[R] normalité test using over identifying moment conditions
Hi, I'm trying to interpret the following results, with respect to normality test using the over identifying moment conditions where returns have normal distribution with parameter mu,sd and i have 4 moment conditions E[r-mu/sd]=0 E[(r-mu)^2/sd-1]=0 E[(r-mu)^3/sd^3]=0 E[(r-mu)^4/sd^4-3]=0 output.. gel(g = g, x = returns, tet0 = c(f3$estimate[1], f3$estimate[2])) Type of GEL: EL Coefficients: Estimate Std. Error t value Pr(|t|) mean -0.01168 0.05614-0.20805 0.83519 sd 1.77591 0.0396544.79218 0.0 Lambdas: Estimate Std. Error t valuePr(|t|) Lambda[1] -0.097430.03912-2.490280.01276 Lambda[2]0.657280.0244326.905050.0 Lambda[3]0.032470.01304 2.489610.01279 Lambda[4] -0.109540.00407 -26.904230.0 Over-identifying restrictions tests: degrees of freedom is 2 statistics p-value LR test 2.3341e+02 2.0730e-51 LM test 7.2417e+02 5.5954e-158 J test 7.2417e+025.5954e-158 Convergence code for the coefficients: 0 Convergence code for the lambdas: 0 does the J-test p-value rejecting the null E[g(theta,x)]=0, and which moment condition is true under normality [[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.
[R] ks.test and wilcoxon.test results differ from other stat.packages
Probably, it's an obvious info, but I have not found anything in R FAQ concerning this feature/bug. The results of ks.test and wilcoxon.test (in the Mann-Whitney version, paired = 'FALSE') don't coincide with the results from the other statistical packages, e.g. Statistica, Medcalc, and (as for MW test) from the numerous online MW tests. E.g. Statistica p-value=0.0435353 Medcalc p-value=0.0435354 R p-value=0.04635 If I want to obtain result of test once, it does not matter. But what should I do if I want to perform Monte-Carlo simulations and I need in 1 or even 100 000 p-values and then will build some distribution and then use results of Statictica? Whtever, the descrepancy bothers. Are there some alternative packages for non-parametric statistics that produce results comparable with the other program packages? Or, probably, there is exists some environment variable to perform calculations in more common way? Thank you! Sasha. __ 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] difftime() out by 1 hour
I have a problem with results from difftime being 1 hour different than
expected. 2 examples are given below:
datetime - matrix(data=rbind(c(2012-03-31 21:00:00, 2012-04-01 00:00:00,
2012-04-01 03:00:00, 2012-04-01 06:00:00),
c(2012-10-06 21:00:00, 2012-10-07 00:00:00, 2012-10-07
03:00:00, 2012-10-07 06:00:00)), ncol=4, nrow=2)
for (row in 1:2) {
x - datetime[row,]
x1=x[1:length(x)-1]
x2=x[2:length(x)]
xd=difftime(x2,x1)
print (x)
print (xd)
}
[1] 2012-03-31 21:00:00 2012-04-01 00:00:00 2012-04-01 03:00:00
2012-04-01 06:00:00
Time differences in hours
[1] 3 4 3
attr(,tzone)
[1]
[1] 2012-10-06 21:00:00 2012-10-07 00:00:00 2012-10-07 03:00:00
2012-10-07 06:00:00
Time differences in hours
[1] 3 2 3
attr(,tzone)
[1]
Accurate results would be 3 3 3 (hours) in both cases.
The problem occurs randomly, but only around midnight.
All the dates between 2012-04-01 06:00:00 and 2012-10-06 21:00:00 are fine.
I have seen postings about problems with POSIXlt and POSIXct accuracy
because of truncation rather than rounding of floating point representations.
The response was that it was expected behaviour.
So my question is this expected behaviour, have I used difftime incorrectly,
and if not, is there a set of date/time functions and classes that use
integer representation to give more accurate results? I am new to R and I may
well be using the wrong set of functions.
Garry
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[R] Filter according to the latest data
Hello together, i have a data.frame, like this one: No. Change Date A 123 final2013-01-15 B 123 error 2013-01-16 C 123 bug fixed 2013-01-17 D 111 final2013-01-12 and now a want a new data.frame which includes only the newest entry for each number. The solution look like this: No. Change Date C 123 bug fixed 2013-01-17 D 111 final2013-01-12 is there any way to filter my data.frame to the latest data, perhabs max? Thanks. Mat -- View this message in context: http://r.789695.n4.nabble.com/Filter-according-to-the-latest-data-tp4657248.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.
Re: [R] difftime() out by 1 hour
Hello,
Here is the result I get using your script:
sessionInfo()
R version 2.15.2 (2012-10-26)
Platform: x86_64-suse-linux-gnu (64-bit)
[1] 2012-03-31 21:00:00 2012-04-01 00:00:00 2012-04-01 03:00:00
[4] 2012-04-01 06:00:00
Time differences in hours
[1] 3 3 3
attr(,tzone)
[1]
[1] 2012-10-06 21:00:00 2012-10-07 00:00:00 2012-10-07 03:00:00
[4] 2012-10-07 06:00:00
Time differences in hours
[1] 3 3 3
attr(,tzone)
[1]
Regards,
Pascal
Le 01/02/2013 16:27, Boots a écrit :
I have a problem with results from difftime being 1 hour different than
expected. 2 examples are given below:
datetime - matrix(data=rbind(c(2012-03-31 21:00:00, 2012-04-01 00:00:00, 2012-04-01
03:00:00, 2012-04-01 06:00:00),
c(2012-10-06 21:00:00, 2012-10-07 00:00:00, 2012-10-07
03:00:00, 2012-10-07 06:00:00)), ncol=4, nrow=2)
for (row in 1:2) {
x - datetime[row,]
x1=x[1:length(x)-1]
x2=x[2:length(x)]
xd=difftime(x2,x1)
print (x)
print (xd)
}
[1] 2012-03-31 21:00:00 2012-04-01 00:00:00 2012-04-01 03:00:00 2012-04-01
06:00:00
Time differences in hours
[1] 3 4 3
attr(,tzone)
[1]
[1] 2012-10-06 21:00:00 2012-10-07 00:00:00 2012-10-07 03:00:00 2012-10-07
06:00:00
Time differences in hours
[1] 3 2 3
attr(,tzone)
[1]
Accurate results would be 3 3 3 (hours) in both cases.
The problem occurs randomly, but only around midnight.
All the dates between 2012-04-01 06:00:00 and 2012-10-06 21:00:00 are fine.
I have seen postings about problems with POSIXlt and POSIXct accuracy
because of truncation rather than rounding of floating point representations.
The response was that it was expected behaviour.
So my question is this expected behaviour, have I used difftime incorrectly,
and if not, is there a set of date/time functions and classes that use
integer representation to give more accurate results? I am new to R and I may
well be using the wrong set of functions.
Garry
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Re: [R] use name (not values!) of a dataframe inside a funktion
Hi
Maybe others can help you but here is my comment. I already use R for many
years and never used such construction. Objects in global environment shall not
be modified by functions it is a bad practice. Imagine you have some data frame
you prepared and controlled in many steps and use some function from a package.
If this function scrambles object without warning and the result cannot be
easily repaired I would be tempted to curse the function with many nasty four
letter words.
When I want to change some data by a function I use
fff - function(x, no=2) {
x[,no]-factor(x[, no])
x
}
dfb.f - fff(dfb)
In this case I will end with old object dfb and new object dfb.f. Of course it
is possible to use
dfb - fff(dfb)
and in this case dfb object is changed
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Winfried Moser
Sent: Thursday, January 31, 2013 3:35 PM
To: r-help
Subject: [R] use name (not values!) of a dataframe inside a funktion
Dear Listers,
can anyone help me, please.
Since several days i try to figure out, how to assign values, vectors,
functions etc to variables with dynamically generated names inside of
functions.
Sometimes I succeed, but the success is rather arbitrary, it seems. up
to now i don't fully understand, why things like get, assign, - etc
do sometimes work, and sometimes not.
here's one of my daily examples, i am stuck with: Example 1 does work,
but example 2 doesn't?
How kann i tell R, that i want it to expand the string dfb to
dfb[,2]
inside the function.
In the end i want the function to change the second variable of the
dataframe dfb permanently to factor (not just inside the function).
Thanks in advance!
Winfried
Example 1:
dfa - data.frame(a=c(1:4),b=c(1:4))
dfa[,2] - factor(dfa[,2])
is.factor(dfa[,2])
TRUE
Example 2:
dfb - data.frame(a=c(1:4),b=c(1:4))
f.fact - function(x) {x[,2] - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
FALSE
PS: I tried a whole lot of other things like, ...
I really don't know where to keep on searching.
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x)[,2] - factor(get(x)[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x[,2]) - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x)[,2] - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {assign(x[,2], factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
Ungültiges erstes Argument
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {quote(x)[,2], factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Unerwartetes ','
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {
name - paste0(quote(x),[,2])
assign(name, factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
FALSE
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {
name - paste0(get(x),[,2])
assign(name, factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
Falsche Anzahl von Dimensionen
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {
name - paste0(x,[,2])
assign(name, factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
Falsche Anzahl von Dimensionen
ächz ...
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Re: [R] Transforming 4x3 data frame into 2 column df in R
Dear Gundala, Try as.data.frame.table(foo) HTH, Jorge.- On Fri, Feb 1, 2013 at 5:00 PM, Gundala Viswanath wrote: I have the following data frame: foo w x y z n 1.51550092 1.4337572 1.2791624 1.1771230 q 0.09977303 0.8173761 1.6123402 0.1510737 r 1.17083866 1.2469347 0.8712135 0.8488029 What I want to do is to change it into : newdf 1 nw 1.51550092 2 q w 0.09977303 3 r w 1.17083866 4 nx 1.43375725 5 q x 0.81737606 6 r x 1.24693468 7 n y 1.27916241 8 q y 1.61234016 9 r y 0.87121353 10 nz 1.17712302 11 q z 0.15107369 12 rz 0.84880292 Whtat's the way to do it? - G.V. [[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.
Re: [R] facet plot
Hi: Try DF - structure(list(Gene = structure(1:5, .Label = c(NM_001001130, NM_001001144, NM_001001152, NM_001001160, NM_001001176 ), class = factor), T0h = c(68L, 0L, 79L, 1L, 0L), T0.25h = c(95L, 1L, 129L, 1L, 0L), T0.5h = c(56L, 4L, 52L, 2L, 0L), T1h = c(43L, 0L, 50L, 0L, 0L), T2h = c(66L, 1L, 24L, 0L, 0L), T3h = c(62L, 1L, 45L, 1L, 0L), T6h = c(68L, 1L, 130L, 0L, 0L), T12h = c(90L, 4L, 154L, 0L, 0L), T24h = c(63L, 1L, 112L, 0L, 0L), T48h = c(89L, 2L, 147L, 1L, 0L)), .Names = c(Gene, T0h, T0.25h, T0.5h, T1h, T2h, T3h, T6h, T12h, T24h, T48h), class = data.frame, row.names = c(1, 2, 3, 4, 5)) library(reshape2) library(ggplot2) # stack the time-related variables by Gene Dfm - melt(DF, id = Gene) # extract the numeric time values Dfm$time - as.numeric(gsub([a-zA-Z], , as.character(Dfm$variable))) ggplot(Dfm, aes(x = time, y = value)) + geom_point() + facet_wrap(~ Gene) Dennis On Thu, Jan 31, 2013 at 2:37 AM, Jose Iparraguirre jose.iparragui...@ageuk.org.uk wrote: Hi Robin, You didn't provide the list with a clear structure of your data, but I hope I understood it properly. If not, it'd be easy for you to adapt what follows. I guess your data looks like this: Gene T0h T0.25h T0.5h T1h T2h T3h T6h T12h T24h T48h 1 NM_001001130 68 9556 43 66 62 68 90 63 89 2 NM_0010011440 1 40111412 3 NM_001001152 7912952 50 24 45 130 154 112 147 4 NM_0010011601 1 20010001 5 NM_0010011760 0 00000000 ... If this is the case, first reshape your data (say, gene.df) to a long format: Gene Time 1 NM_001001130 68 2 NM_0010011440 3 NM_001001152 79 4 NM_0010011601 5 NM_0010011760 6 NM_0010011771 7 NM_0010011780 8 NM_0010011790 9 NM_001001182 4691 ... Then, you create the ggplot: sp - ggplot(gene.df, aes(x=Gene, y=Time)) + geom_point(shape=1) And use the function facet_wrap(): sp + facet_wrap( ~ Gene,ncol=3) You can set a different number of columns and a different shape for the markers, of course. Hope this helps. Regards, José José Iparraguirre Chief Economist Age UK -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Robin Mjelle Sent: 31 January 2013 09:06 To: r-help@r-project.org Subject: [R] facet plot Hi all, I want to plot a facet plot with column names as x and column values as y. One plot for each row. here is part of my dataset: Gene T0h T0.25h T0.5h T1h T2h T3h T6h T12h T24h T48h NM_001001130 68 95 56 43 66 62 68 90 63 89 NM_001001144 0 1 4 0 1 1 1 4 1 2 NM_001001152 79 129 52 50 24 45 130 154 112 147 NM_001001160 1 1 2 0 0 1 0 0 0 1 NM_001001176 0 0 0 0 0 0 0 0 0 0 NM_001001177 1 3 2 3 0 1 1 0 2 0 NM_001001178 0 0 0 0 0 0 0 0 0 0 NM_001001179 0 0 0 0 0 0 0 0 0 0 NM_001001180 1 0 0 1 0 0 0 0 0 0 NM_001001181 350 539 424 470 441 451 554 553 419 548 NM_001001182 4691 6458 5686 6761 5944 4486 6030 5883 6009 7552 NM_001001183 22 23 12 21 25 17 34 40 33 32 NM_001001184 138 147 111 100 54 61 61 77 57 99 NM_001001185 2 3 4 3 4 5 9 2 4 4 NM_001001186 231 337 187 168 148 178 275 319 181 279 [[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. Wrap Up and Run 10k is back! Also, new for 2013 – 2km intergenerational walks at selected venues. So recruit a buddy, dust off the trainers and beat the winter blues by signing up now: http://www.ageuk.org.uk/10k Milton Keynes | Oxford | Sheffield | Crystal Palace | Exeter | Harewood House, Leeds | Tatton Park, Cheshire | Southampton | Coventry Age UK Improving later life http://www.ageuk.org.uk --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual
Re: [R] Nested loop and output help
Hi
see inline
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of staysafe23
Sent: Friday, February 01, 2013 1:01 AM
To: r-help@r-project.org
Subject: [R] Nested loop and output help
Hello Everyone,
My name is Thomas and I have been using R for one week. I recently
found your site and have been able to search the archives of posts.
This has given me some great information that has allowed me to craft
an initial design to an inquiry I would like to make into the breakdown
of McNemar's test. I have read an intro to R manual and the posting
guides and hope I am not violating them with this post. If so I will
re-ask my question in the proper format.
I have succeeded in writing a loop to vary one condition of my inquiry
but I am unable to understand how to vary the remaining three
conditions, each with its own for loop. Each time I try to do so I fail
miserably. Here is my current code :
n - seq(5,10,by=1)
for(i in n) {
z1 - rnorm(n,mean=400, sd=70)
z2 - rnorm(n,mean=450, sd=90)
cor - runif(1,min=0.4, max=0.6)
X - z1
Y = cor*z1+(1-cor)*z2
mat1 - cbind(X,Y)
dev1 - sample(-40:40, 1, replace=T)
cut1 - mean(X) + dev1
dev2 - sample(12:54, 1, replace=T)
cut2 - mean(X) + dev1 + dev2
X2 - X-cut1
Y2 - Y-cut2
c3 - cor(X2,Y2)
mat2 -cbind(X2,Y2)
a11 - ifelse( X cut1 Y cut2, 1, 0)
a12 - ifelse( X cut1 Y = cut2, 1, 0)
a21 - ifelse( X = cut1 Y cut2, 1, 0)
a22 - ifelse( X = cut1 Y = cut2, 1, 0)
mat3 -matrix(c(sum(a11),sum(a21), sum(a12),sum(a22)), nrow = 2)
mat4 -matrix(c(sum(a11),sum(a22), sum(a12),sum(a21)), nrow = 2)
out3a - mcnemar.test(mat3, correct=FALSE)
out3b - mcnemar.test(mat3, correct=TRUE)
out4a - chisq.test(mat4, correct = FALSE)
out4b - chisq.test(mat4, correct = TRUE)
print(mat1)
print(mat2)
print(cut1)
print(cut2)
print(mat3)
print(out3a)
print(out3b)
print(mat4)
print(out4a)
print(out4b)
}
.
Use list structure for keeping such results.
lll-list()
for(j in 1:5) {
lll[[j]] - list()
for( i in 1:3) lll[[j]][[i]]-rnorm(10)
}
after population of a list you can print it as a whole or only part. Here is an
example with your code.
n - seq(5,10,by=1)
lll - vector(mode = list, length = 10)
names(lll) - c(mat1, mat2, mat3, mat4, cut1, cut2, out3a,
out3b, out4a, out4b)
n - seq(5,10,by=1)
for(i in n) {
z1 - rnorm(n,mean=400, sd=70)
z2 - rnorm(n,mean=450, sd=90)
cor - runif(1,min=0.4, max=0.6)
X - z1
Y = cor*z1+(1-cor)*z2
lll[[mat1]] - cbind(X,Y)
dev1 - sample(-40:40, 1, replace=T)
lll[[cut1]] - mean(X) + dev1
dev2 - sample(12:54, 1, replace=T)
lll[[cut2]] - mean(X) + dev1 + dev2
X2 - X-cut1
Y2 - Y-cut2
c3 - cor(X2,Y2)
lll[[mat2]] -cbind(X2,Y2)
a11 - ifelse( X cut1 Y cut2, 1, 0)
a12 - ifelse( X cut1 Y = cut2, 1, 0)
a21 - ifelse( X = cut1 Y cut2, 1, 0)
a22 - ifelse( X = cut1 Y = cut2, 1, 0)
lll[[mat3]] -matrix(c(sum(a11),sum(a21), sum(a12),sum(a22)), nrow = 2)
lll[[mat4]] -matrix(c(sum(a11),sum(a22), sum(a12),sum(a21)), nrow = 2)
lll[[out3a]] - mcnemar.test(mat3, correct=FALSE)
lll[[out3b]] - mcnemar.test(mat3, correct=TRUE)
lll[[out4a]] - chisq.test(mat4, correct = FALSE)
lll[[out4b]] - chisq.test(mat4, correct = TRUE)
}
print(lll)
Other than the sample size of the random draws I would like to nest for
loops for cor, dev1, and dev2 according to the following sequences
respectively:
cor - seq(-0.5,0.5, by=0.25)
do not use floating points in cycle.
better to use
for (k in 1:n) {
cc - cor[k]
make computation with cc
}
dev1 - seq(-100,100, by=10)
dev2 - seq(12,54, by=10)
.
After doing so I would like to put each matrix and their respective
tests into a text file so that I can examine the results. I would like
to put the results in a .txt file each time the loop finishes one case.
I would like to append this text file with subsequent matrices and
results rendered by each iteration of the nested for loop. I have seen
some very nice examples of output that R can render. I would like to
simply display each matrix and their tests.
maybe R2HTML or latex in Hmisc package can
Regards
Petr
Thank you to all the teachers and students on this forum. The only
reason I have been able to craft this inquiry is due to the questions
and answers I have found through searching the archive. Thank you
kindly for your assistance and for freely sharing your knowledge.
Best wishes,
Thomas
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Re: [R] difftime() out by 1 hour
Hi
The same with me on Windows
Most probably an issue of daylight savings time setting.
Sys.timezone()
[1] CET
Regards
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Pascal Oettli
Sent: Friday, February 01, 2013 9:18 AM
To: Boots
Cc: R Help
Subject: Re: [R] difftime() out by 1 hour
Hello,
Here is the result I get using your script:
sessionInfo()
R version 2.15.2 (2012-10-26)
Platform: x86_64-suse-linux-gnu (64-bit)
[1] 2012-03-31 21:00:00 2012-04-01 00:00:00 2012-04-01 03:00:00
[4] 2012-04-01 06:00:00
Time differences in hours
[1] 3 3 3
attr(,tzone)
[1]
[1] 2012-10-06 21:00:00 2012-10-07 00:00:00 2012-10-07 03:00:00
[4] 2012-10-07 06:00:00
Time differences in hours
[1] 3 3 3
attr(,tzone)
[1]
Regards,
Pascal
Le 01/02/2013 16:27, Boots a écrit :
I have a problem with results from difftime being 1 hour different
than expected. 2 examples are given below:
datetime - matrix(data=rbind(c(2012-03-31 21:00:00, 2012-04-01
00:00:00, 2012-04-01 03:00:00, 2012-04-01 06:00:00),
c(2012-10-06 21:00:00, 2012-10-07 00:00:00,
2012-10-07 03:00:00, 2012-10-07 06:00:00)), ncol=4, nrow=2)
for (row in 1:2) {
x - datetime[row,]
x1=x[1:length(x)-1]
x2=x[2:length(x)]
xd=difftime(x2,x1)
print (x)
print (xd)
}
[1] 2012-03-31 21:00:00 2012-04-01 00:00:00 2012-04-01 03:00:00
2012-04-01 06:00:00
Time differences in hours
[1] 3 4 3
attr(,tzone)
[1]
[1] 2012-10-06 21:00:00 2012-10-07 00:00:00 2012-10-07 03:00:00
2012-10-07 06:00:00
Time differences in hours
[1] 3 2 3
attr(,tzone)
[1]
Accurate results would be 3 3 3 (hours) in both cases.
The problem occurs randomly, but only around midnight.
All the dates between 2012-04-01 06:00:00 and 2012-10-06 21:00:00
are fine.
I have seen postings about problems with POSIXlt and POSIXct
accuracy because of truncation rather than rounding of floating point
representations.
The response was that it was expected behaviour.
So my question is this expected behaviour, have I used difftime
incorrectly, and if not, is there a set of date/time functions and
classes that use integer representation to give more accurate results?
I am new to R and I may well be using the wrong set of functions.
Garry
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[R] Cumulative Incidence Function and Pseudovalue
Hi, I want to write own R functions for cumulative incidence function and also for the pseudovalue of the cumulative incidence function. Can you help me? Tas. [[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.
[R] Was confused with options(error = expression(NULL)) in example(stop)
In example for function 'stop' in R, there is options(error = expression(NULL)) with comment # don't stop on stop(.) Use with CARE! I was interested, wanted to know how don't stop on stop(.) was. So, I tried it. Typing example(stop) at the R prompt and pressing ENTER give this. example(stop) stop options(error = expression(NULL)) stop # don't stop on stop(.) Use with CARE! stop stop iter - 12 stop if(iter 10) stop(too many iterations) Error in eval(expr, envir, enclos) : too many iterations R still stops on stop(.). That confused me. Then, I tried manually. But I still couldn't get options(error = expression(NULL)) worked. Everything seemed to be as usual. I gave up. Some time later, I saw somewhere in the internet that options(error = expression(NULL)) worked in batch mode. Ah, running from R CMD BATCH. It had no effect in interactive session. I hadn't thought of it before. sessionInfo() R version 2.15.2 (2012-10-26) Platform: i386-w64-mingw32/i386 (32-bit) locale: [1] LC_COLLATE=English_United States.1252 [2] LC_CTYPE=English_United States.1252 [3] LC_MONETARY=English_United States.1252 [4] LC_NUMERIC=C [5] LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base loaded via a namespace (and not attached): [1] tools_2.15.2 __ 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] Using eigen() for extracting only few major eigenpairs
Thanks a lot, I'll test this package very soon.
As it seems general purpose (ie not specifically fitted to the square
symmetric context), I hope the advantage over the standard routine for
symmetric matrices remains significant.
Pierrick Bruneau
CRP Gabriel Lippmann
On Thursday, January 31, 2013, Mark Leeds wrote:
hi: the irlba package does what you're looking for.
On Thu, Jan 31, 2013 at 3:32 AM, Pierrick Bruneau
pbrun...@gmail.comjavascript:_e({}, 'cvml', 'pbrun...@gmail.com');
wrote:
Hi everyone,
I am using eigen() to extract the 2 major eigenpairs from a large real
square symmetric matrix. The procedure is already rather efficient, but
becomes somehow slow for real time needs with moderately large matrices
(few thousand lines).
The R implementation statically extracts all eigenvalues (and optionally
associated eigenvectors). I heard about optimizations of the eigen
decomposition when only few eigenpairs are needed : did somebody already
care about this problem (through a contributed package for example) ? Or
do
I have to directly try to mess around with the LAPACK library (and
contribute the package myself afterwards) ?
Thanks by advance for your help,
Pierrick Bruneau
CRP Gabriel Lippmann
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[R] how to setdiff on lists of lists
Dear List, I have a data structure like: aa [[1]] [1] 1 2 3 [[2]] [1] 4 5 6 bb [[1]] [1] 3 [[2]] [1] 5 I would like to set differences between aa and bb and get as result another list of lists like: res [[1]] [1] 1 2 [[2]] [1] 4 6 I am trying: lapply(aa, setdiff, bb) but I simply get aa back as result. Could you please point me in the right direction about the combination of functions that I need to accomplish this task? Best regards, Simone -- Simone Gabbriellini, PhD PostDoc@DISI, University of Bologna mobile: +39 340 39 75 626 email: simone.gabbriell...@unibo.it home: www.digitaldust.it DigitalBrains srl Amministratore mobile: +39 340 39 75 626 email: simone.gabbriell...@digitalbrains.it home: www.digitalbrains.it __ 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 setdiff on lists of lists
Hi you have something wrong in your R session, works for me. aa-list(c(1:3), c(4:6)) aa [[1]] [1] 1 2 3 [[2]] [1] 4 5 6 bb-list(3,5) bb [[1]] [1] 3 [[2]] [1] 5 lapply(aa, setdiff, bb) [[1]] [1] 1 2 [[2]] [1] 4 6 Regards Petr -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Simone Gabbriellini Sent: Friday, February 01, 2013 11:16 AM To: r-help@r-project.org Subject: [R] how to setdiff on lists of lists Dear List, I have a data structure like: aa [[1]] [1] 1 2 3 [[2]] [1] 4 5 6 bb [[1]] [1] 3 [[2]] [1] 5 I would like to set differences between aa and bb and get as result another list of lists like: res [[1]] [1] 1 2 [[2]] [1] 4 6 I am trying: lapply(aa, setdiff, bb) but I simply get aa back as result. Could you please point me in the right direction about the combination of functions that I need to accomplish this task? Best regards, Simone -- Simone Gabbriellini, PhD PostDoc@DISI, University of Bologna mobile: +39 340 39 75 626 email: simone.gabbriell...@unibo.it home: www.digitaldust.it DigitalBrains srl Amministratore mobile: +39 340 39 75 626 email: simone.gabbriell...@digitalbrains.it home: www.digitalbrains.it __ 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] ks.test and wilcoxon.test results differ from other stat.packages
Impossible to say w/o a reproducible example, but to start with let me suggest looking at the exact= (both functions) and correct= (wilcox.test) arguments. Experience shows that some change of the default settings allows you to reproduce results from other software (and the help pages will explain you what the settings do; by that, you might also learn what your other tools actually calculate if you have no documentation at hand). HTH, Michael -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Alexander Favorov Sent: Freitag, 1. Februar 2013 09:00 To: r-help@r-project.org Subject: [R] ks.test and wilcoxon.test results differ from other stat.packages Probably, it's an obvious info, but I have not found anything in R FAQ concerning this feature/bug. The results of ks.test and wilcoxon.test (in the Mann-Whitney version, paired = 'FALSE') don't coincide with the results from the other statistical packages, e.g. Statistica, Medcalc, and (as for MW test) from the numerous online MW tests. E.g. Statistica p-value=0.0435353 Medcalc p-value=0.0435354 R p-value=0.04635 If I want to obtain result of test once, it does not matter. But what should I do if I want to perform Monte-Carlo simulations and I need in 1 or even 100 000 p-values and then will build some distribution and then use results of Statictica? Whtever, the descrepancy bothers. Are there some alternative packages for non-parametric statistics that produce results comparable with the other program packages? Or, probably, there is exists some environment variable to perform calculations in more common way? Thank you! Sasha. __ 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.
[R] how to
Dear List,
I have a list of lists and on each list I want to apply a function
from the igraph package, graph.knn. I need to calculate graph.knn for
each list on a graph g, then retrieve one of the result called $knn
and calculate its mean. The non-R style code looks like this:
for(i in listOfLists){
print(mean(graph.knn(g, i)$knn))
}
Is there any way to convert this in a one-liner? I have tried to
figure it out with lapply() or mapply() but with no success.
Thank you in advance for your help.
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University of Bologna
mobile: +39 340 39 75 626
email: simone.gabbriell...@unibo.it
home: www.digitaldust.it
DigitalBrains srl
Amministratore
mobile: +39 340 39 75 626
email: simone.gabbriell...@digitalbrains.it
home: www.digitalbrains.it
__
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Re: [R] help on proportions
Hi Srini, This is a statistics question, not a question about R, so this may not be the best place to ask. Try posting at http://stats.stackexchange.com/ or another statistics help list. Best, Ista On Thu, Jan 31, 2013 at 11:11 PM, Srinivas Iyyer srini_iyyer_...@yahoo.com wrote: Hi: Apologies for asking the following question. As this may sound very basic and stupid for this forum , I honestly do not know how to solve it and I do not have a teacher who can help me understand. I have list of genes (200) that are involved in a particular process and I call this as a ProcSet. From an independent experiment I found that out of 10,000 genes, 1500 are significant and I call these1500 genes as ResultSet. The intersection of ResultSet and ProcSet are 80 genes. That means 40% of ProcSet are significant. How do I calculate that 40% is significant and more than I expect by chance given ResultSet and 10,000 genes I evaluated in the experiment. What I have: n = 200 (ProcSet) p = 0.4 N = 1500 (ResultSet) N1 =10,000 Pn = 0.15 What kind of test will help me know that 0.4 is significant given 0.15. Any suggestions will greatly help me. Thank you. Srini __ 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] Question on plotCI function
There are many plotCI functions in many different packages... which one are you referring to? Also please construct a reproducible example illustrating your problem. Best, Ista On Thu, Jan 31, 2013 at 11:58 PM, li li hannah@gmail.com wrote: Hi all, In my plotCI function, the argument x is chosen to be seq(0.05, 0.95, by=0.05). However, when I make the plot, the plot has the x coordinate goes 1:19. Does anyone know how to make the x coordinate to be (0,0.5, 0.1, ..., 0.95). Thank you. Hanna [[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. __ 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
Hi
I do not see any problem with your code. *apply functions are also hidden
cycles and there shall not be substantial improvement in speed.
Why you do not want to use for cycle?
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Simone Gabbriellini
Sent: Friday, February 01, 2013 12:54 PM
To: r-help@r-project.org
Subject: [R] how to
Dear List,
I have a list of lists and on each list I want to apply a function from
the igraph package, graph.knn. I need to calculate graph.knn for each
list on a graph g, then retrieve one of the result called $knn and
calculate its mean. The non-R style code looks like this:
for(i in listOfLists){
print(mean(graph.knn(g, i)$knn))
}
Is there any way to convert this in a one-liner? I have tried to figure
it out with lapply() or mapply() but with no success.
Thank you in advance for your help.
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University of Bologna
mobile: +39 340 39 75 626
email: simone.gabbriell...@unibo.it
home: www.digitaldust.it
DigitalBrains srl
Amministratore
mobile: +39 340 39 75 626
email: simone.gabbriell...@digitalbrains.it
home: www.digitalbrains.it
__
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.
__
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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] Filter according to the latest data
try this:
x - read.table(text = No. Change Date
+ A 123 final2013-01-15
+ B 123 error 2013-01-16
+ C 123 'bug fixed' 2013-01-17
+ D 111 final2013-01-12
+ , header = TRUE
+ , as.is = TRUE
+ )
do.call(rbind, lapply(split(x, x$No.), function(.sec){
+ .sec[which(.sec$Date == max(.sec$Date))[1L], ]
+ }))
No.Change Date
111 111 final 2013-01-12
123 123 bug fixed 2013-01-17
On Fri, Feb 1, 2013 at 3:04 AM, Mat matthias.we...@fnt.de wrote:
Hello together,
i have a data.frame, like this one:
No. Change Date
A 123 final2013-01-15
B 123 error 2013-01-16
C 123 bug fixed 2013-01-17
D 111 final2013-01-12
and now a want a new data.frame which includes only the newest entry for
each number.
The solution look like this:
No. Change Date
C 123 bug fixed 2013-01-17
D 111 final2013-01-12
is there any way to filter my data.frame to the latest data, perhabs max?
Thanks.
Mat
--
View this message in context:
http://r.789695.n4.nabble.com/Filter-according-to-the-latest-data-tp4657248.html
Sent from the R help mailing list archive at Nabble.com.
__
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
__
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Re: [R] how to setdiff on lists of lists
You can also try 'mapply' aa-list(c(1:3), c(4:6)) bb-list(3,5) mapply(setdiff, aa, bb) [,1] [,2] [1,]14 [2,]26 On Fri, Feb 1, 2013 at 5:16 AM, Simone Gabbriellini simone.gabbriell...@gmail.com wrote: Dear List, I have a data structure like: aa [[1]] [1] 1 2 3 [[2]] [1] 4 5 6 bb [[1]] [1] 3 [[2]] [1] 5 I would like to set differences between aa and bb and get as result another list of lists like: res [[1]] [1] 1 2 [[2]] [1] 4 6 I am trying: lapply(aa, setdiff, bb) but I simply get aa back as result. Could you please point me in the right direction about the combination of functions that I need to accomplish this task? Best regards, Simone -- Simone Gabbriellini, PhD PostDoc@DISI, University of Bologna mobile: +39 340 39 75 626 email: simone.gabbriell...@unibo.it home: www.digitaldust.it DigitalBrains srl Amministratore mobile: +39 340 39 75 626 email: simone.gabbriell...@digitalbrains.it home: www.digitalbrains.it __ 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. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. __ 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
Hi,
Yes I know it works, but I'd like to assign the results like this:
V(g)$meanknn - ONELINER
Where V(g) elencates all the nodes in my graph...
Thanks,
Simone
Sent from my iPhone. Please excuse brevity and odd typos
On 01/feb/2013, at 13:31, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
I do not see any problem with your code. *apply functions are also hidden
cycles and there shall not be substantial improvement in speed.
Why you do not want to use for cycle?
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Simone Gabbriellini
Sent: Friday, February 01, 2013 12:54 PM
To: r-help@r-project.org
Subject: [R] how to
Dear List,
I have a list of lists and on each list I want to apply a function from
the igraph package, graph.knn. I need to calculate graph.knn for each
list on a graph g, then retrieve one of the result called $knn and
calculate its mean. The non-R style code looks like this:
for(i in listOfLists){
print(mean(graph.knn(g, i)$knn))
}
Is there any way to convert this in a one-liner? I have tried to figure
it out with lapply() or mapply() but with no success.
Thank you in advance for your help.
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University of Bologna
mobile: +39 340 39 75 626
email: simone.gabbriell...@unibo.it
home: www.digitaldust.it
DigitalBrains srl
Amministratore
mobile: +39 340 39 75 626
email: simone.gabbriell...@digitalbrains.it
home: www.digitalbrains.it
__
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.
[R] Converting Date to Unix Time
Dear All, I am struggling with a simple problem. I did some online research, but I am still banging my head against the floor. I read a csv file into a data frame and I have a columns with entries like 2012-04-13 00:23:45 2012-04-22 07:53:09 etc.. Each line posted above is the content of a single cell. How can I convert that into Unix time? Many thanks Lorenzo __ 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] Transforming 4x3 data frame into 2 column df in R
newdf-data.frame(k1=rep(row.names(foo),ncol(foo)),stack(foo)) --- On Fri, 1/2/13, Gundala Viswanath gunda...@gmail.com wrote: From: Gundala Viswanath gunda...@gmail.com Subject: [R] Transforming 4x3 data frame into 2 column df in R To: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch Date: Friday, 1 February, 2013, 11:30 AM I have the following data frame: foo w x y z n 1.51550092 1.4337572 1.2791624 1.1771230 q 0.09977303 0.8173761 1.6123402 0.1510737 r 1.17083866 1.2469347 0.8712135 0.8488029 What I want to do is to change it into : newdf 1 n w 1.51550092 2 q w 0.09977303 3 r w 1.17083866 4 n x 1.43375725 5 q x 0.81737606 6 r x 1.24693468 7 n y 1.27916241 8 q y 1.61234016 9 r y 0.87121353 10 n z 1.17712302 11 q z 0.15107369 12 r z 0.84880292 Whtat's the way to do it? - G.V. [[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.
[R] How does this function print, why is n1 which equals 1 printed as 2?
Windows 7, R 2.12.1
Colleagues,
I am trying to understand the n.for.2means function. The code below is a copy
of the function (renamed to n.for.2means.js). I have inserted a single line of
code towards the bottom of the function which uses the cat function to print
the value of n1. You will note the value (preceded by stars) is printed as 1.
The function (1) prints a lot of output without any instructions in the
function to print anything (other than the cat statement I added), and when it
prints (2) reports the value of n1 to be 2!.
I have two questions, (i) how is the function printing when there is no code to
print and (ii) how is n1 which equals 1 being reported as 2? I suspect there is
something fundamental about R that I don't know.
Thank you for the help.
John
library(epicalc)
n.for.2means.js - function (mu1, mu2, sd1, sd2, ratio = 1, alpha = 0.05, power
= 0.8)
{
n1 - (sd1^2 + sd2^2/ratio) * (qnorm(1 - alpha/2) - qnorm(1 - power))^2/(mu1
- mu2)^2
n1 - round(n1)
n2 - ratio * n1
if (length(alpha) == 1) {
alpha1 - NULL
}
else {
alpha1 - alpha
}
if (length(power) == 1) {
power1 - NULL
}
else {
power1 - power
}
if (length(ratio) == 1) {
ratio1 - NULL
}
else {
ratio1 - ratio
}
table1 - cbind(mu1, mu2, sd1, sd2, n1, n2, alpha1, power1,
ratio1)
colnames(table1)[colnames(table1) == alpha1] - alpha
colnames(table1)[colnames(table1) == power1] - power
colnames(table1)[colnames(table1) == ratio1] - n2/n1
table1 - as.data.frame(table1)
cat(**n1***=,n1,\n)
returns - list(mu1 = mu1, mu2 = mu2, sd1 = sd1, sd2 = sd2,
alpha = alpha, n1 = n1, n2 = n2, power = power, ratio =
ratio,
table = table1)
class(returns) - c(n.for.2means, list)
returns
}
n.for.2means.js(0,8,10/6,10/6)
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for the sole use of the
intended recipient(s) and may contain confidential and privileged information.
Any unauthorized use, disclosure or distribution is prohibited. If you are not
the intended recipient, please contact the sender by reply email and destroy
all copies of the original message.
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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.
[R] expand.grid on contents of a list
Hello! I have a list of variable length. One example is: X=vector(list,3) X[[1]]=1:2 X[[2]]=1:2 X[[3]]=1:2 How could I run expand.grid on the elements of X so that the results would be the same as expand.grid(1:2,1:2,1:2)? Thank you! Dimitri -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ [[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.
Re: [R] Converting Date to Unix Time
Mr Isella, On 1 February 2013 05:37, Lorenzo Isella lorenzo.ise...@gmail.com wrote: How can I convert that into Unix time? format.POSIXct(dateCol, '%s'); -- H -- Sent from my mobile device Envoyait de mon portable __ 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 does this function print, why is n1 which equals 1 printed as 2?
On 13-02-01 8:47 AM, John Sorkin wrote:
Windows 7, R 2.12.1
Colleagues,
I am trying to understand the n.for.2means function. The code below is a copy
of the function (renamed to n.for.2means.js). I have inserted a single line of
code towards the bottom of the function which uses the cat function to print
the value of n1. You will note the value (preceded by stars) is printed as 1.
The function (1) prints a lot of output without any instructions in the
function to print anything (other than the cat statement I added), and when it
prints (2) reports the value of n1 to be 2!.
I have two questions, (i) how is the function printing when there is no code to
print and (ii) how is n1 which equals 1 being reported as 2? I suspect there is
something fundamental about R that I don't know.
I haven't run the code, but presumably it's just the usual auto
printing. Unless a function sets the result to be invisible, results
of functions are printed by calling print() after they are returned.
Since your function returns something with class
c(n.for.2means, list)
R will look for a print method for those classes, and use it to print
the result. There's no print.list method in standard R; it just uses
the default. But there's probably a print.n.for.2means function in the
package (possibly not exported; you might need
epicalc:::print.n.for.2means to see it). There might be a print.list
function there instead or as well.
Duncan Murdoch
Thank you for the help.
John
library(epicalc)
n.for.2means.js - function (mu1, mu2, sd1, sd2, ratio = 1, alpha = 0.05, power
= 0.8)
{
n1 - (sd1^2 + sd2^2/ratio) * (qnorm(1 - alpha/2) - qnorm(1 -
power))^2/(mu1 - mu2)^2
n1 - round(n1)
n2 - ratio * n1
if (length(alpha) == 1) {
alpha1 - NULL
}
else {
alpha1 - alpha
}
if (length(power) == 1) {
power1 - NULL
}
else {
power1 - power
}
if (length(ratio) == 1) {
ratio1 - NULL
}
else {
ratio1 - ratio
}
table1 - cbind(mu1, mu2, sd1, sd2, n1, n2, alpha1, power1,
ratio1)
colnames(table1)[colnames(table1) == alpha1] - alpha
colnames(table1)[colnames(table1) == power1] - power
colnames(table1)[colnames(table1) == ratio1] - n2/n1
table1 - as.data.frame(table1)
cat(**n1***=,n1,\n)
returns - list(mu1 = mu1, mu2 = mu2, sd1 = sd1, sd2 = sd2,
alpha = alpha, n1 = n1, n2 = n2, power = power, ratio =
ratio,
table = table1)
class(returns) - c(n.for.2means, list)
returns
}
n.for.2means.js(0,8,10/6,10/6)
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for the sole use of the
intended recipient(s) and may contain confidential and privileged information.
Any unauthorized use, disclosure or distribution is prohibited. If you are not
the intended recipient, please contact the sender by reply email and destroy
all copies of the original message.
__
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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.
__
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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] how to
Hi
So why you do not assign it inside a cycle? As I said lapply does the same as
for cycle does (maybe in some more compact manner).
Regards
Petr
-Original Message-
From: Simone Gabbriellini [mailto:simone.gabbriell...@gmail.com]
Sent: Friday, February 01, 2013 2:29 PM
To: PIKAL Petr
Cc: r-help@r-project.org
Subject: Re: [R] how to
Hi,
Yes I know it works, but I'd like to assign the results like this:
V(g)$meanknn - ONELINER
Where V(g) elencates all the nodes in my graph...
Thanks,
Simone
Sent from my iPhone. Please excuse brevity and odd typos
On 01/feb/2013, at 13:31, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
I do not see any problem with your code. *apply functions are also
hidden cycles and there shall not be substantial improvement in speed.
Why you do not want to use for cycle?
Petr
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Simone Gabbriellini
Sent: Friday, February 01, 2013 12:54 PM
To: r-help@r-project.org
Subject: [R] how to
Dear List,
I have a list of lists and on each list I want to apply a function
from the igraph package, graph.knn. I need to calculate graph.knn
for
each list on a graph g, then retrieve one of the result called $knn
and calculate its mean. The non-R style code looks like this:
for(i in listOfLists){
print(mean(graph.knn(g, i)$knn))
}
Is there any way to convert this in a one-liner? I have tried to
figure it out with lapply() or mapply() but with no success.
Thank you in advance for your help.
Simone
--
Simone Gabbriellini, PhD
PostDoc@DISI, University of Bologna
mobile: +39 340 39 75 626
email: simone.gabbriell...@unibo.it
home: www.digitaldust.it
DigitalBrains srl
Amministratore
mobile: +39 340 39 75 626
email: simone.gabbriell...@digitalbrains.it
home: www.digitalbrains.it
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PLEASE do read the posting guide http://www.R-project.org/posting-
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reproducible code.
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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] expand.grid on contents of a list
Oops, it was easier than I thought: expand.grid(X) Dimitri On Fri, Feb 1, 2013 at 8:48 AM, Dimitri Liakhovitski dimitri.liakhovit...@gmail.com wrote: Hello! I have a list of variable length. One example is: X=vector(list,3) X[[1]]=1:2 X[[2]]=1:2 X[[3]]=1:2 How could I run expand.grid on the elements of X so that the results would be the same as expand.grid(1:2,1:2,1:2)? Thank you! Dimitri -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ [[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.
Re: [R] How does this function print, why is n1 which equals 1 printed as 2?
On Feb 1, 2013, at 7:47 AM, John Sorkin wrote:
Windows 7, R 2.12.1
Colleagues,
I am trying to understand the n.for.2means function. The code below
is a copy of the function (renamed to n.for.2means.js). I have
inserted a single line of code towards the bottom of the function
which uses the cat function to print the value of n1. You will note
the value (preceded by stars) is printed as 1.
The function (1) prints a lot of output without any instructions in
the function to print anything (other than the cat statement I
added), and when it prints (2) reports the value of n1 to be 2!.
I have two questions, (i) how is the function printing when there is
no code to print and (ii) how is n1 which equals 1 being reported as
2? I suspect there is something fundamental about R that I don't know.
1) on my machine the output from the cat() call is:
**n1***= 1
2) All of the output without any instructions in the function to
print anything is just the value of the list object from the
function. Unless you return values using the `invisible` function, any
user define function executed at the console will print its value.
That is standard interactive session behavior. So one gets after the
the cat output:
$mu1
[1] 0
$mu2
[1] 8
$sd1
[1] 1.67
$sd2
[1] 1.67
$alpha
[1] 0.05
$n1
[1] 1
$n2
[1] 1
$power
[1] 0.8
$ratio
[1] 1
$table
mu1 mu2 sd1 sd2 n1 n2
1 0 8 1.67 1.67 1 1
attr(,class)
[1] n.for.2means list
--
David.
Thank you for the help.
John
library(epicalc)
n.for.2means.js - function (mu1, mu2, sd1, sd2, ratio = 1, alpha =
0.05, power = 0.8)
{
n1 - (sd1^2 + sd2^2/ratio) * (qnorm(1 - alpha/2) - qnorm(1 -
power))^2/(mu1 - mu2)^2
n1 - round(n1)
n2 - ratio * n1
if (length(alpha) == 1) {
alpha1 - NULL
}
else {
alpha1 - alpha
}
if (length(power) == 1) {
power1 - NULL
}
else {
power1 - power
}
if (length(ratio) == 1) {
ratio1 - NULL
}
else {
ratio1 - ratio
}
table1 - cbind(mu1, mu2, sd1, sd2, n1, n2, alpha1, power1,
ratio1)
colnames(table1)[colnames(table1) == alpha1] - alpha
colnames(table1)[colnames(table1) == power1] - power
colnames(table1)[colnames(table1) == ratio1] - n2/n1
table1 - as.data.frame(table1)
cat(**n1***=,n1,\n)
returns - list(mu1 = mu1, mu2 = mu2, sd1 = sd1, sd2 = sd2,
alpha = alpha, n1 = n1, n2 = n2, power = power,
ratio = ratio,
table = table1)
class(returns) - c(n.for.2means, list)
returns
}
n.for.2means.js(0,8,10/6,10/6)
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:15}}
__
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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] help on proportions
Hello Srini, It sounds as if you are attempting to establish a prior probability and compare it to the posterior probability -- a perfect candidate for bayesian analysis. I would simply do a search for 'bayesian analysis of gene expression data' -- there are a number of statistical packages that are available. A number of R packages are available as well as a software package from Yale: http://www.yale.edu/townsend/Software/BAGELTutorial.html Hope this helps, James On Thu, Jan 31, 2013 at 11:11 PM, Srinivas Iyyer srini_iyyer_...@yahoo.comwrote: Hi: Apologies for asking the following question. As this may sound very basic and stupid for this forum , I honestly do not know how to solve it and I do not have a teacher who can help me understand. I have list of genes (200) that are involved in a particular process and I call this as a ProcSet. From an independent experiment I found that out of 10,000 genes, 1500 are significant and I call these1500 genes as ResultSet. The intersection of ResultSet and ProcSet are 80 genes. That means 40% of ProcSet are significant. How do I calculate that 40% is significant and more than I expect by chance given ResultSet and 10,000 genes I evaluated in the experiment. What I have: n = 200 (ProcSet) p = 0.4 N = 1500 (ResultSet) N1 =10,000 Pn = 0.15 What kind of test will help me know that 0.4 is significant given 0.15. Any suggestions will greatly help me. Thank you. Srini __ 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. -- *James C. Whanger* * * *It ain't what you don't know that gets you into trouble. It's what you know for sure that just ain't so. Mark Twain* [[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.
Re: [R] expand.grid on contents of a list
Hi, expand.grid(X) # Var1 Var2 Var3 #1111 #2211 #3121 #4221 #5112 #6212 #7122 #8222 expand.grid(1:2,1:2,1:2) # Var1 Var2 Var3 #1111 #2211 #3121 #4221 #5112 #6212 #122 #8222 A.K. - Original Message - From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Friday, February 1, 2013 8:48 AM Subject: [R] expand.grid on contents of a list Hello! I have a list of variable length. One example is: X=vector(list,3) X[[1]]=1:2 X[[2]]=1:2 X[[3]]=1:2 How could I run expand.grid on the elements of X so that the results would be the same as expand.grid(1:2,1:2,1:2)? Thank you! Dimitri -- Dimitri Liakhovitski gfk.com http://marketfusionanalytics.com/ [[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. __ 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] obtainl survival curves for single strata
Stephen, I can almost but not quite get my arms around what you are asking. A bit more detail would help. Like cph.approve = what kind of object, generated by function __ But, let me make a guess: cph is the result of coxph, and the model has both covariates and a strata You want predictioned survival curves, more than 1, of the type covariates = a, b,c, strata=1 covariates = d,e, f, strata=2, ... for arbitrary covariates and strata. Now, the predicted survival curves for different strata are different lengths. The survfit.coxph routine gets around this by being verbose: it expects you to give it covariate sets, and returns all of the strata for each covariate. This allows it to give a compact result. You can always do: newpred - survfit(cox-model-fit, newdata=something) newpred[5,17] # survival curve for the 5th strata, covariates from the 17th row of newdata But, you want to put the results into a matrix, for some pre-specifed set of time points. Take it one step further. mytimepoints - seq(0, 5*365, by=30) # every 30 days z - summary(newpred[5,17], time=mytimepoints, extend=TRUE)$surv The summary.survfit function's time argument was originally written for people who only wanted to print certain time points, but it works as well for those who only want to extract certain ones. It correctly handles the fact that the curve is a step function. Terry Therneau On 02/01/2013 05:00 AM, r-help-requ...@r-project.org wrote: What is the syntax to obtain survival curves for single strata on many subjects? I have a model based on Surv(time,response) object, so there is a single row per subject and no start,stop and no switching of strata. The newdata has many subjects and each subject has a strata and the survival based on the subject risk and the subject strata is needed. If I do newpred- survfit(cph.approve,new=newapp,se=F) I get all strata for every subject. Attempting newpred- survfit(cph.approve,new=newapp,id=CertId,se=F) Error in survfit.coxph(cph.approve, new = newapp, id = CertId, se = F) : The individual option is only valid for start-stop data newpred- survfit(cph.approve,new=newapp,indi=T,se=F) Error in survfit.coxph(cph.approve, new = newapp, indi = T, se = F) : The individual option is only valid for start-stop data Please, advise if obtaining a single strata for a basic (time,response) model is possible? Due to differing lengths of the surv for different strata this will not go in a wide data.frame without padding. Thanks everybody and have a great day. Stephen B __ 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] Question on plotCI function
Thanks so much for the reply, Ista. I used plotrix library. Here is my example: xx - seq(0.05, 0.95, by=0.05) lower - c(-2.896865, -2.728416, -2.642574, -2.587724, -2.548672, -2.518994, -2.495409, -2.476031, -2.459662, -2.445513, -2.433014, -2.421739, -2.411344, -2.401536, -2.392040, -2.382571, -2.372786, -2.362198, -2.349891) upper - c(2.311539, 2.372006, 2.423280, 2.469220, 2.511851, 2.552421, 2.591797, 2.630657, 2.669579, 2.709135, 2.749928, 2.792670, 2.838268, 2.887976, 2.943683, 3.008502, 3.088240, 3.195954, 3.373528) library(plotrix) plotCI(xx,ui=upper,li=lower,err=y,pch=NA, xlab=, ylab=, ylim=c(-5, 5), slty=solid,scol=blue, lwd=2) My question is how can I change the xllim to be c(0,1) which corresponds to the xx values. Hanna 2013/2/1 Ista Zahn istaz...@gmail.com There are many plotCI functions in many different packages... which one are you referring to? Also please construct a reproducible example illustrating your problem. Best, Ista On Thu, Jan 31, 2013 at 11:58 PM, li li hannah@gmail.com wrote: Hi all, In my plotCI function, the argument x is chosen to be seq(0.05, 0.95, by=0.05). However, when I make the plot, the plot has the x coordinate goes 1:19. Does anyone know how to make the x coordinate to be (0,0.5, 0.1, ..., 0.95). Thank you. Hanna [[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.htmlhttp://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.
Re: [R] ks.test and wilcoxon.test results differ from other stat.packages
Hello, As for the KS test, the op might also want to look at Simard, L'Ecuyer (2011) http://www.jstatsoft.org/v39/i11 for an account on the different algorithms available, and their accuracy. Apparently, R uses different algorithms according to the test in question. See the details section of ?ks.test. See also the references there, such as Marsaglia, Tsang, Wang (2003) http://www.jstatsoft.org/v08/i18/ Hope this helps, Rui Barradas Em 01-02-2013 11:24, Meyners, Michael escreveu: Impossible to say w/o a reproducible example, but to start with let me suggest looking at the exact= (both functions) and correct= (wilcox.test) arguments. Experience shows that some change of the default settings allows you to reproduce results from other software (and the help pages will explain you what the settings do; by that, you might also learn what your other tools actually calculate if you have no documentation at hand). HTH, Michael -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Alexander Favorov Sent: Freitag, 1. Februar 2013 09:00 To: r-help@r-project.org Subject: [R] ks.test and wilcoxon.test results differ from other stat.packages Probably, it's an obvious info, but I have not found anything in R FAQ concerning this feature/bug. The results of ks.test and wilcoxon.test (in the Mann-Whitney version, paired = 'FALSE') don't coincide with the results from the other statistical packages, e.g. Statistica, Medcalc, and (as for MW test) from the numerous online MW tests. E.g. Statistica p-value=0.0435353 Medcalc p-value=0.0435354 R p-value=0.04635 If I want to obtain result of test once, it does not matter. But what should I do if I want to perform Monte-Carlo simulations and I need in 1 or even 100 000 p-values and then will build some distribution and then use results of Statictica? Whtever, the descrepancy bothers. Are there some alternative packages for non-parametric statistics that produce results comparable with the other program packages? Or, probably, there is exists some environment variable to perform calculations in more common way? Thank you! Sasha. __ 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. __ 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] Relative Risk in logistic regression
Dear colleagues, I have 2 points: One opinion and one question. 1) In one paper in a peer-reviewed journal, I read about the idea of using a logit regression as a surrogate for the log-binomial, just adding the numerator to the denominator ... It’s tempting to immediately get the RR instead of OR ... I tried it and I think it's a bad idea, the confidence intervals dramatically inflated! Any opinions? 2) What would be the criteria for selection of link - functions for binary data? Usually I use the logit - just for simplest interpretation of parameters. Using logit, probit, log-log, and log-log, I get identical values of the maximum-likelihood, Pearson statistics, overdispersion parameter, etc. However, the regression coefficients and its standard errors are different (for logit b is the maximum, for the probit – min., for log-log C- log-log are between them). LRs close, but the maximum has the log-log. Wald criterion - the maximum for the probit. ?What are the interpretations for regression parameters (except logit)??? Ivan, IPAE RAS -- View this message in context: http://r.789695.n4.nabble.com/Relative-Risk-in-logistic-regression-tp4657040p4657297.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.
Re: [R] Question on plotCI function
You haven't given any y values to plotCI. Try, for example, plotCI(xx, (lower+upper)/2, ui=upper, (etc) What you got was in effect plot( seq(along=xx), xx ) which is standard behavior for plot() when no y values are supplied. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 2/1/13 7:21 AM, li li hannah@gmail.com wrote: Thanks so much for the reply, Ista. I used plotrix library. Here is my example: xx - seq(0.05, 0.95, by=0.05) lower - c(-2.896865, -2.728416, -2.642574, -2.587724, -2.548672, -2.518994, -2.495409, -2.476031, -2.459662, -2.445513, -2.433014, -2.421739, -2.411344, -2.401536, -2.392040, -2.382571, -2.372786, -2.362198, -2.349891) upper - c(2.311539, 2.372006, 2.423280, 2.469220, 2.511851, 2.552421, 2.591797, 2.630657, 2.669579, 2.709135, 2.749928, 2.792670, 2.838268, 2.887976, 2.943683, 3.008502, 3.088240, 3.195954, 3.373528) library(plotrix) plotCI(xx,ui=upper,li=lower,err=y,pch=NA, xlab=, ylab=, ylim=c(-5, 5), slty=solid,scol=blue, lwd=2) My question is how can I change the xllim to be c(0,1) which corresponds to the xx values. Hanna 2013/2/1 Ista Zahn istaz...@gmail.com There are many plotCI functions in many different packages... which one are you referring to? Also please construct a reproducible example illustrating your problem. Best, Ista On Thu, Jan 31, 2013 at 11:58 PM, li li hannah@gmail.com wrote: Hi all, In my plotCI function, the argument x is chosen to be seq(0.05, 0.95, by=0.05). However, when I make the plot, the plot has the x coordinate goes 1:19. Does anyone know how to make the x coordinate to be (0,0.5, 0.1, ..., 0.95). Thank you. Hanna [[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.htmlhttp://www.r-project.org/post ing-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. __ 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] Summary of data for each year
Here is another option using plyr: library(plyr) creek - read.csv(creek.csv) library(ggplot2) creek[1:10,] colnames(creek) - c(date,flow) creek$date - as.Date(creek$date, %m/%d/%Y) ddply(creek,year,summarise,MED=median(flow),MEAN=mean(flow),SD=sd(flow),MIN=min(flow)) Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA http://www.fws.gov/redbluff/rbdd_jsmp.aspx From: Pascal Oettli kri...@ymail.com To: Janesh Devkota janesh.devk...@gmail.com Cc: r-help@r-project.org Sent: Thursday, January 31, 2013 11:52 PM Subject: Re: [R] Summary of data for each year Hello, One possibility is: creek - read.csv(creek.csv) colnames(creek) - c(date,flow) creek$date - as.Date(creek$date, %m/%d/%Y) creek - within(creek, year - format(date, '%Y')) with(creek, aggregate(flow, by=list(year=year), summary)) HTH, Pascal Le 01/02/2013 16:32, Janesh Devkota a écrit : Hello All, I have a data with two columns. In one column it is date and in another column it is flow data. I was able to read the data as date and flow data. I used the following code: creek - read.csv(creek.csv) library(ggplot2) creek[1:10,] colnames(creek) - c(date,flow) creek$date - as.Date(creek$date, %m/%d/%Y) The link to my data is https://www.dropbox.com/s/eqpena3nk82x67e/creek.csv Now, I want to find the summary of each year. I want to especially know mean, median, maximum etc. Thanks. Janesh [[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. __ 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.
[R] Range difference of plot two arrays in one plot
Hello all,
When I tried to plot the following two arrays in one figure with the following:
x = c(0,0,0,10,20,30)
y = c(40,50,60,70,80,90)
plot(x, type='o', ylim=c(min(x),max(x)))
par(new=T)
plot(y, type='l', ylim=c(min(y),max(y)))
Found that the first points and last points from those two arrays are
overlapping together, but the value 30 is not equal to 90. How could I draw
those two arrays in one plot with their real values shown in the plot? i.e., x
is in the lower part of the plot and y is in the upper part of the plot.
Thanks very much!
Cheers,
Rebecca
--
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[R] [R-pkgs] Version 1.2 of pa package available on CRAN
Dear useRs, Version 1.2 of the R package 'pa' is now available on CRAN. This package provides tools for conducting equity portfolio attribution. Two methods included in the package are the Brinson Method and a regression-based analysis. The new version allows users to view attribution results for sub-categories based on the Brinson Method. URL â http://cran.at.r-project.org/web/packages/pa/index.html It also includes a vignette at http://cran.at.r-project.org/web/packages/pa/vignettes/pa.pdf. Best regards, Yang Lu Yang Lu '14 SU 2896 Paresky Center Williams College, MA (413)884-4847 [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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] Filter according to the latest data
library(sqldf) k1-data.frame(ID=LETTERS[1:4], No=c(rep(123,3),111), Change=c(final,error,bug fixed,final), Date=c(2013-01-15,2013-01-16,2013-01-17,2013-01-12)) k1$Date=as.Date(as.character(k1$Date),tz=UK) sqldf(select * from k1 group by No having max(Date)) --- On Fri, 1/2/13, Mat matthias.we...@fnt.de wrote: From: Mat matthias.we...@fnt.de Subject: [R] Filter according to the latest data To: r-help@r-project.org Date: Friday, 1 February, 2013, 1:34 PM Hello together, i have a data.frame, like this one: No. Change Date A 123 final 2013-01-15 B 123 error 2013-01-16 C 123 bug fixed 2013-01-17 D 111 final 2013-01-12 and now a want a new data.frame which includes only the newest entry for each number. The solution look like this: No. Change Date C 123 bug fixed 2013-01-17 D 111 final 2013-01-12 is there any way to filter my data.frame to the latest data, perhabs max? Thanks. Mat -- View this message in context: http://r.789695.n4.nabble.com/Filter-according-to-the-latest-data-tp4657248.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. [[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.
[R] Error: package/namespace load failed for ‘arm’
Hi! I get the following error messages when trying to run the package 'arm'. library(arm) Error : Functions found when exporting methods from the namespace arm which are not S4 generic: fixef, ranef Error: package/namespace load failed for arm My R version is: version _ platform i386-w64-mingw32 arch i386 os mingw32 system i386, mingw32 status major 2 minor 15.2 year 2012 month 10 day26 svn rev61015 language R version.string R version 2.15.2 (2012-10-26) nickname Trick or Treat When running the install for 'arm' it looks like the install was successful and 'arm' shows up in my package list and has created a folder in the library (C:\Users\Pernille\Documents\R\win-library\2.15\arm) install.packages(arm) Installing package(s) into C:/Users/Pernille/Documents/R/win-library/2.15 (as lib is unspecified) trying URL ' http://cran.rstudio.com/bin/windows/contrib/2.15/arm_1.6-01.02.zip' Content type 'application/zip' length 237391 bytes (231 Kb) opened URL downloaded 231 Kb package arm successfully unpacked and MD5 sums checked I have tried uninstalling and reinstalling 'arm'. I recently updated R from version 2.15.0 to 2.15.2 but 'arm' ran with the same error messages in both versions. It has never worked on my computer. I have tried in both 64 and 32 bit R. Hope someone can help me, I much appreciate it! Pernille Heelsberg [[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.
[R] x-axis labelling
Hi Folks! I have a question regarding an issue on which I´ve read a few threads but can´t resolve -- Labelling every other tick mark on the x-axis using factors. Here´s an example: db-data.frame(date=as.factor(c(1:50)),var=rnorm(50)) barplot(db$var,names.arg=levels(db$date),las=2,cex.axis=0.8,cex=0.8) This labels all the tickmarks in my barplot, which is the default behavior. I would like to label every other tick mark in the barplot -- 1,3,5,7 and so on. Thank you very much for your help! All the best, A -- Alfredo Tello (http://alfredotello.com) Sustainable Aquaculture Group Institute of Aquaculture University of Stirling Scotland, UK. [[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.
[R] Are there a read and write functions for data in the vowpal_wabbit format?
Has anyone implemented a reader and writer for data in the vowpal wabbit format? It's a format for representing sparse data in ASCII files, see https://github.com/JohnLangford/vowpal_wabbit - Roy [[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.
Re: [R] Filter according to the latest data
Hi, Perhaps, (#Untested) do.call(rbind,lapply(split(dat1,dat1$No),function(x) tail(x,1))) #or library(plyr) ddply(dat1,.(No), function(x) x[nrow(x),]) A.K. - Original Message - From: Mat matthias.we...@fnt.de To: r-help@r-project.org Cc: Sent: Friday, February 1, 2013 3:04 AM Subject: [R] Filter according to the latest data Hello together, i have a data.frame, like this one: No. Change Date A 123 final 2013-01-15 B 123 error 2013-01-16 C 123 bug fixed 2013-01-17 D 111 final 2013-01-12 and now a want a new data.frame which includes only the newest entry for each number. The solution look like this: No. Change Date C 123 bug fixed 2013-01-17 D 111 final 2013-01-12 is there any way to filter my data.frame to the latest data, perhabs max? Thanks. Mat -- View this message in context: http://r.789695.n4.nabble.com/Filter-according-to-the-latest-data-tp4657248.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. __ 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] Doing mixed stock analysis with predict function using DFA
Dear R Help Mailing List Members, I have some baseline data (attached) on which I plan to run a DFA in R to find out the best predictor variables that best allow discrimination between the different stocks. I then need to apply this DFA function to the mixed data set (attached), which contains all the predictor variables of the baseline data set, but does not contain a stock variable, as their origin is unknown. I am trying to classify the mixed stock fish back to their origin stock using the DFA function generated from the baseline data. My question is whether it is possible to use the predict function with the generated DFA function and provide the mixed data set as the newdata argument within this to generate the classification, even though the mixed data set does not contain the stock variable? Unfortunately, the mixstock package that has recently become available is not capable of dealing with continuous variables at this time, so I am unable to use it. I would be grateful for any guidance you may be able to provide. Thank you, Melanie Melanie Zölck (Zoelck) PhD Candidate Galway-Mayo Institute of Technology Marine and Freshwater Research Centre Commercial Fisheries Research Group Department of Life Science Dublin Road Galway Republic of Ireland E-mail: mzoe...@hotmaill.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.
Re: [R] use name (not values!) of a dataframe inside a funktion
It is strongly discouraged in R to have functions that change data values
in the global workspace (or any location other than their local
environment).
The usual procedure in R is to have your function return a modified version
of the object and the user then decides what to do with it. They can
assign it back to the same original object so that there is still only one
copy and it has changed (but the user made that decision, not the
programmer), or they can save it to a different name and not lose the
original.
If you really want to change the original copy (and there are sometimes
when the exception to the rule makes sense) then you can either use
environments (which don't copy on modify) or use macros instead of
functions. Given your examples I would look at the macro approach first.
There is a 'defmacro' function in the 'gtools' package and the reference
on the help page for 'defmacro' leads to the original R news (now R
Journal) article describing the use of macros in R (definitely read this if
you are considering this approach).
On Thu, Jan 31, 2013 at 7:34 AM, Winfried Moser winfried.mo...@gmail.comwrote:
Dear Listers,
can anyone help me, please.
Since several days i try to figure out, how to assign values, vectors,
functions etc to variables with dynamically generated names inside of
functions.
Sometimes I succeed, but the success is rather arbitrary, it seems. up to
now i don't fully understand, why things like get, assign, - etc do
sometimes work, and sometimes not.
here's one of my daily examples, i am stuck with: Example 1 does work, but
example 2 doesn't?
How kann i tell R, that i want it to expand the string dfb to dfb[,2]
inside the function.
In the end i want the function to change the second variable of the
dataframe dfb permanently to factor (not just inside the function).
Thanks in advance!
Winfried
Example 1:
dfa - data.frame(a=c(1:4),b=c(1:4))
dfa[,2] - factor(dfa[,2])
is.factor(dfa[,2])
TRUE
Example 2:
dfb - data.frame(a=c(1:4),b=c(1:4))
f.fact - function(x) {x[,2] - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
FALSE
PS: I tried a whole lot of other things like, ...
I really don't know where to keep on searching.
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x)[,2] - factor(get(x)[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x[,2]) - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {get(x)[,2] - factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Object 'x' nicht gefunden
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {assign(x[,2], factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
Ungültiges erstes Argument
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {quote(x)[,2], factor(x[,2])}
f.fact(dfb)
is.factor(dfb[,2])
Unerwartetes ','
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {
name - paste0(quote(x),[,2])
assign(name, factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
FALSE
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {
name - paste0(get(x),[,2])
assign(name, factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
Falsche Anzahl von Dimensionen
dfb - data.frame(a=c(1,2,3,4),b=c(1,2,3,4))
f.fact - function(x) {
name - paste0(x,[,2])
assign(name, factor(x[,2]))}
f.fact(dfb)
is.factor(dfb[,2])
Falsche Anzahl von Dimensionen
ächz ...
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Re: [R] glm poisson and quasipoisson
I am sure, that this is not a pure Poisson! Huge overdispersion! You get inflated confidence intervals! (although, the point estimates of the regression coefficients stay the same) Try to look for the causes of overdispersion! It may be geteroscedastisity? What is the nature of the response, is it the positive integers? Perhaps in your model still missing something important predictors? Or just you can try the Gamma or log-normal distr. Ivan, IPAE UB RAS -- View this message in context: http://r.789695.n4.nabble.com/glm-poisson-and-quasipoisson-tp4657234p4657310.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.
Re: [R] Relative Risk in logistic regression
Ivan, In reference to your part 2), in 1989 Li and Duan published a paper where they examined the effect of using the wrong link function ( http://projecteuclid.org/DPubS?service=UIversion=1.0verb=Displayhandle=euclid.aos/1176347254). The short version is that they found that in common models (lm and glm and others) and when the x-variables meet certain conditions, then the estimates of the slopes will change by a multiplicative constant as will the variance co-variance matrix. Many of the tests are still well behaved in this condition. The link functions you mentioned are all very similar to each other, so the impact of using a wrong one will be very minor. On Fri, Feb 1, 2013 at 8:33 AM, Ivan-K k...@ipae.uran.ru wrote: Dear colleagues, I have 2 points: One opinion and one question. 1) In one paper in a peer-reviewed journal, I read about the idea of using a logit regression as a surrogate for the log-binomial, just adding the numerator to the denominator ... Its tempting to immediately get the RR instead of OR ... I tried it and I think it's a bad idea, the confidence intervals dramatically inflated! Any opinions? 2) What would be the criteria for selection of link - functions for binary data? Usually I use the logit - just for simplest interpretation of parameters. Using logit, probit, log-log, and log-log, I get identical values of the maximum-likelihood, Pearson statistics, overdispersion parameter, etc. However, the regression coefficients and its standard errors are different (for logit b is the maximum, for the probit min., for log-log C- log-log are between them). LRs close, but the maximum has the log-log. Wald criterion - the maximum for the probit. ?What are the interpretations for regression parameters (except logit)??? Ivan, IPAE RAS -- View this message in context: http://r.789695.n4.nabble.com/Relative-Risk-in-logistic-regression-tp4657040p4657297.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. -- Gregory (Greg) L. Snow Ph.D. 538...@gmail.com [[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.
Re: [R] glm poisson and quasipoisson
On Thu, Jan 31, 2013 at 2:13 PM, Wim Kreinen wkrei...@gmail.com wrote: Hello, I have a question about modelling via glm. I think you are way off track. Either the data, glm, or both, are not what you think they are. I have a dataset skn300.tab - structure(list(n = 1:97, freq = c(0L, 0L, 0L, 0L, 1L, 7L, 40L, 100L, 276L, 543L, 952L, 1414L, 1853L, 2199L, 2435L, 2270L, 2042L, 1679L, 1386L, 1108L, 922L, 792L, 642L, 597L, 453L, 424L, 370L, 297L, 278L, 218L, 208L, 172L, 174L, 149L, 124L, 98L, 98L, 67L, 78L, 67L, 46L, 34L, 31L, 42L, 34L, 21L, 28L, 18L, 18L, 18L, 10L, 19L, 6L, 9L, 10L, 6L, 6L, 5L, 3L, 9L, 4L, 3L, 4L, 5L, 2L, 6L, 4L, 2L, 2L, 3L, 3L, 0L, 0L, 0L, 0L, 2L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 2L, 1L, 0L, 0L, 0L, 0L, 2L, 0L, 0L, 0L, 1L), kum = c(0L, 0L, 0L, 0L, 1L, 8L, 48L, 148L, 424L, 967L, 1919L, L, 5186L, 7385L, 9820L, 12090L, 14132L, 15811L, 17197L, 18305L, 19227L, 20019L, 20661L, 21258L, 21711L, 22135L, 22505L, 22802L, 23080L, 23298L, 23506L, 23678L, 23852L, 24001L, 24125L, 24223L, 24321L, 24388L, 24466L, 24533L, 24579L, 24613L, 24644L, 24686L, 24720L, 24741L, 24769L, 24787L, 24805L, 24823L, 24833L, 24852L, 24858L, 24867L, 24877L, 24883L, 24889L, 24894L, 24897L, 24906L, 24910L, 24913L, 24917L, 24922L, 24924L, 24930L, 24934L, 24936L, 24938L, 24941L, 24944L, 24944L, 24944L, 24944L, 24944L, 24946L, 24947L, 24947L, 24947L, 24947L, 24947L, 24947L, 24948L, 24948L, 24948L, 24949L, 24951L, 24952L, 24952L, 24952L, 24952L, 24952L, 24954L, 24954L, 24954L, 24954L, 24955L)), .Names = c(n, freq, kum), row.names = c(NA, -97L), class = data.frame) that looks like as if it where poisson distributed (actually I would appreciate that) but it isnt plot(skn300.tab) My guess, we are looking at the pdf and cdf (maybe even of a Poisson process), but not at any data that lends itself to a (generalized) linear model. Consult a statistician, post on stackexchange, read about regression, or better define your actual R problem here, demonstrating this is not homework - see the posting guide. Cheers because mean unequals var. mean (x) [1] 901.7827 var (x) [1] 132439.3 Anyway, I tried to model it via poisson and quasipoisson. Actually, just to get an impression how glm works. But I dont know how to interprete the data. Of course this is the case because my knowledge concerning logistic regressions is rather limited. Hoping there is somebody with mercy I would like to understand which parameters are important, e.g. which paramter might give me a hint that a poisson model is a bad idea. For hints concerning some tutorials about reading glm-output I would appreciate as well. Thanks Wim skn300.glmp - glm (freq~n, data=skn300.tab, family=poisson) summary (skn300.glmp) Call: glm(formula = freq ~ n, family = poisson, data = skn300.tab) Deviance Residuals: Min 1Q Median 3Q Max -51.332 -9.383 -6.599 -3.959 55.111 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 7.2374375 0.0093285 775.8 2e-16 *** n -0.0539424 0.0003699 -145.8 2e-16 *** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (Dispersion parameter for poisson family taken to be 1) Null deviance: 71731 on 96 degrees of freedom Residual deviance: 37383 on 95 degrees of freedom AIC: 37800 Number of Fisher Scoring iterations: 6 skn300.glmq - glm (freq~n, data=skn300.tab, family=quasipoisson) summary (skn300.glmq) Call: glm(formula = freq ~ n, family = quasipoisson, data = skn300.tab) Deviance Residuals: Min 1Q Median 3Q Max -51.332 -9.383 -6.599 -3.959 55.111 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 7.237438 0.186381 38.831 2e-16 *** n -0.053942 0.007391 -7.298 8.8e-11 *** --- Signif. codes: 0 *** 0.001 ** 0.01 * 0.05 . 0.1 1 (Dispersion parameter for quasipoisson family taken to be 399.1874) Null deviance: 71731 on 96 degrees of freedom Residual deviance: 37383 on 95 degrees of freedom AIC: NA Number of Fisher Scoring iterations: 6 dput (skn300.tab) structure(list(n = 1:97, freq = c(0L, 0L, 0L, 0L, 1L, 7L, 40L, 100L, 276L, 543L, 952L, 1414L, 1853L, 2199L, 2435L, 2270L, 2042L, 1679L, 1386L, 1108L, 922L, 792L, 642L, 597L, 453L, 424L, 370L, 297L, 278L, 218L, 208L, 172L, 174L, 149L, 124L, 98L, 98L, 67L, 78L, 67L, 46L, 34L, 31L, 42L, 34L, 21L, 28L, 18L, 18L, 18L, 10L, 19L, 6L, 9L, 10L, 6L, 6L, 5L, 3L, 9L, 4L, 3L, 4L, 5L, 2L, 6L, 4L, 2L, 2L, 3L, 3L, 0L, 0L, 0L, 0L, 2L, 1L, 0L, 0L, 0L, 0L, 0L, 1L, 0L, 0L, 1L, 2L, 1L, 0L, 0L, 0L, 0L, 2L, 0L, 0L, 0L, 1L), kum = c(0L, 0L, 0L, 0L, 1L, 8L, 48L, 148L, 424L, 967L, 1919L, L, 5186L, 7385L, 9820L, 12090L, 14132L, 15811L, 17197L, 18305L, 19227L, 20019L, 20661L, 21258L, 21711L, 22135L, 22505L, 22802L, 23080L, 23298L, 23506L, 23678L, 23852L, 24001L,
[R] Correcting and Adding to Data Frame
Running on Slackware here with R-2.15.2. There is a data frame I need to edit to correct mis-spellings and to add a row. I used the edit() command with emacs but could not find the mis-spelled strings nor figure out how to add another row. The data frame is stored in /usr/lib/R/library/bio.infer/data/itis.ttable.rda and I load it after invoking the bio.infer library. Running the bio.infer get.taxonomic() function brings up a dialog box asking me to correct 5 entries; four are mis-spellings that I correct. When I close the dialog box the function returns a message that one taxon is not in the table: bcnt - get.taxonomic(emapben) The following taxa are not in ITIS: RADOTANYPUS Adding this taxon to the table seems to make no difference: itis.ttable - rbind(itis.ttable, data.frame(PHYLUM = ARTHROPODA, SUBPHYLUM = NA, SUPERCLASS = NA, CLASS = INSECTA, SUBCLASS = NA, INFRACLASS = NA, SUPERORDER = NA, ORDER = DIPTERA, SUBORDER = NEMATOCERA, INFRAORDER = CULICOMORPHA, SUPERFAMILY = CHIRONOMOIDAE, FAMILY = CHIRONOMIDAE, SUBFAMILY = TANYPODINAE, TRIBE = NA, SUBTRIBE = NA, GENUS = RADOTANYPUS, TAXON = NA)) bcnt - get.taxonomic(emapben) The following taxa are not in ITIS: RADOTANYPUS When I edit the names in the get.taxonomic() dialog box they seem to not be corrected in the data frame as they also are displayed the second time. And, using rbind() to add the missing row does not seem to take. What steps have I missed here, and how should I make these corrections and the addition? Rich __ 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] Range difference of plot two arrays in one plot
One method is to follow correct usage of base graphics and only use the plot
function once for each plot. To overlay, use a function like points or lines.
Another approach is to use lattice graphics or ggplot2 and give the data to
higher-level plot routines that know how to plot multiple groups of data
together.
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Yuan, Rebecca rebecca.y...@bankofamerica.com wrote:
Hello all,
When I tried to plot the following two arrays in one figure with the
following:
x = c(0,0,0,10,20,30)
y = c(40,50,60,70,80,90)
plot(x, type='o', ylim=c(min(x),max(x)))
par(new=T)
plot(y, type='l', ylim=c(min(y),max(y)))
Found that the first points and last points from those two arrays are
overlapping together, but the value 30 is not equal to 90. How could I
draw those two arrays in one plot with their real values shown in the
plot? i.e., x is in the lower part of the plot and y is in the upper
part of the plot.
Thanks very much!
Cheers,
Rebecca
--
This message, and any attachments, is for the intended
r...{{dropped:5}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Range difference of plot two arrays in one plot
Hello David,
When I used the same ylim for both plots, I got what I need:
x = c(0,0,0,10,20,30)
y = c(40,50,60,70,80,90)
plot(x, type='o', ylim=c(min(x,y),max(x,y)))
par(new=T)
plot(y, type='l', ylim=c(min(x,y),max(x,y)))
Thanks,
Rebecca
-Original Message-
From: David Winsemius [mailto:dwinsem...@comcast.net]
Sent: Friday, February 01, 2013 11:39 AM
To: Yuan, Rebecca
Subject: Re: [R] Range difference of plot two arrays in one plot
On Feb 1, 2013, at 10:28 AM, Yuan, Rebecca wrote:
Hello all,
When I tried to plot the following two arrays in one figure with the
following:
x = c(0,0,0,10,20,30)
y = c(40,50,60,70,80,90)
plot(x, type='o', ylim=c(min(x),max(x)))
par(new=T)
plot(y, type='l', ylim=c(min(y),max(y)))
Found that the first points and last points from those two arrays are
overlapping together, but the value 30 is not equal to 90. How could I
draw those two arrays in one plot with their real values shown in the
plot? i.e., x is in the lower part of the plot and y is in the upper
part of the plot.
Use the same ylim for each plot?
--
David Winsemius, MD
Alameda, CA, USA
--
This message, and any attachments, is for the intended r...{{dropped:2}}
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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] Transforming 4x3 data frame into 2 column df in R
Hi, library(reshape) foo$id-row.names(foo) melt(foo,id.var=id) id variable value #1 n w 1.51550092 #2 q w 0.09977303 #3 r w 1.17083866 #4 n x 1.43375720 #5 q x 0.81737610 #6 r x 1.24693470 #7 n y 1.27916240 #8 q y 1.61234020 #9 r y 0.87121350 #10 n z 1.17712300 #11 q z 0.15107370 #12 r z 0.84880290 A.K. - Original Message - From: Gundala Viswanath gunda...@gmail.com To: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch Cc: Sent: Friday, February 1, 2013 1:00 AM Subject: [R] Transforming 4x3 data frame into 2 column df in R I have the following data frame: foo w x y z n 1.51550092 1.4337572 1.2791624 1.1771230 q 0.09977303 0.8173761 1.6123402 0.1510737 r 1.17083866 1.2469347 0.8712135 0.8488029 What I want to do is to change it into : newdf 1 n w 1.51550092 2 q w 0.09977303 3 r w 1.17083866 4 n x 1.43375725 5 q x 0.81737606 6 r x 1.24693468 7 n y 1.27916241 8 q y 1.61234016 9 r y 0.87121353 10 n z 1.17712302 11 q z 0.15107369 12 r z 0.84880292 Whtat's the way to do it? - G.V. [[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. __ 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] CFA with lavaan or with SEM
Not sure if you are aware of the OpenMx SEM package (http://openmx.psyc.virginia.edu/). It's a very full-featured structural equation modeling package. __ 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] Summary of data for each year
Hi, You could use: creek - read.csv(creek.csv,sep=\t) colnames(creek) - c(date,flow) creek$date - as.Date(creek$date, %m/%d/%Y) creek1 - within(creek, year - format(date, '%Y')) library(data.table) creek2- data.table(creek1) creek2[,list(MEAN=.Internal(mean(flow)),MEDIAN=median(flow),MAX=max(flow),MIN=min(flow)),by=list(year)] # year MEAN MEDIAN MAX MIN #1: 1999 0.6365604 0.47695 7.256 0.3187 #2: 2000 0.2819057 0.20810 2.380 0.1370 #3: 2001 0.2950348 0.22260 2.922 0.1769 #4: 2002 0.5345666 0.21190 14.390 0.1279 #5: 2003 1.0351742 0.71730 10.150 0.3492 #6: 2004 0.9691180 0.65240 11.710 0.4178 #7: 2005 1.2338066 0.72790 17.720 0.4722 #8: 2006 0.5458652 0.42820 3.351 0.2651 #9: 2007 0.6331271 0.40410 9.629 0.2784 #10: 2008 0.8792396 0.64770 4.596 0.4131 #11: 2009 0.8465300 0.59450 6.383 0.3877 A.K. - Original Message - From: Janesh Devkota janesh.devk...@gmail.com To: r-help@r-project.org Cc: Sent: Friday, February 1, 2013 2:32 AM Subject: [R] Summary of data for each year Hello All, I have a data with two columns. In one column it is date and in another column it is flow data. I was able to read the data as date and flow data. I used the following code: creek - read.csv(creek.csv) library(ggplot2) creek[1:10,] colnames(creek) - c(date,flow) creek$date - as.Date(creek$date, %m/%d/%Y) The link to my data is https://www.dropbox.com/s/eqpena3nk82x67e/creek.csv Now, I want to find the summary of each year. I want to especially know mean, median, maximum etc. Thanks. Janesh [[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. __ 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] obtainl survival curves for single strata
Dr. Therneau,
You are correct about the fit:
(cph.approve -
coxph(Surv(fundterm,resp)~I(CommitAmt/1e5)+res+CommitedRate+dayflag+mth+strata(termfac),
data=dfa,
subset=(HedgeDate2012-11-15 p1!=FixedOpen), method=efron))
However the output from survfit has individuals in each column and the strata
are stacked so, in order to use that I have to subset the exact rows for the
strata I need even though the strata is provided in the newdata argument.
Based on the help file it seems like the function should be able to use the
strata info and return a single strata per subject. I am pasting my current
code, which is not fast due to calls to reshape. If sth similar can be achieved
by calling the compiled code it should run much faster allowing 100k subjects
to be done in 2-3 min.
To use survfit as is, I would need to achieve subsetting in a for loop (going
through columns), which is even slower than reshape. In my old code I processed
subjects one by one in a for loop and that was much slower than grouping all
subjects from the same strata together as in the code below.
surv.approve - survfit(cph.approve)
b1 - c(1,cumsum(surv.approve$strata)+1)
b2 - cumsum(surv.approve$strata)
b1 - b1[-length(b1)]
stratbins -
data.frame(termfac=as.integer(substring(names(b2),9,9)),start=b1,finish=b2)
stratbins
termfac start finish
1 1 1 93
2 294187
3 3 188282
4 4 283372
5 5 373462
6 6 463553
7 7 554633
8 8 634695
9 9 696789
strats - sort(unique(newapp$termfac))
for (jj in strats){
cat('strata ',jj,'\n')
block - newapp[newapp$termfac==jj,]
surv -
surv.approve$surv[stratbins[stratbins$termfac==jj,start]:stratbins[stratbins$termfac==jj,finish]]
risk - predict(cph.approve,new=block,type=risk,ref=sample)
newsurv - outer(surv,risk,^)
days -
as.Date(outer(surv.approve$time[stratbins[stratbins$termfac==jj,start]:stratbins[stratbins$termfac==jj,finish]],
block$HedgeDate,+))
fund - t(t(newsurv)*block$CommitAmt)
fund - rbind(block$CommitAmt,fund)
fund - -diff(fund)
fund - as.data.frame(t(fund))
fund$acct - as.integer(rownames(fund))
ncols - ncol(fund)-1
fundlong -
reshape(fund,dir=long,varying=list(colnames(fund)[1:ncols]),idvar=acct,timevar=daycnt)
fundlong - fundlong[order(fundlong$acct,fundlong$daycnt),]
days - matrix(days,nrow=nrow(days))
days - t(days)
days - cbind(fund[,acct],days)
days - as.data.frame(days)
colnames(days)[1] - acct
daylong -
reshape(days,dir=long,varying=list(colnames(days)[2:(ncols+1)]),idvar=acct,timevar=dt)
daylong - daylong[order(daylong$acct,daylong$dt),]
daylong$V2 - as.Date(daylong$V2,origin = 1970-01-01)
fundlong$dt - as.character(daylong$V2)
sqlSave(ch1,fundlong,tmpfund)
sqlQuery(ch1, insert into RRfund (acct,dt,fund,daycnt) select
acct,dt,V1,daycnt from tmpfund)
sqlQuery(ch1,drop table tmpfund)
}
Output looks like:
acctdt funddaycnt
3623963 2012-11-16 00:00:00 472.99329489343 1
3623963 2012-11-17 00:00:00 5842.482289437712
3623963 2012-11-18 00:00:00 7807.171026727333
3623963 2012-11-19 00:00:00 7244.017697003454
3623963 2012-11-20 00:00:00 7073.831095927985
3623963 2012-11-21 00:00:00 8745.915155128846
3623963 2012-11-22 00:00:00 9473.871528069497
3623963 2012-11-23 00:00:00 12627.88904229168
3623963 2012-11-24 00:00:00 19598.56847130979
3623963 2012-11-25 00:00:00 56094.581213680210
3623963 2012-11-26 00:00:00 25690.439183149 11
3623963 2012-11-27 00:00:00 25420.991525671412
3623963 2012-11-28 00:00:00 30322.375086543413
3623963 2012-11-29 00:00:00 18651.113684675814
3623963 2012-11-30 00:00:00 20291.452806729215
Stephen
-Original Message-
From: Terry Therneau [mailto:thern...@mayo.edu]
Sent: Friday, February 01, 2013 10:11 AM
To: r-help@r-project.org; Bond, Stephen
Subject: Re: obtainl survival curves for single strata
Stephen,
I can almost but not quite get my arms around what you are asking. A bit
more detail
would help. Like
cph.approve = what kind of object, generated by function __
But, let me make a guess:
cph is the result of coxph, and the model has both covariates and a strata
You want predictioned survival curves, more than 1, of the type covariates
= a, b,c,
strata=1 covariates = d,e, f, strata=2, ... for arbitrary covariates and
strata.
Now, the predicted survival curves for different strata are different lengths.
The survfit.coxph routine gets around this by being verbose: it expects you to
give it
covariate sets, and returns
all of the strata for each covariate. This allows it to give a compact result.
You can
always do:
newpred - survfit(cox-model-fit, newdata=something)
newpred[5,17] # survival curve for the 5th strata, covariates from
[R] heatmap tile size question
Hello List I was wondering if it is possible to make the individual 'tiles' in a heatmap larger. Often when I plot heatmaps and want to label the rows with eg gene names I either have to shrink the text or leave it out altogether as it becomes so small as to be unreadable. I wondered if there was a way to make the 'tiles' deeper and therefore allow more room for the row labels. I apologise if this is not entirely clear but the toy code below might illustrate the problem. library(hgu133plus2.db) # fake up some data testMat - matrix(rnorm(4500, 4, 2), 150, 30) testMat[,15:30] - testMat[,15:30]*4 # clear differnce geneNames - as.list(hgu133plus2SYMBOL) # get some gene names rownames(testMat) - stack(sample(geneNames, 150))[,1] # relabel the matrix heatmap(testMat) # row labels overlap Best Iain [[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.
Re: [R] Fastest way to compare a single value with all values in one column of a data frame
I've compared the solutions.
*Solution 1:*
myf - function( df1, df2 ){
cond - df2$a min(df1$a)
if( cond )
{
idx - which( df1$a == min(df1$a) )
df1[idx, ] - df2[1, ]
}
df1
}
# On a larger example,
set.seed(4530)
tst - data.frame(item = 1:1000,a = rnorm(1000),b = rnorm(1000)) # large
data frame
u-tst
system.time(
for(i in 1:10){
y-data.frame(item=(1000+i),a=rnorm(1),b=rnorm(1)) # small data frame,
every time new
u - myf(u, y)
})
Took me about 31.90 sec
*Solution 2:*
set.seed(4530)
x - data.frame(item = 1:1000,a = rnorm(1000),b = rnorm(1000)) # large data
frame
system.time(
for(i in 1:10){
y-data.frame(item=(1000+i),a=rnorm(1),b=rnorm(1)) # small data frame,
every time new
u[intersect(which(u$a y$a),which.min(u$a)),] - y
})
The solution is correct (despite warnings) but took longer - about 48.84
sec.
Dimitri
On Wed, Jan 30, 2013 at 3:27 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
In realy, values in a will be not integers, but numeric. They will never
be identical, but it could be that they are pretty close - I don't know
after how many points after the comma matter.
Dimitri
On Wed, Jan 30, 2013 at 2:06 PM, arun smartpink...@yahoo.com wrote:
Hi,
Any chance x$a to have the same number repeated?
If `Item` and `a` are unique, I guess both the solutions should work.
set.seed(1851)
x-
data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:45,20,replace=F),b=sample(20:50,20,replace=F),stringsAsFactors=F)
y- data.frame(item=z,a=3,b=10,stringsAsFactors=F)
x[intersect(which(x$a y$a),which.min(x$a)),]
# item a b
#17c 1 48
x[x$a==which.min(x$a[x$ay$a]),]
# item a b
#17c 1 48
#or
x[x$a%in%which.min(x$a[x$ay$a]),]
# item a b
#17c 1 48
x[x$a%in%which.min(x$a[x$ay$a]),]-y
tail(x)
# item a b
#15q 45 30
#16g 10 23
#17z 3 10
#18r 15 39
#19l 18 45
#20t 35 33
#However, if `item` column is unique, but `a` is not, then the one I
mentioned previously arise.
set.seed(1851)
x1-
data.frame(item=sample(letters[1:20],20,replace=F),a=sample(1:10,20,replace=T),b=sample(20:50,20,replace=F),stringsAsFactors=F)
y1- data.frame(item=z,a=3,b=10,stringsAsFactors=F)
x1[intersect(which(x1$a y1$a),which.min(x1$a)),]
# item a b
#3s 1 41
x1[x1$a==which.min(x1$a[x1$ay1$a]),]
# item a b
#3 s 1 41
#11h 1 46
#17c 1 48
x1[x1$a==which.min(x1$a[x1$ay1$a]),]- y1
A.K.
From: Dimitri Liakhovitski dimitri.liakhovit...@gmail.com
To: arun smartpink...@yahoo.com
Cc: R help r-help@r-project.org; Jessica Streicher
j.streic...@micromata.de
Sent: Wednesday, January 30, 2013 1:49 PM
Subject: Re: [R] Fastest way to compare a single value with all values in
one column of a data frame
Sorry - I should have clarified:
My identifiers (in column item) will always be unique. In other words,
one entry in column item will never be repeated - neither in x nor in y.
Dimitri
On Wed, Jan 30, 2013 at 1:27 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Thank you, everyone! I'll try to test those different approaches. Really
appreciate your help!
Dimitri
On Wed, Jan 30, 2013 at 11:03 AM, arun smartpink...@yahoo.com wrote:
HI,
Sorry, my previous solution doesn't work.
This should work for your dataset:
set.seed(1851)
x-
data.frame(item=sample(letters[1:5],20,replace=TRUE),a=sample(1:15,20,replace=TRUE),b=sample(20:30,20,replace=TRUE),stringsAsFactors=F)
y- data.frame(item=f,a=3,b=10,stringsAsFactors=F)
x[x$a%in%which.min(x[x$ay$a,]$a),]- y #if there are multiple minimum
values
set.seed(1241)
x1-
data.frame(item=sample(letters[1:10],1e4,replace=TRUE),a=sample(1:30,1e4,replace=TRUE),b=sample(1:100,1e4,replace=TRUE),stringsAsFactors=F)
y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F)
length(x1$a[x1$a==1])
#[1] 330
system.time({x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1})
# user system elapsed
# 0.000 0.000 0.001
length(x1$a[x1$a==1])
#[1] 0
#For some reason, it is not working when the multiple number of minimum
values some value
set.seed(1241)
x1-
data.frame(item=sample(letters[1:10],1e5,replace=TRUE),a=sample(1:30,1e5,replace=TRUE),b=sample(1:100,1e5,replace=TRUE),stringsAsFactors=F)
y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F)
length(x1$a[x1$a==1])
#[1] 3404
x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1
length(x1$a[x1$a==1])
#[1] 3404 #not getting replaced
#However, if I try:
set.seed(1241)
x1-
data.frame(item=sample(letters[1:10],1e6,replace=TRUE),a=sample(1:5000,1e6,replace=TRUE),b=sample(1:100,1e6,replace=TRUE),stringsAsFactors=F)
y1- data.frame(item=f,a=3,b=10,stringsAsFactors=F)
length(x1$a[x1$a==1])
#[1] 208
system.time(x1[x1$a%in%which.min(x1[x1$ay1$a,]$a),]- y1)
#user system elapsed
# 0.124 0.016 0.138
length(x1$a[x1$a==1])
#[1] 0
#Tried Jessica's solution:
set.seed(1851)
x-
[R] Change default order of colors line types
Dear R users, I'd like to change the default order of colors line types. Especially I am using ggplot2 and using color Set1. In Set1, the default color order is red, blue, green, violet,.. ect. However, I want to put red in fourth (not first). Likewise, I want to change the order of default linetype. I want to put solid line in fourth. How can I do thses? R code to draw the graph is qplot(variable, power, data = m.powers, colour = method, linetype=method, ylab = Power) + geom_line(aes(group = method), ylim = c(0,1)) + scale_colour_brewer(palette=Set1) Thank you, __ 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] Loading a list into the environment
R-helpers: Say I have a list: myvariables - list(a=1:10,b=20) Is there a way to load the list components into the environment as variables based on the component names? i.e. by applying this theoretical function to myvariables I would have the variables a and b loaded into the environment without having to explicitly define them. --j -- Jonathan A. Greenberg, PhD Assistant Professor Global Environmental Analysis and Remote Sensing (GEARS) Laboratory Department of Geography and Geographic Information Science University of Illinois at Urbana-Champaign 607 South Mathews Avenue, MC 150 Urbana, IL 61801 Phone: 217-300-1924 http://www.geog.illinois.edu/~jgrn/ AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307, Skype: jgrn3007 [[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.
[R] Armadillo error in R extension
Is there anyway with some experience in using armadillo in R C++ extensions?
My problem is the following:
I programmed a function in a header looking like
#include armadillo
inline arma::vec foo(input) {
... do something
return an arma::vec object
}
compiling this via R CMD INSTALL packagename (PKG_CXXFLAGS =
-I/folder/of/armadillo and armadillo_bits in my package)
I get the following error
error: expected initializer before 'foo'
The exact line and word is where inline arma::vec ends.
Does anyone know what this error could be?
Best
Simon
__
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] Change default order of colors line types
Dear R users, I'd like to change the default order of colors line types. Especially I am using ggplot2 and using color Set1. In Set1, the default color order is red, blue, green, violet,.. ect. However, I want to put red in fourth (not first). Likewise, I want to change the order of default linetype. I want to put solid line in fourth. How can I do thses? R code to draw the graph is qplot(variable, power, data = m.powers, colour = method, linetype=method, ylab = Power) + geom_line(aes(group = method), ylim = c(0,1)) + scale_colour_brewer(palette=Set1) Thank you, __ 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] order function
Hi I'm having a simple issue with the order function. When I use the following code my data is not ordered correctly output[order(output[,3]),] Namebeta pval 881 9.09303277751237 0.000100253350483199 74026.40553461638365 0.00010228641631914 4879 -8.88509881106217 0.000103251645995887 However when I export the data and sort it in excel I see the following: Name beta pval pval 25037 -5.70737 2.48E-07 34294 -19.6931 1.04E-05 36002 -12.2478 1.63E-05 Any suggestions on how I can get this sort to work properly on data in scientifict format? .kripa [[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.
[R] order function
Hi I'm having a simple issue with the order function. When I use the following code my data is not ordered correctly output[order(output[,3]),] Namebeta pval 881 9.09303277751237 0.000100253350483199 74026.40553461638365 0.00010228641631914 4879 -8.88509881106217 0.000103251645995887 However when I export the data and sort it in excel I see the following: Name beta pval pval 25037 -5.70737 2.48E-07 34294 -19.6931 1.04E-05 36002 -12.2478 1.63E-05 Any suggestions on how I can get this sort to work properly on data in scientifict format? .kripa [[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.
Re: [R] cumulative sum by group and under some criteria
I am new to this mailing list. Is everything I posted seen for everyone in the mailing list? or just you can see the post? On Fri, Feb 1, 2013 at 11:33 AM, arun kirshna [via R] ml-node+s789695n465731...@n4.nabble.com wrote: HI, Could you show the modified code and also str(dataset)? A.K. -- If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657316.html To unsubscribe from cumulative sum by group and under some criteria, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4657074code=WmpvYW5uYTIwMTNAZ21haWwuY29tfDQ2NTcwNzR8LTE3NTE1MDA0MzY= . NAMLhttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewerid=instant_html%21nabble%3Aemail.namlbase=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespacebreadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml -- View this message in context: http://r.789695.n4.nabble.com/cumulative-sum-by-group-and-under-some-criteria-tp4657074p4657319.html Sent from the R help mailing list archive at Nabble.com. [[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.
Re: [R] Nested loop and output help
Thank you very much Petr,
I believe I have fixed my inquiry to not use floating points in my cycle as
you pointed out and to use the list structure to keep my results. I am
still at a loss as to how to run the multiple loops. I have tried quite a
few different strategies but my failure seems to illustrate that my
understanding of how the loops will run is nonexistent.
I would like to simultaneously let the following 4 things vary:
z1 - rnorm(ss,mean=400, sd=70) and z2 - rnorm(ss,mean=450, sd=90) by ss
- seq(5,9,by=1) which yields 5 6 7 8 9
r - cc by cc - seq(-0.5,0.5, by=0.25) which yields -0.50 -0.25 0.00 0.25
0.50
dev1 - oo1 by oo1 - seq(-10,10, by=5) which yields -10 -5 0 5 10
dev2 - oo2 by oo2 - seq(0,20, by=5) which yields 0 5 10 15 20
I tried to run the loops that would vary each of these above conditions
with the looped code attached below and failed very badly.
Thank you Petr and all,
Best,
Thomas
###SINGLE RUN
CODE
lll - vector(mode = list, length = 10)
names(lll) - c(mat1, mat2, mat3, mat4, cut1, cut2, out3a,
out3b, out4a, out4b)
z1 - rnorm(20,mean=400, sd=70)
z2 - rnorm(20,mean=450, sd=90)
cor - runif(1,min=0.4, max=0.6)
X - z1
Y = cor*z1+(1-cor)*z2
lll[[mat1]] - cbind(X,Y)
dev1 - sample(-40:40, 1, replace=T)
lll[[cut1]] - mean(X) + dev1
dev2 - sample(12:54, 1, replace=T)
lll[[cut2]] - mean(X) + dev1 + dev2
X2 - X-lll[[cut1]]
Y2 - Y-lll[[cut2]]
c3 - cor(X2,Y2)
lll[[mat2]] -cbind(X2,Y2)
a11 - ifelse( X lll[[cut1]] Y lll[[cut2]], 1, 0)
a12 - ifelse( X lll[[cut1]] Y = lll[[cut2]], 1, 0)
a21 - ifelse( X = lll[[cut1]] Y lll[[cut2]], 1, 0)
a22 - ifelse( X = lll[[cut1]] Y = lll[[cut2]], 1, 0)
lll[[mat3]] -matrix(c(sum(a11),sum(a21), sum(a12),sum(a22)), nrow = 2)
lll[[mat4]] -matrix(c(sum(a11),sum(a22), sum(a12),sum(a21)), nrow = 2)
lll[[out3a]] - mcnemar.test(lll[[mat3]], correct=FALSE)
lll[[out3b]] - mcnemar.test(lll[[mat3]], correct=TRUE)
lll[[out4a]] - chisq.test(lll[[mat4]], correct = FALSE)
lll[[out4b]] - chisq.test(lll[[mat4]], correct = TRUE)
print(lll)
capture.output(print(lll), file = C:/Chi_Square_fix/temp.txt, append =
TRUE)
##LOOPED
CODE#
lll - vector(mode = list, length = 10)
names(lll) - c(mat1, mat2, mat3, mat4, cut1, cut2, out3a,
out3b, out4a, out4b)
ss - seq(5,9,by=1)
cc - seq(-0.5,0.5, by=0.25)
oo1 - seq(-10,10, by=5)
oo2 - seq(0,20, by=5)
for(i in ss) {
for (j in cc) {
for (k in oo1) {
for (l in oo2) {
z1 - rnorm(ss,mean=400, sd=70)
z2 - rnorm(ss,mean=450, sd=90)
r - cc
X - z1
Y = r*z1+(1-r)*z2
lll[[mat1]] - cbind(X,Y)
dev1 - oo1
lll[[cut1]] - mean(X) + dev1
dev2 - oo2
lll[[cut2]] - mean(X) + dev1 + dev2
X2 - X-lll[[cut1]]
Y2 - Y-lll[[cut2]]
c3 - cor(X2,Y2)
lll[[mat2]] -cbind(X2,Y2)
a11 - ifelse( X lll[[cut1]] Y lll[[cut2]], 1, 0)
a12 - ifelse( X lll[[cut1]] Y = lll[[cut2]], 1, 0)
a21 - ifelse( X = lll[[cut1]] Y lll[[cut2]], 1, 0)
a22 - ifelse( X = lll[[cut1]] Y = lll[[cut2]], 1, 0)
lll[[mat3]] -matrix(c(sum(a11),sum(a21), sum(a12),sum(a22)), nrow = 2)
lll[[mat4]] -matrix(c(sum(a11),sum(a22), sum(a12),sum(a21)), nrow = 2)
lll[[out3a]] - mcnemar.test(lll[[mat3]], correct=FALSE)
lll[[out3b]] - mcnemar.test(lll[[mat3]], correct=TRUE)
lll[[out4a]] - chisq.test(lll[[mat4]], correct = FALSE)
lll[[out4b]] - chisq.test(lll[[mat4]], correct = TRUE)
print(lll)
capture.output(print(lll), file = C:/Chi_Square_fix/temp.txt, append =
TRUE)
}
}
}
}
On Feb 1, 2013 2:01 AM, PIKAL Petr petr.pi...@precheza.cz wrote:
Hi
see inline
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of staysafe23
Sent: Friday, February 01, 2013 1:01 AM
To: r-help@r-project.org
Subject: [R] Nested loop and output help
Hello Everyone,
My name is Thomas and I have been using R for one week. I recently
found your site and have been able to search the archives of posts.
This has given me some great information that has allowed me to craft
an initial design to an inquiry I would like to make into the breakdown
of McNemar's test. I have read an intro to R manual and the posting
guides and hope I am not violating them with this post. If so I will
re-ask my question in the proper format.
I have succeeded in writing a loop to vary one condition of my inquiry
but I am unable to understand how to vary the remaining three
conditions, each with its own for loop. Each time I try to do so I fail
miserably. Here is my current code :
n - seq(5,10,by=1)
for(i in n) {
z1 - rnorm(n,mean=400, sd=70)
z2 - rnorm(n,mean=450, sd=90)
cor - runif(1,min=0.4, max=0.6)
X - z1
Y = cor*z1+(1-cor)*z2
mat1 - cbind(X,Y)
dev1 - sample(-40:40, 1, replace=T)
cut1 - mean(X) + dev1
dev2 - sample(12:54, 1, replace=T)
cut2 - mean(X) + dev1 + dev2
X2 - X-cut1
Y2 - Y-cut2
Re: [R] order function
On 13-02-01 12:15 PM, Kripa R wrote: Hi I'm having a simple issue with the order function. When I use the following code my data is not ordered correctly output[order(output[,3]),] Namebeta pval 881 9.09303277751237 0.000100253350483199 74026.40553461638365 0.00010228641631914 4879 -8.88509881106217 0.000103251645995887 That's correctly ordered in the pval column. However when I export the data and sort it in excel I see the following: Name beta pval pval 25037 -5.70737 2.48E-07 34294 -19.6931 1.04E-05 36002 -12.2478 1.63E-05 That is a different dataset, also correctly ordered in the third column. Excel seems to have added a 4th heading for some reason. Any suggestions on how I can get this sort to work properly on data in scientifict format? You need to explain what is wrong before we can fix it. Duncan Murdoch .kripa [[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. __ 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] Loading a list into the environment
On Fri, Feb 1, 2013 at 5:24 PM, Jonathan Greenberg j...@illinois.edu wrote: R-helpers: Say I have a list: myvariables - list(a=1:10,b=20) Is there a way to load the list components into the environment as variables based on the component names? i.e. by applying this theoretical function to myvariables I would have the variables a and b loaded into the environment without having to explicitly define them. ?list2env -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.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.
Re: [R] Armadillo error in R extension
Look up the terribly wonderful RcppArmadillo package.
MW
On Feb 1, 2013, at 8:38 PM, Simon Zehnder szehn...@uni-bonn.de wrote:
Is there anyway with some experience in using armadillo in R C++ extensions?
My problem is the following:
I programmed a function in a header looking like
#include armadillo
inline arma::vec foo(input) {
... do something
return an arma::vec object
}
compiling this via R CMD INSTALL packagename (PKG_CXXFLAGS =
-I/folder/of/armadillo and armadillo_bits in my package)
I get the following error
error: expected initializer before 'foo'
The exact line and word is where inline arma::vec ends.
Does anyone know what this error could be?
Best
Simon
__
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] Armadillo error in R extension
On 1 February 2013 at 21:38, Simon Zehnder wrote:
| Is there anyway with some experience in using armadillo in R C++ extensions?
Yes, sure -- we wrote (and use) a package RcppArmadillo that provides very
easy access to Armadillo via Rcpp.
| My problem is the following:
|
| I programmed a function in a header looking like
|
| #include armadillo
|
| inline arma::vec foo(input) {
|
| ... do something
|
| return an arma::vec object
| }
|
| compiling this via R CMD INSTALL packagename (PKG_CXXFLAGS =
-I/folder/of/armadillo and armadillo_bits in my package)
|
| I get the following error
|
| error: expected initializer before 'foo'
|
| The exact line and word is where inline arma::vec ends.
|
| Does anyone know what this error could be?
We do. I would like to invite to subscribe to the rcpp-devel list, to peruse
its list archives, to study the RcppArmadillo examples (which cover all this)
and, if you still have questions, pose a small reproducible example on the
rcpp-devel list.
Also, we recently opened the Rcpp Gallery (http://gallery.rcpp.org) which
has a set of RcppArmadillo related posts you can read via the tags
http://gallery.rcpp.org/tags/armadillo/
Hope this helps, Dirk
--
Dirk Eddelbuettel | e...@debian.org | http://dirk.eddelbuettel.com
__
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] Loading a list into the environment
Hello,
Something like this?
myfun - function(x, envir = .GlobalEnv){
nm - names(x)
for(i in seq_along(nm))
assign(nm[i], x[[i]], envir)
}
myvariables - list(a=1:10,b=20)
myfun(myvariables)
a
b
Hope this helps,
Rui Barradas
Em 01-02-2013 22:24, Jonathan Greenberg escreveu:
R-helpers:
Say I have a list:
myvariables - list(a=1:10,b=20)
Is there a way to load the list components into the environment as
variables based on the component names? i.e. by applying this theoretical
function to myvariables I would have the variables a and b loaded into the
environment without having to explicitly define them.
--j
__
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] Change default order of colors line types
Sure. Start by reading [1] and take its advice to heart. Then study the many ways to create factors with values ordered in sync with the color ordering. The color ordering can be set using the default, using any of the various color picker functions, or manually. [1] http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Soyeon Kim yunni0...@gmail.com wrote: Dear R users, I'd like to change the default order of colors line types. Especially I am using ggplot2 and using color Set1. In Set1, the default color order is red, blue, green, violet,.. ect. However, I want to put red in fourth (not first). Likewise, I want to change the order of default linetype. I want to put solid line in fourth. How can I do thses? R code to draw the graph is qplot(variable, power, data = m.powers, colour = method, linetype=method, ylab = Power) + geom_line(aes(group = method), ylim = c(0,1)) + scale_colour_brewer(palette=Set1) Thank you, __ 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.
[R] mountain lion install, error message
Hi, I have been trying to install R on my mac which is running mountain lion. It is partially working, but one error message I am getting is: Error in function () : object '.activeModel' not found. I cannot find anything on this by googling. It appears, for example, when I try to 'add observation statistics to data' under the 'models' menu in R commander. I wondered if anyone might have any idea what this could mean, and how I might try and fix my installation. I have installed the various recommended additional packages, and uninstalled and re-installed R around 5 times now. I have had exactly the same performance after each installation. I'm rather at a loss at where to turn next to get things working properly. Thanks. Claire. __ 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 calculating p-values
I am trying to figure out how to calculate p-values for the difference in prevalence of a risk factor between men and women. For example, I find that 277 out of 710 male patients and 125 out of 305 female patients have obesity, what is the p-value for their difference? If there is a package that can calculate this in bulk, I would appreciate to learn about it! Thank you [[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.
Re: [R] Help calculating p-values
This is very basic stuff, so I think your main problem is statistical, not R related. Try posting on stats.stackexchange.com to get statistical advice or do some reading about differences in proportions, contingency tables, and the like. Incidentally, off the top of my head, I'd say there's no evidence that the proportions are different. See if I'm right when you do the correct calculation. -- Bert On Fri, Feb 1, 2013 at 3:45 PM, Jin Choi jin.hj.c...@gmail.com wrote: I am trying to figure out how to calculate p-values for the difference in prevalence of a risk factor between men and women. For example, I find that 277 out of 710 male patients and 125 out of 305 female patients have obesity, what is the p-value for their difference? If there is a package that can calculate this in bulk, I would appreciate to learn about it! Thank you [[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. -- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm __ 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] mountain lion install, error message
Perhaps you would benefit from joining R-sig-mac, and/or perusing their archives. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. clairehewson@zen zen11...@zen.co.uk wrote: Hi, I have been trying to install R on my mac which is running mountain lion. It is partially working, but one error message I am getting is: Error in function () : object '.activeModel' not found. I cannot find anything on this by googling. It appears, for example, when I try to 'add observation statistics to data' under the 'models' menu in R commander. I wondered if anyone might have any idea what this could mean, and how I might try and fix my installation. I have installed the various recommended additional packages, and uninstalled and re-installed R around 5 times now. I have had exactly the same performance after each installation. I'm rather at a loss at where to turn next to get things working properly. Thanks. Claire. __ 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] mountain lion install, error message
Dear Claire, This error is due to a bug in the previous version of the Rcmdr package; it is fixed in the current version on CRAN, version 1.9-4. Simply reinstall the Rcmdr package from CRAN and verify that you have the correct version. I hope this helps, John -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of clairehewson@zen Sent: Friday, February 01, 2013 6:29 PM To: r-help@r-project.org Subject: [R] mountain lion install, error message Hi, I have been trying to install R on my mac which is running mountain lion. It is partially working, but one error message I am getting is: Error in function () : object '.activeModel' not found. I cannot find anything on this by googling. It appears, for example, when I try to 'add observation statistics to data' under the 'models' menu in R commander. I wondered if anyone might have any idea what this could mean, and how I might try and fix my installation. I have installed the various recommended additional packages, and uninstalled and re-installed R around 5 times now. I have had exactly the same performance after each installation. I'm rather at a loss at where to turn next to get things working properly. Thanks. Claire. __ 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] mountain lion install, error message
Dear Jeff, As I just explained to Claire, this error is caused by a bug in an earlier version of the Rcmdr package; the bug affects the Rcmdr on all platforms, not just Mac OS X, and is fixed in the current version 1.9-4 of the Rcmdr on CRAN. Best, John -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jeff Newmiller Sent: Friday, February 01, 2013 8:14 PM To: clairehewson@zen; r-help@r-project.org Subject: Re: [R] mountain lion install, error message Perhaps you would benefit from joining R-sig-mac, and/or perusing their archives. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. clairehewson@zen zen11...@zen.co.uk wrote: Hi, I have been trying to install R on my mac which is running mountain lion. It is partially working, but one error message I am getting is: Error in function () : object '.activeModel' not found. I cannot find anything on this by googling. It appears, for example, when I try to 'add observation statistics to data' under the 'models' menu in R commander. I wondered if anyone might have any idea what this could mean, and how I might try and fix my installation. I have installed the various recommended additional packages, and uninstalled and re-installed R around 5 times now. I have had exactly the same performance after each installation. I'm rather at a loss at where to turn next to get things working properly. Thanks. Claire. __ 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. __ 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] x-axis labelling
On Feb 1, 2013, at 9:30 AM, Alfredo Tello wrote: Hi Folks! I have a question regarding an issue on which I´ve read a few threads but can´t resolve -- Labelling every other tick mark on the x-axis using factors. Here´s an example: Why not: db-data.frame(date=as.factor(c(1:50)),var=rnorm(50)) barplot(db$var,names.arg=ifelse( rep( c(TRUE,FALSE), length=length(levels(db$date) ) ), levels(db$date), ),las=2,cex.axis=0.8,cex=0.8) barplot(db$var,names.arg=ifelse( c(TRUE,FALSE), levels(db$date), },las=2,cex.axis=0.8,cex=0.8) -- David. This labels all the tickmarks in my barplot, which is the default behavior. I would like to label every other tick mark in the barplot -- 1,3,5,7 and so on. Thank you very much for your help! All the best, A -- Alfredo Tello (http://alfredotello.com) Sustainable Aquaculture Group Institute of Aquaculture University of Stirling Scotland, UK. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] repeating autocovariate functions
Hi there, Just wondering why my post was rejected? cheersRachel Subject: repeating autocovariate functions From: r-help-ow...@r-project.org To: moy...@hotmail.com Date: Sat, 2 Feb 2013 02:56:27 +0100 Message rejected by filter rule match --Forwarded Message Attachment-- Date: Fri, 1 Feb 2013 17:56:14 -0800 From: moy...@hotmail.com To: r-help@r-project.org Subject: repeating autocovariate functions Hi there, I would like to repeat an autocovariate term calculation using 30 different neighborhood sizes. Then I would like to run a nominal logistic regression on the generated autocovariate values and their respective neighborhood sizes to see which would be most appropriate to use in the final calculation of my autocovariate term. I have a matrix of x,y values: x y 174.7173-35.967 174.7166-35.9649 174.7174-35.968 174.7418-35.9678 174.741 -35.9672 174.7395-35.9671 (and 150 more) To calculate the autocovariate terms my code is as follows: library(spdep) Taranga - read.csv(file.choose()) #contains xy coordinates xy - cbind(Taranga$x, Taranga$y) acinvb - autocov_dist(Taranga$burrows.pa, xy, nbs=1,zero.policy = TRUE, type=inverse) datfrm - data.frame(autocov=acinvb, nbs=1) write.table(dfrm, file = 'results.csv',sep=,,row.names = FALSE) acinva - autocov_dist(Taranga$burrows.pa, xy, nbs=100,zero.policy = TRUE, type=inverse) dfrm -data.frame(autocov=acinva, nbs=100) names(datfrm) - NULL write.table(datfrm, file = 'results.csv',sep=,,append=TRUE, row.names = FALSE, col.names=FALSE) I want to repeat this function, each time adding 100 to nbs Then I will run a model from the resulting results.csv thanks in advance for any insight! Rachel -- View this message in context: http://r.789695.n4.nabble.com/repeating-autocovariate-functions-tp4657353.html Sent from the R help mailing list archive at Nabble.com. [[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.
Re: [R] Help calculating p-values
This smacks of homework to me. cheers, Rolf Turner On 02/02/2013 12:45 PM, Jin Choi wrote: I am trying to figure out how to calculate p-values for the difference in prevalence of a risk factor between men and women. For example, I find that 277 out of 710 male patients and 125 out of 305 female patients have obesity, what is the p-value for their difference? If there is a package that can calculate this in bulk, I would appreciate to learn about it! __ 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] Choice of text for intermediate level R programming course
The Subject line mostly says it. I'm designing it as a semester-long, 3 hours per week, course that takes in students who got the basics of R in stats classes, but don't really know how to program in it. Translation: if their own examples don't look enough like examples from previous work, they're stumped. Does anybody have a text for an intermediate R course (but not too highbrow, no total guide to object oriented programming, and class s3 vs. s4 objects), they would recommend. I read 24 books some ielementary some advance not many intermediate, and I have two I like but for quite difference reasons. I won't name them so as not to prejudice anybody. It needs to be English language and I greatly prefer a good index like \LaTeX\ produces by the makeindex() prograrm. Regards, William Grove -- William M. Grove, Ph.D. Psychology Department University of Minnesota My public encryption key.is: -BEGIN PGP PUBLIC KEY BLOCK- Version: GnuPG v1.2.2 (MingW32) mQGiBE+tS9ERBACCTYW1uclqHbUvBvP92rUN8zOlMWe2zr3NfRp6ELNyYa60H5PK +Xx/vbHGWF0MXhGLLVFMiRNBRomJ4S30Ify4exTg6vWpPrVsmDvyKLmYilWSOuw1 0yoimDdr4DphGQSKqVD/O/tLUhO1wyEKJI89qtJmFG4usnn7f/Bvt4cRZwCgq76W LRhGATk0E2GYl0n+2UeFU+8D/ipLpFYNJEkz3y492/9orimBsxsAdnXmK5bpOBG5 pQaoFuvvPMtkUYhVNYUqMyjJniM3F6mnv7OX52sHHcYn89ScO9S+WtWAYsekwTTE xb1wvwB9u7PoGDC46tihXhTl533oUcMFGMwlVKvFZ4zpWYo5W3i0DmOAtjcMPzl2 4wcfA/0fgbUe2JIkPTf6MFMQTFwpNhCGeDvTfbcC5YCCcAXWcnUT/GKFsvRT0Wje i/A4PMopJkfLRreM1GSZ6JY0cOL3U071M6YhVmldrL4X74OUAjfdo8mYosxe/qTe Ji0s6jg/oiXgY/hHLZzDE+AtQ6LRi0/1tpms2Qw9/7pXs8HHErQ0V2lsbGlhbSBN LiBHcm92ZSAoR1BHIFB1YmxpYyBLZXkpIDxncm92ZTAwMUB1bW4uZWR1PohbBBMR AgAbBQJPrUvRBgsJCAcDAgMVAgMDFgIBAh4BAheAAAoJEG2MPb65SVkFnNcAnRCi URO4Oy9dk5Kb6j/1yN1iElTJAKCnlwfwOt0J8jf2nJ78sNkiZNjM6bkCDQRPrUvT EAgAg52fj9/7k6BuJgmY+ynKm4rwWaRW6VjrnS5VI+luTmEkQi/w9Vz6dOmCJWlz VLsm7mEW0tvB9VwZbkrttoWnJOmqzRWNJxlRlj78jRq6yN16JZNhbPGDSbjL/+JV LzlhMetUpuTYPziuatSr7C2RxYYRd5pabgd3G/XowqSBtyD3jrViJ8CqcZlMQGfJ BRoiyBDlVEOO+aJ/KGLinfbhyz06tR50BpXFFlxBN1m+DInRLF6zRkjZUO+OyFO6 Acy0SM4htfetEI0H0+X1w/Cw8e2Ew48Wwsve+USRr+OLXqwkgBP/jTNRko6cUI5m LiXjkSXYdR0JonTzgUSh66TDBwADBQf/TR/xC5P5Vs6X5OPu3iHsKRp/dGVpL33D KBH2Ofw3ps48kQ65la/uMW+fZDvwZJukTIJL4dq5OGjxnwznj3hkZJZWLknRFDx+ ByoumskJXPENphqlnpDV+W2stNwoSY3f5gO9xTQSGeoqqLB6+Xwe4CDZo/wOuhMA nCXQeIJxiH3GsCnxkZY9e3rw9HdcE/6NbvKGZ/OOLMdorASG9eSuyQZDgKN7jXZ3 2GPbppaRjzUWuRGau1RBv9V0QMY5C0Z1c51njCHuR0cgcItad69IcgaujlC2kouu C/MW/VOA5rfGhvBBI7nr26MjhxwHUYWQw6i/KgzYwJ+X43cEBCoITIhGBBgRAgAG BQJPrUvTAAoJEG2MPb65SVkFDLoAn36dv6dBZyIrTD+lF3y1lXO/R1OHAJ0Sn0bi 0TdH/wwxUEFeq5eCMfoZTg== =91Ch -END PGP PUBLIC KEY BLOCK- [[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.
Re: [R] cumulative sum by group and under some criteria
Hi,
Saw your reply on Nabble:
#Your code:
library(zoo)
res1- do.call(rbind,lapply(lapply(split(d,list(d$m1,d$n1)),function(x)
{x$cterm1_p0L[x$Qm=c11]- cumsum(x$term1_p0[x$Qm=c11]);
x$cterm1_p0H[x$Qn=c12]- cumsum(x$term1_p0[x$Qn=c12]);
x$cterm1_p1L[x$Qm=c11]- cumsum(x$term1_p1[x$Qm=c11]);
x$cterm1_p1H[x$Qm=c12]- cumsum(x$term1_p1[x$Qn=c12]); #Check this line Qm
and Qn
x}),function(x) {x$cterm1_p0L-na.locf(x$cterm1_p0L,na.rm=F);
x$cterm1_p0H-na.locf(x$cterm1_p0H,na.rm=F);
x$cterm1_p1L-na.locf(x$cterm1_p1L,na.rm=F);
x$cterm1_p1H-na.locf(x$cterm1_p1H,na.rm=F);x}))
#should be:
colnames(d)-c(m1,n1,x1,y1,Fmm, Fnn, Qm, Qn, term1_p0,
term1_p1)
res1- do.call(rbind,lapply(lapply(split(d,list(d$m1,d$n1)),function(x)
{x$cterm1_p0L[x$Qm=c11]- cumsum(x$term1_p0[x$Qm=c11]);
x$cterm1_p0H[x$Qn=c12]- cumsum(x$term1_p0[x$Qn=c12]);
x$cterm1_p1L[x$Qm=c11]- cumsum(x$term1_p1[x$Qm= c11]);
x$cterm1_p1H[x$Qn=c12]- cumsum(x$term1_p1[x$Qn= c12]);
x}),function(x) {x$cterm1_p0L-na.locf(x$cterm1_p0L,na.rm=F);
x$cterm1_p0H-na.locf(x$cterm1_p0H,na.rm=F);
x$cterm1_p1L-na.locf(x$cterm1_p1L,na.rm=F);
x$cterm1_p1H-na.locf(x$cterm1_p1H,na.rm=F);x}))
row.names(res1) - 1:nrow(res1)
res1[,11:14][is.na(res1[,11:14])]- 0
res1[,11:14][is.na(res1[,11:14])]- 0
res1
# m1 n1 x1 y1 Fmm Fnn Qm Qn term1_p0 term1_p1 cterm1_p0L
cterm1_p0H cterm1_p1L cterm1_p1H
#1 2 2 0 0 0.00 0.00 1.000 1.000 0.8145062500 0.40960 0.0e+00
0.00 0.0 0.0
#2 2 2 0 1 0.00 0.60 1.000 0.400 0.0857375000 0.20480 0.0e+00
0.00 0.0 0.0
#3 2 2 0 2 0.00 1.00 1.000 0.000 0.0022562500 0.02560 0.0e+00
0.0022562500 0.0 0.02560
#4 2 2 1 0 0.61 0.00 0.695 0.695 0.0857375000 0.20480 0.0e+00
0.0022562500 0.0 0.02560
#5 2 2 1 1 0.61 0.62 0.390 0.380 0.009025 0.10240 0.0e+00
0.0022562500 0.0 0.02560
#6 2 2 1 2 0.63 1.00 0.370 0.000 0.0002375000 0.01280 0.0e+00
0.0024937500 0.0 0.03840
#7 2 2 2 0 1.00 0.00 0.500 0.500 0.0022562500 0.02560 0.0e+00
0.0024937500 0.0 0.03840
#8 2 2 2 1 1.00 0.67 0.165 0.165 0.0002375000 0.01280 2.37500e-04
0.0027312500 0.01280 0.05120
#9 2 2 2 2 1.00 1.00 0.000 0.000 0.062500 0.00160 2.43750e-04
0.0027375000 0.01440 0.05280
#10 3 2 0 0 0.00 0.00 1.000 1.000 0.7737809375 0.32768 0.0e+00
0.00 0.0 0.0
#11 3 2 0 1 0.00 0.65 1.000 0.350 0.0814506250 0.16384 0.0e+00
0.00 0.0 0.0
#12 3 2 0 2 0.00 1.00 1.000 0.000 0.0021434375 0.02048 0.0e+00
0.0021434375 0.0 0.02048
#13 3 2 1 0 0.67 0.00 0.665 0.665 0.1221759375 0.24576 0.0e+00
0.0021434375 0.0 0.02048
#14 3 2 1 1 0.60 0.64 0.400 0.360 0.0128606250 0.12288 0.0e+00
0.0021434375 0.0 0.02048
#15 3 2 1 2 0.66 1.00 0.340 0.000 0.0003384375 0.01536 0.0e+00
0.0024818750 0.0 0.03584
#16 3 2 2 0 0.71 0.00 0.645 0.645 0.0064303125 0.06144 0.0e+00
0.0024818750 0.0 0.03584
#17 3 2 2 1 0.69 0.66 0.325 0.325 0.0006768750 0.03072 0.0e+00
0.0024818750 0.0 0.03584
#18 3 2 2 2 0.64 1.00 0.360 0.000 0.178125 0.00384 0.0e+00
0.0024996875 0.0 0.03968
#19 3 2 3 0 1.00 0.00 0.500 0.500 0.0001128125 0.00512 0.0e+00
0.0024996875 0.0 0.03968
#20 3 2 3 1 1.00 0.74 0.130 0.130 0.118750 0.00256 1.18750e-05
0.0025115625 0.00256 0.04224
#21 3 2 3 2 1.00 1.00 0.000 0.000 0.003125 0.00032 1.21875e-05
0.0025118750 0.00288 0.04256
#22 2 3 0 0 0.00 0.00 1.000 1.000 0.7737809375 0.32768 0.0e+00
0.00 0.0 0.0
#23 2 3 0 1 0.00 0.60 1.000 0.400 0.1221759375 0.24576 0.0e+00
0.00 0.0 0.0
#24 2 3 0 2 0.00 0.65 1.000 0.350 0.0064303125 0.06144 0.0e+00
0.00 0.0 0.0
#25 2 3 0 3 0.00 1.00 1.000 0.000 0.0001128125 0.00512 0.0e+00
0.0001128125 0.0 0.00512
#26 2 3 1 0 0.77 0.00 0.615 0.615 0.0814506250 0.16384 0.0e+00
0.0001128125 0.0 0.00512
#27 2 3 1 1 0.60 0.62 0.400 0.380 0.0128606250 0.12288 0.0e+00
0.0001128125 0.0 0.00512
#28 2 3 1 2 0.61 0.72 0.390 0.280 0.0006768750 0.03072 0.0e+00
0.0001128125 0.0 0.00512
#29 2 3 1 3 0.65 1.00 0.350 0.000 0.118750 0.00256 0.0e+00
0.0001246875 0.0 0.00768
#30 2 3 2 0 1.00 0.00 0.500 0.500 0.0021434375 0.02048 0.0e+00
0.0001246875 0.0 0.00768
#31 2 3 2 1 1.00 0.58 0.210 0.210 0.0003384375 0.01536 0.0e+00
0.0001246875 0.0 0.00768
#32 2 3 2 2 1.00 0.60 0.200 0.200 0.178125 0.00384 1.78125e-05
0.0001425000 0.00384 0.01152
#33 2 3 2 3 1.00 1.00 0.000
Re: [R] cumulative sum by group and under some criteria
HI,
I think this should be more correct:
maxN-9
c11-0.2
c12-0.2
p0L-0.05
p0H-0.05
p1L-0.20
p1H-0.20
d - structure(list(m1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3),
n1 = c(2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3,
3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2), x1 = c(0,
0, 0, 1, 1, 1, 2, 2, 2, 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2,
2, 0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3), y1 = c(0, 1, 2, 0,
1, 2, 0, 1, 2, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1,
2, 0, 1, 2, 0, 1, 2, 0, 1, 2), Fmm = c(0, 0, 0, 0.7, 0.59,
0.64, 1, 1, 1, 0, 0, 0, 0, 0.63, 0.7, 0.74, 0.68, 1, 1, 1,
1, 0, 0, 0, 0.62, 0.63, 0.6, 0.63, 0.6, 0.68, 1, 1, 1), Fnn = c(0,
0.64, 1, 0, 0.51, 1, 0, 0.67, 1, 0, 0.62, 0.69, 1, 0, 0.54,
0.62, 1, 0, 0.63, 0.73, 1, 0, 0.63, 1, 0, 0.7, 1, 0, 0.7,
1, 0, 0.58, 1), Qm = c(1, 1, 1, 0.65, 0.45, 0.36, 0.5, 0.165,
0, 1, 1, 1, 1, 0.685, 0.38, 0.32, 0.32, 0.5, 0.185, 0.135,
0, 1, 1, 1, 0.69, 0.37, 0.4, 0.685, 0.4, 0.32, 0.5, 0.21,
0), Qn = c(1, 0.36, 0, 0.65, 0.45, 0, 0.5, 0.165, 0, 1, 0.38,
0.31, 0, 0.685, 0.38, 0.32, 0, 0.5, 0.185, 0.135, 0, 1, 0.37,
0, 0.69, 0.3, 0, 0.685, 0.3, 0, 0.5, 0.21, 0), term1_p0 = c(0.81450625,
0.0857375, 0.00225625, 0.0857375, 0.009025, 0.0002375, 0.00225625,
0.0002375, 6.25e-06, 0.7737809375, 0.1221759375, 0.0064303124999,
0.0001128125, 0.081450625, 0.012860625, 0.000676875, 1.1875e-05,
0.0021434375, 0.0003384375, 1.78125e-05, 3.125e-07, 0.7737809375,
0.081450625, 0.0021434375, 0.1221759375, 0.012860625, 0.0003384375,
0.0064303124999, 0.000676875, 1.78125e-05, 0.0001128125,
1.1875e-05, 3.125e-07), term1_p1 = c(0.4096, 0.2048, 0.0256,
0.2048, 0.1024, 0.0128, 0.0256, 0.0128, 0.0016, 0.32768,
0.24576, 0.06144, 0.00512, 0.16384, 0.12288, 0.03072, 0.00256,
0.02048, 0.01536, 0.00384, 0.00032, 0.32768, 0.16384, 0.02048,
0.24576, 0.12288, 0.01536, 0.06144, 0.03072, 0.00384, 0.00512,
0.00256, 0.00032)), .Names = c(m1, n1, x1, y1, Fmm,
Fnn, Qm, Qn, term1_p0, term1_p1), row.names = c(NA,
33L), class = data.frame)
library(zoo)
lst1- split(d,list(d$m1,d$n1))
res2-do.call(rbind,lapply(lst1[lapply(lst1,nrow)!=0],function(x){
x[,11:14]-NA;
x[,11:12][x$Qm=c11,]-cumsum(x[,9:10][x$Qm=c11,]);
x[,13:14][x$Qn=c12,]-cumsum(x[,9:10][x$Qn=c12,]);
colnames(x)[11:14]- c(cterm1_P0L,cterm1_P1L,cterm1_P0H,cterm1_P1H);
x1-na.locf(x);
x1[,11:14][is.na(x1[,11:14])]-0;
x1}))
row.names(res2)- 1:nrow(res2)
res2
# m1 n1 x1 y1 Fmm Fnn Qm Qn term1_p0 term1_p1 cterm1_P0L
cterm1_P1L cterm1_P0H cterm1_P1H
#1 2 2 0 0 0.00 0.00 1.000 1.000 0.8145062500 0.40960 0.00
0.0 0.00 0.0
#2 2 2 0 1 0.00 0.64 1.000 0.360 0.0857375000 0.20480 0.00
0.0 0.00 0.0
#3 2 2 0 2 0.00 1.00 1.000 0.000 0.0022562500 0.02560 0.00
0.0 0.0022562500 0.02560
#4 2 2 1 0 0.70 0.00 0.650 0.650 0.0857375000 0.20480 0.00
0.0 0.0022562500 0.02560
#5 2 2 1 1 0.59 0.51 0.450 0.450 0.009025 0.10240 0.00
0.0 0.0022562500 0.02560
#6 2 2 1 2 0.64 1.00 0.360 0.000 0.0002375000 0.01280 0.00
0.0 0.0024937500 0.03840
#7 2 2 2 0 1.00 0.00 0.500 0.500 0.0022562500 0.02560 0.00
0.0 0.0024937500 0.03840
#8 2 2 2 1 1.00 0.67 0.165 0.165 0.0002375000 0.01280 0.0002375000
0.01280 0.0027312500 0.05120
#9 2 2 2 2 1.00 1.00 0.000 0.000 0.062500 0.00160 0.0002437500
0.01440 0.0027375000 0.05280
#10 3 2 0 0 0.00 0.00 1.000 1.000 0.7737809375 0.32768 0.00
0.0 0.00 0.0
#11 3 2 0 1 0.00 0.63 1.000 0.370 0.0814506250 0.16384 0.00
0.0 0.00 0.0
#12 3 2 0 2 0.00 1.00 1.000 0.000 0.0021434375 0.02048 0.00
0.0 0.0021434375 0.02048
#13 3 2 1 0 0.62 0.00 0.690 0.690 0.1221759375 0.24576 0.00
0.0 0.0021434375 0.02048
#14 3 2 1 1 0.63 0.70 0.370 0.300 0.0128606250 0.12288 0.00
0.0 0.0021434375 0.02048
#15 3 2 1 2 0.60 1.00 0.400 0.000 0.0003384375 0.01536 0.00
0.0 0.0024818750 0.03584
#16 3 2 2 0 0.63 0.00 0.685 0.685 0.0064303125 0.06144 0.00
0.0 0.0024818750 0.03584
#17 3 2 2 1 0.60 0.70 0.400 0.300 0.0006768750 0.03072 0.00
0.0 0.0024818750 0.03584
#18 3 2 2 2 0.68 1.00 0.320 0.000 0.178125 0.00384 0.00
0.0 0.0024996875 0.03968
#19 3 2 3 0 1.00 0.00 0.500 0.500 0.0001128125 0.00512 0.00
0.0 0.0024996875 0.03968
#20 3 2 3 1 1.00 0.58 0.210 0.210 0.118750 0.00256 0.00
0.0 0.0024996875 0.03968
#21 3 2 3 2 1.00 1.00 0.000 0.000 0.003125 0.00032 0.003125
0.00032 0.002500 0.04000
#22 2 3 0 0 0.00 0.00 1.000
Re: [R] repeating autocovariate functions
The short answer is ... because it was from Nabble. The longer answer is that the filters needed to be tightened on Nabble postings because of a an autobot spammer and the moderation queue was being swamped with cases. If you send your questions as email (from your email client) to r-help@r-project.org you should have fewer problems. (you might want to learn to post in plain text, too.) -- David On Feb 1, 2013, at 7:58 PM, rachel buxton wrote: Hi there, Just wondering why my post was rejected? cheers Rachel Subject: repeating autocovariate functions From: r-help-ow...@r-project.org To: moy...@hotmail.com Date: Sat, 2 Feb 2013 02:56:27 +0100 Message rejected by filter rule match --Forwarded Message Attachment-- Date: Fri, 1 Feb 2013 17:56:14 -0800 From: moy...@hotmail.com To: r-help@r-project.org Subject: repeating autocovariate functions Hi there, I would like to repeat an autocovariate term calculation using 30 different neighborhood sizes. Then I would like to run a nominal logistic regression on the generated autocovariate values and their respective neighborhood sizes to see which would be most appropriate to use in the final calculation of my autocovariate term. I have a matrix of x,y values: x y 174.7173-35.967 174.7166-35.9649 174.7174-35.968 174.7418-35.9678 174.741 -35.9672 174.7395-35.9671 (and 150 more) To calculate the autocovariate terms my code is as follows: library(spdep) Taranga - read.csv(file.choose()) #contains xy coordinates xy - cbind(Taranga$x, Taranga$y) acinvb - autocov_dist(Taranga$burrows.pa, xy, nbs=1,zero.policy = TRUE, type=inverse) datfrm - data.frame(autocov=acinvb, nbs=1) write.table(dfrm, file = 'results.csv',sep=,,row.names = FALSE) acinva - autocov_dist(Taranga$burrows.pa, xy, nbs=100,zero.policy = TRUE, type=inverse) dfrm -data.frame(autocov=acinva, nbs=100) names(datfrm) - NULL write.table(datfrm, file = 'results.csv',sep=,,append=TRUE, row.names = FALSE, col.names=FALSE) I want to repeat this function, each time adding 100 to nbs Then I will run a model from the resulting results.csv thanks in advance for any insight! Rachel -- View this message in context: http://r.789695.n4.nabble.com/repeating-autocovariate-functions-tp4657353.html Sent from the R help mailing list archive at Nabble.com. David Winsemius, MD Alameda, CA, USA __ 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.
