) in System affect differently Response in a significant way. I am basing my analysis on this URL: https://stats.idre.ucla.edu/r/dae/logit-regression/ This is the result of my analysis: fit <- glm(Response ~ System, data = scrd, family = "binomial") summary(fit) Call: glm(formula = Response ~ System, family = "binomial", data = scrd) Deviance Residuals: Min 1Q Median 3Q Max -2.8840 0.1775 0.2712 0.2712 0.5008 Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept)3.2844 0.2825 11.626 < 2e-16 *** SystemB -1.2715 0.3379 -3.763 0.000168 *** SystemC0.8588 0.4990 1.721 0.085266 . --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 411.26 on 1023 degrees of freedom Residual deviance: 376.76 on 1021 degrees of freedom AIC: 382.76 Number of Fisher Scoring iterations: 6 Following this analysis I perform the wald test in order to understand whether there is an overall effect of System: library(aod) wald.test(b = coef(fit), Sigma = vcov(fit), Terms = 1:3) Wald test: -- Chi-squared test: X2 = 354.6, df = 3, P(> X2) = 0.0 The chi-squared test statistic of 354.6, with 3 degrees of freedom is associated with a p-value < 0.001 indicating that the overall effect of System is statistically significant. Now I check whether there are differences between the coefficients using again the wald test: # Here difference between system B and C: l <- cbind(0, 1, -1) wald.test(b = coef(fit), Sigma = vcov(fit), L = l) Wald test: -- Chi-squared test: X2 = 22.3, df = 1, P(> X2) = 2.3e-06 # Here difference between system A and C: l <- cbind(1, 0, -1) wald.test(b = coef(fit), Sigma = vcov(fit), L = l) Wald test: -- Chi-squared test: X2 = 12.0, df = 1, P(> X2) = 0.00052 # Here difference between system A and B: l <- cbind(1, -1, 0) wald.test(b = coef(fit), Sigma = vcov(fit), L = l) Wald test: -- Chi-squared test: X2 = 58.7, df = 1, P(> X2) = 1.8e-14 My understanding is that from this analysis I can state that the three systems lead to a significantly different Response. Am I right? If so, how should I report the results of this analysis? What is the correct way? Thanks in advance Best wishes FJ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistrasse 52 building W 34 20246 Hamburg Phone: +49 (0) 40 7410 - 58243 Fax: +49 (0) 40 7410 - 57790 Web: www.uke.de/imbe -- _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg | www.uke.de Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Marya Verdel _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Genuine relative paths with R
Dear Olivier, you may find the rprojroot package useful, see https://github.com/r-lib/rprojroot and a discussion https://gist.github.com/jennybc/362f52446fe1ebc4c49f#file-2014-10-12_stop-working-directory-insanity-md Cheers Am 06.10.2018 um 13:48 schrieb Olivier GIVAUDAN: Dear R users, I would like to work with genuine relative paths in R for obvious reasons: if I move all my scripts related to some project as a whole to another location of my computer or someone else's computer, if want my scripts to continue to run seamlessly. What I mean by "genuine" is that it should not be necessary to hardcode one single absolute path (making the code obviously not "portable" - to another place - anymore). For the time being, I found the following related posts, unfortunately never conclusive or even somewhat off-topic: https://stackoverflow.com/questions/1815606/rscript-determine-path-of-the-executing-script https://stackoverflow.com/questions/47044068/get-the-path-of-current-script/47045368 http://r.789695.n4.nabble.com/Script-auto-detecting-its-own-path-td2719676.html So I found 2 workarounds, more or less satisfactory: 1. Either create a variable "ScriptPath" in the first lines of each of my R scripts and run a batch (or shell, etc.) to replace every single occurrence of "ScriptPath <-" by "ScriptPath <- [Absolute path of the R script]" in all the R scripts located in the folder (and possibly subfolders) of the batch file. 2. Or create an R project file with RStudio and use the package "here" to get the absolute path of the R project file and put all the R scripts related to this project in the R project directory, as often recommended. But I am really wondering why R doesn't have (please tell me if I'm wrong) this basic feature as many other languages have it (batch, shell, C, LaTeX, SAS with macro-variables, etc.)? Do you know whether the language will have this kind of function in a near future? What are the obstacles / what is the reasoning for not having it already? Do you know other workarounds? Best regards, Olivier [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistrasse 52 building W 34 20246 Hamburg Phone: +49 (0) 40 7410 - 58243 Fax: +49 (0) 40 7410 - 57790 Web: www.uke.de/imbe -- _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg | www.uke.de Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Martina Saurin (komm.) _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregate and list elements of variables in data.frame
Hi,
if you are willing to use dplyr, you can do all in one line of code:
library(dplyr)
df<-data.frame(id=1:10,A=c(123,345,123,678,345,123,789,345,123,789))
df%>%group_by(unique_A=A)%>%summarise(list_id=paste(id,collapse=", "))->r
cheers
Am 06.06.2018 um 10:13 schrieb Massimo Bressan:
> #given the following reproducible and simplified example
>
> t<-data.frame(id=1:10,A=c(123,345,123,678,345,123,789,345,123,789))
> t
>
> #I need to get the following result
>
> r<-data.frame(unique_A=c(123, 345, 678,
> 789),list_id=c('1,3,6,9','2,5,8','4','7,10'))
> r
>
> # i.e. aggregate over the variable "A" and list all elements of the variable
> "id" satisfying the criteria of having the same corrisponding value of "A"
> #any help for that?
>
> #so far I've just managed to "aggregate" and "count", like:
>
> library(sqldf)
> sqldf('select count(*) as count_id, A as unique_A from t group by A')
>
> library(dplyr)
> t%>%group_by(unique_A=A) %>% summarise(count_id = n())
>
> # thank you
>
>
> [[alternative HTML version deleted]]
>
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistrasse 52
building W 34
20246 Hamburg
Phone: +49 (0) 40 7410 - 58243
Fax: +49 (0) 40 7410 - 57790
Web: www.uke.de/imbe
--
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg | www.uke.de
Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Joachim Prölß, Martina Saurin (komm.)
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Re: [R] Plot in real unit (1:1)
How about this:
in2mm<-25.4 # scale factor to convert inches to mm
pdf("test.pdf",width=8.3,height=11.7)
pin<-par("pin")
plot(c(0,pin[1]*in2mm),c(0,pin[2]*in2mm), type="n", xaxs="i", yaxs="i")
lines(c(10,10),c(0,10))
text(11,5,"1 cm", adj=0)
lines(c(0,40),c(20,20))
text(20,24,"4 cm")
polygon(c(50,50,70,70),c(50,70,70,50))
text(60,60,"2x2 cm")
dev.off()
cheers
Am 06.06.2018 um 16:00 schrieb Christian Brandstätter:
> Dear List,
>
> Is it possible to plot in R in "real" units? I would like to draw a
> plot on A4 paper, where 1 plot unit would be a mm in reality. Is
> something like that possible? I would also like to be able to scale the
> plot in x and y direction.
> Background: For a project I would have to draw around 65 fast sketches
> of elevation courves.
>
> Copied from here, due to no answer: https://stackoverflow.com/questions
> /50606797/plot-in-real-units-mm
>
> Thank you!
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistrasse 52
building W 34
20246 Hamburg
Phone: +49 (0) 40 7410 - 58243
Fax: +49 (0) 40 7410 - 57790
Web: www.uke.de/imbe
--
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg | www.uke.de
Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Joachim Prölß, Martina Saurin (komm.)
_
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Re: [R] Problem with regression line
Hi Anne,
I would suggest to change the linear model to lm(BloodPressure~Age), as
this model makes more sense in biological means (you would assume that
age influences pressure, not vice versa) and also obeys the statistical
assumption of weak exogeneity, that age can be measured without error,
at least compared to error-prone bp measures.
Cheers
Am 18.04.2018 um 16:07 schrieb Gerrit Eichner:
> Hi, Anne,
>
> assign Age and Bloodpressure in the correct order
> to the axes in your call to plot as in:
>
> plot(y = Age, x = BloodPressure)
> abline(SimpleLinearReg1)
>
>
> Hth -- Gerrit
>
> -
> Dr. Gerrit Eichner Mathematical Institute, Room 212
> gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
> Tel: +49-(0)641-99-32104 Arndtstr. 2, 35392 Giessen, Germany
> http://www.uni-giessen.de/eichner
> -
>
>
> Am 18.04.2018 um 15:26 schrieb CHATTON Anne via R-help:
>> Hello,
>>
>> I am trying to graph a regression line using the followings:
>>
>> Age <- c(39, 47, 45, 47, 65, 46, 67, 42, 67, 56, 64, 56, 59, 34, 42,
>> 48, 45,
>> 17, 20, 19, 36, 50, 39, 21, 44, 53, 63, 29, 25, 69)
>> BloodPressure <- c(144, 220, 138, 145, 162, 142, 170, 124, 158, 154, 162,
>> 150, 140, 110, 128, 130, 135, 114, 116, 124, 136, 142, 120, 120, 160,
>> 158,
>> 144, 130, 125, 175)
>> SimpleLinearReg1=lm(Age ~ BloodPressure)
>> summary(SimpleLinearReg1)
>> eq = paste0("y = ", round(coeff[2],1), "*x ", round(coeff[1],1))
>> plot(Age, BloodPressure, pch = 16, cex = 1.3, col = "blue",
>> main = eq, xlab = "Age (Year)", ylab = "Blood Pressure (mmHg)")
>> abline(SimpleLinearReg1, col="red")
>> mean(Age)
>> mean(BloodPressure)
>> abline(h=142.53, col="green")
>> abline(v=45.13, col="green")
>>
>> But I cannot get the regression line. Can anybody tell me what's wrong
>> with the codes ? I would appreciate very much. Thanks for any help.
>>
>> Anne
>>
>> [[alternative HTML version deleted]]
>>
>> __________
>> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
>
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistrasse 52
building W 34
20246 Hamburg
Phone: +49 (0) 40 7410 - 58243
Fax: +49 (0) 40 7410 - 57790
Web: www.uke.de/imbe
--
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg | www.uke.de
Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Joachim Prölß, Martina Saurin (komm.)
_
SAVE PAPER - THINK BEFORE PRINTING
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Re: [R] Calculate LC50
Have a look at the drc-package. btw. a linear model does not fit your data very well. Cheers. Am 22.02.2018 um 08:20 schrieb AbouEl-Makarim Aboueissa: > Dear All: good morning > > > I need helps with the calculation of the *LC50* from the data below > > > x<-c(0,0.3,0.7,1,4,10) > y<-c(100,86,65,51.3,19.2,7.4) > > yxreg<-lm(y~x) > > > any help will be highly appreciated. > > > with many thanks > abou > __ > > > *AbouEl-Makarim Aboueissa, PhD* > > *Professor of Statistics* > > *Department of Mathematics and Statistics* > *University of Southern Maine* > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistrasse 52 building W 34 20246 Hamburg Phone: +49 (0) 40 7410 - 58243 Fax: +49 (0) 40 7410 - 57790 Web: www.uke.de/imbe -- _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg | www.uke.de Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Martina Saurin (komm.) _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] outlining (highlighting) pixels in ggplot2
just add limits=c(-zmax1,zmax1) to the scale_fill_gradient2-call hth Am 20.12.2017 um 18:21 schrieb Morway, Eric: > I apprecaite the guidance Eik, that works great! I'm also wondering if > you have any pointers for how I might stretch the color scale so that > the max and min values are the same? Right now, the min is -0.064 and > the max is something closer to 0.04. As you can see in what I sent, I > tried adding: > > zmax1 = max(abs(m1)) > > ggplot(..., autoscale = FALSE, zmin = -1 * zmax1, zmax = zmax1) + ... > > to ggplot, but I'm either using the wrong arguments or have not added > them to the correct spot. I haven't been able to find the fix for this > and would very much appreciate any further guidance you can offer for > this last fix. -- Eik Vettorazzi Universitätsklinikum Hamburg-Eppendorf Institut für Medizinische Biometrie und Epidemiologie Martinistraße 52 Gebäude W 34 20246 Hamburg Telefon: +49 (0) 40 7410 - 58243 Fax: +49 (0) 40 7410 - 57790 Web: www.uke.de/imbe -- _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg | www.uke.de Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Martina Saurin (komm.) _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] outlining (highlighting) pixels in ggplot2
Hi Eric,
you can use an annotate-layer, eg
ind<-which(sig>0,arr.ind = T)
ggplot(m1.melted, aes(x = Month, y = Site, fill = Concentration), autoscale
= FALSE, zmin = -1 * zmax1, zmax = zmax1) +
geom_tile() +
coord_equal() +
scale_fill_gradient2(low = "darkred",
mid = "white",
high = "darkblue",
midpoint = 0)
+annotate("rect",ymin=ind[,"row"]-.5,ymax=.5+ind[,"row"],
xmin=-.5+ind[,"col"],xmax=.5+ind[,"col"],colour="red", size=.5,
linetype=1, fill=NA)
Cheers
Am 20.12.2017 um 01:32 schrieb Morway, Eric:
> library(ggplot2)
> library(RColorBrewer)
> library(reshape2)
>
> m1 <- matrix(c(
> -0.0024, -0.0031, -0.0021, -0.0034, -0.0060, -1.00e-02, -8.47e-03, -0.0117,
> -0.0075, -0.0043, -0.0026, -0.0021,
> -0.0015, -0.0076, -0.0032, -0.0105, -0.0107, -2.73e-02, -3.37e-02, -0.0282,
> -0.0149, -0.0070, -0.0046, -0.0039,
> -0.0121, -0.0155, -0.0203, -0.0290, -0.0330, -3.19e-02, -1.74e-02, -0.0103,
> -0.0084, -0.0180, -0.0162, -0.0136,
> -0.0073, -0.0053, -0.0050, -0.0058, -0.0060, -4.38e-03, -2.21e-03, -0.0012,
> -0.0026, -0.0026, -0.0034, -0.0073,
> -0.0027, -0.0031, -0.0054, -0.0069, -0.0071, -6.28e-03, -2.88e-03, -0.0014,
> -0.0031, -0.0037, -0.0030, -0.0027,
> -0.0261, -0.0223, -0.0216, -0.0293, -0.0327, -3.17e-02, -1.77e-02, -0.0084,
> -0.0059, -0.0060, -0.0120, -0.0157,
> 0.0045, 0.0006, -0.0031, -0.0058, -0.0093, -9.20e-03, -6.76e-03,
> -0.0033, 0.0002, 0.0045, 0.0080, 0.0084,
> -0.0021, -0.0018, -0.0020, -0.0046, -0.0080, -2.73e-03, 7.43e-04, 0.0004,
> -0.0010, -0.0017, -0.0022, -0.0024,
> -0.0345, -0.0294, -0.0212, -0.0194, -0.0192, -2.25e-02, -2.05e-02, -0.0163,
> -0.0179, -0.0213, -0.0275, -0.0304,
> -0.0034, -0.0038, -0.0040, -0.0045, -0.0059, -1.89e-03, 6.99e-05, -0.0050,
> -0.0114, -0.0112, -0.0087, -0.0064,
> -0.0051, -0.0061, -0.0052, -0.0035, 0.0012, -7.41e-06, -3.43e-03, -0.0055,
> -0.0020, 0.0016, -0.0024, -0.0069,
> -0.0061, -0.0068, -0.0089, -0.0107, -0.0104, -7.65e-03, 2.43e-03, 0.0008,
> -0.0006, -0.0014, -0.0021, -0.0057,
> 0.0381, 0.0149, -0.0074, -0.0302, -0.0550, -6.40e-02, -5.28e-02, -0.0326,
> -0.0114, 0.0121, 0.0367, 0.0501,
> -0.0075, -0.0096, -0.0123, -0.0200, -0.0288, -2.65e-02, -2.08e-02, -0.0176,
> -0.0146, -0.0067, -0.0038, -0.0029,
> -0.0154, -0.0162, -0.0252, -0.0299, -0.0350, -3.40e-02, -2.51e-02, -0.0172,
> -0.0139, -0.0091, -0.0119, -0.0156),
> nrow = 15, ncol = 12, byrow=TRUE,
> dimnames = list(rev(c("TH1", "IN1", "IN3", "GL1", "LH1", "ED9", "TC1",
> "TC2", "TC3", "UT1", "UT3", "UT5", "GC1", "BC1", "WC1")),
> c(format(seq(as.Date('2000-10-01'),
> as.Date('2001-09-30'), by='month'), "%b"
>
> # palette definition
> palette <- colorRampPalette(c("darkblue", "blue", "white", "red",
> "darkred"))
>
> # find max stretch value
> zmax1 = max(abs(m1))
>
> m1.melted <- melt(m1)
> names(m1.melted) <- c('Site','Month', 'Concentration')
>
> # Set up an example matrix with binary code for which results (pixels) are
> significant
> set.seed(4004)
> sig <- matrix(round(abs(rnorm(15*12)/3)), nrow = 15, ncol = 12)
>
> ggplot(m1.melted, aes(x = Month, y = Site, fill = Concentration), autoscale
> = FALSE, zmin = -1 * zmax1, zmax = zmax1) +
> geom_tile() +
> coord_equal() +
> scale_fill_gradient2(low = "darkred",
>mid = "white",
>high = "darkblue",
>midpoint = 0)
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistrasse 52
building W 34
20246 Hamburg
Phone: +49 (0) 40 7410 - 58243
Fax: +49 (0) 40 7410 - 57790
Web: www.uke.de/imbe
--
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg | www.uke.de
Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Joachim Prölß, Martina Saurin (komm.)
_
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Re: [R] YNT: ggtern and bquote...
t; information by a person other than the intended recipient is unauthorized and > may be illegal. In this case, please immediately notify the sender and delete > it from your system. Anadolu University does not guarantee the accuracy or > completeness of any information included in this message. Therefore, by any > means Anadolu University is not responsible for the content of the message, > and the transmission, reception, storage, and use of the information. The > opinions expressed in this message only belong to the sender of it and may > not reflect the opinions of Anadolu University. > > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistrasse 52 building W 34 20246 Hamburg Phone: +49 (0) 40 7410 - 58243 Fax: +49 (0) 40 7410 - 57790 Web: www.uke.de/imbe -- _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg | www.uke.de Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Martina Saurin (komm.) _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] unique identifier for number sequence
cumsum(c(x[1],pmax(0,diff(x*x Am 23.11.2015 um 15:59 schrieb PIKAL Petr: > Dear all > > I have a vector ones and zeroes like that > x<-c(rep(0,5), rep(1,5), rep(0,10), rep(1,8)) > > and I need to get result like that > x.i<-c(rep(0,5), rep(1,5), rep(0,10), rep(2,8)) > > It means I need an unique identifier for each sequence of ones. > > It probably can be done by rle, cumsum and some fiddling with data but maybe > there is some clever way which I overlooked. > > Cheers > Petr > > > > Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou > určeny pouze jeho adresátům. > Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně > jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze > svého systému. > Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email > jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat. > Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či > zpožděním přenosu e-mailu. > > V případě, že je tento e-mail součástí obchodního jednání: > - vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, > a to z jakéhokoliv důvodu i bez uvedení důvodu. > - a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; > Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany > příjemce s dodatkem či odchylkou. > - trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným > dosažením shody na všech jejích náležitostech. > - odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost > žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně > pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu > případně osobě, kterou adresát zastupuje, předloženy nebo jejich existence je > adresátovi či osobě jím zastoupené známá. > > This e-mail and any documents attached to it may be confidential and are > intended only for its intended recipients. > If you received this e-mail by mistake, please immediately inform its sender. > Delete the contents of this e-mail with all attachments and its copies from > your system. > If you are not the intended recipient of this e-mail, you are not authorized > to use, disseminate, copy or disclose this e-mail in any manner. > The sender of this e-mail shall not be liable for any possible damage caused > by modifications of the e-mail or by delay with transfer of the email. > > In case that this e-mail forms part of business dealings: > - the sender reserves the right to end negotiations about entering into a > contract in any time, for any reason, and without stating any reasoning. > - if the e-mail contains an offer, the recipient is entitled to immediately > accept such offer; The sender of this e-mail (offer) excludes any acceptance > of the offer on the part of the recipient containing any amendment or > variation. > - the sender insists on that the respective contract is concluded only upon > an express mutual agreement on all its aspects. > - the sender of this e-mail informs that he/she is not authorized to enter > into any contracts on behalf of the company except for cases in which he/she > is expressly authorized to do so in writing, and such authorization or power > of attorney is submitted to the recipient or the person represented by the > recipient, or the existence of such authorization is known to the recipient > of the person represented by the recipient. > __ > R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg | www.uke.de Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] doubt with Odds ratio - URGENT HELP NEEDED
>>> >>>> I can’t get the OR and OR CI :( >>>> >>>> >>>> *DATA:* >>>> >>>> >>>> >>>> >>>> >>>> >>>> Best, >>>> >>>> RO >>>> >>>> >>>> >>>> >>>> Atenciosamente, >>>> Rosa Oliveira >>>> >>>> -- >>>> >>>> >>>> >>>> >>>> >>>> >>>> Rosa Celeste dos Santos Oliveira, >>>> >>>> E-mail:rosit...@gmail.com <http://gmail.com> >>>> <mailto:rosit...@gmail.com> >>>> Tlm: +351 939355143 >>>> Linkedin: https://pt.linkedin.com/in/rosacsoliveira >>>> >>>> >>>> "Many admire, few know" >>>> Hippocrates >>>> >>>>> On 23 Sep 2015, at 12:02, Michael Dewey <li...@dewey.myzen.co.uk >>>>> <mailto:li...@dewey.myzen.co.uk> >>>>> <mailto:li...@dewey.myzen.co.uk>> wrote: >>>>> >>>>> Dear Rosa >>>>> >>>>> It would help if you posted the error messages where they occur so >>>>> that we can see which of your commands caused which error. However see >>>>> comment inline below. >>>>> >>>>> On 22/09/2015 22:17, Rosa Oliveira wrote: >>>>>> Dear all, >>>>>> >>>>>> >>>>>> I’m trying to compute Odds ratio and OR confidence interval. >>>>>> >>>>>> I’m really naive, sorry for that. >>>>>> >>>>>> >>>>>> I attach my data and my code. >>>>>> >>>>>> I’m having lots of errors: >>>>>> >>>>>> 1. Error in data.frame(tas1 = tas.data$tas_d2, tas2 = >>>>>> tas.data$tas_d3, tas3 = tas.data$tas_d4, : >>>>>> arguments imply differing number of rows: 90, 0 >>>>>> >>>>> >>>>> At least one of tas_d2, tas_d3, tas_d4 does not exist >>>>> >>>>> I suggest fixing that one and hoping the rest go away. >>>>> >>>>>> 2. Error in data.frame(tas = c(unlist(tas.data[, -8:-6])), time = >>>>>> rep(c(0:4), : >>>>>> arguments imply differing number of rows: 630, 450, 0 >>>>>> >>>>>> 3. Error: object 'tas.data.long' not found >>>>>> >>>>>> 4. Error in data.frame(media = c(mean.dead, mean.alive), >>>>>> standarderror = c(se.dead, : >>>>>> arguments imply differing number of rows: 14, 10 >>>>>> >>>>>> 5. Error in ggplot(summarytas, aes(x = c(c(1:5), c(1:5)), y = mean, >>>>>> colour = discharge)) : >>>>>> object 'summarytas' not found >>>>>> >>>>>> 6. Error in summary(glm(tas.data[, 6] ~ tas.data[, 4], family = >>>>>> binomial(link = probit))) : >>>>>> error in evaluating the argument 'object' in selecting a method for >>>>>> function 'summary': Error in eval(expr, envir, enclos) : y values >>>>>> must be 0 <= y <= 1 >>>>>> >>>>>> 7. Error in wilcox.test.default(pred[obs == 1], pred[obs == 0], >>>>>> alternative = "great") : >>>>>> not enough (finite) 'x' observations >>>>>> In addition: Warning message: >>>>>> In is.finite(x) & apply(pred, 1, f) : >>>>>> longer object length is not a multiple of shorter object length >>>>>> >>>>>> >>>>>> and off course I’m not getting OR. >>>>>> >>>>>> Nonetheless all this errors, I think I have not been able to compute >>>>>> de code to get OR and OR confidence interval. >>>>>> >>>>>> >>>>>> Can anyone help me please. It’s really urgent. >>>>>> >>>>>> PLEASE >>>>>> >>>>>> THE CODE: >>>>>> >>>>>> the hospital outcome is discharge. >>>>>> >>>>>> require(gdata
Re: [R] Help with vectors!
how about this:
match(VAS,unique(VAS))
#or, preserving given order
match(VAS,c("White","Yellow","Green","Black"))
Cheers.
Am 05.09.2015 um 23:14 schrieb Dan D:
> # your data
> VAS<-c("Green","Green","Black","Green","White","Yellow","Yellow","Black","Green","Black")
>
> # declare the new vector
> New_Vector<-numeric(length(VAS))
>
> # brute force:
> New_Vector[VAS=="White"]<-1
> New_Vector[VAS=="Yellow"]<-2
> New_Vector[VAS=="Green"]<-3
> New_Vector[VAS=="Black"]<-4
>
> # a little more subtle
> cols<-c("White","Yellow","Green","Black")
> for (i in 1:length(cols)) New_Vector[VAS==cols[i]]<-i
>
> # and a general approach (that may give a different indexing, but can be
> used for any array)
> for (i in 1:length(unique(VAS))) New_Vector[VAS==unique(VAS)[i]]<-i
> cbind(1:length(unique(VAS)),unique(VAS)) # a decoding key for the color
> index
>
>
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Help-with-vectors-tp4711801p4711895.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg | www.uke.de
Vorstandsmitglieder: Prof. Dr. Burkhard Göke (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Joachim Prölß, Rainer Schoppik
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Re: [R] Logistic Regression with 200K features in R?
it is simply because you can't do a regression with more predictors than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms - terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
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_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden),
Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik
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Re: [R] Logistic Regression with 200K features in R?
I thought so (with all the limitations due to collinearity and so on),
but actually there is a limit for the maximum size of an array which is
independent of your memory size and is due to the way arrays are
indexed. You can't create an object with more than 2^31-1 = 2147483647
elements.
https://stat.ethz.ch/pipermail/r-help/2007-June/133238.html
cheers
Am 12.12.2013 12:34, schrieb Romeo Kienzler:
ok, so 200K predictors an 10M observations would work?
On 12/12/2013 12:12 PM, Eik Vettorazzi wrote:
it is simply because you can't do a regression with more predictors than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms -
terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Besuchen Sie uns auf: www.uke.de
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden),
Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik
_
SAVE PAPER - THINK BEFORE PRINTING
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Logistic Regression with 200K features in R?
thanks Duncan for this clarification.
A double precision matrix with 2e11 elements (as the op wanted) would
need about 1.5 TB memory, that's more than a standard (windows 64bit)
computer can handle.
Cheers.
Am 12.12.2013 13:00, schrieb Duncan Murdoch:
On 13-12-12 6:51 AM, Eik Vettorazzi wrote:
I thought so (with all the limitations due to collinearity and so on),
but actually there is a limit for the maximum size of an array which is
independent of your memory size and is due to the way arrays are
indexed. You can't create an object with more than 2^31-1 = 2147483647
elements.
https://stat.ethz.ch/pipermail/r-help/2007-June/133238.html
That post is from 2007. The limits were raised considerably when R
3.0.0 was released, and it is now 2^48 for disk-based operations, 2^52
for working in memory.
Duncan Murdoch
cheers
Am 12.12.2013 12:34, schrieb Romeo Kienzler:
ok, so 200K predictors an 10M observations would work?
On 12/12/2013 12:12 PM, Eik Vettorazzi wrote:
it is simply because you can't do a regression with more predictors
than
observations.
Cheers.
Am 12.12.2013 09:00, schrieb Romeo Kienzler:
Dear List,
I'm quite new to R and want to do logistic regression with a 200K
feature data set (around 150 training examples).
I'm aware that I should use Naive Bayes but I have a more general
question about the capability of R handling very high dimensional
data.
Please consider the following R code where mygenestrain.tab is a 150
by 20 matrix:
traindata - read.table('mygenestrain.tab');
mylogit - glm(V1 ~ ., data = traindata, family = binomial);
When executing this code I get the following error:
Error in terms.formula(formula, data = data) :
allocMatrix: too many elements specified
Calls: glm ... model.frame - model.frame.default - terms -
terms.formula
Execution halted
Is this because R can't handle 200K features or am I doing something
completely wrong here?
Thanks a lot for your help!
best Regards,
Romeo
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Besuchen Sie uns auf: www.uke.de
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden),
Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik
_
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Solving a normal distribution pnorm for q
Assuming r, sd1 and sd2 as known(fixed) components, this works: X - pnorm(q=q,sd=sd1)*r+pnorm(q=q,sd=sd2)*(1-r) uniroot(function(x)pnorm(q=x,sd=sd1)*r+pnorm(q=x,sd=sd2)*(1-r)-X,c(-1e6,1e6)) Cheers Am 12.12.2013 15:58, schrieb Johannes Radinger: Thanks for the fast response. Of course I totally overlooked qnorm as I had a more complex task in my head. I wanted to reverse following equation: r=0.67 q=-150 sd1=100 sd2=1000 X - pnorm(q=q,sd=sd1)*r+pnorm(q=q,sd=sd2)*(1-r) Maybe its mathematically really easy, but somehow I don't get it how to do reverse and provide X and get q especially with the presence of r respectively as weighting factors for the two distributions. /J On Thu, Dec 12, 2013 at 3:09 PM, Dániel Kehl ke...@ktk.pte.hu wrote: Looks like homework. Try ?qnorm Feladó: r-help-boun...@r-project.org [r-help-boun...@r-project.org] ; meghatalmaz#243;: Johannes Radinger [johannesradin...@gmail.com] Küldve: 2013. december 12. 14:56 To: R help Tárgy: [R] Solving a normal distribution pnorm for q Hi, I found follwowing example of pnorm here: http://www.r-tutor.com/elementary-statistics/probability-distributions/normal-distribution Problem Assume that the test scores of a college entrance exam fits a normal distribution. Furthermore, the mean test score is 72, and the standard deviation is 15.2. What is the percentage of students scoring 84 or more in the exam? Solution pnorm(84, mean=72, sd=15.2, lower.tail=FALSE) [1] 0.21492 That is straight forward, however what if I want to know the score the best 30% students are reaching at least. So I know the solution of pnorm but want to know its q. How can that be achieved in R? Any suggestions? /Johannes [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Besuchen Sie uns auf: www.uke.de _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] adding tables
How about this: t2 - table(c(10,11,12,13)) t1 -table(c(1,1,2,4,5)) t12-(rbind(as.data.frame(t1),as.data.frame(t2))) xtabs(Freq~.,t12) it works for overlapping tables t2 - table(c(10,11,12,13,1,2)) t1 -table(c(1,1,2,4,5,11,12)) t12-(rbind(as.data.frame(t1),as.data.frame(t2))) xtabs(Freq~.,t12) and it will work for two-dimensional tables as well: t2 - table(c(10,11,12,13),letters[rep(1:2,each=2)]) t1 -table(c(1,1,2,4,5),letters[rep(c(1,3),length.out=5)]) t12a-(rbind(as.data.frame(t1),as.data.frame(t2))) xtabs(Freq~.,t12a) cheers. Am 10.12.2013 00:59, schrieb Ross Boylan: Can anyone recommend a good way to add tables? Ideally I would like t1 - table(x1) t2 - table(x2) t1+t2 It t1 and t2 have the same levels this works fine, but I need something that will work even if they differ, e.g., t1 1 2 4 5 2 1 1 1 t2 - table(c(10, 11, 12, 13)) t1+t2 # apparently does simple vector addition 1 2 4 5 3 2 2 2 whereas I want 1 2 4 5 10 11 12 13 2 1 1 1 1 1 1 1 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Besuchen Sie uns auf: www.uke.de _ Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Christian Gerloff (Vertreter des Vorsitzenden), Prof. Dr. Dr. Uwe Koch-Gromus, Joachim Prölß, Rainer Schoppik _ SAVE PAPER - THINK BEFORE PRINTING __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Automatically Remove Aliased Terms from a Model
Hi Thorn,
it is not entirely clear (at least for me) what you want to accomplish.
an easy and fail safe way of extracting used terms in a (g)lm-object is
names(model.frame(l))
if you want to extract terms to finally select a model, have a look at
drop1 and/or MASS::dropterm
Hth
Am 28.10.2013 17:19, schrieb Thaler,Thorn,LAUSANNE,Applied Mathematics:
Dear all,
I am trying to implement a function which removes aliased terms from a model.
The challenge I am facing is that with alias I get the aliased coefficients
of the model, which I have to translate into the terms from the model
formula. What I have tried so far:
--8--
d - expand.grid(a = 0:1, b=0:1)
d$c - (d$a + d$b) %% 2
d$y - rnorm(4)
d - within(d, {a - factor(a); b - factor(b); c - factor(c)})
l - lm(y ~ a * b + c, d)
removeAliased - function(mod) {
## Retrieve all terms in the model
X - attr(mod$terms, term.label)
## Get the aliased coefficients
rn - rownames(alias(mod)$Complete)
## remove factor levels from coefficient names to retrieve the terms
regex.base - unique(unlist(lapply(mod$model[, sapply(mod$model,
is.factor)], levels)))
aliased - gsub(paste(regex.base, $, sep = , collapse = |), ,
gsub(paste(regex.base, :, sep = , collapse = |), :, rn))
uF - formula(paste(. ~ ., paste(aliased, collapse = -), sep = -))
update(mod, uF)
}
removeAliased(l)
--8--
This function works in principle, but this workaround with removing the
factor levels is just, well, a workaround which could cause problems in some
circumstances (when the name of a level matches the end of another variable,
when I use a different contrast and R names the coefficients differently etc.
- and I am not sure which other cases I am overlooking).
So my question is whether there are some more intelligent ways of doing what
I want to achieve? Is there a function to translate a coefficient of a LM
back to the term, something like:
termFromCoef(a1) ## a1
termFromCoef(a1:b1) ## a:b
With this I could simply translate the rownames from alias into the terms
needed for the model update.
Thanks for your help.
Kind Regards,
Thorn Thaler
NRC Lausanne
Applied Mathematics
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--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Besuchen Sie uns auf: www.uke.de
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Joachim Prölß, Rainer Schoppik
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How can I use a script l (LaTeX \ell) in mathematical annotation of plots?
Hi Byron,
You may have a look at the tikzDevice package, it solves the aesthetic
issue of font mixing.
You can sweave the following example
\documentclass{article}
\usepackage{tikz}
\begin{document}
\SweaveOpts{concordance=TRUE}
\begin{figure}[!h]
\centering
fig1, echo=FALSE, fig=FALSE, results=hide=
library(tikzDevice)
plotfile-simpleEx.tex
tikz(plotfile, width =3.5 , height =3.5)
plot(1, ,type=n)
text(1,1,$\\ell$ within ordinary \\LaTeX text)
dev.off()
@
\input{\Sexpr{plotfile}}
\caption{Test TeX annotation}
\end{figure}
\end{document}
cheers
Am 24.10.2013 18:44, schrieb Byron Dom:
Thanks. That works for me too.
I had worked that out based on Brian's response to my original post.
If you read thru my response to him in detail, you'll see:
Here are a couple of examples using it that worked:
plot(1:10,xlab=\u2113)
plot(1:10,xlab=\u2113(\u2113 + 1))
From: Eik Vettorazzi e.vettora...@uke.de
To: Byron Dom byron_...@yahoo.com; r help r-help@r-project.org
Cc: rip...@stats.ox.ac.uk rip...@stats.ox.ac.uk
Sent: Thursday, October 24, 2013 2:50 AM
Subject: Re: [R] How can I use a script l (LaTeX \ell) in mathematical
annotation of plots?
this works for me:
plot(1,main=\u2113)
cheers
Am 24.10.2013 01:39, schrieb Byron Dom:
Original post: On 13/10/2013 18:53, Byron Dom wrote:
Due to convention a script l - $$\ell$$ (LaTeX \ell) is used to
represent a certain quantity in something I'm working on. I'm
unable to figure out how to use it in R. It's not included in the
list on ?plotmath.
Can anyone tell me how to use it? Its unicode is U+2113. This
page has a list of various encodings of it:
http://www.fileformat.info/info/unicode/char/2113/encoding.htm.
Is there a way to include it by using one of these encodings somehow?
-
On 13/10/2013 22:06 Prof Brian Ripley responded:
What do you want to do with it? plotmath is about plotting, but you
have not otherwise mentioned that, let alone the device on which you
want to plot.
Read the help for plotmath: on some plot devices, just use \u2113. It
is not AFAICS in the Adobe symbol encoding.
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:
+44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
-
My response:
Thanks I worked out how to do it based on your mention of u2113.
See below.
I'm sorry if my information wasn't complete enough. I assumed that
what I said, combined with the subject (How can I use a script l
(LaTeX \ell) in
mathematical annotation of plots?) would have been
enough.
I just wanted to be able to do this on the default device, which
displays plots within the R session window. I have a simple path
to go from there to .png form, which I include in LaTeX documents
and for other purposes. I am using R version 3.0.1 in Windows 7,
for which I believe the default plot device is windows.
An example of the kind of thing I wanted to do is to include
ylab = expression(hat(gamma)) (which is equivalent to the LaTex
\hat{\gamma}}) among the base-graphics plot() arguments. That
works. In LaTeX a script l is produced with \ell, but
something like ylab = expression(ell) doesn't work in R.
I wanted to be able to do the equivalent of that, but to obtain ℓ.
Here are a couple of examples using it that worked:
plot(1:10,xlab=\u2113)
plot(1:10,xlab=\u2113(\u2113 + 1))
The only (slight) problem with this is the minor aesthetic issue
that the ℓ one gets this way is in an obviously different font
from what one gets using the LaTeX \lambda command/symbol.
Strangely, earlier, when I tried
plot(1:10,xlab=expression(symbol(\u2113)))
it did the same thing as
plot(1:10,xlab=expression(lambda))
So I got a lower-case greek lambda - λ, rather than ℓ
(script l).
When I used the unicode representation for lowercase
lambda - λ - as follows
plot(1:10,xlab=expression(symbol(\u03bb)))
I got this for an x-axis label: Y+03BB. On the other hand,
the following did work
plot(1:10,xlab=\u03bb),
giving me an x-axis label of lambda - λ.
[[alternative HTML version deleted]]
__
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and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Besuchen Sie uns auf: www.uke.de
Re: [R] How can I use a script l (LaTeX \ell) in mathematical annotation of plots?
this works for me:
plot(1,main=\u2113)
cheers
Am 24.10.2013 01:39, schrieb Byron Dom:
Original post: On 13/10/2013 18:53, Byron Dom wrote:
Due to convention a script l - $$\ell$$ (LaTeX \ell) is used to
represent a certain quantity in something I'm working on. I'm
unable to figure out how to use it in R. It's not included in the
list on ?plotmath.
Can anyone tell me how to use it? Its unicode is U+2113. This
page has a list of various encodings of it:
http://www.fileformat.info/info/unicode/char/2113/encoding.htm.
Is there a way to include it by using one of these encodings somehow?
-
On 13/10/2013 22:06 Prof Brian Ripley responded:
What do you want to do with it? plotmath is about plotting, but you
have not otherwise mentioned that, let alone the device on which you
want to plot.
Read the help for plotmath: on some plot devices, just use \u2113. It
is not AFAICS in the Adobe symbol encoding.
--
Brian D. Ripley, ripley at stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel:
+44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
-
My response:
Thanks I worked out how to do it based on your mention of u2113.
See below.
I'm sorry if my information wasn't complete enough. I assumed that
what I said, combined with the subject (How can I use a script l
(LaTeX \ell) in
mathematical annotation of plots?) would have been
enough.
I just wanted to be able to do this on the default device, which
displays plots within the R session window. I have a simple path
to go from there to .png form, which I include in LaTeX documents
and for other purposes. I am using R version 3.0.1 in Windows 7,
for which I believe the default plot device is windows.
An example of the kind of thing I wanted to do is to include
ylab = expression(hat(gamma)) (which is equivalent to the LaTex
\hat{\gamma}}) among the base-graphics plot() arguments. That
works. In LaTeX a script l is produced with \ell, but
something like ylab = expression(ell) doesn't work in R.
I wanted to be able to do the equivalent of that, but to obtain ℓ.
Here are a couple of examples using it that worked:
plot(1:10,xlab=\u2113)
plot(1:10,xlab=\u2113(\u2113 + 1))
The only (slight) problem with this is the minor aesthetic issue
that the ℓ one gets this way is in an obviously different font
from what one gets using the LaTeX \lambda command/symbol.
Strangely, earlier, when I tried
plot(1:10,xlab=expression(symbol(\u2113)))
it did the same thing as
plot(1:10,xlab=expression(lambda))
So I got a lower-case greek lambda - λ, rather than ℓ
(script l).
When I used the unicode representation for lowercase
lambda - λ - as follows
plot(1:10,xlab=expression(symbol(\u03bb)))
I got this for an x-axis label: Y+03BB. On the other hand,
the following did work
plot(1:10,xlab=\u03bb),
giving me an x-axis label of lambda - λ.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Besuchen Sie uns auf: www.uke.de
_
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Joachim Prölß, Rainer Schoppik
_
SAVE PAPER - THINK BEFORE PRINTING
__
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Re: [R] Jul 26, 2013; 12:34am
Dear Frank, you can use grconvertY and grconvertX to convert from user coordinates to inches etc and vice versa. E.g. lines(rep(.25*1500,2),grconvertY(grconvertY(par(usr)[3],user,inches)+c(0,.1),from=inches,to=user)) should produce a tick of 0.1 inches height, regardless of the actual plot height. Hope this helps. Am 26.07.2013 13:57, schrieb Frank Harrell: Thanks Rich and Jim and apologies for omitting the line x - c(285, 43.75, 94, 150, 214, 375, 270, 350, 41.5, 210, 30, 37.6, 281, 101, 210) But the fundamental problem remains that vertical spacing is not correct unless I waste a lot of image space at the top. Frank -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] measuring distances between colours?
Hi John,
i would propose a one-liner for the hexcode transformation:
hex2dec-function(hexnums)sapply(strtoi(hexnums,16L),function(x)x%/%256^(2:0)%%256)
#instead of
hexnumerals - 0:15
names(hexnumerals) - c(0:9, LETTERS[1:6])
hex2decimal - function(hexnums){
hexnums - strsplit(hexnums, )
decimals - matrix(0, 3, length(hexnums))
decimals[1, ] - sapply(hexnums, function(x)
sum(hexnumerals[x[1:2]] * c(16, 1)))
decimals[2, ] - sapply(hexnums, function(x)
sum(hexnumerals[x[3:4]] * c(16, 1)))
decimals[3, ] - sapply(hexnums, function(x)
sum(hexnumerals[x[5:6]] * c(16, 1)))
decimals
}
#some tests
cols-c(AA, 002200, 99, 00, BB00BB, 00)
cols-sub(^#,,toupper(cols))
#actually 'toupper' is not needed for hex2dec
#check results
hex2decimal(cols)
hex2dec(cols)
#it is not only shorter ocde, but even faster.
cols.test-sprintf(%06X,sample(0:(256^3),10))
system.time(hex2decimal(cols.test))
# User System verstrichen
# 3.540.003.61
system.time(hex2dec(cols.test))
# User System verstrichen
# 0.530.000.53
cheers.
Am 30.05.2013 14:13, schrieb John Fox:
Dear r-helpers,
I'm interested in locating the named colour that's closest to an arbitrary
RGB colour. The best that I've been able to come up is the following, which
uses HSV colours for the comparison:
r2c - function(){
hexnumerals - 0:15
names(hexnumerals) - c(0:9, LETTERS[1:6])
hex2decimal - function(hexnums){
hexnums - strsplit(hexnums, )
decimals - matrix(0, 3, length(hexnums))
decimals[1, ] - sapply(hexnums, function(x)
sum(hexnumerals[x[1:2]] * c(16, 1)))
decimals[2, ] - sapply(hexnums, function(x)
sum(hexnumerals[x[3:4]] * c(16, 1)))
decimals[3, ] - sapply(hexnums, function(x)
sum(hexnumerals[x[5:6]] * c(16, 1)))
decimals
}
colors - colors()
hsv - rgb2hsv(col2rgb(colors))
function(cols){
cols - sub(^#, , toupper(cols))
dec.cols - rgb2hsv(hex2decimal(cols))
colors[apply(dec.cols, 2, function(dec.col)
which.min(colSums((hsv - dec.col)^2)))]
}
}
rgb2col - r2c()
I've programmed this with a closure so that hsv gets computed only once.
Examples:
rgb2col(c(AA, 002200, 99, 00, BB00BB, #00))
[1] darkred darkgreen blue4 darkgreen magenta3 darkgreen
rgb2col(c(00, #00))
[1] darkgoldenrod cyan4
Some of these colour matches, e.g., #00 - darkgreen seem poor to me.
Even if the approach is sound, I'd like to be able to detect that there is no
sufficiently close match in the vector of named colours. That is, can I
establish a maximum acceptable distance in the HSV (or some other) colour
space?
I vaguely recall a paper or discussion concerning colour representation in R
but can't locate it.
Any suggestions would be appreciated.
John
John Fox
Sen. William McMaster Prof. of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
http://socserv.mcmaster.ca/jfox/
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Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
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Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Prof. Dr. Dr. Uwe
Koch-Gromus, Astrid Lurati (Kommissarisch), Joachim Prölß, Matthias Waldmann
(Kommissarisch)
Bitte erwägen Sie, ob diese Mail ausgedruckt werden muss - der Umwelt zuliebe.
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Re: [R] How to call an object given a string?
Hi Rui, how about this sapply(ls(),get) cheers Am 29.04.2013 13:07, schrieb Rui Esteves: Hello, This is very basic and very frustrating. Suppose this: A=5 B=5 C=10 ls() A B C I would like this xpto() 5 5 10 How can I do xpto()? Thanks Rui [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] all.vars for nested expressions
Hi Felix,
I thought, this could be an easy task for substitute, and the following
works as expected:
all.vars(substitute(expression(tp/a),list(a=expression(fn+tp
# [1] tp fn
But (of course)
all.vars(substitute(sen,list(a=a)))
does not yield the desired result, and I can't figure out, how to set up
as.name, bquote, eval, deparse etc to do the task properly.
Instead, my approach is a recursive call to all.vars
xall.help-function(x){
#check if there is an object with name x
if(exists(x)) lapply(all.vars(get(x)),xall.help) else x}
xall.vars-function(x){
if (!is.character(x)) x-paste(substitute(x))
#for convenience put in a single vecotr
#xall.help returns a 'parsed tree'
unique(unlist(xall.help(x)))
}
#example
fn-expression(n1+n2)
a - expression(fn+tp)
sen - expression(tp/a)
xall.vars(sen)
# [1] tp n1 n2
cheers.
Am 29.04.2013 13:33, schrieb flxms:
Dear R fellows,
Assume I define
a - expression(fn+tp)
sen - expression(tp/a)
Now I'd like to know, which variables are necessary for calculating sen
all.vars(sen)
This results in a vector c(tp,a). But I'd like all.vars to evaluate the
sen-object down to the ground level, which would result in a vector
c(tp,fn) (because a was defined as fn+tp). In other words, I'd like
all.vars to expand the a-object (and all other downstream objects). I am
looking for a solution, that works with much more levels. This is just a
very simple example.
I'd appreciate any suggestions how to do that very much!
Thanks in advance,
Felix
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander
Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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Re: [R] Frequency Block Chart
Hi Angelo, you might have a look at this http://lmdvr.r-forge.r-project.org/figures/figures.html?chapter=06;figure=06_15;theme=stdBW;code=right Cheers Am 23.04.2013 12:42, schrieb Angelo Scozzarella Tiscali: Hi, friends. I'd like to illustrate the relationship between two categorical variables with a block chart (a three-dimensional bar chart) with the height of the blocks being proportional to the frequencies. Is there a way to do it with R? Thanks in advance Angelo [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to search a list that contains multiple dissimilar vectors?
Hi Eric,
something like the following might me a starter
(index-lapply(alist,function(x)which(x==x.search)))
cheers.
Am 28.03.2013 08:40, schrieb Eric Rupley:
Dear All,
This is a simple question, but I'm stumped about the simplest way to search a
list object such as the following:
This randomish snippet:
n - c(round(runif(round(runif(1,1,10),0),1,10),0))
alist - new(list)
for (i in seq_along(n)) {
alist[[i]] - c(round(runif(round(runif(1,1,10),0),1,10),0))
}
names(alist) - sample(letters[1:length(n)])
rm(n);c(alist)
...produces something like this:
$d
[1] 4
$b
[1] 3 5 3
$a
[1] 2 5 7 3 10 3 4 9 9
$c
[1] 6 3 7 4 5 10 8 10 3
My question is how does one search the list for a given value, in a most
compressed set of commands, in order two return two separate indices: a) the
index of the list element(s) containing the value, and b) the index of the
matching value(s) within the vector.
Right now, I'm writing cumbersome loops to iterate though the elements, but
there must be a simple, effective method to which I have not found a
reference.
Many thanks in advance, and apologies if I have overlooked a reference
passage.
Best,
Eric
--
Eric Rupley
University of Michigan, Museum of Anthropology
1109 Geddes Ave, Rm. 4013
Ann Arbor, MI 48109-1079
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--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander
Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] urgent: question concerning data manipulation
There’s more than one way to skin a cat, here is another mm-model.matrix(~personId+law+0,testdata) merge(testdata,aggregate(mm[,-1],list(personId=mm[,personId]),max)) cheers Am 04.03.2013 16:44, schrieb David Studer: Hello everyone! Does anyone of you know how I could solve the following problem. I guess, it is not a very difficult question, but I simply lack of the right idea: I have a dataset containing data of convictions. This dataset contains 4 columns: - personId: individual number that identifies the offender - law: law which has been violated - article: article which has been violated # Testdata: personId-c(1,1,2,2,2,2,2,3,4,4) law-c(SVG, SVG, StGB, StGB, SVG, AuG, StGB, SVG, StGB, AuG) article-c(10, 10, 123, 122, 10, 40, 126, 10, 111, 40) testdata-data.frame(personId, law, article) Now I'd like to create three additional dummy-coded columns for each law (SVG, StGB, AuG). For each offender (all offenders have the same personId) it should be checked, whether there are any violations against the three laws. If there are any violations against SVG (for example), then in all rows of this offender the column SVG should have the value 1 (otherwise 0). For example offender 2 has once violated against law SVG therefore his four entries should have the value 1 at the column SVG. I hope you can understand my problem. I'd really appreciate any hints and solutions! Thank you! David [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problems with line types in plots saved as PDF files
Hi Ian,
besides encouraging alternative pdf viewers, you just could thin out
your interpolation grid, such as
b = seq(-10, 10, 0.02) #instead of 0.0002
and everything goes fine even with AcroRead XI.
Cheers
Am 20.02.2013 01:09, schrieb Ian Renner:
Hi,
I am trying to save a plot as a PDF with different line types. In the example
below, R displays the plot correctly with 2 curves in solid lines, 2 curves
in dashed lines and 1 curve as a dotted line. However, when I save the image
as PDF (as attached), the dashed lines become solid. This also happens if I
use the pdf command directly (by removing the # symbols).
Is there any way around this? Saving the image as a JPEG or PNG file works
fine but the image quality is not desirable.
b.hat = 6
a.1 = -12
a.2 = 0
a.3 = 200
b = seq(-10, 10, 0.0002)
l = a.1*(b - b.hat)^2 + a.2*(b - b.hat) + a.3
lambda = 20
p = -lambda*abs(b)
pen.like = l + p
y.min = 3*min(p)
y.max = max(c(l, p, pen.like))
#pdf(file = TestPlot.pdf, 6, 6)
#{
plot(b, l, type = l, ylim = c(y.min, y.max), lwd = 2, xlab =
expression(beta), ylab = , col = green, yaxt = n, xaxt = n)
points(b, p, type = l, lty = dotted, lwd = 2, col = red)
points(b, pen.like, type = l, lwd = 2, lty = dashed, col = green)
axis(1, at = c(0))
axis(2, at = c(0))
lambda.hat = which.max(pen.like)
lambda.glm = which(b == b.hat)
points(b[lambda.glm], l[lambda.glm], pch = 16, cex = 1.5)
points(b[lambda.hat], l[lambda.hat], pch = 17, cex = 1.5)
b.hat = -3
a.1 = -1.5
a.2 = 0
a.3 = 120
l = a.1*(b - b.hat)^2 + a.2*(b - b.hat) + a.3
pen.like = l + p
points(b, l, type = l, lwd = 2, col = blue)
points(b, pen.like, type = l, lwd = 2, lty = dashed, col = blue)
lambda.hat = which.max(pen.like)
lambda.glm = which(b == b.hat)
points(b[lambda.glm], l[lambda.glm], pch = 16, cex = 1.5)
points(b[lambda.hat], l[lambda.hat], pch = 17, cex = 1.5)
abline(h = 0)
abline(v = 0)
#}
#dev.off()
Thanks,
Ian
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--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander
Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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Re: [R] subsetting data file by intoducing a second file
Hi Ozgul, the interesting part is the select parameter in subset: subset(bg,select=c_bg[,1]) cheers Am 12.02.2013 15:29, schrieb Ozgul Inceoglu: Hello, I have a very data matrix and I have a file which has the names that I need to subset. However I cannot manage to subset the main file. ANy idea? bg - read.table (file.choose(), header=T, row.names) bg Otu00022 Otu00029 Otu00039 Otu00042 Otu00101 Otu00105 Otu00125 Otu00131 Otu00137 Otu00155 Otu00158 Otu00172 Otu00181 Otu00185 Otu00190 Otu00209 Otu00218 Gi20Jun11 0.0012170 0.0012170 0.0000 0 0.00121700000 0.0012170 0.001217 Gi40Jun11 0.000 0.000 0.0000 0 0.0000000 0.000 0.00 Gi425Jun11 0.000 0.000 0.0000 0 0.0000000 0.000 0.00 Gi45Jun11 0.000 0.000 0.00151300 0 0.0000000 0.000 0.00 Gi475Jun11 0.000 0.000 0.0000 0 0.0000000 0.000 0.00 Gi50Jun11 0.000 0.000 0.0000 0 0.0000000 0.000 0.00 ... #second file which has the names that I want to subset c_bg [,1] [1,] Otu0128 [2,] Otu0218 [3,] Otu0034 [4,] Otu0257 [5,] Otu0212 [6,] Otu0279 [7,] Otu0318 [8,] Otu0266 [9,] Otu0056 ... #by using the c_bg name file, I would like to subset bg file g1-subset(bg,colnames(bg) %in% (c_bg)) # this returns me the all the column names in bg file. Thank you, Ö __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to draw confidence interval lines of a fitted curve of polynominal regression
Hi Elaine you can do all at one go using the ggplot2-package library(ggplot2) qplot(age,weight)+ geom_smooth(method=lm,formula=y~poly(x,2)) Have a look at ?geom_smooth when you want to incooperate customized model predictions cheers. Am 07.02.2013 02:31, schrieb Elaine Kuo: Hello, I drew a plot of weight and height of people and fitted it with a polynominal regression x^2. (using curve()) Now I would like to draw the confidence interval line for the fitted curve. Please kindly advise the code for the purpose. Thank you. Elaine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] weighing proportion of rowSums in dataframe
Hi Alain, here is a one-liner for a df without the rowSum column df-data.frame(id=c(x01,x02,x03,x04,x05,x06),a=c(1,2,NA,4,5,6),b=c(2,4,6,8,10,NA),c=c(NA,3,9,12,NA,NA)) (df$wSum-apply(sweep(df[,-1],1,rowSums(df[,-1],na.rm=T),/),1,function(x)sum(x*w,na.rm=T))) hth. Am 06.02.2013 14:17, schrieb D. Alain: Dear R-List, I am sure there must be a very simple way to do this - I just do not know how... This is what I want to do: #my dataframe df-data.frame(id=c(x01,x02,x03,x04,x05,x06),a=c(1,2,NA,4,5,6),b=c(2,4,6,8,10,NA),c=c(NA,3,9,12,NA,NA),sum=c(3,9,15,24,15,6)) ida b c sum 1 x01 1 2NA 3 2 x02 2 4 3 9 3 x03 NA 6 9 15 4 x04 4 81224 5 x05 510 NA 15 6 x06 6 NA NA 6 #my weights w-c(10.5,9,12.8) #now I want to calculate the proportion of the rowsum = sum of every other number in the row, that is df[1,2]/df[1,5], df[1,3]/df[1,5], ... #e.g. df.prop id a b c sum 1 x01 0.33 0.66 NA3 2 x02 0.22 0.44 0.33 9 3 x03 NA 0.4 0.6 15 4 x04 0.16 0.33 0.524 5 x05 0.33 0.66 NA 15 6 x06 1 NA NA 6 #and then calculate a rowsum, were each column is weighed by its associated factor in the vector w, i.e. all entries in column df$a should be weighed by the factor 10.5, df$b by 9 and df$c by 12.8 and then summed up to a weighed rowsum (wrowsum=10.5*0.33+9*0.66). Any suggestions how to achieve this in a simple way with dataframes that have 9000 rows and 44 columns (so I cannot do this row by row) - sorry, if this is to easy a question. Thank you very much in advance! Best wishes Alain [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Omitting repeated occurrence in a string
Hi Christopher,
what is the rule to omit ah which is also repeated in Text?
The following might be a start:
Text - ahsgdvasgAbcabcsdahj
#finds first repetion of substrings of length 2 or more, here ah
gsub((?i)([a-z]{2,})(.*)\\1,\\1\\2,Text,perl=T)
#finds all repetions of substrings of length 3 or more, here Abc
gsub((?i)([a-z]{3,})(.*)\\1,\\1\\2,Text,perl=T)
#finds only subsequent repetions of substrings of length 2 or more
gsub((?i)([a-z]{2,})\\1,\\1,Text,perl=T)
hth.
Am 06.02.2013 17:46, schrieb Christofer Bogaso:
Hello again,
I was looking for some way on How to delete repeated appearance in a
String. Let say I have following string:
Text - ahsgdvasgAbcabcsdahj
Here you see Abc appears twice. But I want to keep only 1
occurrence. Therefore I need that:
Text_result - ahsgdvasgAbcsdahj (i.e. the first one).
Can somebody help me if it is possible using some R function?
Thanks and regards,
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
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Re: [R] merge 2 data.frames
Hi Mat, just have a look at ?rbind cheers. Am 06.02.2013 15:55, schrieb Mat: Hello together, i have probably a easy question, but how can i sum 2 data.frames among each other. I have 2 data.frames which look like this one: Cu.No. place level 1 123London A 2111 Paris B Cu.No. place level 1 333Berlin C 2444 MadridA and now i want this data in the same data.frame. like this one: Cu.No. place level 1 123London A 2111 Paris B 3 333Berlin C 4444 MadridA how can i do this. Thanks. Mat -- View this message in context: http://r.789695.n4.nabble.com/merge-2-data-frames-tp4657702.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] reshape help
Hi Robert, how about this? tmp-read.table(text=ID Dx Anausea Adiabetes Akidney_failure Aheart_attack Afever Bfever Bpneumonia Bheart_attack Bnausea Bcough Ckidney_failure Cnausea Cfoot_pain,header=T) data.frame(cbind(ID=levels(tmp$ID),(table(tmp$ID,tmp$Dx hth. Am 04.02.2013 09:16, schrieb Robert Strother: Dear R users - I have a list of patient identifiers and diagnoses from inpatient admissions. I would like to reorganize the list, presently in a long format to a wide format in reshape, but in the absence of a time element, I am uncertain how to do this - any help greatly appreciated. ID Dx Anausea Adiabetes Akidney failure Aheart attack Afever Bfever Bpneumonia Bheart attack Bnausea Bcough Ckidney failure Cnausea Cfoot pain I want this to be IDnausea diabetes kidney_failure heart_attack fever pneumonia A 11 1 1 1 0 B 00 0 1 1 1 R. Matthew Strother, MD Clinical Pharmacology Medical Oncology [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R plot like candlestick
Hi Denis, there is no if, only how in R ;) how about this: rmail2-read.table(textConnection(item, min, int_1, int_2, max a, 2.5, 3, 4, 5.5 b, 2, 3.5, 4, 4.5 c, 3.5, 4, 4.5, 5),header=T,sep=,) with(rmail2,symbols(item, (int_1+ int_2)/2, boxplots=cbind(.25, int_2-int_1, int_1-min,max-int_2,0),inches=F,ylim=range(rmail2[,-1]),xaxt=n,ylab=)) axis(1,seq_along(rmail2$item),labels=rmail2$item) hth. Am 28.01.2013 22:26, schrieb Denis Francisci: Hi all, I'm new on this list so I greet all. My question is: does exist in R a plot similar to candlestick plot but not based on xts (time series)? I have to plot a range of 4 value: for every item I have min value, max value and 2 intermediate values. I would like plot this like a candlestick, i.e. with a box between 2 intermediate values and 1 segment between box and min value and a segment between box and max value. Candlestick plot is provided by quantmod package, but it is very specific for financial purpose and it uses time series for x axis. I need a simpler method for plotting my data that are stored in a table like this: item, min, int_1, int_2, max a, 2.5, 3, 4, 5.5 b, 2, 3.5, 4, 4.5 c, 3.5, 4, 4.5, 5 . Does anyone know if there is an R-plot for this purpose? Thank you very much, D. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] a confidence interval from an ANOVA model
Hi Erin, how about this: dta-read.table(textConnection(trt resp A 23 A 28 A 37 A 30 B 37 B 44 B 31 B 35 C 42 C 47 C 52 C 38),header=T) summary(ano-aov(resp~trt,dta)) ano0-update(ano,.~.+0) confint(ano0) # and the corresponding means are coef(ano0) # or use the effects package library(effects) summary(allEffects(ano)) hth. Am 20.01.2013 03:11, schrieb Erin Hodgess: Dear R People: I have a data set with trt and resp: trt resp A 23 A 28 A 37 A 30 B 37 B 44 B 31 B 35 C 42 C 47 C 52 C 38 And I did read.table, constructed an aov object and ran the summary on the aov object. I also ran the TukeyHSD function. All is well so far. My question is: is there a way to put together a confidence interval from the aov object on a particular mean, please? It seems like there should be. Thanks, Have a great night/day, Sincerely, Erin -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Hello R User
Hi Bibek, how about this? dta-read.table(textConnection(ID Time 1 3 1 6 1 7 1 10 1 16 2 12 2 18 2 19 2 25 2 28 2 30),header=T) dta$delta-with(dta,ave(Time,ID,FUN=function(x)c(0,diff(x dta hth. Am 14.12.2012 16:51, schrieb bibek sharma: Hello R User, In the sample data given below, time is recorded for each id subsequently. For the analysis, for each id, I would like to set 1st recorded time to zero and thereafter find the difference from previous time. I.e. for ID==1, I would like to see Time=0,3,1,3,6. This needs to be implemented to big data set. Any suggestions are much appreciated! Thanks, Bibek IDTime 1 3 1 6 1 7 1 10 1 16 2 12 2 18 2 19 2 25 2 28 2 30 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] error in IF condition with factor evaluation
Hi Edoardo,
there is a difference between comparisons and assignments, both
semantically as well as in R syntax: == vs = or -, latter being
more obvious an assignment.
This is the source of your error.
But to change the labels of a factor object, it is easier to do sth like
at-factor(1:5,labels=letters[1:5])
at
levels(at)[3]-xyz
at #check
hth
Am 23.11.2012 10:42, schrieb edoardo baldoni:
Cam anyone tell me why the condition x[i] == DISCONECTED looks like
producing an NA instead of TRUE/FALSE
I would like to rename DISCONNECTED those factors inside the variable
dataset$STATUS.x that are named DISCONECTED
thank you
summary(dataset$STATUS.x)
ACTIVE DISCONECTED PENDING SUSPENDED TERMINATED
158869 16918130288565 47233
NA's
6
class(dataset$STATUS.x)
[1] factor
fff = function(x) {
+ for (i in 1:length(x)){
+ if (x[i] == DISCONECTED) {
+ x[i] == DISCONNECTED
+ } else x[i] == x[i]
+ }
+ return(x)
+ }
r = fff(dataset$STATUS.x)
Error in if (x[i] == DISCONECTED) { :
missing value where TRUE/FALSE needed
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Martin Zeitz (Vorsitzender), Dr. Alexander
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Re: [R] How can I improve an ugly, dumb hack
Hi Bert, maybe I'm missing the point, but dd-cbind(d,m) does 1, 2 and 3 as desired: n - data.frame(nx=letters[7:9], ny = 7:9) str(cbind(d,m,n)) t-letters[7:9] str(cbind(d,t)) cheers. Am 06.09.2012 18:03, schrieb Bert Gunter: Hi Folks: Here's the situation: m - cbind(x=letters[1:3], y = letters[4:6]) m x y [1,] a d [2,] b e [3,] c f ## m is a 2 column character matrix d - data.frame(a=1:3,b=4:6) d$c - m d a b c.x c.y 1 1 4 a d 2 2 5 b e 3 3 6 c f ## But please note (as was remarked in a thread here a couple of months ago) ncol(d) [1] 3 ## d is a ** 3 ** column data frame Now what I wish to do is programmatically convert d to a 4 column frame with names c(a,b,x,y). Of course: 1. The column classes/modes must be preserved (character going to factor and numeric remaining numeric). 2. I assume that I do not know a priori which of d's components/columns are matrices and which are vectors. 3. There may be many more columns which are vectors or matrix than just the three in this little example. I can easily and sensibly accomplish these 3 tasks, but the problem is that I run afoul of data frame column naming procedures in doing so, about which the data.frame Help page says rather enigmatically: How the names of the data frame are created is complex, and the rest of this paragraph is only the basic story. Indeed! (This, of course, is shorthand for Go look at the source if you want to know! ) Anyway, AFAICT from the Help, any simple approach to conversion using data.frame results in c.x and c.y for the names of the last two columns. I **can** get what I want by explicitly constructing the vector of names via the following ugly hack; my question is, can it be improved? dd - do.call(data.frame,d) dd a b c.x c.y 1 1 4 a d 2 2 5 b e 3 3 6 c f ncol(dd) [1] 4 cnames - sapply(d,colnames) cnames $a NULL $b NULL $c [1] x y names(dd) - unlist(ifelse(sapply(cnames,is.null),names(d),cnames)) ##Yuck! dd a b x y 1 1 4 a d 2 2 5 b e 3 3 6 c f Cheers to all, Bert -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] define subset argument for function lm as variable?
Hi Rainer, I got an error while replicating your data.frame construction. But this worked for me ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) Cheers Am 21.08.2012 17:11, schrieb Rainer M Krug: On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] define subset argument for function lm as variable?
Josh's solution should be fine. just one more note, trt0 may not work as intended since trt is a factor in your example. Just check subset(dat,trt0) wich is just 'dat'. Am 21.08.2012 17:41, schrieb Rainer M Krug: On 21/08/12 17:35, Eik Vettorazzi wrote: Hi Rainer, I got an error while replicating your data.frame construction. But this worked for me ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) dat - data.frame( group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) Sorry - to much garbage in my workspace left. Just add: rm(ctl) rm(trt) after the definition lm(weight ~ group, data=dat, subset=trt0) this obviously works, but I would like to have: subst - trt0 lm(weight ~ group, data=dat, subset=subst) Sorry about this, Rainer Cheers Am 21.08.2012 17:11, schrieb Rainer M Krug: On 21/08/12 16:57, Bert Gunter wrote: ?? I do not groc what you mean. ... subset == subs would work fine in your lm call. So unless someone else does get it, you may need to elaborate. OK - here is an example: dat - data.frame( ctl = c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14), trt = c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69), group = gl(2,10,20, labels=c(Ctl,Trt)), weight = c(ctl, trt) ) lm(weight ~ group, data=dat, subset=trt0) subst - trt0 ### here I get the obvious error: Error: object 'trt' not found # I want to use: lm(weight ~ group, data=dat, subset=subst) In general, ?substitute, ?bquote, and ?quote are useful to avoid immediate evaluation of calls, but I don't know if that's relevant to what you want here. Looks promising from the help, but I don't get it to work. Rainer -- Bert On Tue, Aug 21, 2012 at 7:44 AM, Rainer M Krug r.m.k...@gmail.com wrote: Hi I want to do a series of linear models, and would like to define the input arguments for lm() as variables. I managed easily to define the formula arguments in a variable, but I also would like to have the subset in a variable. My reasoning is, that I have the subset in the results object. So I wiould like to add a line like: subs - dead==FALSE recTreat==FALSE which obviously does not work as the expression is evaluated immediately. Is is it possible to do what I want to do here, or do I have to go back to use dat - subset(dat, dead==FALSE recTreat==FALSE) ? dat - loadSPECIES(SPECIES) feff - height~pHarv*year # fixed effect in the model reff - ~year|plant # random effect in the model, where year is the dat.lme - lme( fixed = feff, # fixed effect in the model data = dat, random = reff, # random effect in the model correlation = corAR1(form=~year|plant), # subset = dead==FALSE recTreat==FALSE, # na.action = na.omit ) dat.lm - lm( formula = feff, # fixed effect in the model data = dat, subset = dead==FALSE recTreat==FALSE, na.action = na.omit ) Thanks, Rainer -- Rainer M. Krug, PhD (Conservation Ecology, SUN), MSc (Conservation Biology, UCT), Dipl. Phys. (Germany) Centre of Excellence for Invasion Biology Stellenbosch University South Africa Tel : +33 - (0)9 53 10 27 44 Cell: +33 - (0)6 85 62 59 98 Fax : +33 - (0)9 58 10 27 44 Fax (D):+49 - (0)3 21 21 25 22 44 email: rai...@krugs.de Skype: RMkrug __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Chopping a vector up into smaller vectors
Hi Stephen,
how about this:
chop-function(x,counts){
stopifnot(sum(counts)==length(x))
split(9:1,unlist(mapply(rep,seq_along(counts),counts)))
}
chop(9:1,c(3,2,4))
cheers
Am 02.08.2012 12:29, schrieb Stephen Eglen:
Anyone got a neat way to chop a vector up into smaller subvectors?
This is what I have now, which seems inelegant:
chop - function(v, counts) {
stopifnot(sum(counts)==length(v))
end - cumsum(counts)
beg - c(1, 1+end[-length(end)])
begend - cbind(beg, end)
apply(begend, 1, function(x) v[x[1]:x[2]])
}
chop(9:1, c(3,2,4))
[[1]]
[1] 9 8 7
[[2]]
[1] 6 5
[[3]]
[1] 4 3 2 1
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--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
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Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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Re: [R] Polygon shaded area
Hi Jose, how about this example(polygon) I think the second one is pretty much what you want. cheers. Am 02.08.2012 14:33, schrieb Jose Narillos de Santos: Hi all, I have two vectors (columns) called efinal and efinal 2. I want to plot them on the same plot and draw a shaded area beween the two lines using function polygon I have tried all but I don ´t understand the polygon area, can you help me with examples? plot(efinal,type=l,ylim=range(min(efinal2), max(efinal)),ylab=,main=plot) par(new=T) plot(efinal2,type=l,ylim=range(min(efinal2), max(efinal)),ylab=) The efinal are two temporal series I attach? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Polygon shaded area
Hi Jose, this should work (but I think you need a deeper understanding, what plot(efinal) does, see ?plot): plot(efinal,ylim=range(c(efinal,efinal2)),type=n,ylab=) xv-seq_along(efinal) polygon(c(xv,rev(xv)),c(efinal,rev(efinal2)),col=red) cheers Am 02.08.2012 16:36, schrieb Jose Narillos de Santos: Hi I attach my function. No error message (it seems a line appears but nothing similar to examples(polygon)) 2012/8/2 Eik Vettorazzi e.vettora...@uke.de mailto:e.vettora...@uke.de Hi Jose, how about this example(polygon) I think the second one is pretty much what you want. cheers. Am 02.08.2012 14:33, schrieb Jose Narillos de Santos: Hi all, I have two vectors (columns) called efinal and efinal 2. I want to plot them on the same plot and draw a shaded area beween the two lines using function polygon I have tried all but I don ´t understand the polygon area, can you help me with examples? plot(efinal,type=l,ylim=range(min(efinal2), max(efinal)),ylab=,main=plot) par(new=T) plot(efinal2,type=l,ylim=range(min(efinal2), max(efinal)),ylab=) The efinal are two temporal series I attach? __ R-help@r-project.org mailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 tel:%2B%2B49%2F40%2F7410-58243 F ++49/40/7410-57790 tel:%2B%2B49%2F40%2F7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] apply function over same column of all objects in a list
I still try to figure out, what you are finally asking for, but maybe
res-sapply(list4,function(x)max(x[[1]]$coconut))
names(res)-c(mango, banana, pineapple)
res
max(res)
is worth a try?
Especially i did not get the point of doing something like
x[which.max(x)] because this is in any case just max(x)
if you would like to plug in different functions or want to select
different columns,
fun-max # or min, median etc
sapply(lapply(list4,[[,1),fun)
could be a start, where lapply(list4,[[,1) extracts the first object
in each sublist - which is in your case a data.frame with just one
column ('coconut'), where 'fun' can be applied to. So to address
'coconut' explicitly, just nest this lapply approach:
sapply(lapply(lapply(list4,[[,1),[[,coconut),fun)
#or use an anonymous extractor function
sapply(lapply(list4,function(x)x[[1]][[coconut]]),fun)
cheers
Am 02.08.2012 16:09, schrieb gail:
Hi,
my dataset is the result of the function density on another set of data.
It refers to that data in its variable call, though since all the results
are actually reproduced (except that I've removed all rows bar 10), I am not
sure why R still needs it. But I've understood now why your code doesn't
work on my data: your sublists are still just dataframes, whereas my
sublists contain all sorts of vectors (2 numerical vectors and 5 other
things). I've changed your data to make it look a bit more like mine, and
you'll see that the code doesn't work.
mango - list(data.frame(coconut=1:6), y=c(4,5,3,2,1,8),
bw=c(4,18,12,3,4,9), nn=paste0(a,1:6))
banana - list(data.frame(coconut=c(1,2,18,16,15)), y=c(4,5,9,2,1),
bw=c(4,18,22,3,4), nn=paste0(a,1:5))
pineapple - list(data.frame(coconut=c(4,6,9)), y=c(8,24,12),
bw=c(14,31,36), nn=paste0(a,1:3))
list4 - list(mango, banana, pineapple)
b - list()
for(i in 1:3){
b[[i]] - list()
b[[i]] - lapply (list4[[i]]$coconut, FUN=function(x) x[which.max(x)])
}
b
summary( b)
What I want to end up with is the following:
coconut
mango 6
banana 18
pineapple 9
(that's for the individual sublists)
and
coconut
18
(that's for the list as a whole).
Hope this is becoming a bit clearer ...
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--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] apply function over same column of all objects in a list
Hi Arun,
if you name the data.frame in the list, e.g.
mango - list( y=c(4,5,3,2,1,8), bw=c(4,18,12,3,4,9),
nn=paste0(a,1:6),target=data.frame(coconut=1:6))
banana - list(target=data.frame(coconut=c(1,2,18,16,15)), y=c(4,5,9,2,1),
bw=c(4,18,22,3,4), nn=paste0(a,1:5))
pineapple - list(target=data.frame(coconut=c(4,6,9)), y=c(8,24,12),
bw=c(14,31,36), nn=paste0(a,1:3))
list6-list(mango,banana,pineapple)
then
sapply(lapply(list5,[[,target),max)
still works.
Regards
Am 02.08.2012 20:21, schrieb arun kirshna [via R]:
Hi Elk,
I tried to test with another case where coconut is in different position in
the sublist.
mango - list( y=c(4,5,3,2,1,8), bw=c(4,18,12,3,4,9),
nn=paste0(a,1:6),data.frame(coconut=1:6))
banana - list(data.frame(coconut=c(1,2,18,16,15)), y=c(4,5,9,2,1),
bw=c(4,18,22,3,4), nn=paste0(a,1:5))
pineapple - list(data.frame(coconut=c(4,6,9)), y=c(8,24,12),
bw=c(14,31,36), nn=paste0(a,1:3))
list5-list(mango,banana,pineapple)
#Your code
fun-max # or min, median etc
sapply(lapply(list5,[[,1),fun)
[1] 8 18 9
# Here, first number should be 6, instead it did the max of y from mango.
#Then, I tried second solution of yours
sapply(lapply(lapply(list5,[[,1),[[,coconut),fun)
#Error in FUN(X[[1L]], ...) : subscript out of bounds
#Third solution
sapply(lapply(list5,function(x)x[[1]][[coconut]]),fun)
#Error in x[[1]][[coconut]] : subscript out of bounds
#I tried with another way to come up with the result (not as elegant)
b3-do.call(c,list5)
names(b3)[4]-mango
names(b3)[5]-banana
names(b3)[9]-pineapple
b4-data.frame(key=names(b3),value=unlist(lapply(b3,FUN=function(x)
max(x
key1-c(mango,banana,pineapple)
b5-subset(b4, b4$key%in% key1)
b5
#key value
#4 mango 6
#5banana18
#9 pineapple 9
b6-within(b5,{value-as.numeric(as.character(value))})
max(b6$value)
#[1] 18
A.K.
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Re: [R] help with a regression problem
Hi,
maybe working with a data.frame in long format is an option - then you
can use e.g. lmList and so on up to mixed models, depending on your
final goals of analyses (e.g. check for differential slopes).
vmat-matrix(c(X1,X2,X3,X4,Y1,Y2,Y3,Y4),nrow=2,byrow=T)
aa.l-reshape(aa,idvar=ID,direction=long,varying=vmat,v.names=c(X,Y))
library(nlme)
(ll-lmList(Y~X|ID,aa.l,na.action=na.omit))
summary(ll)
cheers.
Am 01.08.2012 15:06, schrieb R Heberto Ghezzo, Dr:
Hello,
I have a big data frame where consecutive time dates and corresponding
observed values for each subject (ID) are on a line. I want to compute the
linear slope for each subject. I would like to use apply but I do
not know how to express the corresponding function. An example using a loop
follows
#
# create dummy data set There are missing values
a - c(1,2,3,4, 1,1,1,1, 2,2,3,3, 3,4,NA,4, 5,5,5,5,
2.1,2.2,2.3,2.4, 2.3,2.4,2.6,2.6, 2.5,2.6,2.9,3,
2.6,NA,3.2,4)
a - matrix(a, nr=4)
aa - as.data.frame(a)
names(aa) - c(ID,X1,X2,X3,X4,Y1,Y2,Y3,Y4)
#
# I want the regression coefficientes of the Y on the X for each ID
#
sl - rep(NA,4)
for(i in 1:4) {
x1 - a[i,2:5]
y1 - a[i,6:9]
sl[i] - lm(y1 ~ x1)$coef[2]
}
sl
#
# I would like to use apply on the data.frame aa but with which function?
#
sl - apply(aa,1,FUN) # FUN = ??
#
Thanks for any help
R.Heberto Ghezzo Ph.D.
Montreal - Canada
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Re: [R] Optimize a function with Discrete inputs
Hi,
not sure if that is what you are looking for, but have a look at
cmb-t(combn(c(0,3,5,8),2)) #get all pairs of combinations
cbind(cmb,apply(cmb,1,diff)) #for each pair, get the difference
cheers
Am 01.08.2012 12:29, schrieb loyolite270:
Hi
I need to optimize the below function:
a=function(x){
A=x[1]
B=x[2]
C=B-A
return(C)
}
I need to optimize the above function such that x can be any combination of
these number (0,3,5,8) of vector length 2
(i.e) x can be (3,0), (5,0), (8,0), (3,5), (3,8), (5,8), .. etc
can someone please help me solve this problem ?
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Re: [R] Variables in a Tabular form. easily saved in a txt file
Hi Alex, sprintf with left alignment might be a start: cat(sprintf(%-7s| %-12d| %-5d,paste(City,1:3,sep=),(12345)%/%10^c(1:3),c(2,5,54433)),sep=\n) #cat has also a file option, see ?cat. but acutally your given output doesn't look like a conventional table but much more like this: cat(sprintf(paste(%-7s| %-,12:14,d| %-5d,sep=),paste(City,1:3,sep=),(12345)%/%10^c(1:3),c(2,5,54433)),sep=\n) cheers. Am 26.07.2012 13:29, schrieb Alaios: Dear all, I would like to save few variable-names with their values in a tabular form, with that I mean that files can be printed easily in R in a tabular form and also saved in a ascii file that when one opens it see also the variables in a nice tabular format. IS that possible? Below a small example of how should look results in R and in a txt file. Postal Code | Superb City1 | 2134 | 2 City2 | 254| 5 City3 | 12 | 54433 I am ok if you can give me some hard coded example to try. Regards Alex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] package memisc: recode examples
Hi Marion, the Hmisc package has also a function called recode, maybe you loaded this package after memisc so the memisc version was masked (you should have got a warning message about that). Or perhaps you wrote you own recode function? Try find(recode,mode=function) to see all search list entries. You can access the desired function regardless of the place in the search path by memisc::recode(...) hth. Am 24.07.2012 13:53, schrieb Marion Wenty: Dear people, Yesterday I looked at the recode command in the memisc package and ran the following example stated in the manual: x - as.item(sample(1:6,20,replace=TRUE), labels=c( a=1, b=2, c=3, d=4, e=5, f=6)) print(x) f - as.factor(x) f recode(f, A=c(a,b), B=c(c,d),otherwise=copy ) I also used this command for my dataset and it work - I didn't get an error message. Today I have tried several times running this example, also with my dataset, and always got the same error message (in German): Fehler in recode(f, A = c(a, b), B = c(c, d), otherwise = copy) : unbenutzte(s) Argument(e) (A = c(a, b), B = c(c, d), otherwise = copy) (Error in recode ... unused argument(s) ...) I also couldn't find anything in the internet. I don't understand why this is not working anymore. Does anyone know what the problem is? Thank you very much for your help in advance! Marion [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Maintaining Column names while writing csv file.
Hi Vincy, have you checked names(df2) and names(df_new) because by default 'data.frame' checks the column names to ensure that they are syntactically valid variable names and 1w and 2w aren't, so an X is prepended (see ?data.frame and ?make.names). compare tst.fail-data.frame(a=1:3,1a=1:3,1b=1:3) names(tst.fail) #and tst-data.frame(a=1:3,1a=1:3,1b=1:3,check.names = F) names(tst) this works for me: tst2-data.frame(a=1:3,w1=4:6,w3=4:6) write.csv(merge(tst,tst2),file=testr.csv) hth. Am 19.07.2012 08:55, schrieb Vincy Pyne: Dear R helpers, I have one trivial problem while writing an output file in csv format. I have two dataframes say df1 and df2 which I am reading from two different csv files. df1 has column names as date, r1, r2, r3 while the dataframe df2 has column names as date, 1w, 2w. (the dates in both the date frames are identical also no of elements in each column are equal say = 10). I merge these dataframes as df_new = merge(df1, df2, by = date, all = T) So my new data frame has columns as date, r1, r2, r3, 1w, 2w However, if I try to write this new dataframe as a csv file as write.csv(data.frame(df_new), 'df_new.csv', row.names = FALSE) The file gets written, but when I open the csv file, the column names displayed are as date, r1, r2, r3, X1w, X2w My original output file has about 200 columns so it is not possible to write column names individually. Also, I can't change the column names since I am receiving these files from external source and need to maintain the column names. Kindly guide Regards Vincy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Finding the last value before a certain date
Hi Robert, how about this (assuming your data.frame is ordered by date): tmp-read.table(textConnection(datey 1 2010-09-27 1356 2 2010-10-04 1968 3 2010-10-11 2602 4 2010-10-17 3116 5 2010-10-24 3496 6 2010-10-31 3958),header=T,colClasses=c(date=Date)) tmp[max(which(tmp$dateas.Date(2010-10-06))),y] hth. Am 19.07.2012 09:42, schrieb Robert Latest: Hello all, I have a dataframe that looks like this: head(df) datey 1 2010-09-27 1356 2 2010-10-04 1968 3 2010-10-11 2602 4 2010-10-17 3116 5 2010-10-24 3496 6 2010-10-31 3958 I need a function that, given any date, returns the y value corresponding to the given date or the last day before the given date. Example: Input: as.Date(2010-10-06). Output: 1968 (because the last value is from 2010-10-04) I've been tinkering with this for an hour now, without success. I think the solution is either surprisingly complicated or surprisingly simple. Thanks, robert __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with using apply for dataframe
Hi Marion, as stated in the help file ?apply coerces a data.frame object into an array. Since an array has only one type of data, this coercion turns all your variables into strings (because this data type can hold all information given without loss). If it happens that your data.frame consists only of numeric variables, then this type suffices for the array as well, and things like sum or mean can be applied row/columnwise. btw: apply(mydataframe,2,function(x)sum(x)) doesn't work either (for the same reason). Cheers Am 19.07.2012 11:48, schrieb Marion Wenty: Dear people, I am including an example of a dataframe: mydataframe-data.frame(X=c(1:4),total_bill=c(16.99,10.34,21.01,23.68),tip=c(1.01,1.66,3.50,3.31),sex=c(Male,Male,Male,Female)) When I use the sapply function getting the information about the factors works: sapply(mydataframe,function(x)is.factor(x)) X total_billtipsex FALSE FALSE FALSE TRUE But if I use the apply function it doesn't work: apply(mydataframe,2,function(x)is.factor(x)) X total_billtipsex FALSE FALSE FALSE FALSE I don't understand why, because I have used the apply function for dataframes before e.g. using sum(x) instead of is.factor(x) and it did work. Could anyone help me with this? Thank you very much for your help in advance! Marion [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] duplicate data between two data frames according to row names
Hi Jeff, looks like a job for ?rbind and ?merge merge(rbind(DF1,DF2),DF3) hth Am 18.07.2012 10:21, schrieb jeff6868: Hi everybody. I'll first explain my problem and what I'm trying to do. Admit this example: I'm working on 5 different weather stations. I have first in one file 3 of these 5 weather stations, containing their data. Here's an example of this file: DF1 - data.frame(station=c(ST001,ST004,ST005),data=c(5,2,8)) And my two other stations in this other data.frame: DF2 - data.frame(station=c(ST002,ST003),data=c(3,7)) I would like to add geographical coordinates of these weather stations inside these two data.frames, according to the number of the weather station. All of my geographical coordinates for each of the 5 weather stations are inside another data frame: DF3 - data.frame(station=c(ST001,ST002,ST003,ST004,ST005),lat=c(40,41,42,43,44),lon=c(1,2,3,4,5)) My question is: how can I put automatically these geographical coordinates inside my first 2 data frames, according to the number of the weather station? For this example, the first two data frames DF1 and DF2 should become: DF1 - data.frame(station=c(ST001,ST004,ST005),lat=c(40,43,44),lon=c(1,4,5),data=c(5,2,8)) and DF2 - data.frame(station=c(ST002,ST003),lat=c(41,42),lon=c(2,3),data=c(3,7)) I need to automatize this method because my real dataset contains 70 weather stations, and each file contains other (or same sometimes) stations , but each station can be found in the list of the coordinates file (DF3). Is there any way or any function able to do this kind of thing? Thank you very much! -- View this message in context: http://r.789695.n4.nabble.com/duplicate-data-between-two-data-frames-according-to-row-names-tp4636845.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] contour
Hi Akhil, in addition to Rui's post, here is a solution using lattice graphics. tmp-read.table(textConnection(x1 x2 x3 0 12 0 21 0 35 1 14 1 22 1 33),header=T) library(lattice) contourplot(x3~x1*x2,tmp) hth. Am 18.07.2012 11:23, schrieb Akhil dua: Hello Everyone I have the data long format and I want to draw the contour plot with it x1 x2 x3 0 12 0 21 0 35 1 14 1 22 1 33 when I am using contour(x1,x2,x3,col=heat.colors) or fill.contour its giving me an error that increasing x and y expected So please tell me what is the right function to draw contour when the data is not ordered and you cant order it. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to create table with file link in Rd File
Hi, I think \tabular does what you want, see see http://cran.r-project.org/doc/manuals/R-exts.html#Lists-and-tables hth. Am 17.07.2012 14:45, schrieb purushothaman: Hi, i need to create table like this --- module name class namefunction name fun description --- CSV Functions CSVoperations http://readcsv.hml Read csv used to read input csv file Thanks B.Purushothaman -- View this message in context: http://r.789695.n4.nabble.com/how-to-create-table-with-file-link-in-Rd-File-tp4636740p4636742.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] I need to get function name from R file
Hi,
how about this:
sandbox-new.env()
sys.source(test.R,envir=sandbox)
#all objects defined in test.r
(tstf-ls(env=sandbox))
#just the functions
tstf[sapply(tstf, function(n) exists(n, envir = sandbox, mode =
function, inherits = FALSE))]
hth
Eik
Am 11.07.2012 15:22, schrieb purushothaman:
Hi,
i need to get functions name from R file
example
test.R file having 2 functions like
add-function()
{
a+b
}
sub-function()
{
a-b
}
how to get function name from test.R file
i need output is functions are add,sub.
Thanks
B.Purushothaman
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--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Import single variables from SPSS
Hi Marion, package memisc does what you want, just have a look at ?importers. Hth Eik Am 05.07.2012 15:05, schrieb Marion Wenty: Dear all, I have got a very big SPSS dataframe and I would like to just import one or a few variables (the dataframe has got a lot of colums). I used the package foreign but I couldn't find anything in there for my problem. Thank you very much for your help in advance. Marion [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] inverse binomial in R
Hi Anna,
you are at the upper end of the support of the respective binomial
distribution when asking for invbinomial(n,n,p). The probability of
seeing n or less successes in n trials is 1, regardless of the
underlying success rate. It cannot be less, so asking for
invbinomial(50,50, 0.6) makes no sense at all - neither does the result
of stata.
You may insert some exception handling in the function by throwing a
warning or error, when this special case occurs - or even return 0 to
cope the erroneous stata result.
cheers
Am 19.06.2012 11:39, schrieb anna freni sterrantino:
Hi Duncan and Rlist,
I've notice a different behaviour in the invbinomial
you suggest me and invbinomial in stata.
invbinomial(50,50, 0.4)
Error in uniroot(function(x) pbinom(k, n, x) - p, c(0, 1)) :
f() values at end points not of opposite sign
invbinomial(50,50, 0.6)
Error in uniroot(function(x) pbinom(k, n, x) - p, c(0, 1)) :
f() values at end points not of opposite sign
while stata
gen p3=invbinomial(50,50, 0.4)
. display p3
0
. gen p4=invbinomial(50,50, 0.6)
. display p4
0
Thanks
Cheers
Anna
Da: Duncan Murdoch murdoch.dun...@gmail.com
Cc: Rcran help r-help@r-project.org
Inviato: Giovedì 31 Maggio 2012 15:32
Oggetto: Re: [R] inverse binomial in R
On 12-05-31 9:10 AM, anna freni sterrantino wrote:
Hello!
I'm having some trouble
trying to replicate in R a Stata function
invbinomial(n,k,p)
Domain n: 1 to 1e+17
Domain k: 0 to n - 1
Domain p: 0 to 1 (exclusive)
Range:0 to 1
Description: returns the inverse of the cumulative binomial; i.e.,
it
returns the probability of success on one trial
such
that the probability of observing floor(k) or
fewer
successes in floor(n) trials is p.
I've found some hints on the web like
http://rwiki.sciviews.org/doku.php?id=guides:tutorials:regression:table
I tried to replicate using qbinom
the results obtained in
invbinomial(10,5, 0.5)
.54830584
but with no success.
I don't think base R has a function like that, though some contributed
package probably does. If you're writing it yourself you'd need to use
uniroot or some other solver, e.g
invbinomial - function(n, k, p) {
uniroot(function(x) pbinom(5, 10, x) - p, c(0, 1))
}
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20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
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Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
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Re: [R] Day or Month difference between dates???
Hi Tammy, I suspect that the first 4 digits compose the year, so with Year_Month-c(2010*100+9:11,2011*100+1:2) #check composition Year_Month ym-(Year_Month%/%100*12+Year_Month%%100) ym-ym[1] should give the difference in months. Differences in Days are a bit trickier, since not all months have an equal number of days. Something like xd-as.Date(paste(Year_Month%/%100,Year_Month%%100,01,sep=-)) xd-xd[1] deals with that. Cheers Am 01.06.2012 13:31, schrieb Tammy Ma: HI, R-Users: I got a questions. have been struggling so long time I have this data: m1$Year_Month 201009 201010 201011 201101 201102 min(m1$Year_Month) 201009 I want to calculate the following two answers, how do I program it? difference in Month? [1] 0 1 2 4 5 difference in Days? 0 31 61 Thank you in advance!!! Tammy [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] \Sexpr{}
Hi Hua,
try
options(SweaveSyntax=SweaveSyntaxNoweb)
before sweaving your file (see A.14 in the FAQ section of the manual).
In most cases R2HTML interferes with Sweave, causing your problem
hth.
Am 02.04.2012 15:12, schrieb Liang, Hua:
Did anyone know whether the problem it doesn't evaluate \Sexpr{} in Sweave
has been fixed? I checked r help page and realized it was asked a couple of
years ago. But I didn't find solutions.
Thanks for your help in advance!
Hua
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--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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Re: [R] Test Normality
Hi Sindy, you might try Snows penultimate normality test from the TeachingDemos package. But read the help file carefully. http://www.inside-r.org/packages/cran/TeachingDemos/docs/SnowsPenultimateNormalityTest cheers. Am 28.03.2012 02:32, schrieb Sindy Carolina Lizarazo: Good Night I made different test to check normality and multinormality in my dataset, but I don´t know which test is better. To verify univariate normality I checked: shapiro.test, cvm.test, ad.test, lillie.test, sf.test or jaque.bera.test and To verify multivariate normal distribution I use mardia, mvShapiro.Test, mvsf, mshapiro.test, mvnorm.e. I have a dataset with almost 1000 data and 9 variables, in both cases the result is non-normality. For this reason, I transformed data with bcPower function and I want to check normality again. I really appreciate your help. Thanks. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot points using circles filled half in red and half in blue.
Hi Alan, on an UTF-8 locale you can use \u25D6 and \u25D7, but you have to plot group a=4 twice. On Windows in a non-utf locale you can use a special Windows font: windowsFonts(wdg2=windowsFont(Wingdings 2)) plot(0:1,0:1) strh-strheight(x)/2.02 points(x = 0.5, y = 0.75+strh/2,pch=¼,family=wdg2,col=red) points(x = 0.5, y = 0.75-strh/2,pch=½,family=wdg2,col=green) strw-strwidth(º) points(.4-strw/2,.7,pch=º,family=wdg2,col=red) points(.4+strw/2,.7,pch=»,family=wdg2,col=green) Cheers Am 28.03.2012 04:49, schrieb alan: I want to plot many points and want to use circles. The filling color depends on variable a. if a=1, then not fill if a=2 then fill with red, if a=3 then fill with blue, if a=4, fill half with red and half with blue. Can anyone tell me how to plot the case a=4? Thanks a lot __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What are the color's name in heat.color(5)
Hi, just one more note, col2rgb works for color codes as well: (HC.m - col2rgb(heat.colors(5))) and to convert named colors to color codes, something like tabl-apply(col2rgb(colors()),2,function(x)do.call(rgb,c(as.list(x),alpha=255,maxColorValue=255))) names(tabl)-colors() tabl should work. cheers Am 28.03.2012 15:51, schrieb David Winsemius: On Mar 28, 2012, at 9:08 AM, Kevin Wright wrote: They have hex RGB values instead of names: R heat.colors(5) [1] #FFFF #FF5500FF #FFAA00FF #00FF #80FF Kevin The first one is the same as 'red' as can be seen by parsing the RGB values above: col2rgb(red) [,1] red255 green0 blue 0 HC.m matrix( c( strtoi( substr(heat.colors(5), 2,3), 16L), strtoi(substr(heat.colors(5), 4,5), 16L), strtoi(substr(heat.colors(5), 6,7), 16L) ), nrow=3, byrow=TRUE) HC.m [,1] [,2] [,3] [,4] [,5] [1,] 255 255 255 255 255 [2,]0 85 170 255 255 [3,]0000 128 One of the others 'yellow' is also in named colors: which( apply(col2rgb(colors()) , 2 , function(x) all( x == c(HC.m[,2]))) ) integer(0) which( apply(col2rgb(colors()) , 2 , function(x) all( x == c(HC.m[,3]))) ) integer(0) which( apply(col2rgb(colors()) , 2 , function(x) all( x == c(HC.m[,1]))) ) [1] 552 553 which( apply(col2rgb(colors()) , 2 , function(x) all( x == c(HC.m[,4]))) ) [1] 652 653 which( apply(col2rgb(colors()) , 2 , function(x) all( x == c(HC.m[,5]))) ) integer(0) colors()[552: 553] [1] red red1 colors()[652:653] [1] yellow yellow1 -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SPSS R-Menu for Ordinal Factor Analysis
Hi Till, you need a plug-in for SPSS to run R, which is bound to a fixed R version. SPSS 19 uses R2.10 and SPSS 20 uses R2.12 (both not up-to-date R versions). So you have to download this plugin (after registering on IBM Support) for your spss version and the according R version - or start with a brand new R2.14 ;) Cheers. Am 26.03.2012 13:32, schrieb Till Below: Dear all, I am trying to conduct an enhanced version of factor analysis with a SPSS interface that allows to use R. This approach has been suggested in the recent article: Basto, M. and J.M. Pereira An SPSS R-Menu for Ordinal Factor Analysis. Journal of Statistical Software 46, pp. 1-29. My variables are ordinal-type and the tool of Basto allows to run polychoric correlations in the SPSS environment with the R essentials ans Plugin. I used R 2.10 version for spss 20 (I only have access to spss 20). I loaded the required packages. When running the factor analysis in spss with the plugin I receive the error message posted below. Aparently, the tool of Basto runs only with spss 19. Since the packages and the plugin are installed I think it must be a problem of the spss version. Could someone suggest me a solution for the problem? I am not familiar with R, and so I would prefer to run the factor analysis in spss. Is there any possebility to run it with spss V. 20? Best regards and compliments for your work, Till Below (PhD student, Humboldt Universität zu Berlin, Germany) SPSS output with error message: GET FILE='C:\X. DATASET NAME DatenSet1 WINDOW=FRONT. DATASET ACTIVATE DatenSet1. SAVE OUTFILE='C:\XX /COMPRESSED. *Mário Basto, José Manuel Pereira, IPCA *Required: SPSS 19 and R Integration Plugin *R Packages required: psych, polycor, GPArotation, nFactors, corpcor, ICS. set printback off. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Best way to compute the difference between two levels of a factor ?
Hi Sylvain, assuming your data frame is ordered by ID and TIME, how about this aggregate(cbind(X,Y)~ID,data, function(x)(x[2]-x[1])) #or doing this for all but the first 2 columns of data: aggregate(data[,-(1:2)],by=list(data$ID), function(x)(x[2]-x[1])) cheers. Am 21.03.2012 09:48, schrieb wphantomfr: Dear R-help Members, I am wondering if anyone think of the optimal way of computing for several numeric variable the difference between 2 levels of a factor. To be clear let's generate a simple data frame with 2 numeric variables collected for different subjects (ID) and 2 levels of a TIME factor (time of evaluation) data=data.frame(ID=c(AA,AA,BB,BB,CC,CC),TIME=c(T1,T2,T1,T2,T1,T2),X=rnorm(6,10,2.3),Y=rnorm(6,12,1.9)) ID TIME X Y 1 AA T1 9.959540 11.140529 2 AA T2 12.949522 9.896559 3 BB T1 9.039486 13.469104 4 BB T2 10.056392 14.632169 5 CC T1 8.706590 14.939197 6 CC T2 10.799296 10.747609 I want to compute for each subject and each variable (X, Y, ...) the difference between T2 and T1. Until today I do it by reshaping my dataframe to the wide format (the columns are then ID, X.T1, X.T2, Y.T1,Y.T2) and then compute the difference between successive columns one by one : data$Xdiff=data$X.T2-data$X.T1 data$Ydiff=data$Y.T2-data$Y.T1 ... but this way is probably not optimal if the difference has to be computed for a large number of variables. How will you handle it ? Thanks in advance Sylvain Clément __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] constant function
Hi Chris, how about this: constant-function(x) rep(7,length(x)) curve(constant,-1000,100) cheers. Am 07.03.2012 06:41, schrieb Chris Waggoner: Is it possible to define a constant function (for example the zero function) in R? I found the following unsatisfying hack to create the seven function: constant - stepfun( x = -100:100, y = rep(7, 201) ) This is mathematically unsatisfying because the domain is artificially restricted to [-100,100] rather than (-Inf,Inf). Additionally, unless I use pch=n there are ugly dots on the plot (and again this feels hackish). Any suggestions on what else to do? Thanks in advance. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Aggregate with Function List ?
Hi Michael,
something like the following might be a starting point for aggregating
data using arbitrary lists of functions:
(it is lacking a method for data.frame objects)
maggregate-function(...)UseMethod(maggregate)
maggregate.default-function(x, FUN, ...){
tmp-lapply(FUN,function(fct)aggregate(x,FUN=fct,...))
names(tmp)-sapply(substitute(FUN), deparse)[-1]
r2-data.frame(tmp[[1]][,1],sapply(tmp,[,-1))
names(r2)[1]-names(tmp[[1]])[1]
r2
}
maggregate.formula-function(formula, data, FUN, ..., subset, na.action
= na.omit){
tmp-lapply(FUN,function(fct)aggregate(formula,data,fct,...,na.action =
na.action))
names(tmp)-sapply(substitute(FUN), deparse)[-1]
r2-data.frame(tmp[[1]][,1],sapply(tmp,[,-1))
names(r2)[1]-names(tmp[[1]])[1]
r2
}
#using formula method
maggregate(Sepal.Length~Species,data=iris,FUN=c(mean,median,sd))
maggregate(Sepal.Length~Species,data=iris,FUN=list(mean,quantile))
#check if parameters are passed to quantile
maggregate(Sepal.Length~Species,iris,FUN=list(mean,quantile),probs=c(.25,.5,.75))
maggregate(iris$Sepal.Length,by=list(iris$Species),FUN=list(mean,quantile))
Cheers
Am 23.02.2012 19:41, schrieb Michael Karol:
R Experts
I wish to tabulate into one data frame statistics summarizing
concentration data. The summary is to include mean, standard
deviation, median, min and max. I wish to have summaries by Dose, Day
and Time. I can do this by calling aggregate once for each of the
statistics (mean, standard deviation, median, min and max) and then
execute 4 merges to merging the 5 data frames into one. (Example
aggregate code for mean only is shown below.)
Can someone show me the coding to do this as one command, rather than
5 calls to aggregate and 4 merges. In other words, in essence, I'd like
to present to FUN = a list of functions, so all the summary stats come
back in one data frame. Your assistance is appreciated. Thank you.
MeansByDoseDayTime - aggregate(as.double(DF$Concentration), by =
list(DF$Dose, DF$Day, DF$Time), FUN = mean, trim = 0, na.rm = T,
weights=NULL)
Regards,
Michael
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--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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Re: [R] Loop
Hi Florian,
'yet that doesn't work' is an improper question on this list, see the
posting guide.
Besides that, something like
set-vector(mode = list, length = 10)
for (i in 1:10){
set[[i]] - complete(imp,i)}
or saving some typing
set-lapply(1:10,function(i)complete(imp,i))
should work.
cheers
Am 22.02.2012 11:32, schrieb Florian Weiler:
Dear all,
I have a (probably very basic) question. I am imputing data with the mice
package, using 10 chains. I can then write out the 10 final values of the
chains simply by
name1 - complete(imp, 1)
:
:
name10 - complete(imp,10)
Not a big deal, I just wanted to do that in a little loop as follows:
for (i in 1:10){
set[i] - complete(imp,i)}
Yet that doesn't work, I also tried other things like:
for (i in 1:10){
set[[i]] - complete(imp,i)}
Again, no success. It only saves a couple of lines of code, but there must
be an easy solution, right?
Thanks,
Florian
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Re: [R] Using substitute in nested function calls
Hi Sebastian,
how about this:
mycurve - function (expr) {
do.call(curve,list(substitute(expr),-100,100))
}
mycurve(x^2)
mycurve(sin(x/20))
mycurve(x+x^2-x^3)
cheers
Am 21.02.2012 06:28, schrieb Sebastian Kranz:
Dear List members,
I really, like the feature that one can call R functions with
mathematical expressions, e.g.
curve(x^2, 0, 1)
I wonder, how I can construct in a simple way a function like
mycurve = function (expr) {...}
such that that a call
mycurve(x^2)
has the same effect as the call
curve(x^2, -100,100)
Below is some code that works, but it seems much to complicated: it
first substitutes and deparses the expression, creates a string with the
new function call and then parses and evaluates the string again. Does
anybody know a simpler, more elegant solution?
mycurve = function(expr) {
# The following attempt does not work
# curve(substitute(expr),-100,100)
# Transform original expression to a string
org.expr = deparse(substitute(expr))
# Construct a string, parse it and evaluates it
eval(parse(text=paste(curve(,org.expr,,-100,100),sep=)))
}
mycurve(x^2)
Best regards,
Sebastian
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T ++49/40/7410-58243
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Re: [R] Run function several times changing only one argument - without a loop
Hi Marion, you can either use any of the *apply-functions or vectorize your function (which internally uses mapply): par(las=1) par(mar=c(5,13,4,2)) barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F) #using sapply invisible(sapply((1:9)*10,function(x)axis(2,pos=x,tick=T, tcl=F, labels=F,col=white))) #using Vectorize barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F) vaxis-Vectorize(axis,pos) invisible(vaxis(2,pos=(1:9)*10, tick=T, tcl=F, labels=F,col=white)) Cheers! Am 20.02.2012 11:04, schrieb Marion Wenty: Dear people, I created a plot which looks like this: Ee1-matrix(c(88,86,74,62,41),ncol=5) colnames(Ee1)-c(Lehrer,Lehrerinnen,Klassenkollegen,Klassenkolleginnen,Geschwister) par(las=1) par(mar=c(5,13,4,2)) barplot(Ee1,horiz=T,col=grey85,border=NA,xlim=c(0,100),axes=F) axis(2,pos=10, tick=T, tcl=F, labels=F,col=white) axis(2,pos=20, tick=T, tcl=F, labels=F,col=white) axis(2,pos=30, tick=T, tcl=F, labels=F,col=white) axis(2,pos=40, tick=T, tcl=F, labels=F,col=white) axis(2,pos=50, tick=T, tcl=F, labels=F,col=white) axis(2,pos=60, tick=T, tcl=F, labels=F,col=white) axis(2,pos=70, tick=T, tcl=F, labels=F,col=white) axis(2,pos=80, tick=T, tcl=F, labels=F,col=white) axis(2,pos=90, tick=T, tcl=F, labels=F,col=white) Now I would like to shorten the whole thing - namely use only one step to create the 9 axes without having to use a loop. In general, I would be interested if there is a way to use a function several times changing only one argument, without having to use a loop. Does anyone know how to do that. Marion [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Toggle cASE
Hi Antonio, how about this: txt2 - useRs may fly into JFK or laGuardia chartr(a-zA-Z, A-Za-z, txt2) Cheers. Am 05.12.2011 08:11, schrieb Tonio: Hello R-help list, I am looking for way to toggle the case of the characters like a flip-flop; that is from ''Hello'' to hELLO or vice versa. I know that there are a number of functions like casefold, tolower, toupper, etc. but these functions change the case in an uniform way. Thanks in advance, Antonio Rivero Ostoic Antonio Rivero Ostoic PhD Student, Department of Leadership and Strategy From 1 Jul until 31 Dec 2011 Tel.+61 3 8344 4300 Fax +61 3 9347 6618 Email j...@sdu.dk Addr. The University of Melbourne, Victoria 3010 Australia - UNIVERSITY OF SOUTHERN DENMARK Sdr. Stationsvej 28 · DK-4200 Slagelse · Denmark · Tel. +45 6550 1000 · www.sdu.dk [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] query for xlim/ylim of current plot
Hi Lukas, does par(usr) help? Cheers. Am 31.10.2011 11:34, schrieb eldor ado: der R-listers, does anyone of you know whether there is a function which returns the xlim/ylim ranges of the current plot (i would like to use coordinates to place some text via a function, regardless of the x/y lims of the plot). best regards, lukas kohl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] textplot in layout
Hi Ben,
maybe mtext is of more help here?
par(mar=c(7,3,3,3))
plot(year,rate,main='main',sub='sub')
mtext('test',cex=1,side=1,line=5)
box()
cheers
Am 25.10.2011 15:26, schrieb Ben quant:
Hello,
Someone (Erik) recently posted about putting text on a plot. That thread
didn't help. I'd like to put text directly below the 'sub' text (with no
gap). The code below is the best I can do. Note the large undesirable gap
between 'sub' and 'test'. I'd like the word 'test' to be just below the top
box() boarder (directly below 'sub').
year - c(2000 , 2001 , 2002 , 2003 , 2004)
rate - c(9.34 , 8.50 , 7.62 , 6.93 , 6.60)
op - par(no.readonly = TRUE)
on.exit(par(op))
layout(matrix(c(1,2), 2, 1, byrow = TRUE),heights=c(8,1))
par(mar=c(5,3,3,3))
plot(year,rate,main='main',sub='sub')
library(gplots)
par(mar=c(0,0,0,0),new=F)
textplot('test',valign='top',cex=1)
box()
Note: I'd rather solve it with textplot. If not, my next stop is
grid.text(). Also, the text I am plotting with textplot is much longer so a
multiple line text plot would solve my next issue (of which I have not
looked into yet). Lastly, layout is not necessary. I just used it because I
thought it would do what I wanted.
Thanks,
Ben
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--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
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Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
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Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
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Re: [R] Plotting text?
Hi Kevin, this should be read as halign=center - so is it a typo just in your mail or in your program as well? apart from that, the rules at the bottom lines of every post on this list also apply here: what have you tried and what went wrong? Cheers Am 24.10.2011 14:48, schrieb Kevin Burton: Thank you. This works pretty well. I am having some trouble with the text on the left-hand side getting cut off. I have tried haling=center with not luck. How can I make the text seem wider than it really is to avoid the truncation. It is only a few characters but still it is annoying. Thank you. -Original Message- From: Eik Vettorazzi [mailto:e.vettora...@uke.de] Sent: Saturday, October 22, 2011 7:06 AM To: rkevinbur...@charter.net Cc: r-help@r-project.org Subject: Re: [R] Plotting text? Hi Kevin, have a look at ?textplot from the gplots-package. cheers Am 22.10.2011 02:26, schrieb rkevinbur...@charter.net: I noticed that the text() command adds text to a plot. Is there a way to either make the plot blank or add text to a blank sheet. I would like to plot a page that contains just text, no plot lines, labels, etc. Suggestions? Kevin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] splitting a string into words preserving blanks (using regex)
Hi Mark, here is a way using gsub to insert a split marker and strsplit. strsplit(gsub(([[:alnum:]]+),|\\1|,c( somewords to split ))[[1]] cheers Am 24.10.2011 15:46, schrieb Mark Heckmann: I would like to split a string into words at its blanks but also to preserve all blanks. Example: c( somewords to split ) should become c( , some,, words, , to , , split, ) I was not able to achieve this via strsplit() . But I am not familiar with regular expressions. Is there an easy way to do that using e.g. regex and strsplit? Thanks Mark ––– Mark Heckmann Blog: www.markheckmann.de R-Blog: http://ryouready.wordpress.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Plotting text?
Hi Kevin, have a look at ?textplot from the gplots-package. cheers Am 22.10.2011 02:26, schrieb rkevinbur...@charter.net: I noticed that the text() command adds text to a plot. Is there a way to either make the plot blank or add text to a blank sheet. I would like to plot a page that contains just text, no plot lines, labels, etc. Suggestions? Kevin [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to remove all objects except the sequence
Hi Sergio,
how about this:
rm(list=setdiff(ls(),z))
cheers.
Am 20.10.2011 11:00, schrieb Sergio René Araujo Enciso:
Dear All:
I would like to know if there is plausible way to say to R to remove all
elements in the memory but the sequence. I have a code which makes a loop,
and what I want is after the programme has performed all the operation over
every ith element, to remove all the objects, expect the sequence
parameter. I included the option rm(list=ls(all=TRUE)), but obviously that
removes the sequence as well. I know that I can give the names of the
objects to remove, but instead of doing so I would like to tell R remove all
but the sequence. Is there a way for doing so? The reason for removing all
the objects after each operation is saving some memory, as the operation I
am doing involves some bootstrapping, after the loop reaches a certain ith
element, the operations start to be really slow. So I want to faster the
loop by removing the objects and free memory after every operation.
Below is my code:
setwd(C:\\Dokumente und Einstellung\\.)
library(tsDyn)
z-(1:1000)### sequence parameter
sink(prueba.txt)
for (i in seq(z))
{
P1-read.csv(2R_EQ_P_R1.csv)
P2-read.csv(2R_EQ_P_R2.csv)
c-data.frame(P1[i],P2[i])
c.t-ts(c)*-1
try(print(z[i]))
try(SeoTest-TVECM.SeoTest(c.t, lag=1, beta=1, trim=0.1, nboot=100))
try(summary(SeoTest))
rm(list=ls(all=TRUE)) ### Here I want to erase all but z
}
best,
Sergio René
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--
Eik Vettorazzi
Department of Medical Biometry and Epidemiology
University Medical Center Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
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Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
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Re: [R] how to plot two surfaces with lattice::wireframe
Hi Carl,
I have no idea what z or f(z) are, but maybe outer will help you:
wireframe(outer(seq(0,5,length.out=50),seq(2,4,length.out=40),function(x,y)sin(x*y)))
cheers.
Am 13.10.2011 23:37, schrieb Carl Witthoft:
Hi all,
I'd like to plot the Real and Imaginary parts of some f(z) as two
different surfaces in wireframe (the row/column axes are the real and
imag axes). I know I can do it by, roughly speaking, something like
plotz - expand.grid(x={range of Re(z)}, y={range of Im(z), groups=1:2)
plotz$func-c(Re(f(z),Im(f(z))
wireframe(func~x*y,data=plotz,groups=groups)
But that seems like a clunky way to go, especially if I happen to have
started out with a nice matrix of the f(z) values.
So, is there some simpler way to write the formula in wireframe? I
envision, for a matrix of complex values zmat, pseudocode:
wireframe(c(Re(zmat),Im(zmat), groups=1:2)
-- and yes, I'm fully aware that without a connection between the
'groups' variable and zmat, this won't work as written.
All suggestions (including read the help file for {some lattice func I
didn't know about} ) greatfully accepted.
Carl
--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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Re: [R] Remove specific rows in a matrix/data.frame
Hi Syrvn, how about this dtf-read.table(textConnection(Letter Number a 1 a 1 b 1 b 0 c 0 c 1 d 0 d 0),header=T) aggregate(Number~Letter,data=dtf,max) cheers. Am 13.10.2011 18:42, schrieb syrvn: Hi, imagine the following matrix/data.frame Letter Number a 1 a 1 b 1 b 0 c 0 c 1 d 0 d 0 If the numbers for two identical letters are also identical then I want to remove either the first or the second row of that letter. If for a letter the numbers are 1 and 0 I want to remove the row with the 0. That means if the code works I would and up with the following matrix/data.frame Letter Number a 1 b 1 c 1 d 1 Many thanks, Syrvn -- View this message in context: http://r.789695.n4.nabble.com/Remove-specific-rows-in-a-matrix-data-frame-tp3902149p3902149.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Remove specific rows in a matrix/data.frame
Hi, just put it in the formula: aggregate(Number ~ Letter+Test,data=dtf,max) cheers Am 13.10.2011 19:30, schrieb syrvn: Hello again, dtf-read.table(textConnection(Letter Test Number a b 1 a b 1 b b 1 b b 0 c b 0 c b 1 d b 0 d b 0),header=T) aggregate(Number ~ Letter,data=dtf,max) how can I adjust this solution that the results also includes Test? I tried: aggregate(Number ~ Letter,data=dtf,max,by=list(Letter, Test, Number)) But it breaks with the following error message: Error in aggregate.data.frame(mf[1L], mf[-1L], FUN = FUN, ...) : arguments must have same length -- View this message in context: http://r.789695.n4.nabble.com/Remove-specific-rows-in-a-matrix-data-frame-tp3902149p3902286.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Parallel processing for R loop
Hi Sandeep, still missing an answer? Perhaps you cross check your post with the rules of the posting guide and find what is missing at all here. Anyway, depending on your OS, package multicore, snow/snowfall may fit your needs - but you have to re-formulate your loop using adequate multicore *apply-functions. hth Am 11.10.2011 14:13, schrieb Sandeep Patil: I have an R script that consists of a for loop that repeats a process for many different files. I want to process this parallely on machine with multiple cores, is there any package for it ? Thanks -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to test if two C statistics are significantly different?
Hi Yujie, there is still a lot of work in progress, I think. As http://faculty.washington.edu/heagerty/Software/SurvROC/RisksetROC/risksetROCdiscuss.pdf states: [...] for inference and variance estimation, we now suggest bootstrapping [...]. Recently I catched a glimpse on roc.test from the pROC package, they implemented, amongst others, a bootstrap algorithm - maybe this is a start for your own work? Hth. Am 10.10.2011 21:35, schrieb Yujie Wang: Hey all, In order to test if a marker is a risk factor, I built two models (using cox proportional hazard model). One model included this marker, and the other is not. Then, I use R package risksetROC to test how much predictive value did the marker add to this model. I get two C statistics by analyzing the linear predictors of the two models into this package. The qustion is How to test if two C statistics are significantly different? Your help will be greatly appreciated! Yujie [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] need help on read.spss
Hi, if you specify to.data.frame=T, then use.missings is implictly set to T as well, which causes different results for (user-defined) missing values. cheers. Am 11.10.2011 12:07, schrieb Smart Guy: Hi, I have one doubt about one of the parameter of 'read.spss()' from 'foreign' package. Here is the syntax :- read.spss ( file, use.value.labels = TRUE, to.data.frame = FALSE, max.value.labels = Inf, trim.factor.names = FALSE, trim_values = TRUE, reencode = NA, use.missings = to.data.frame ) In above syntax when I pass *'to.data.frame= FALSE*' it gives me missing values from SPSS file (that I try to read using read.spss() ). But when I pass '*to.data.frame = TRUE*' then its not giving me missing values. And need to get missing values. According to read.spss() documentation *to.data.frame : return a data frame?* I am curious to know, if we pass *'to.data.frame = TRUE*' , is it going to cause some issue or effect something? I didn't understand the read.spss() documentation correctly. Please explain. Thanks in Advance -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] anova.rq {quantreg) - Why do different level of nesting changes the P values?!
Hi Tal, you are comparing different things. The Details-section of ?anova.qr states test the hypothesis that smaller models are adequate relative to the largest specified model. So in your first anova you compare fit2 with fit1 and fit0, in your second attempt fit1 with fit0, so you have different base-models to compare with, and consequently different p-values. hth. Am 06.10.2011 10:05, schrieb Tal Galili: Hello dear R help members. I am trying to understand the anova.rq, and I am finding something which I can not explain (is it a bug?!): The example is for when we have 3 nested models. I run the anova once on the two models, and again on the three models. I expect that the p.value for the comparison of model 1 and model 2 would remain the same, whether or not I add a third model to be compared with. However, the P values change, and I do not understand why. Here is an example code (following with it's input): data(barro) fit0 - rq(y.net ~ lgdp2 + fse2 , data = barro) fit1 - rq(y.net ~ lgdp2 + fse2 + gedy2 , data = barro) fit2 - rq(y.net ~ lgdp2 + fse2 + gedy2 + Iy2 , data = barro) anova(fit0,fit1,fit2, R = 1000) anova(fit0,fit1, R = 1000) Output: data(barro) fit0 - rq(y.net ~ lgdp2 + fse2 , data = barro) fit1 - rq(y.net ~ lgdp2 + fse2 + gedy2 , data = barro) fit2 - rq(y.net ~ lgdp2 + fse2 + gedy2 + Iy2 , data = barro) anova(fit0,fit1,fit2, R = 1000) Quantile Regression Analysis of Deviance Table Model 1: y.net ~ lgdp2 + fse2 + gedy2 + Iy2 Model 2: y.net ~ lgdp2 + fse2 + gedy2 Model 3: y.net ~ lgdp2 + fse2 Df Resid Df F valuePr(F) 1 1 156 29.494 2.110e-07 *** 2 2 156 18.194 7.901e-08 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 anova(fit0,fit1, R = 1000) Quantile Regression Analysis of Deviance Table Model 1: y.net ~ lgdp2 + fse2 + gedy2 Model 2: y.net ~ lgdp2 + fse2 Df Resid Df F value Pr(F) 1 1 157 3.9532 0.04852 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 sessionInfo() R version 2.13.1 (2011-07-08) Platform: i386-pc-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=Hebrew_Israel.1255 LC_CTYPE=Hebrew_Israel.1255 [3] LC_MONETARY=Hebrew_Israel.1255 LC_NUMERIC=C [5] LC_TIME=Hebrew_Israel.1255 attached base packages: [1] splines stats graphics grDevices utils datasets methods base other attached packages: [1] rms_3.3-1Hmisc_3.8-3 survival_2.36-9 [4] colorspace_1.1-0 quantreg_4.71SparseM_0.89 [7] PerformanceAnalytics_1.0.3.2 xts_0.8-2zoo_1.7-4 [10] reporttools_1.0.6xtable_1.5-6 loaded via a namespace (and not attached): [1] cluster_1.14.0 grid_2.13.1 lattice_0.19-33 tools_2.13.1 Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] converting commas to points
Hi Anna, have a look at ?write.csv2, which deals with the Excel conventions for CSV in German locale. cheers Am 06.10.2011 17:39, schrieb Anna Lee: Hello everyone! I work with a german excell version which uses commas instead of points for seperating decimal places. R work with points so in order to be able to save my excell tables without changing the commas to points, whenever I load a table I type in: read.table(..., dec = ,) only R puts the points into the wron places. For excample excell has a cell with the number: 0,09 so what R does, it writes the number as 0.9 which is wrong and makes my calculations become useless. Does anyone know this problem? Maby I made a mistake somewhere? I would be glad about your answers! Cheers, Anna -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with regexp
Hi Jannis,
just use the backreferences in gsub, see ?gsub, - replacement
test - c('filename_1_def.pdf', 'filename_2_abc.pdf')
gsub(.*_([A-z]+)\\.pdf, \\1, test)
hth.
Am 05.10.2011 13:56, schrieb Jannis:
Dear list memebers,
I am stuck with using regular expressions.
Imagine I have a vector of character strings like:
test - c('filename_1_def.pdf', 'filename_2_abc.pdf')
How could I use regexpressions to extract only the 'def'/'abc' parts of these
strings?
Some try from my side yielded no results:
testresults - grep('(?=filename_[[:digit:]]_).{1,3}(?=.pdf)', perl = TRUE,
value = TRUE)
Somehow I seem to miss some important concept here. Until now I always used
nested sub expressions like:
testresults - sub('.pdf$', '', sub('^filename_[[:digit:]]_', '' , test))
but this tends to become cumbersome and I was wondering whether there is a
more elegant way to do this?
Thanks for any help
Jannis
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and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr.
Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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Re: [R] unload a library while testing?
Hi Rainer, for better or worse unlibrary actually is done by detach in R, ?detach #first example cheers Am 05.10.2011 15:04, schrieb Rainer M Krug: Hi I am testing a package, and after I make changes, I have to close R and open R again to load the new version (same version number) of the package I am working on. So my question: is there a function which removes a package, i.e library(myPackage) Package is loaded unlibrary(myPackage) package is not loaded any more Thanks, Rainer -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Guido Sauter (Vertreter des Vorsitzenden), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] String manipulation with regexpr, got to be a better way
Hi Chris,
why not using routines for dates
dates - c(09/10/2003, 10/22/2005)
format(strptime(dates,format=%m/%d/%Y),%Y)
or take just the last 4 chars from dates
gsub(.*([0-9]{4})$,\\1,dates)
cheers
Am 29.09.2011 16:23, schrieb Chris Conner:
Help-Rs,
I'm doing some string manipulation in a file where I converted a string date
in mm/dd/ format and returned the date .
I've used regexpr (hat tip to Gabor G for a very nice earlier post on this
function) in steps (I've un-nested the code and provided it and an example of
what I did below. My question is: is there a more efficient way to do this.
Specifically is there a way to use regexpr or some other string function to
return not the first instance, but the 2nd (or for that matter 3rd, 4th or
5th instance) of a certain string?
#first find the first occurence of / and create a variable for this
firstslash - unlist(regexpr(/, dates, fixed = TRUE)) #then use frist/ to
cut the string field into an intermediate variable e.g., from 1/1/2008 to
1/2008. step1 - substr( dates, (firstslash + 1), nchar(dates) ) #then
repeat steps 1 and 2...there's got to be a better way step2 -
unlist(regexpr(/, step1, fixed = TRUE)) #then use step2 to cut string into
final product e.g., from 1/2008 to 2008. final - substring(step1,step2 + 1,
nchar(step1) )
Thx!
C
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Jörg F. Debatin (Vorsitzender), Dr. Alexander
Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Plotting Lines Through multiple groups
Hi Clara, are there repeated measures of an animal on the same day? Otherwise I did not get the point of the level of aggregation. Using lattice you can do sth like xyplot(Cort~Day,groups=Animal,type=c(p,a),data=df) but with your given data this just connects the raw points hth. Am 28.09.2011 08:03, schrieb clara_eco: Hi I have data in the following format Cort Day Animal 230 1 273 1 240 2 271 2 342 2 303 2 244 2 200 3 241 3 282 3 344 3 etc. It is measured across time(day) however no every individual is measured the same number of times. All I want to do is plot the Raw data and then run a line connecting the data points of each individual animal, for the example above there would be three lines as there are three animals. I've tried, gplots, lattice and car but I'm not really figuring it out, any help would be greatly appreciated! Thanks Clara -- View this message in context: http://r.789695.n4.nabble.com/Plotting-Lines-Through-multiple-groups-tp3850099p3850099.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Jörg F. Debatin (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with predict and lines in plotting binomial glm
Hi Anina, predict.glm returns predicted probabilities, when used with type=response, so you have to either scale the probs to the number of trials for any x or you plot probs from start: par(mfcol=c(1,2)) plot(x, successes) lines(x, (successes+failures)*predict(glm1, type= response), lwd=2) plot(x, successes/(successes+failures)) lines(x, predict(glm1, type= response), lwd=2) If you use predict with new data, please name the rows accordingly, when you actually want to use them: newdf - data.frame(seq(min(x), max(x), length=20)) #you would expect 20 predictions, but *surprise* you get length(predict(glm2, type=response, newdata= newdf)) #52! newdf - data.frame(x=seq(min(x), max(x), length=20)) length(predict(glm2, type=response, newdata= newdf)) #ok And you should use the appropriate x in lines as well: plot(x,successes/(successes+failures)) lines(newdf$x, predict(glm2, type=response, newdata= newdf),col=red) Cheers Am 21.09.2011 09:55, schrieb Heystek, A, Me 15418...@sun.ac.za: Problems with predict and lines in plotting binomial glm Dear R-helpers I have found quite a lot of tips on how to work with glm through this mailing list, but still have a problem that I can't solve. I have got a data set of which the x-variable is count data and the y-variable is proportional data, and I want to know what the relationship between the variables are. The data was overdispersed (the residual deviance is much larger than the residual degrees of freedom) therefore I am using the quasibinomial family, thus the y-variable is a matrix of successes and failures (20 trials for every sample, thus each y-variable row counts up to 20). x - c(1200, 1200, 1200, 1200, 1200, 1200, 1200, 1200, 1800, 1800, 1800, 1800, 1800, 1800, 1800, 1800, 1800, 2400, 2400, 2400, 2400, 2400, 2400, 2400, 3000, 3000, 3600, 3600, 3600, 3600, 4200, 4200, 4800, 4800, 5400, 6600, 6600, 7200, 7800, 7800, 8400, 8400, 8400, 9000, 9600, 10200, 13200, 18000, 20400, 24000, 25200, 36600) successes - c(6, 16, 11, 14, 11, 16, 13, 13, 14, 16, 12, 12, 11, 15, 12, 9, 7, 7, 17, 15, 13, 9, 9, 12, 14, 8, 9, 16, 7, 9, 14, 11, 8, 8, 13, 6, 16, 11, 9, 7, 9, 8, 4, 14, 7, 3, 3, 9, 12, 8, 4, 6) failures - c(14, 4, 9, 6, 9, 4, 7, 7, 6, 4, 8, 8, 9, 5, 8, 11, 13, 13, 3, 5, 7, 11, 11, 8, 6, 12, 11, 4, 13, 11, 6, 9, 12, 12, 7, 14, 4, 9, 11, 13, 11, 12, 16, 6, 13, 17, 17, 11, 8, 12, 16, 14) y - cbind(successes, failures) data - data.frame(y, x) glm1 - glm(y ~ x, family= quasibinomial, data= data) glm2 - glm(y ~ log(x), family=quasibinomial, data= data) # residual deviance is lower with log transformed x-value plot(x, successes) lines(x, predict(glm1, type= response), lwd=2) Firstly, because of the skewed distribution of the x variable I am not sure whether it should be log transformed or not. When I do log transform it, the residual deviance and the p-value for the slope is lower. Either way, the lines command does not plot any line and neither does it give any error messages. On some of my other data it plots a line way below all the data points. From what I can gather, the predict function as it is now uses the fitted values because no newdata argument is specified. I want the line to be predicted from the same x-values. I tried two ways of adding the newdata argument: ## a data.frame using the original x-values lines(x, predict(glm2, type= response, newdata= as.data.frame(x))) ## or a data.frame with values (the same length as y) from the range of x values newdf - data.frame(seq(min(x), max(x), length=52)) lines(x, predict(glm2, type=response, newdata= newdf)) Only the second option plotted a line once, but then I could never get it to do the same again on a new plot even though I used the same variables and same code. Thank you very much for your time and patient Anina Heystek BSc Honours student Department of Botany and Zoology University of Stellenbosch, Stellenbosch, South Africa 15418...@sun.ac.za [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Jörg F. Debatin (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https
Re: [R] pasting elements of one character vector together
Hi marion, just transpose the matrix: mm-matrix(LETTERS[1:20],nrow=5) paste(t(mm),collapse=) or - if you want the result seperated by rows apply(mm,1,paste,collapse=) cheers Am 20.09.2011 11:55, schrieb Marion Wenty: I have another question concerning the paste command: now instead of a vector I would like to paste the elements of a matrix together, which works in the same: Mypastedmatrix - paste(Mymatrix,collapse=) My problem now is that the program does this BY COLUMN, but I would like to have the elements pasted together BY ROW. Could anybody help me with this? Marion 2011/9/19 Marion Wenty marion.we...@gmail.com hello michael and dimitris, yes, this was it! now i understand this collapse-argument. thank you very much. marion 2011/9/19 Dimitris Rizopoulos d.rizopou...@erasmusmc.nl Try this: object - c(Hello, World) paste(object, collapse = ) I hope it helps. Best, Dimitris On 9/19/2011 3:58 PM, Marion Wenty wrote: hello, i am familiar with the paste command with which i can paste for exaple: object- Hello paste(object,World) now i would like to be able to paste all the elements of the same vector together e.g: object- c(Hello,World) getting as a result also: Hello World. Does anyone know the solution to this problem? Thank you very much in advance. Marion [[alternative HTML version deleted]] __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 Web: http://www.erasmusmc.nl/**biostatistiek/http://www.erasmusmc.nl/biostatistiek/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Jörg F. Debatin (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Possible or not possible: serif axis labels with plotmath [but everything else sans serif]?
Hi Jan Marius, using the tikzDevice-package, nearly everything is possible (at least, all what can be done in LaTeX). cheers Am 19.09.2011 11:58, schrieb Hofert Jan Marius: Dear expeRts, I it possible to have serif labels in the following plot? x - 1:10 y - x plot(x, y, type=b, xlab=expression(x[1]), ylab=expression(x[2])) I know that one can use pdf(, family=serif), but then also the axis tick marks are printed in serif font. Apart from the fact that it may not look nice, I'm just interested if one can have serif axis labels but everything else in sans serif (default). Cheers, Marius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Jörg F. Debatin (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Possible or not possible: serif axis labels with plotmath [but everything else sans serif]?
Hi Jan Marius, I had a lot of positive experience with tkiz, but yes, handling large datasets can push it to the limits. Anyway, in your case, isn't it as simple as plot(x, y, type=b,xlab=,ylab=) title(xlab=expression(x[1]), ylab=expression(x[2]),family=serif) cheers Am 19.09.2011 14:51, schrieb Hofert Jan Marius: Dear Eik, although possible in this case, tikzDevice is certainly not a general solution to all kinds of problems :-) I used it for quite some time before I gave up: I had a simple bar plot, the bars being black. This already caused errors like TeX capacity exceeded ... and I obtained these a lot. In fact, enlarging the TeX capacity (not trivial but possible) did not solve these issues. That's why I gave up on this package [although, clearly, the idea of full TeX support is totally appealing -- that's why I looked at the package in the first place]. Cheers, Marius On 2011-09-19, at 14:38 , Eik Vettorazzi wrote: Hi Jan Marius, using the tikzDevice-package, nearly everything is possible (at least, all what can be done in LaTeX). cheers Am 19.09.2011 11:58, schrieb Hofert Jan Marius: Dear expeRts, I it possible to have serif labels in the following plot? x - 1:10 y - x plot(x, y, type=b, xlab=expression(x[1]), ylab=expression(x[2])) I know that one can use pdf(, family=serif), but then also the axis tick marks are printed in serif font. Apart from the fact that it may not look nice, I'm just interested if one can have serif axis labels but everything else in sans serif (default). Cheers, Marius __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Jörg F. Debatin (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus ETH Zurich Dr. Marius Hofert RiskLab, Department of Mathematics HG E 65.2 Rämistrasse 101 8092 Zurich Switzerland Phone +41 44 632 2423 marius.hof...@math.ethz.ch http://www.math.ethz.ch/~hofertj -- Eik Vettorazzi Department of Medical Biometry and Epidemiology University Medical Center Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 -- Pflichtangaben gemäß Gesetz über elektronische Handelsregister und Genossenschaftsregister sowie das Unternehmensregister (EHUG): Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts; Gerichtsstand: Hamburg Vorstandsmitglieder: Prof. Dr. Jörg F. Debatin (Vorsitzender), Dr. Alexander Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Replace a for loop with a function
Hi Iris,
maybe I misinterpret this, but I think in the end it all comes down to
sample(1:3,50,prob=c(0.5,0.15,0.35),replace=T))
cheers
Am 19.09.2011 17:16, schrieb Eekhout, I.:
Hi all,
I would like to replace the for loop in the code below with a function
to improve the speed and to make the script more efficient.
The loop creates a vector of integers (x) with the probability of f for
each integer.
The length of f is variable, but sums to 1.
I tried to use a function with optional arguments which did not work.
Here is the code:
f - data.matrix(c(0.5,0.15,0.35))
u - runif(50)
x - data.matrix(rep(1, n))
fc - 0
for(i in 1:length(f)) {
fc - fc + f[i]
cf - ifelse(ufc,1,0)
x - x + cf
}
x
Can anyone help me with this translation?
Thanks in advance,
Iris
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--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
--
Pflichtangaben gemäß Gesetz über elektronische Handelsregister und
Genossenschaftsregister sowie das Unternehmensregister (EHUG):
Universitätsklinikum Hamburg-Eppendorf; Körperschaft des öffentlichen Rechts;
Gerichtsstand: Hamburg
Vorstandsmitglieder: Prof. Dr. Jörg F. Debatin (Vorsitzender), Dr. Alexander
Kirstein, Joachim Prölß, Prof. Dr. Dr. Uwe Koch-Gromus
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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