[R] Decomposing tests of interaction terms in mixed-effects models
Dear R colleagues, a friend and I are trying to develop a modest workflow for the problem of decomposing tests of higher-order terms into interpretable sets of tests of lower order terms with conditioning. For example, if the interaction between A (3 levels) and C (2 levels) is significant, it may be of interest to ask whether or not A is significant at level 1 of C and level 2 of C. The textbook response seems to be to subset the data and perform the tests on the two subsets, but using the RSS and DF from the original fit. We're trying to answer the question using new explanatory variables. This approach (seems to) work just fine in aov, but not lme. For example, ## # Build the example dataset set.seed(101) Block - gl(6, 6, 36, lab = paste(B, 1:6, sep = )) A - gl(3, 2, 36, lab = paste(A, 1:3, sep = )) C - gl(2, 1, 36, lab = paste(C, 1:2, sep = )) example - data.frame(Block, A, C) example$Y - rnorm(36)*2 + as.numeric(example$A)*as.numeric(example$C)^2 + 3 * rep(rnorm(6), each=6) with(example, interaction.plot(A, C, Y)) # lme require(nlme) anova(lme(Y~A*C, random = ~1|Block, data = example)) summary(aov(Y ~ Error(Block) + A*C, data = example)) # Significant 2-way interaction. Now we would like to test the effect # of A at C1 and the effect of A at C2. Construct two new variables # that decompose the interaction, so an LRT is possible. example$AC - example$AC1 - example$AC2 - interaction(example$A, example$C) example$AC1[example$C == C1] - A1.C1 # A is constant at C1 example$AC2[example$C == C2] - A1.C2 # A is constant at C2 # lme fails (as does lmer) lme(Y ~ AC1 + AC, random = ~ 1 | Block, data = example) # but aov seems just fine. summary(aov(Y ~ Error(Block) + AC1 + AC, data = example)) ## AC was not significant, so A is not significant at C1 summary(aov(Y ~ Error(Block) + AC2 + AC, data = example)) ## AC was significant, so A is significant at C2 ## Some experimentation suggests that lme doesn't like the 'partial ## confounding' approach that we are using, rather than the variables ## that we have constructed. lme(Y ~ AC, random = ~ 1 | Block, data = example) lme(Y ~ A + AC, random = ~ 1 | Block, data = example) ## Are we doing anything obviously wrong? Is there another approach to the goal that we are trying to achieve? Many thanks, Andrew -- Andrew Robinson Department of Mathematics and StatisticsTel: +61-3-8344-6410 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 http://www.ms.unimelb.edu.au/~andrewpr http://blogs.mbs.edu/fishing-in-the-bay/ __ 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] about the 95%CI around the median...
Dear people I've learnt that by using the boxplot.stats command in the grDevices library I can get the 5-number summaries of a boxplot, plus other important information, like the confidence interval around the median. I'm interested in knowing the actual formula to used in that package to calculate that confidence interval. Can someone help me with this? Cheers, Fernando __ 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] about the 95%CI around the median...
See ?fivenum in the stats package. If you just type stats::fivenum you will get the code. The crucial calculations are in the last few lines. Simon. On Mon, 2008-08-04 at 16:19 +0930, Fernando Marmolejo Ramos wrote: Dear people I've learnt that by using the boxplot.stats command in the grDevices library I can get the 5-number summaries of a boxplot, plus other important information, like the confidence interval around the median. I'm interested in knowing the actual formula to used in that package to calculate that confidence interval. Can someone help me with this? Cheers, Fernando __ 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. -- Simon Blomberg, BSc (Hons), PhD, MAppStat. Lecturer and Consultant Statistician Faculty of Biological and Chemical Sciences The University of Queensland St. Lucia Queensland 4072 Australia Room 320 Goddard Building (8) T: +61 7 3365 2506 http://www.uq.edu.au/~uqsblomb email: S.Blomberg1_at_uq.edu.au Policies: 1. I will NOT analyse your data for you. 2. Your deadline is your problem. The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. - John Tukey. __ 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] graph
Hello, Is ?segments what you are looking for ? François -Message d'origine- De : [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] De la part de elyakhlifi mustapha Envoyé : vendredi 1 août 2008 17:15 À : r-help@r-project.org Objet : [R] graph Hello. I don't know how to do to ouput segments between points like the chart attached. Can you help me please? Thanks. .yahoo.fr __ 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] Major difference in multivariate analyses SPSS and R
Dear colleagues, I know SPSS can not compute linear mixed models. I used 'R' before for computing multivariate analyses. But, I never encountered such a major difference in outcome between SPSS and 'R': In SPSS the Pearson correlation between variable 1 and variable 2 is 31% p0.001. In SPSS binary logistic regression gives us an Odds Ratio (OR)=4.9 (95% CI 2.7-9.0), p0.001, n=338. OR lower upper gender 1,120 0,565 2,221 age 0,985 0,956 1,015 variable 2 4,937 2,698 9,032 In R multilevel logistic regression using statistical package 'lmer' gives us an Odds Ratio=10.2 (95% CI 6.3-14), p=0.24, n=338, groups: group 1, 98; group 2 84. OR lower upper gender 2,295-2,840 7,430 age 0,003-70,047 70,054 variable 2 10,176 6,295 14,056 The crosstabs gives us: variable A Var B 0 1 0 156 108 1 17 57 Would somebody know how it is possible that in SPSS we get p0.001 and in R we get p=0.24? And, in 'R' the 95% CI of the Odds Ratio is 6.2-14.1. Why is the p-value=0.24? Thanks, Ronald [[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] Sweave and ggplot2
Dear Sorn,
It's hard to guess what your problem is, as you don't provide any sample
code. My guess is that the graphics are empty. Did you use
print(qplot(...)) or just qplot(). The latter won't work. You need
print(qplot(...))
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens [EMAIL PROTECTED]
Verzonden: maandag 4 augustus 2008 6:55
Aan: r-help@r-project.org
Onderwerp: [R] Sweave and ggplot2
Hi all,
I've been trying to run Sweave with R code embedded - using the ggplot2
package and in particular the qplot command. There appears to be a
problem
in Sweave not picking up that qplot is a function. Has anybody else
tried
to use qplot in Sweave and have you been successful? Any help would be
very much appreciated.
Kind Regards,
Sorn
Notice:
This email and any attachments may contain information\ ...{{dropped:10}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Decomposing tests of interaction terms in mixed-effects models
Andrew Robinson wrote: Dear R colleagues, a friend and I are trying to develop a modest workflow for the problem of decomposing tests of higher-order terms into interpretable sets of tests of lower order terms with conditioning. For example, if the interaction between A (3 levels) and C (2 levels) is significant, it may be of interest to ask whether or not A is significant at level 1 of C and level 2 of C. The textbook response seems to be to subset the data and perform the tests on the two subsets, but using the RSS and DF from the original fit. We're trying to answer the question using new explanatory variables. This approach (seems to) work just fine in aov, but not lme. For example, ## # Build the example dataset set.seed(101) Block - gl(6, 6, 36, lab = paste(B, 1:6, sep = )) A - gl(3, 2, 36, lab = paste(A, 1:3, sep = )) C - gl(2, 1, 36, lab = paste(C, 1:2, sep = )) example - data.frame(Block, A, C) example$Y - rnorm(36)*2 + as.numeric(example$A)*as.numeric(example$C)^2 + 3 * rep(rnorm(6), each=6) with(example, interaction.plot(A, C, Y)) # lme require(nlme) anova(lme(Y~A*C, random = ~1|Block, data = example)) summary(aov(Y ~ Error(Block) + A*C, data = example)) # Significant 2-way interaction. Now we would like to test the effect # of A at C1 and the effect of A at C2. Construct two new variables # that decompose the interaction, so an LRT is possible. example$AC - example$AC1 - example$AC2 - interaction(example$A, example$C) example$AC1[example$C == C1] - A1.C1 # A is constant at C1 example$AC2[example$C == C2] - A1.C2 # A is constant at C2 # lme fails (as does lmer) lme(Y ~ AC1 + AC, random = ~ 1 | Block, data = example) # but aov seems just fine. summary(aov(Y ~ Error(Block) + AC1 + AC, data = example)) ## AC was not significant, so A is not significant at C1 summary(aov(Y ~ Error(Block) + AC2 + AC, data = example)) ## AC was significant, so A is significant at C2 ## Some experimentation suggests that lme doesn't like the 'partial ## confounding' approach that we are using, rather than the variables ## that we have constructed. lme(Y ~ AC, random = ~ 1 | Block, data = example) lme(Y ~ A + AC, random = ~ 1 | Block, data = example) ## Are we doing anything obviously wrong? Is there another approach to the goal that we are trying to achieve? This looks like a generic issue with lme/lmer, in that they are not happy with rank deficiencies in the design matrix. Here's a dirty trick which you are not really supposed to know about because it is hidden inside the stats namespace: M - model.matrix(~AC1+AC, data=example) M2 - stats:::Thin.col(M) summary(lme(Y ~ M2 - 1, random = ~ 1 | Block, data = example) (Thin.col(), Thin.row(), and Rank() are support functions for anova.mlm() et al., but maybe we should document them and put them out in the open.) -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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] about the 95%CI around the median...
On Mon, 2008-08-04 at 17:00 +1000, Simon Blomberg wrote: See ?fivenum in the stats package. If you just type stats::fivenum you will get the code. The crucial calculations are in the last few lines. That will only give the code to calculate the five number summary, but Fernando wants to know how the confidence interval is calculated in boxplot.stats. To see the code just type boxplot.stats followed by return at the command line in R. The relevant line is: conf - if (do.conf) stats[3] + c(-1.58, 1.58) * iqr/sqrt(n) Which is working on the median (stats[3]). Details of this computation are in ?boxplot.stats which should have been Fernando's first port of call. Two works are cited in support of the calculation with full references in the References section of that help page. HTH G Simon. On Mon, 2008-08-04 at 16:19 +0930, Fernando Marmolejo Ramos wrote: Dear people I've learnt that by using the boxplot.stats command in the grDevices library I can get the 5-number summaries of a boxplot, plus other important information, like the confidence interval around the median. I'm interested in knowing the actual formula to used in that package to calculate that confidence interval. Can someone help me with this? Cheers, Fernando __ 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] Unexpected nls behaviour: Solved
Hi Everyone,
I'd omitted the non-optional 'parameters' argument to selfStart. Making this
change to SSbatch gives the same (successful) result from the two calls to
nls.
SSbatch-selfStart(
model=function(Batch, Coeffs)
{
Coeffs[Batch]
}
,initial=function(mCall, data, LHS)
{
# Estimate coefficients as mean of each batch
xy - sortedXyData(mCall[[Batch]], LHS, data)
Batch - data[[as.character(mCall[[Batch]])]]
# check Batch is successive integers starting at 1
if ((min(xy$x) !=1) | (any(diff(xy$x)!=1))) stop(
Batch is not a successive integers sequence)
Lval - list(xy$y)
names(Lval) - mCall[Coeffs]
Lval
}
, parameters = c(Coeffs)
)
Sorry for wasting anyones time.
Keith Jewell
-
Keith Jewell [EMAIL PROTECTED] wrote in message news:...
Hi everyone,
I thought that for a selfStart function, these two should be exactly
equivalent
nls(Aform, DF)
nls(Aform, DF, start=getInitial(Aform, DF))
but in this example that is not the case in R (although it is in S-plus
V6.2)
--
SSbatch-selfStart(
model=function(Batch, Coeffs)
{
Coeffs[Batch]
}
,initial=function(mCall, data, LHS)
{
# Estimate coefficients as mean of each batch
xy - sortedXyData(mCall[[Batch]], LHS, data)
Batch - data[[as.character(mCall[[Batch]])]]
# check Batch is successive integers starting at 1
if ((min(xy$x) !=1) | (any(diff(xy$x)!=1))) stop(
Batch is not a successive integers sequence)
Lval - list(xy$y)
names(Lval) - mCall[Coeffs]
Lval
}
)
DF - data.frame(A=c(0.9, 1.1, 1.9, 2.0, 2.1, 2.9, 3.0),
Batch=c(1,1,2,2,2,3,3))
Aform - formula(A~SSbatch(Batch,cA))
nls(Aform, DF, start=getInitial(Aform, DF))
nls(Aform, DF)
Don't ask why I'd want such a silly selfStart, that's a long story.
I guess wherever I would have used nls(Aform, DF)
I could use nls(Aform, DF, start=getInitial(Aform, DF))
but that seems clumsy.
Can anyone point out my mistake? Or is this a limitation of nls in R (I
hesitate to use the b*g word).
Thanks in advance,
Keith Jewell
--
I don't think it's relevant but, for completeness:
sessionInfo()
version 2.7.0 (2008-04-22)
i386-pc-mingw32
locale:
LC_COLLATE=English_United Kingdom.1252;LC_CTYPE=English_United
Kingdom.1252;LC_MONETARY=English_United
Kingdom.1252;LC_NUMERIC=C;LC_TIME=English_United Kingdom.1252
attached base packages:
[1] stats graphics grDevices datasets tcltk utils methods
base
other attached packages:
[1] xlsReadWrite_1.3.2 svSocket_0.9-5 svIO_0.9-5 R2HTML_1.58
svMisc_0.9-5 svIDE_0.9-5
loaded via a namespace (and not attached):
[1] tools_2.7.0 VGAM_0.7-7
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Re: [R] How to format the output file just the way I want ?
I think I have found the answer myself.
If you have something better, please write it.
Pierre8r
My R code :
---
library(quantmod)
library(xts)
Lines -
2008.07.01,02:00,1.5761,1.5766,1.5760,1.5763,65
2008.07.01,02:15,1.5762,1.5765,1.5757,1.5761,95
2008.07.01,02:30,1.5762,1.5765,1.5758,1.5759,58
2008.07.01,02:45,1.5758,1.5758,1.5745,1.5746,91
2008.07.01,03:00,1.5745,1.5751,1.5744,1.5744,117
2008.07.01,03:15,1.5742,1.5742,1.5727,1.5729,100
2008.07.01,03:30,1.5730,1.5736,1.5730,1.5736,61
2008.07.01,03:45,1.5735,1.5740,1.5735,1.5739,55
quotes - read.csv(textConnection(Lines), header=FALSE)
x - as.xts(quotes[,-(1:2)],
as.POSIXct(paste(quotes[,1],quotes[,2]),format='%Y.%m.%d %H:%M'))
colnames(x) - c('Open','High','Low','Close','Volume')
x$Open - sprintf(%5.04f, x$Open)
x$High - sprintf(%5.04f, x$High)
x$Low - sprintf(%5.04f, x$Low)
x$Close - sprintf(%5.04f, x$Close)
x$Volume - sprintf(%.0f, x$Volume)
write.table(x,file = K:\\OutputFile.csv, quote=FALSE, col.names=FALSE,
row.names=format(index(x),%Y.%m.%d,%H:%M), sep=,)
--
View this message in context:
http://www.nabble.com/How-to-format-the-output-file-just-the-way-I-want---tp18790926p18808276.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] source a script file straight from a subversion repository
SM == Steven McKinney [EMAIL PROTECTED] on Fri, 1 Aug 2008 18:38:54 -0700 writes: SM Thanks to Duncan Murdoch and Marc Schwartz for their SM excellent help. SM As we don't yet have the apache http: interface to svn SM running yet, 'svn export' is the access mechanism. yes. Note that svn.r-project.org for the R sources (and some recommended packages' ones) *only* runs the apache secure http interface 'https:/.' for security reasons. It has the big advantage that there exist no (!) user logins on that machine at all, and hence no compromised user ssh information can be misused to break in. *and* you can use source(...)! Martin Maechler, ETH Zurich __ 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] xyplot strip=function for two conditioning variables
Checkout this one: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/82452.html On Mon, Aug 4, 2008 at 6:24 AM, Henning Wildhagen [EMAIL PROTECTED] wrote: Dear list, for a data structure like in df: set.seed(100) Treatment-rep(c(Nitrogen,Carbon, Sulfur),each=9) week-rep(c(1,5,9),3,each=3) genes-rep(c(18s, EF1b, NR),9) copies-rnorm(27, 100,40) df-data.frame(Treatment,week,genes,copies) i wrote this code for a xyplot: library(lattice) PLOT-xyplot(data=df, copies~week|Treatment+genes,type=c(b,g)) I would like to change PLOT in a way, that the strips for Treatment are displayed only once at the top of each of the three columns. Additionally i would like to strips for genes at the top of each panel. I tried to adopt some code i found in the helplist using strip=function... but i did not manage to get what i would like. Can anyone give me advice? Thanks, Henning -- [[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.
[R] R and emacs
I wrote a program truc.r with emacs In emacs, I start the buffer R, then I eval the buffer truc.r (C-c C-b) All is correct and I have my results. But, when I return to Console R, and make: source (truc.r), I obtain an error: Erreur dans source(truc.r) : invalid multibyte character in parser at line 1 If I copy, line by line the instructions of the emacs buffer and put them in R console, all is correct. Thanks Jacques Vernin __ 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] xyplot strip=function for two conditioning variables
Dear list, for a data structure like in df: set.seed(100) Treatment-rep(c(Nitrogen,Carbon, Sulfur),each=9) week-rep(c(1,5,9),3,each=3) genes-rep(c(18s, EF1b, NR),9) copies-rnorm(27, 100,40) df-data.frame(Treatment,week,genes,copies) i wrote this code for a xyplot: library(lattice) PLOT-xyplot(data=df, copies~week|Treatment+genes,type=c(b,g)) I would like to change PLOT in a way, that the strips for Treatment are displayed only once at the top of each of the three columns. Additionally i would like to strips for genes at the top of each panel. I tried to adopt some code i found in the helplist using strip=function... but i did not manage to get what i would like. Can anyone give me advice? Thanks, Henning -- [[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] R and emacs
pir2.jv wrote: I wrote a program truc.r with emacs In emacs, I start the buffer R, then I eval the buffer truc.r (C-c C-b) All is correct and I have my results. But, when I return to Console R, and make: source (truc.r), I obtain an error: Erreur dans source(truc.r) : invalid multibyte character in parser at line 1 If I copy, line by line the instructions of the emacs buffer and put them in R console, all is correct. Sounds like encoding/locale confusion. Presumably, you have an accented character in the file somewhere, but in a different encoding to what source() expects. What is your platform? Notice also that emacs sometimes switches encodings on the fly. (Remember to follow up to the R-help, not to me. On some platforms, there are things I Just Don't Know, and it is good practice anyway to keep things on the list.) Thanks Jacques Vernin __ 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. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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] Exporting data to a text file
Hi John I don't get an error message but a warning write.table(myclara$clustering,cluster.dat,append=TRUE) Warning message: In write.table(myclara$clustering, cluster.dat, append = TRUE) : appending column names to file Here it is the output of str(myclara), it looks strange to me. I think clustering are integers and data are real numbers str(myclara) List of 10 $ sample: chr [1:56] 32356 33277 43230 52386 ... $ medoids : num [1:8, 1:14] 7.888 12.019 5.427 0.725 17.688 ... ..- attr(*, dimnames)=List of 2 .. ..$ : chr [1:8] 109056 98194 56959 109806 ... .. ..$ : chr [1:14] lon lat sst01 sst02 ... $ i.med : int [1:8] 20482 16158 5137 20722 48599 56033 68028 64308 $ clustering: Named int [1:75459] 1 1 1 1 1 1 1 1 1 1 ... ..- attr(*, names)= chr [1:75459] 12296 12297 12298 12299 ... $ objective : num 3.22 $ clusinfo : num [1:8, 1:4] 15055 9474 5164 13702 11340 ... ..- attr(*, dimnames)=List of 2 .. ..$ : NULL .. ..$ : chr [1:4] size max_diss av_diss isolation $ diss :Classes 'dissimilarity', 'dist' atomic [1:1540] 1.11 6.54 4.62 3.30 4.32 ... .. ..- attr(*, Size)= int 56 .. ..- attr(*, Metric)= chr euclidean .. ..- attr(*, Labels)= chr [1:56] 32356 33277 43230 52386 ... $ call : language clara(x = mydata, k = 8) $ silinfo :List of 3 ..$ widths : num [1:56, 1:3] 1 1 1 1 1 1 1 1 2 2 ... .. ..- attr(*, dimnames)=List of 2 .. .. ..$ : chr [1:56] 96250 109056 130058 116317 ... .. .. ..$ : chr [1:3] cluster neighbor sil_width ..$ clus.avg.widths: num [1:8] 0.343 0.355 0.533 0.265 0.308 ... ..$ avg.width : num 0.362 $ data : num [1:75459, 1:14] 8.68 8.72 8.77 8.81 8.86 ... ..- attr(*, dimnames)=List of 2 .. ..$ : chr [1:75459] 12296 12297 12298 12299 ... .. ..$ : chr [1:14] lon lat sst01 sst02 ... - attr(*, class)= chr [1:2] clara partition I can't output the two variables in two different files without any problem. Thanks 2008/8/1 John Kane [EMAIL PROTECTED] try str(myclara) to see what you have - a data frame , matrix etc Are you getting any error messages? I tried your write.table commands and they work okay. --- On Fri, 8/1/08, pacomet [EMAIL PROTECTED] wrote: From: pacomet [EMAIL PROTECTED] Subject: [R] Exporting data to a text file To: r-help@r-project.org Received: Friday, August 1, 2008, 12:49 PM HI R users With clara function I get a data frame (maybe this is not the exact word, I'm new to R) with the following variables: names(myclara) [1] sample medoids i.med clustering objective [6] clusinfo diss call silinfo data I want to export clustering and data to a new text file so I try write.table(myclara$data,cluster.dat) write.table(myclara$clustering,cluster.dat,append=TRUE) Variable data is properly exported but clustering is not appended to the output file. Please, where is the mistake? is it possible to export the two variables in just a sentence? thanks in advance Paco -- _ El ponent la mou, el llevant la plou Usuari Linux registrat: 363952 --- Fotos: http://picasaweb.google.es/pacomet [[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. __ Connect with friends from any web browser - no download required. Try the new Yahoo! Canada Messenger for the Web BETA at http://ca.messenger.yahoo.com/webmessengerpromo.php -- _ El ponent la mou, el llevant la plou Usuari Linux registrat: 363952 --- Fotos: http://picasaweb.google.es/pacomet [[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] greek letters in italic font
Dear HelpeRs,
I am trying to write axis labels with some letters in cursive for later
inclusion in a LaTeX document. The following code does what I want with
latin letters (c cursive and the rest not cursive):
plot(1:10, 1:10, ylab = expression(italic(c)*(MB)))
or
plot(1:10, 1:10, ylab = expression(paste(italic(c)*(MB), / ,
mu*g~L^{-1}))
but it does not work for greek letters:
plot(1:10, 1:10, ylab = expression(italic(rho)*(MB)))
Using text() or mtext() with the option font = 3 does not work either.
I found a way around this by writing rho and then having rho be
changed to the corresponding greek letter in LaTeX by using the PSfrag
package... but it only works for text(), not for mtext() nor xlab/ylab
within the plot() command.
I would be grateful for some ideas.
Many thanks and best regards,
Luis
--
Luis Tercero, M.Sc.
Engler-Bunte-Institut der Universität Karlsruhe (TH)
Bereich Wasserchemie
Engler-Bunte-Ring 1
D-76131 Karlsruhe
Tel. +49 721 608 6381
Fax: +49 721 608 7051
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Re: [R] Plotting ordered nominal data
Many thanks
I was sure it was simple. That was exactly what I wanted.
I should have clarified that I was looking for a box-plot.
Thanks to all who responed.
Sandy
S Ellison wrote:
Sandy,
You can re-order a factor with
df$Eyeball-factor(df$Eyeball, levels=c(Normal, Mild, Moderate,
Severe), ordered=T)
(assuming df is your data frame and that you want an _ordered_ factor;
the latter is not essential to your plots)
Incidentally, NULL isn't a particularly friendly item to find in a
data frame. NULL often implies I'm not here at all while NA says I
exist but I'm a missing value. For an example of when it might matter,
try
length(c(1,2,NULL,3))
#versus
length(c(1,2,NA,3))
Steve E
Sandy Small [1][EMAIL PROTECTED] 01/08/2008 16:21:29
Hi
I'm sure this question has been asked before but I can't find it in the
archives.
I want to plot them in the order Normal, Mild,
Moderate, Severe so that the trend (or not) is obvious.
***
This email and any attachments are confidential. Any use...{{dropped:29}}
References
1. mailto:[EMAIL PROTECTED]
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Re: [R] greek letters in italic font
You could piece it together using the Hershey fonts for the italic rho:
op - par(xpd = NA)
plot(1, type = n, ylab = )
u - par(usr)
text(u[1] - .1 * diff(u[1:2]), 1, \\*r, vfont = c(serif, italic))
par(op)
Now add the rest.
?Hershey
?par
?strwidth
On Mon, Aug 4, 2008 at 7:12 AM, Luis Tercero
[EMAIL PROTECTED] wrote:
Dear HelpeRs,
I am trying to write axis labels with some letters in cursive for later
inclusion in a LaTeX document. The following code does what I want with
latin letters (c cursive and the rest not cursive):
plot(1:10, 1:10, ylab = expression(italic(c)*(MB)))
or
plot(1:10, 1:10, ylab = expression(paste(italic(c)*(MB), / ,
mu*g~L^{-1}))
but it does not work for greek letters:
plot(1:10, 1:10, ylab = expression(italic(rho)*(MB)))
Using text() or mtext() with the option font = 3 does not work either.
I found a way around this by writing rho and then having rho be changed
to the corresponding greek letter in LaTeX by using the PSfrag package...
but it only works for text(), not for mtext() nor xlab/ylab within the
plot() command.
I would be grateful for some ideas.
Many thanks and best regards,
Luis
--
Luis Tercero, M.Sc.
Engler-Bunte-Institut der Universität Karlsruhe (TH)
Bereich Wasserchemie
Engler-Bunte-Ring 1
D-76131 Karlsruhe
Tel. +49 721 608 6381
Fax: +49 721 608 7051
__
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] Decomposing tests of interaction terms in mixed-effects models
On Mon, Aug 04, 2008 at 10:17:38AM +0200, Peter Dalgaard wrote: Andrew Robinson wrote: Dear R colleagues, a friend and I are trying to develop a modest workflow for the problem of decomposing tests of higher-order terms into interpretable sets of tests of lower order terms with conditioning. For example, if the interaction between A (3 levels) and C (2 levels) is significant, it may be of interest to ask whether or not A is significant at level 1 of C and level 2 of C. The textbook response seems to be to subset the data and perform the tests on the two subsets, but using the RSS and DF from the original fit. We're trying to answer the question using new explanatory variables. This approach (seems to) work just fine in aov, but not lme. For example, ## # Build the example dataset set.seed(101) Block - gl(6, 6, 36, lab = paste(B, 1:6, sep = )) A - gl(3, 2, 36, lab = paste(A, 1:3, sep = )) C - gl(2, 1, 36, lab = paste(C, 1:2, sep = )) example - data.frame(Block, A, C) example$Y - rnorm(36)*2 + as.numeric(example$A)*as.numeric(example$C)^2 + 3 * rep(rnorm(6), each=6) with(example, interaction.plot(A, C, Y)) # lme require(nlme) anova(lme(Y~A*C, random = ~1|Block, data = example)) summary(aov(Y ~ Error(Block) + A*C, data = example)) # Significant 2-way interaction. Now we would like to test the effect # of A at C1 and the effect of A at C2. Construct two new variables # that decompose the interaction, so an LRT is possible. example$AC - example$AC1 - example$AC2 - interaction(example$A, example$C) example$AC1[example$C == C1] - A1.C1 # A is constant at C1 example$AC2[example$C == C2] - A1.C2 # A is constant at C2 # lme fails (as does lmer) lme(Y ~ AC1 + AC, random = ~ 1 | Block, data = example) # but aov seems just fine. summary(aov(Y ~ Error(Block) + AC1 + AC, data = example)) ## AC was not significant, so A is not significant at C1 summary(aov(Y ~ Error(Block) + AC2 + AC, data = example)) ## AC was significant, so A is significant at C2 ## Some experimentation suggests that lme doesn't like the 'partial ## confounding' approach that we are using, rather than the variables ## that we have constructed. lme(Y ~ AC, random = ~ 1 | Block, data = example) lme(Y ~ A + AC, random = ~ 1 | Block, data = example) ## Are we doing anything obviously wrong? Is there another approach to the goal that we are trying to achieve? This looks like a generic issue with lme/lmer, in that they are not happy with rank deficiencies in the design matrix. Here's a dirty trick which you are not really supposed to know about because it is hidden inside the stats namespace: M - model.matrix(~AC1+AC, data=example) M2 - stats:::Thin.col(M) summary(lme(Y ~ M2 - 1, random = ~ 1 | Block, data = example) (Thin.col(), Thin.row(), and Rank() are support functions for anova.mlm() et al., but maybe we should document them and put them out in the open.) That is a neat idea, thanks, Peter, but it doesn't quite fit the bill. The summary provides t-tests but I am hoping to find F-tests, otherwise I'm not sure how to efficiently test A (3 levels) at the two levels of C. The anova.lme function doesn't help, sadly: anova(lme(Y ~ M2 - 1, random = ~ 1 | Block, data = example)) numDF denDF F-value p-value M2 625 23.0198 .0001 so I'm still flummoxed! Andrew -- Andrew Robinson Department of Mathematics and StatisticsTel: +61-3-8344-6410 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 http://www.ms.unimelb.edu.au/~andrewpr http://blogs.mbs.edu/fishing-in-the-bay/ __ 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] memory problem - failing to load rgl in R 2.7.1 patched
Hi,
yesterday i had the surprise not to be able to load the package ca on R
2.7.0 saying that cannot find required package rgl although it was there. So
today i've upgraded to 7.2.1. patched and i got the following error:
local({pkg - select.list(sort(.packages(all.available = TRUE)))
+ if(nchar(pkg)) library(pkg, character.only=TRUE)})
Loading required package: rgl
Error : cannot allocate vector of size 2.0 Gb
Error: package 'rgl' could not be loaded
I've up-ed the memory to max which is 4 Gb for my computer and tried again
... same result. I know how to install packages and load them or at
least i thought so. Do you know what is wrong??
sysname release
Windows XP
version nodename
build 2600, Service Pack 2
machine
x86
sessionInfo()
R version 2.7.1 Patched (2008-07-31 r46185)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] e1071_1.5-18 class_7.2-42
loaded via a namespace (and not attached):
[1] tools_2.7.1
Thanks for any advice,
Monica
_
Time for vacation? WIN what you need- enter now!
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and provide commented, minimal, self-contained, reproducible code.
[R] SOLVED - memory problem - failing to load rgl in R 2.7.1 patched
Hi,
Sorry for that . Today R did load both ca and rgl packages with no problems
... the single thing i've done since Friday was to put defrag on drive C, i
even didn't re-start the computer ...actually it was re-started several time
with same weird problem ... so i suppose something was segmented badly on the
hard drive.
Monica
From: [EMAIL PROTECTED]
To: r-help@r-project.org
Subject: [R] memory problem - failing to load rgl in R 2.7.1 patched
Date: Mon, 4 Aug 2008 12:27:31 +
Hi,
yesterday i had the surprise not to be able to load the package ca on R
2.7.0 saying that cannot find required package rgl although it was there. So
today i've upgraded to 7.2.1. patched and i got the following error:
local({pkg - select.list(sort(.packages(all.available = TRUE)))
+ if(nchar(pkg)) library(pkg, character.only=TRUE)})
Loading required package: rgl
Error : cannot allocate vector of size 2.0 Gb
Error: package 'rgl' could not be loaded
I've up-ed the memory to max which is 4 Gb for my computer and tried again
... same result. I know how to install packages and load them or at
least i thought so. Do you know what is wrong??
sysname release
Windows XP
version nodename
build 2600, Service Pack 2
machine
x86
sessionInfo()
R version 2.7.1 Patched (2008-07-31 r46185)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] e1071_1.5-18 class_7.2-42
loaded via a namespace (and not attached):
[1] tools_2.7.1
Thanks for any advice,
Monica
_
Time for vacation? WIN what you need- enter now!
http://www.gowindowslive.com/summergiveaway/?ocid=tag_jlyhm
_
Reveal your inner athlete and share it with friends on Windows Live.
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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] Decomposing tests of interaction terms in mixed-effects models
Andrew Robinson wrote: On Mon, Aug 04, 2008 at 10:17:38AM +0200, Peter Dalgaard wrote: Andrew Robinson wrote: Dear R colleagues, a friend and I are trying to develop a modest workflow for the problem of decomposing tests of higher-order terms into interpretable sets of tests of lower order terms with conditioning. For example, if the interaction between A (3 levels) and C (2 levels) is significant, it may be of interest to ask whether or not A is significant at level 1 of C and level 2 of C. The textbook response seems to be to subset the data and perform the tests on the two subsets, but using the RSS and DF from the original fit. We're trying to answer the question using new explanatory variables. This approach (seems to) work just fine in aov, but not lme. For example, ## # Build the example dataset set.seed(101) Block - gl(6, 6, 36, lab = paste(B, 1:6, sep = )) A - gl(3, 2, 36, lab = paste(A, 1:3, sep = )) C - gl(2, 1, 36, lab = paste(C, 1:2, sep = )) example - data.frame(Block, A, C) example$Y - rnorm(36)*2 + as.numeric(example$A)*as.numeric(example$C)^2 + 3 * rep(rnorm(6), each=6) with(example, interaction.plot(A, C, Y)) # lme require(nlme) anova(lme(Y~A*C, random = ~1|Block, data = example)) summary(aov(Y ~ Error(Block) + A*C, data = example)) # Significant 2-way interaction. Now we would like to test the effect # of A at C1 and the effect of A at C2. Construct two new variables # that decompose the interaction, so an LRT is possible. example$AC - example$AC1 - example$AC2 - interaction(example$A, example$C) example$AC1[example$C == C1] - A1.C1 # A is constant at C1 example$AC2[example$C == C2] - A1.C2 # A is constant at C2 # lme fails (as does lmer) lme(Y ~ AC1 + AC, random = ~ 1 | Block, data = example) # but aov seems just fine. summary(aov(Y ~ Error(Block) + AC1 + AC, data = example)) ## AC was not significant, so A is not significant at C1 summary(aov(Y ~ Error(Block) + AC2 + AC, data = example)) ## AC was significant, so A is significant at C2 ## Some experimentation suggests that lme doesn't like the 'partial ## confounding' approach that we are using, rather than the variables ## that we have constructed. lme(Y ~ AC, random = ~ 1 | Block, data = example) lme(Y ~ A + AC, random = ~ 1 | Block, data = example) ## Are we doing anything obviously wrong? Is there another approach to the goal that we are trying to achieve? This looks like a generic issue with lme/lmer, in that they are not happy with rank deficiencies in the design matrix. Here's a dirty trick which you are not really supposed to know about because it is hidden inside the stats namespace: M - model.matrix(~AC1+AC, data=example) M2 - stats:::Thin.col(M) summary(lme(Y ~ M2 - 1, random = ~ 1 | Block, data = example) (Thin.col(), Thin.row(), and Rank() are support functions for anova.mlm() et al., but maybe we should document them and put them out in the open.) That is a neat idea, thanks, Peter, but it doesn't quite fit the bill. The summary provides t-tests but I am hoping to find F-tests, otherwise I'm not sure how to efficiently test A (3 levels) at the two levels of C. The anova.lme function doesn't help, sadly: anova(lme(Y ~ M2 - 1, random = ~ 1 | Block, data = example)) numDF denDF F-value p-value M2 625 23.0198 .0001 so I'm still flummoxed! Andrew You do have to peek into M2 to know that the test is for whether the last two coefs are zero, but how about M3 - M2[,2:4] M4 - M2[,5:6] anova(lme(Y ~ M3+M4, random = ~ 1 | Block, data = example)) numDF denDF F-value p-value (Intercept) 125 10.66186 0.0032 M3 325 55.31464 .0001 M4 225 1.27591 0.2967 Also, check out estimable() in the gmodels package. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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] Decomposing tests of interaction terms in mixed-effects models
On Mon, Aug 04, 2008 at 02:51:48PM +0200, Peter Dalgaard wrote: Andrew Robinson wrote: On Mon, Aug 04, 2008 at 10:17:38AM +0200, Peter Dalgaard wrote: Andrew Robinson wrote: That is a neat idea, thanks, Peter, but it doesn't quite fit the bill. The summary provides t-tests but I am hoping to find F-tests, otherwise I'm not sure how to efficiently test A (3 levels) at the two levels of C. The anova.lme function doesn't help, sadly: anova(lme(Y ~ M2 - 1, random = ~ 1 | Block, data = example)) numDF denDF F-value p-value M2 625 23.0198 .0001 so I'm still flummoxed! Andrew You do have to peek into M2 to know that the test is for whether the last two coefs are zero, but how about M3 - M2[,2:4] M4 - M2[,5:6] anova(lme(Y ~ M3+M4, random = ~ 1 | Block, data = example)) numDF denDF F-value p-value (Intercept) 125 10.66186 0.0032 M3 325 55.31464 .0001 M4 225 1.27591 0.2967 Marvelous, many thanks, Peter. Also, check out estimable() in the gmodels package. Will do. Actually had done, but will do again. Cheers Andrew -- Andrew Robinson Department of Mathematics and StatisticsTel: +61-3-8344-6410 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 http://www.ms.unimelb.edu.au/~andrewpr http://blogs.mbs.edu/fishing-in-the-bay/ __ 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] about the 95%CI around the median...
Gavin Simpson wrote: On Mon, 2008-08-04 at 17:00 +1000, Simon Blomberg wrote: See ?fivenum in the stats package. If you just type stats::fivenum you will get the code. The crucial calculations are in the last few lines. That will only give the code to calculate the five number summary, but Fernando wants to know how the confidence interval is calculated in boxplot.stats. To see the code just type boxplot.stats followed by return at the command line in R. The relevant line is: conf - if (do.conf) stats[3] + c(-1.58, 1.58) * iqr/sqrt(n) Which is working on the median (stats[3]). Details of this computation are in ?boxplot.stats which should have been Fernando's first port of call. Two works are cited in support of the calculation with full references in the References section of that help page. HTH G I wonder why we don't just use the exact nonparametric confidence interval for the median, which is just as easy to compute. Also, it will be asymmetric if the data are skewed, as it should be. Frank Simon. On Mon, 2008-08-04 at 16:19 +0930, Fernando Marmolejo Ramos wrote: Dear people I've learnt that by using the boxplot.stats command in the grDevices library I can get the 5-number summaries of a boxplot, plus other important information, like the confidence interval around the median. I'm interested in knowing the actual formula to used in that package to calculate that confidence interval. Can someone help me with this? Cheers, Fernando -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] Trouble when I call a function within another
Greetings, I recently discovered R and how to tackle my own functions. I'm blocked on a simple barrier I guess. I call a function that recognizes a string and return the adequate formula in a main function. This is done through a script. When I type the formula in the main function, it works fine. When I attempted to use a second function (with conditional choice) it doesn't work any more. May you guide me? http://www.nabble.com/file/p18811385/annexe.R annexe.R - B. Boulinguiez [EMAIL PROTECTED] Ph.D. in Chemistry Ecole Nationale Supérieure de Chimie de Rennes 35700 Rennes (FRANCE) -- View this message in context: http://www.nabble.com/Trouble-when-I-call-a-function-within-another-tp18811385p18811385.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] Howto Smooth a Curve Created with the Point Function
Hi all, I have this figure: http://docs.google.com/Doc?id=df5zfsj4_103rjt2v4d5 created with the following steps: x [1] 90.4 57.8 77.0 103.7 55.4 217.5 68.1 85.3 152.0 113.0 97.1 89.9 [13] 68.1 83.7 77.4 34.5 104.9 170.3 88.6 88.1 108.8 77.4 85.6 82.7 [25] 81.3 108.0 49.5 71.0 85.7 99.3 203.5 275.9 51.1 84.8 16.5 72.6 [37] 160.5 158.3 136.7 140.0 98.4 116.1 177.0 168.7 208.0 219.3 90.5 119.7 [49] 72.9 138.4 98.6 56.0 151.6 230.0 82.0 65.6 83.0 106.9 141.3 103.9 [61] 76.0 90.6 72.9 85.5 70.5 134.3 129.5 130.9 187.1 131.4 57.7 179.7 [73] 89.7 126.1 124.4 103.2 111.8 143.1 59.4 85.3 122.8 100.0 158.4 188.2 [85] 91.9 90.0 74.5 70.2 105.8 90.0 96.6 109.5 176.8 145.1 126.2 215.9 [97] 103.4 154.3 84.8 112.8 119.8 111.9 107.6 114.3 198.7 98.2 187.6 196.7 [109] 138.5 146.1 75.0 80.0 147.7 171.3 90.1 58.1 109.3 136.6 91.5 93.8 [121] 79.4 71.8 xhist - hist(x,col=blue, xlab=Exp Level,freq=FALSE) alpha - 6.352082 beta - 17.84647 g.pdf - dgamma(x,shape=alpha,scale=beta) points(x,g.pdf,col=red,type='l') ## problem here My question is: Why the command points DOES NOT create smooth red curve as we expected? How can I resolve this problem? I specifically want to use the dgamma function with the learned parameter (alpha,beta). That's why I don't use this command: lines(density(x),add=T, col=red) - Gundala Viswanath Jakarta - Indonesia __ 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] Trouble when I call a function within another
bbouling wrote: Greetings, I recently discovered R and how to tackle my own functions. I'm blocked on a simple barrier I guess. I call a function that recognizes a string and return the adequate formula in a main function. This is done through a script. When I type the formula in the main function, it works fine. When I attempted to use a second function (with conditional choice) it doesn't work any more. May you guide me? http://www.nabble.com/file/p18811385/annexe.R annexe.R You need to show us the output you get, and tell us the output you want, in a reproducible way. Just showing us a function is not enough: we don't know what inputs to give to the function, etc. Duncan Murdoch __ 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] Howto Smooth a Curve Created with the Point Function
Hi Gundala, You have to reorder the points, like below: #(your code ...) g - cbind(x,g.pdf)[order(x),] points(x=[,1],y=g[,2],type='l',col=red) Cheers, gary -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gundala Viswanath Sent: Monday, August 04, 2008 9:50 AM To: [EMAIL PROTECTED] Subject: [R] Howto Smooth a Curve Created with the Point Function Hi all, I have this figure: http://docs.google.com/Doc?id=df5zfsj4_103rjt2v4d5 created with the following steps: x [1] 90.4 57.8 77.0 103.7 55.4 217.5 68.1 85.3 152.0 113.0 97.1 89.9 [13] 68.1 83.7 77.4 34.5 104.9 170.3 88.6 88.1 108.8 77.4 85.6 82.7 [25] 81.3 108.0 49.5 71.0 85.7 99.3 203.5 275.9 51.1 84.8 16.5 72.6 [37] 160.5 158.3 136.7 140.0 98.4 116.1 177.0 168.7 208.0 219.3 90.5 119.7 [49] 72.9 138.4 98.6 56.0 151.6 230.0 82.0 65.6 83.0 106.9 141.3 103.9 [61] 76.0 90.6 72.9 85.5 70.5 134.3 129.5 130.9 187.1 131.4 57.7 179.7 [73] 89.7 126.1 124.4 103.2 111.8 143.1 59.4 85.3 122.8 100.0 158.4 188.2 [85] 91.9 90.0 74.5 70.2 105.8 90.0 96.6 109.5 176.8 145.1 126.2 215.9 [97] 103.4 154.3 84.8 112.8 119.8 111.9 107.6 114.3 198.7 98.2 187.6 196.7 [109] 138.5 146.1 75.0 80.0 147.7 171.3 90.1 58.1 109.3 136.6 91.5 93.8 [121] 79.4 71.8 xhist - hist(x,col=blue, xlab=Exp Level,freq=FALSE) alpha - 6.352082 beta - 17.84647 g.pdf - dgamma(x,shape=alpha,scale=beta) points(x,g.pdf,col=red,type='l') ## problem here My question is: Why the command points DOES NOT create smooth red curve as we expected? How can I resolve this problem? I specifically want to use the dgamma function with the learned parameter (alpha,beta). That's why I don't use this command: lines(density(x),add=T, col=red) - Gundala Viswanath Jakarta - Indonesia __ 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. This message w/attachments (message) may be privileged, confidential or proprietary, and if you are not an intended recipient, please notify the sender, do not use or share it and delete it. Unless specifically indicated, this message is not an offer to sell or a solicitation of any investment products or other financial product or service, an official confirmation of any transaction, or an official statement of Merrill Lynch. Subject to applicable law, Merrill Lynch may monitor, review and retain e-communications (EC) traveling through its networks/systems. The laws of the country of each sender/recipient may impact the handling of EC, and EC may be archived, supervised and produced in countries other than the country in which you are located. This message cannot be guaranteed to be secure or error-free. This message is subject to terms available at the following link: http://www.ml.com/e-communications_terms/. By messaging with Merrill Lynch you consent to the foregoing. __ 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] subset inside a lattice plot using panel.lines
Hi R people
Pulling my hair out here trying to get something very simple to work.
Have a data frame of 774 rows and want to plot first and second half
on same axes with different colours. A variable is present call 'row'
created like this and checked to be OK:
row - seq( len=length( variable1 ) )
...so I just want to plot the two subsets where row = 387 and where
row 387. None of these work properly:
plots both over each other in correct colours
opt_plot2 - function( var_string, c, units ) {
print( xyplot( get(var_string) ~ variable1 | variable2, panel =
function( x, y, ... ) {
panel.xyplot( x, y, type =n, ... )
panel.grid( h=-1, v=-1, lty=3, lwd=1, col=grey )
panel.lines( x, y, col = red, subset = row = 387 )
panel.lines( x, y, col = dark green, subset = row 387 )
},
}
plots both just in red
... panel.lines( x[row = 387], y[row = 387], col = red )
panel.lines( x[row 387], y[row 387], col = dark green )
first - (row = 387)
second - (row 387)
plots both over each other in correct colours
... panel.lines( x, y, col = red, subset = first )
panel.lines( x, y, col = dark green, subset = second )
plots both just in red
... panel.lines( x[first], y[first], col = red )
panel.lines( x[second], y[second], col = dark green )
I'm feeling frustrated and a bit stupid but should this be so
difficult? Any help or tips on what I am doing wrong greatly
appreciated.
TIA
Michael
__
Hopkins Research Touch the Future
__
[[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] subset inside a lattice plot using panel.lines
Hi R people [duplicate - sorry, just posted HTML by mistake]
Pulling my hair out here trying to get something very simple to work.
Have a data frame of 774 rows and want to plot first and second half
on same axes with different colours. A variable is present call 'row'
created like this and checked to be OK:
row - seq( len=length( variable1 ) )
...so I just want to plot the two subsets where row = 387 and where
row 387. None of these work properly:
plots both over each other in correct colours
opt_plot2 - function( var_string, c, units ) {
print( xyplot( get(var_string) ~ variable1 | variable2, panel =
function( x, y, ... ) {
panel.xyplot( x, y, type =n, ... )
panel.grid( h=-1, v=-1, lty=3, lwd=1, col=grey )
panel.lines( x, y, col = red, subset = row = 387 )
panel.lines( x, y, col = dark green, subset = row 387 )
},
}
plots both just in red
... panel.lines( x[row = 387], y[row = 387], col = red )
panel.lines( x[row 387], y[row 387], col = dark
green )
first - (row = 387)
second - (row 387)
plots both over each other in correct colours
... panel.lines( x, y, col = red, subset = first )
panel.lines( x, y, col = dark green, subset = second )
plots both just in red
... panel.lines( x[first], y[first], col = red )
panel.lines( x[second], y[second], col = dark green )
I'm feeling frustrated and a bit stupid but should this be so
difficult? Any help or tips on what I am doing wrong greatly
appreciated.
TIA
Michael
__
Hopkins Research Touch the Future
__
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] subset inside a lattice plot using panel.lines
Hi Michael,
Pulling my hair out here trying to get something very simple to work. ...
I can't quite see what you are trying to do [and I am not sure that you
clearly state it], but you could make things easier and simpler by (1)
creating a factor to identify your groups of rows more cleanly and (2) by
using the groups= argument in lattice.
## Something along these lines
copyDat - originalDat
copyDat$newFac - gl(2, 387, 774, labels=c(A,B))
xyplot(response ~ predictor|maybe.condition, data=copyDat, groups=newFac)
I hope this gives you some ideas.
Mark.
Michael Hopkins wrote:
Hi R people [duplicate - sorry, just posted HTML by mistake]
Pulling my hair out here trying to get something very simple to work.
Have a data frame of 774 rows and want to plot first and second half
on same axes with different colours. A variable is present call 'row'
created like this and checked to be OK:
row - seq( len=length( variable1 ) )
...so I just want to plot the two subsets where row = 387 and where
row 387. None of these work properly:
plots both over each other in correct colours
opt_plot2 - function( var_string, c, units ) {
print( xyplot( get(var_string) ~ variable1 | variable2, panel =
function( x, y, ... ) {
panel.xyplot( x, y, type =n, ... )
panel.grid( h=-1, v=-1, lty=3, lwd=1, col=grey )
panel.lines( x, y, col = red, subset = row = 387 )
panel.lines( x, y, col = dark green, subset = row 387 )
},
}
plots both just in red
... panel.lines( x[row = 387], y[row = 387], col = red )
panel.lines( x[row 387], y[row 387], col = dark
green )
first - (row = 387)
second - (row 387)
plots both over each other in correct colours
... panel.lines( x, y, col = red, subset = first )
panel.lines( x, y, col = dark green, subset = second )
plots both just in red
... panel.lines( x[first], y[first], col = red )
panel.lines( x[second], y[second], col = dark green )
I'm feeling frustrated and a bit stupid but should this be so
difficult? Any help or tips on what I am doing wrong greatly
appreciated.
TIA
Michael
__
Hopkins Research Touch the Future
__
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.
--
View this message in context:
http://www.nabble.com/subset-inside-a-lattice-plot-using-panel.lines-tp18813236p18813818.html
Sent from the R help mailing list archive at Nabble.com.
__
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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] plot() change unit length
I am try to plot a data frame, and encounter some difficulties. my x axis range is from -60 to 80 and unit length is 10 so x axis looks like -60, -40, -20, .., 40,60,80 But I want chang unit length from 10 to 5 -60, -50, -40, .., 60,70,80 Does anyone know how to do it? Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Are there any guis out there, which will allow editing of the graph?
Hi, all, I would like to know if there is any gui interface out there (academic or commercial) that allows one to edit R-language generated graphs (e.g positioning x axis labels.) It would be nice to have something like the user interface of Igor or Origin. I have already used JGR and R-gui. These are good, but they don't allow one to easily edit graphs. I have also tried locator() and the package iplots. Your input is greatly appreciated. Best wishes, Art Roberts University of Washington __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot() change unit length
You can always create the axis yourself: x - seq(-60,80) plot(x, seq_along(x), xaxt='n') axis(1, at=pretty(x,10)) On Mon, Aug 4, 2008 at 12:47 PM, Qian R [EMAIL PROTECTED] wrote: I am try to plot a data frame, and encounter some difficulties. my x axis range is from -60 to 80 and unit length is 10 so x axis looks like -60, -40, -20, .., 40,60,80 But I want chang unit length from 10 to 5 -60, -50, -40, .., 60,70,80 Does anyone know how to do it? Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? __ 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] Are there any guis out there, which will allow editing of the graph?
No. Can't be. Editable graphs require that the graph be produced via code that produces changeable components. All R graphs are essentially static. That said, caveats: graphs drawn via the grid package functionality -- for example lattice graphs -- **are** produced via changeable code. If you read the lattice docs carefully, you'll see that there are a few features there that allow some graph editing. There may be other packages that also have some editing capabilties. R's base graphics also allow a little interaction via identify() and locator(), which can be useful (e.g. for positioning legends). One can also simulate interactivity by recording various components of graph construction and then modifying and redrawing them. But this is just manually doing what you're looking for, so probably a dumb suggestion. While graph editing certainly can be a nice feature, it is very difficult to implement without severely constraining graphing flexibility (IMO, of course). Graphs are very complex beasties, so it's hard to write clean code that allows flexibile editing capabilities. Look at S-Plus's graph editing, which I always found harder to use (and more buggy) than just issuing the commands. (To be fair, it's been some years since I tried). Again, just my 2 bits. Others may well disagree (and perhaps point you to what you seek). Cheers, Bert -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Arthur Roberts Sent: Monday, August 04, 2008 9:51 AM To: [EMAIL PROTECTED] Subject: [R] Are there any guis out there,which will allow editing of the graph? Hi, all, I would like to know if there is any gui interface out there (academic or commercial) that allows one to edit R-language generated graphs (e.g positioning x axis labels.) It would be nice to have something like the user interface of Igor or Origin. I have already used JGR and R-gui. These are good, but they don't allow one to easily edit graphs. I have also tried locator() and the package iplots. Your input is greatly appreciated. Best wishes, Art Roberts University of Washington __ 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] thematic map of USA
My goal is to prepare a thematic map of the US, with states shaded according to their values for a variable of interest. I would like to include an inset for Alaska in the upper left and an inset for Hawaii in the lower left. If possible, I'd like to use Albers conic projection, or something similar. Thus far I have tried using the maps package with its state database (which omits Alaska and Hawaii) and the gmaps package (which places an inset for Alaska in the lower left and uses a projection [Miller cylindrical?] that distorts states along the northern and southern fringe). I would be very grateful for suggestions about alternative approaches. John -- John P. Burkett Department of Environmental and Natural Resource Economics and Department of Economics University of Rhode Island Kingston, RI 02881-0808 USA phone (401) 874-9195 __ 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] Are there any guis out there, which will allow editing of the graph?
On 04/08/2008 12:50 PM, Arthur Roberts wrote: Hi, all, I would like to know if there is any gui interface out there (academic or commercial) that allows one to edit R-language generated graphs (e.g positioning x axis labels.) It would be nice to have something like the user interface of Igor or Origin. I have already used JGR and R-gui. These are good, but they don't allow one to easily edit graphs. I have also tried locator() and the package iplots. Your input is greatly appreciated. I imagine there are a number of commercial packages available that would allow you to edit PDF or PS files, but I haven't used any of them, so I can't recommend one. Duncan Murdoch __ 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] Are there any guis out there, which will allow editing of the graph?
A colleague of mine, quite by accident, discovered that Adobe Illustrator can manipulate plots made by base graphics, and when you do, many pieces of the plot are separate items that can be manipulated with Illustrator. He cuts and pastes from a Quartz window on his Mac, into Illustrator. Apparently Illustrator has two kinds of selection arrows, one of which selects groups of things, the other selects individual things. He confirms that text from R may be changed, colors in polygonal areas may be changed, objects may be moved etc once they are selected. Apparently he saw a notation that this was possible on some R code that went with a Wikipedia entry, and he tried it, and it works. YMMV. Bryan * Bryan Hanson Professor of Chemistry Biochemistry DePauw University, Greencastle IN USA On 8/4/08 1:09 PM, Bert Gunter [EMAIL PROTECTED] wrote: No. Can't be. Editable graphs require that the graph be produced via code that produces changeable components. All R graphs are essentially static. That said, caveats: graphs drawn via the grid package functionality -- for example lattice graphs -- **are** produced via changeable code. If you read the lattice docs carefully, you'll see that there are a few features there that allow some graph editing. There may be other packages that also have some editing capabilties. R's base graphics also allow a little interaction via identify() and locator(), which can be useful (e.g. for positioning legends). One can also simulate interactivity by recording various components of graph construction and then modifying and redrawing them. But this is just manually doing what you're looking for, so probably a dumb suggestion. While graph editing certainly can be a nice feature, it is very difficult to implement without severely constraining graphing flexibility (IMO, of course). Graphs are very complex beasties, so it's hard to write clean code that allows flexibile editing capabilities. Look at S-Plus's graph editing, which I always found harder to use (and more buggy) than just issuing the commands. (To be fair, it's been some years since I tried). Again, just my 2 bits. Others may well disagree (and perhaps point you to what you seek). Cheers, Bert -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Arthur Roberts Sent: Monday, August 04, 2008 9:51 AM To: [EMAIL PROTECTED] Subject: [R] Are there any guis out there,which will allow editing of the graph? Hi, all, I would like to know if there is any gui interface out there (academic or commercial) that allows one to edit R-language generated graphs (e.g positioning x axis labels.) It would be nice to have something like the user interface of Igor or Origin. I have already used JGR and R-gui. These are good, but they don't allow one to easily edit graphs. I have also tried locator() and the package iplots. Your input is greatly appreciated. Best wishes, Art Roberts University of Washington __ 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] Are there any guis out there, which will allow editing of the graph?
Bert Gunter wrote: No. Can't be. Editable graphs require that the graph be produced via code that produces changeable components. All R graphs are essentially static. Well, there's the xfig() device whose output can be edited with xfig This was originally written (by me, for S-PLUS, back in the dark ages, later modified extensively for R by Brian) with the purpose of being able to do things like moving unfortunately placed labels around. In practice, I hardly ever used it because of the writing in sand aspect that affects all such interactive tools. However, xfig is still around and even installable on Windows with a little extra effort. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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] Are there any guis out there, which will allow editing of the graph?
Microsoft Word's graphics editor can edit R graphics saved in metafile format, wmf. That includes x axis labels, etc. On Mon, Aug 4, 2008 at 12:50 PM, Arthur Roberts [EMAIL PROTECTED] wrote: Hi, all, I would like to know if there is any gui interface out there (academic or commercial) that allows one to edit R-language generated graphs (e.g positioning x axis labels.) It would be nice to have something like the user interface of Igor or Origin. I have already used JGR and R-gui. These are good, but they don't allow one to easily edit graphs. I have also tried locator() and the package iplots. Your input is greatly appreciated. Best wishes, Art Roberts University of Washington __ 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] thematic map of USA
On 04/08/2008 1:14 PM, John P. Burkett wrote: My goal is to prepare a thematic map of the US, with states shaded according to their values for a variable of interest. I would like to include an inset for Alaska in the upper left and an inset for Hawaii in the lower left. If possible, I'd like to use Albers conic projection, or something similar. Thus far I have tried using the maps package with its state database (which omits Alaska and Hawaii) and the gmaps package (which places an inset for Alaska in the lower left and uses a projection [Miller cylindrical?] that distorts states along the northern and southern fringe). I would be very grateful for suggestions about alternative approaches. Alaska and Hawaii are in the world map, coded as regions USA:Alaska and Hawaii:Hawaii. I believe you can plot the states data, then add Alaska and Hawaii at locations of your choice afterwards. I don't know how to do multiple regions on the same plot, but here's how to do the components: map(state, projection=albers, parameters=c(25,40)) map(world2Hires, USA:Alaska, projection=albers, parameters=c(55, 70)) map(world2Hires, Hawaii, projection=albers, parameters=c(15, 25)) This needs the mapdata package for world2Hires, and (I think) the mapproj package for the Albers projections. Duncan Murdoch __ 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] Are there any guis out there, which will allow editing of the graph?
Peter Dalgaard wrote: Bert Gunter wrote: No. Can't be. Editable graphs require that the graph be produced via code that produces changeable components. All R graphs are essentially static. Well, there's the xfig() device whose output can be edited with xfig This was originally written (by me, for S-PLUS, back in the dark ages, later modified extensively for R by Brian) Actually, rewritten is more like it. R and S device drivers are rather different beasts, so Brian started from scratch, modifying the postscript device using the fig file specification. with the purpose of being able to do things like moving unfortunately placed labels around. In practice, I hardly ever used it because of the writing in sand aspect that affects all such interactive tools. However, xfig is still around and even installable on Windows with a little extra effort. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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] Sorting a matrix by a column
R 2.6
Windows XP
I have a 100x4 matirx
data-matrix(nrow=100,ncol=4)
I would like to sort the entire matrix by column two, i.e. data[,2]
I looked at the help page for sort() but can not determine how I can use it to
sort a matrix on one of the matrix's columns.
Thanks,
John
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)
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This email message, including any attachments, is for th...{{dropped:6}}
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Re: [R] Sorting a matrix by a column
On Mon, 4 Aug 2008, John Sorkin wrote:
R 2.6
Windows XP
I have a 100x4 matirx
data-matrix(nrow=100,ncol=4)
I would like to sort the entire matrix by column two, i.e. data[,2]
I looked at the help page for sort() but can not determine how I can use it to
sort a matrix on one of the matrix's columns.
Rather, use sort.list, e.g. data[sort.list(data[,2]), ]
There's an example for data frames on ?order.
Thanks,
John
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 th...{{dropped:6}}
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] Sorting a matrix by a column
probably you need order(), e.g.,
data[order(data[, 2]), ]
I hope it helps.
Best,
Dimitris
John Sorkin wrote:
R 2.6
Windows XP
I have a 100x4 matirx
data-matrix(nrow=100,ncol=4)
I would like to sort the entire matrix by column two, i.e. data[,2]
I looked at the help page for sort() but can not determine how I can use it to
sort a matrix on one of the matrix's columns.
Thanks,
John
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:25}}
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[R] Long Range Dependence: Hurst exponent estimation
Dear R Users, Can anyone point me to a package for R vrsion 2.7.1 which implements some Hurst exponent estimation methods ? Thanks in advance, Tolga Generally, this communication is for informational purposes only and it is not intended as an offer or solicitation for the purchase or sale of any financial instrument or as an official confirmation of any transaction. In the event you are receiving the offering materials attached below related to your interest in hedge funds or private equity, this communication may be intended as an offer or solicitation for the purchase or sale of such fund(s). All market prices, data and other information are not warranted as to completeness or accuracy and are subject to change without notice. Any comments or statements made herein do not necessarily reflect those of JPMorgan Chase Co., its subsidiaries and affiliates. This transmission may contain information that is privileged, confidential, legally privileged, and/or exempt from disclosure under applicable law. If you are not the intended recipient, you are hereby notified that any disclosure, copying, distribution, or use of the information contained herein (including any reliance thereon) is STRICTLY PROHIBITED. Although this transmission and any attachments are believed to be free of any virus or other defect that might affect any computer system into which it is received and opened, it is the responsibility of the recipient to ensure that it is virus free and no responsibility is accepted by JPMorgan Chase Co., its subsidiaries and affiliates, as applicable, for any loss or damage arising in any way from its use. If you received this transmission in error, please immediately contact the sender and destroy the material in its entirety, whether in electronic or hard copy format. Thank you. Please refer to http://www.jpmorgan.com/pages/disclosures for disclosures relating to UK legal entities. [[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] FW: Are there any guis out there, which will allow editing of the graph?
Well, just goes to show you how much I know! Glad you were able to get some help. -- Bert -Original Message- From: Arthur Roberts [mailto:[EMAIL PROTECTED] Sent: Monday, August 04, 2008 12:05 PM To: Bert Gunter Subject: Re: [R] Are there any guis out there,which will allow editing of the graph? Dear Bert, I heard from several commentators that you can actually edit Figures by a variety of means. I tried Inkscape and it seems to work pretty well. Best wishes, Art On Aug 4, 2008, at 10:09 AM, Bert Gunter wrote: No. Can't be. Editable graphs require that the graph be produced via code that produces changeable components. All R graphs are essentially static. That said, caveats: graphs drawn via the grid package functionality -- for example lattice graphs -- **are** produced via changeable code. If you read the lattice docs carefully, you'll see that there are a few features there that allow some graph editing. There may be other packages that also have some editing capabilties. R's base graphics also allow a little interaction via identify() and locator(), which can be useful (e.g. for positioning legends). One can also simulate interactivity by recording various components of graph construction and then modifying and redrawing them. But this is just manually doing what you're looking for, so probably a dumb suggestion. While graph editing certainly can be a nice feature, it is very difficult to implement without severely constraining graphing flexibility (IMO, of course). Graphs are very complex beasties, so it's hard to write clean code that allows flexibile editing capabilities. Look at S-Plus's graph editing, which I always found harder to use (and more buggy) than just issuing the commands. (To be fair, it's been some years since I tried). Again, just my 2 bits. Others may well disagree (and perhaps point you to what you seek). Cheers, Bert -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] ] On Behalf Of Arthur Roberts Sent: Monday, August 04, 2008 9:51 AM To: [EMAIL PROTECTED] Subject: [R] Are there any guis out there,which will allow editing of the graph? Hi, all, I would like to know if there is any gui interface out there (academic or commercial) that allows one to edit R-language generated graphs (e.g positioning x axis labels.) It would be nice to have something like the user interface of Igor or Origin. I have already used JGR and R-gui. These are good, but they don't allow one to easily edit graphs. I have also tried locator() and the package iplots. Your input is greatly appreciated. Best wishes, Art Roberts University of Washington __ 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] xyplot strip=function for two conditioning variables
On Mon, Aug 4, 2008 at 3:36 AM, Gabor Grothendieck [EMAIL PROTECTED] wrote: Checkout this one: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/82452.html And there's a wrapper for this in the latticeExtra package: library(latticeExtra) useOuterStrips(xyplot(data=df, copies~week|Treatment+genes,type=c(b,g))) -Deepayan On Mon, Aug 4, 2008 at 6:24 AM, Henning Wildhagen [EMAIL PROTECTED] wrote: Dear list, for a data structure like in df: set.seed(100) Treatment-rep(c(Nitrogen,Carbon, Sulfur),each=9) week-rep(c(1,5,9),3,each=3) genes-rep(c(18s, EF1b, NR),9) copies-rnorm(27, 100,40) df-data.frame(Treatment,week,genes,copies) i wrote this code for a xyplot: library(lattice) PLOT-xyplot(data=df, copies~week|Treatment+genes,type=c(b,g)) I would like to change PLOT in a way, that the strips for Treatment are displayed only once at the top of each of the three columns. Additionally i would like to strips for genes at the top of each panel. I tried to adopt some code i found in the helplist using strip=function... but i did not manage to get what i would like. Can anyone give me advice? Thanks, Henning -- [[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. __ 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] Is there any way to make pretty tables in R to pdf?
Hi, all, All your comments have been very useful. I was wondering if there was a package that can make pretty R tables to pdf. I guess I could use xtable, but I would like something a little more elegant. Your input is greatly appreciated. Best wishes, Art __ 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] subset inside a lattice plot using panel.lines
On Mon, Aug 4, 2008 at 8:44 AM, Mark Difford [EMAIL PROTECTED] wrote:
Hi Michael,
Pulling my hair out here trying to get something very simple to work. ...
I can't quite see what you are trying to do [and I am not sure that you
clearly state it], but you could make things easier and simpler by (1)
creating a factor to identify your groups of rows more cleanly and (2) by
using the groups= argument in lattice.
## Something along these lines
copyDat - originalDat
copyDat$newFac - gl(2, 387, 774, labels=c(A,B))
xyplot(response ~ predictor|maybe.condition, data=copyDat, groups=newFac)
Right, and using 'row' directly should also work (although not with as
good possibilities for annotation):
xyplot(response ~ predictor|maybe.condition, data=copyDat,
groups=(row = 387), type = c(l, g))
If you need a custom panel function, try using panel.groups:
xyplot(response ~ predictor|maybe.condition, data=copyDat,
groups=(row = 387),
panel = panel.superpose,
panel.groups = function(...) {
panel.grid(h = -1, v = -1, lty = 3)
panel.lines(...)
})
-Deepayan
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[R] backslash in character string?
Dear list, After searching many old posts, I can't find the solution to a simple problem. can someone tell me how to create a character string with multiple backslashes, as in: file_dir - c(C:\files\data\) I need to create this string and then paste it to many files names for batch processing in another software program. R won't accept the backslash and removes it. Thanks for any suggestions, zack [[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] an interesting finding on Hurst exponent estimation from fSeries
Dear R Users, I am using code from the following links to do Hurst exponent estimation. link: http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php?rev=1root=rmetricsview=rev file: LongRangeDependence.R Take a look at the following run: x-(cos((1:200)/10)) rsFit(diff(x,15))@hurst$H [1] 1.027420 aggvarFit(diff(x,15))@hurst$H [1] 0.02301331 First of all, can anyone cast a light on the drastically different results (obviously, the R/S fit is just wrong) ? Further, can anyone recommend something which is more stable ? This code appears to be part of fSeries tho it doesnt appear if one installs and loads that package. Instead, one has to source the R code directly from the .R file. Thanks in advance, Tolga Generally, this communication is for informational purposes only and it is not intended as an offer or solicitation for the purchase or sale of any financial instrument or as an official confirmation of any transaction. In the event you are receiving the offering materials attached below related to your interest in hedge funds or private equity, this communication may be intended as an offer or solicitation for the purchase or sale of such fund(s). All market prices, data and other information are not warranted as to completeness or accuracy and are subject to change without notice. Any comments or statements made herein do not necessarily reflect those of JPMorgan Chase Co., its subsidiaries and affiliates. This transmission may contain information that is privileged, confidential, legally privileged, and/or exempt from disclosure under applicable law. If you are not the intended recipient, you are hereby notified that any disclosure, copying, distribution, or use of the information contained herein (including any reliance thereon) is STRICTLY PROHIBITED. Although this transmission and any attachments are believed to be free of any virus or other defect that might affect any computer system into which it is received and opened, it is the responsibility of the recipient to ensure that it is virus free and no responsibility is accepted by JPMorgan Chase Co., its subsidiaries and affiliates, as applicable, for any loss or damage arising in any way from its use. If you received this transmission in error, please immediately contact the sender and destroy the material in its entirety, whether in electronic or hard copy format. Thank you. Please refer to http://www.jpmorgan.com/pages/disclosures for disclosures relating to UK legal entities. [[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] backslash in character string?
Try: file_dir - C:\\files\\data\\ On Mon, Aug 4, 2008 at 5:02 PM, zack holden [EMAIL PROTECTED] wrote: Dear list, After searching many old posts, I can't find the solution to a simple problem. can someone tell me how to create a character string with multiple backslashes, as in: file_dir - c(C:\files\data\) I need to create this string and then paste it to many files names for batch processing in another software program. R won't accept the backslash and removes it. Thanks for any suggestions, zack [[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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] Is there any way to make pretty tables in R to pdf?
Hi Arthur, I was wondering if there was a package that can make pretty R tables to pdf. You got through TeX/LateX, but PDF could be your terminus. Package Hmisc: ? summary.formula and its various arguments and options. You can't get much better. http://cran.za.r-project.org/doc/contrib/Harrell-statcomp-notes.pdf HTH, Mark. Arthur Roberts wrote: Hi, all, All your comments have been very useful. I was wondering if there was a package that can make pretty R tables to pdf. I guess I could use xtable, but I would like something a little more elegant. Your input is greatly appreciated. Best wishes, Art __ 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. -- View this message in context: http://www.nabble.com/Is-there-any-way-to-make-pretty-tables-in-R-to-pdf--tp18818187p18818675.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] Long Range Dependence: Hurst exponent estimation
Thanks Gary. That package is a bit weird. When one installs and loads it up, you don't actually get any of those functions. One has to go to the following link: http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php?rev=1root=rmetricsview=rev and then download and source the file LongRangeDependence.R to get the aggvar and rsFit functions to do Hurst exponent estimation. It appears to be the case that the fSeries package currently does not really have that functionality. Perhaps they meant to put it in there but decided to exclude it in the current release for some reason. I have also cc'd the RMetrics group here to see if they can cast a light on this issue. Thanks, Tolga Ling, Gary \(Electronic Trading\) [EMAIL PROTECTED] 04/08/2008 20:51 To [EMAIL PROTECTED] cc Subject RE: [R] Long Range Dependence: Hurst exponent estimation Hi, you can try the fSeries package. See this doc: http://phase.hpcc.jp/mirrors/stat/R/CRAN/doc/packages/fSeries.pdf -gary -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Monday, August 04, 2008 3:04 PM To: r-help@r-project.org Subject: [R] Long Range Dependence: Hurst exponent estimation Dear R Users, Can anyone point me to a package for R vrsion 2.7.1 which implements some Hurst exponent estimation methods ? Thanks in advance, Tolga Generally, this communication is for informational purposes only and it is not intended as an offer or solicitation for the purchase or sale of any financial instrument or as an official confirmation of any transaction. In the event you are receiving the offering materials attached below related to your interest in hedge funds or private equity, this communication may be intended as an offer or solicitation for the purchase or sale of such fund(s). All market prices, data and other information are not warranted as to completeness or accuracy and are subject to change without notice. Any comments or statements made herein do not necessarily reflect those of JPMorgan Chase Co., its subsidiaries and affiliates. This transmission may contain information that is privileged, confidential, legally privileged, and/or exempt from disclosure under applicable law. If you are not the intended recipient, you are hereby notified that any disclosure, copying, distribution, or use of the information contained herein (including any reliance thereon) is STRICTLY PROHIBITED. Although this transmission and any attachments are believed to be free of any virus or other defect that might affect any computer system into which it is received and opened, it is the responsibility of the recipient to ensure that it is virus free and no responsibility is accepted by JPMorgan Chase Co., its subsidiaries and affiliates, as applicable, for any loss or damage arising in any way from its use. If you received this transmission in error, please immediately contact the sender and destroy the material in its entirety, whether in electronic or hard copy format. Thank you. Please refer to http://www.jpmorgan.com/pages/disclosures for disclosures relating to UK legal entities. [[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. This message w/attachments (message) may be privileged, confidential or proprietary, and if you are not an intended recipient, please notify the sender, do not use or share it and delete it. Unless specifically indicated, this message is not an offer to sell or a solicitation of any investment products or other financial product or service, an official confirmation of any transaction, or an official statement of Merrill Lynch. Subject to applicable law, Merrill Lynch may monitor, review and retain e-communications (EC) traveling through its networks/systems. The laws of the country of each sender/recipient may impact the handling of EC, and EC may be archived, supervised and produced in countries other than the country in which you are located. This message cannot be guaranteed to be secure or error-free. This message is subject to terms available at the following link: http://www.ml.com/e-communications_terms/ . By messaging with Merrill Lynch you consent to the foregoing. Generally, this communication is for informational purposes only and it is not intended as an offer or solicitation for the purchase or sale of any financial instrument or as an official confirmation of any transaction. In the event you are receiving the offering materials attached below related to your interest in hedge funds or private equity, this communication may be intended as an offer or
Re: [R] Is there any way to make pretty tables in R to pdf?
Hi Arthur, Sorry, sent you down the wrong track: this will help you to get there: http://biostat.mc.vanderbilt.edu/twiki/pub/Main/StatReport/summary.pdf Regards, Mark. Arthur Roberts wrote: Hi, all, All your comments have been very useful. I was wondering if there was a package that can make pretty R tables to pdf. I guess I could use xtable, but I would like something a little more elegant. Your input is greatly appreciated. Best wishes, Art __ 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. -- View this message in context: http://www.nabble.com/Is-there-any-way-to-make-pretty-tables-in-R-to-pdf--tp18818187p18818837.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] backslash in character string?
You have to escape every backslash by a second backslash - try this: f - C:\\files\\data\\ # create character string write(f, ) # write to file system(open -t ) # see what happen to the character string file_dir - c(C:\files\data\) Christoph Heibl Systematic Botany Ludwig-Maximilians-Universität München Menzinger Str. 67 D-80638 München GERMANY phone: +49-(0)89-17861-251 e-mail:[EMAIL PROTECTED] http://www.christophheibl.de/ch-home.html SAVE PAPER - THINK BEFORE YOU PRINT [[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] History pruning
On 8/1/08 1:13 PM, Richard M. Heiberger [EMAIL PROTECTED] wrote: I meant 5a 5b 5c. Multiple-line commands are handled correctly. What is is doing is looking for and + prompts. Anything else is removed. When I said 5c) prune any lines that don't have assignment operators I meant to take a sequence like this (to pick a semi-random chunk from my history log): --- df - data.frame(x=2:9, y=(1:8)^2) cor(df) ?cor mad(c(1:9)) ?reshape a - matrix(1:12, nrow=3) b - matrix(2:13, nrow=3) b - matrix(4:15, nrow=3) b - matrix(2:13, nrow=3) c - matrix(4:15, nrow=3) a b c --- And turn it into this: --- df - data.frame(x=2:9, y=(1:8)^2) a - matrix(1:12, nrow=3) b - matrix(2:13, nrow=3) b - matrix(4:15, nrow=3) b - matrix(2:13, nrow=3) c - matrix(4:15, nrow=3) --- Obviously I wouldn't *always* want this performed, but selectively it would be quite nice. Further, if the dependency graph among variable definitions were computable, the sequence could be reduced to this: --- df - data.frame(x=2:9, y=(1:8)^2) a - matrix(1:12, nrow=3) b - matrix(2:13, nrow=3) c - matrix(4:15, nrow=3) --- Note that the starting point of all of this is a sequence of commands (the output of savehistory(), so separating commands from output isn't necessary. I've made a bit of progress on this, hopefully I can get clearance to show my work soon. It would be nice if this could be hooked into ESS for selective pruning or something. -Ken -- Ken Williams Research Scientist The Thomson Reuters Corporation Eagan, MN __ 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] simulate data based on partial correlation matrix
Given four known and fixed vectors, x1,x2,x3,x4, I am trying to generate a fifth vector,z, with specified known and fixed partial correlations. How can I do this? In the past I have used the following (thanks to Greg Snow) to generate a fifth vector based on zero order correlations---however I'd like to modify it so that it can generate a fifth vector with specific partial correlations rather than zero order correlations: # create x1-x4 x1 - rnorm(100, 50, 3) x2 - rnorm(100) + x1/5 x3 - rnorm(100) + x2/5 x4 - rnorm(100) + x3/5 # find current correlations cor1 - cor( cbind(x1,x2,x3,x4) ) cor1 # create 1st version of z z - rnorm(100) # combine in a matrix m1 - cbind( x1, x2, x3, x4, z ) # center and scale m2 - scale(m1) # find cholesky decomp c1 - chol(var(m2)) # force to be independent m3 - m2 %*% solve(c1) # create new correlation matrix: cor2 - cbind( rbind( cor1, z=c(.5,.3,.1,.05) ), z=c(.5,.3,.1,.05,1) ) # create new matrix m4 - m3 %*% chol(cor2) # uncenter and unscale m5 - sweep( m4, 2, attr(m2, 'scaled:scale'), '*') m5 - sweep( m5, 2, attr(m2, 'scaled:center'), '+') ##Check they are equal zapsmall(cor(m5))==zapsmall(cor2) Thanks, ben __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with nested loop for regression
Hi,
I guess my question is really more about the nested for loop construct and
whether it is doing what I intend it to do in my code in the previous post.
I would be grateful if anyone who has used nested loops could let me know if
I am doing something wrong.
Thanks,
rcoder
rcoder wrote:
Hi everyone,
I'm experiencing difficulty getting the results I want when I use a nested
for loop. I have a data set to which I perform some calculations, and then
try to apply a regression over a rolling window. The code runs, but the
regression results I am getting (intercept and slope) are simply the same,
repeated again and again in the results matrix. The regression does not
seem to be obeying the instructions of the nested loop, which intend it to
calculate regression coefficients over a data one row at a time. I have
been struggling with the code for many days now, testing various parts,
and I cannot seem to get the nested loop to work as I want it to. I would
be very grateful for any suggestions. Below is a brief version of my code:
#Code start
library(zoo)
seed.set(1)
Pmat-matrix(rnorm(1000), nrow=100, ncol=100)
maxcol-ncol(Pmat)
maxrow-nrow(Pmat)
startrow-10
period-10
wind-2 #roll window
subdiv-period/wind
rollstart-11 #start roll at period+1
#converting Pmat into ts for rollapply() to work...
PmatTS-ts(Pmat)
Preg-matrix(NA,ncol=maxcol,nrow=2*maxrow)
PmatWt-matrix(NA, nrow=subdiv,ncol=maxcol)
mult_col-matrix(1:5, nrow=5, ncol=1)
#rolling calculations...
for (i in (rollstart):maxrow)
{
#extract the relevant timeframe...
dslice-PmatTS[(i-period):i,]
dslicets-ts(dslice)
#operating on sliced data...
Pmin-rollapply(dslicets, wind, by=wind, min, na.rm=F)
Pmax-rollapply(dslicets, wind, by=wind, max, na.rm=F)
Pmult-Pmin*Pmax#calculating product
tt-time(Pmult)
for (j in 1:5)#1st nested loop
{
PmatWt[j,]-Pmult[j,]*mult_col[j,]
}
#rolling regression analysis...
for (k in 1:maxcol) #2nd nested loop
{
sel_col-PmatWt[,k]
if(!all(is.na(sel_col))) {Preg[,k]-coef(lm(tt~sel_col))}
}
}
#Code End
Thanks,
rcoder
--
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[R] Turning Cross-tab into new data frame
Dear R Experts, I am wondering if anyone has a solution to the following: I am creating some cross-tabulation tables and want to input the results from these tables into a new set of functions. Is there a way to turn the cross-tab table into a new dataset? I haven't yet been able to accomplish this by placing the cross tab function inside a as.matrix or data.frame yet. Any help would be great! Spencer -- Spencer Cohen PhD Student Department of Geography University of Washington Email: [EMAIL PROTECTED] Website: staff.washington.edu/zhuge99 __ 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] Lattice: How to draw curves from given formulae
I have another questions. How can I type specific names into strips of the
resulting plot?
For instance, in the resulting figure from the attached code, instead of
'umbrella(d)', I want have 'UMBRELLA' in the strip.
library(lattice)
flat - function(d) 0 * d
linear - function(d) -(1.65/8) * d
logistic - function(d) 0.015 - 1.73 / (1 + exp(1.2 * (4-d)))
umbrella - function(d) -(1.65/3) * d + (1.65 / 36) * d^2
emax - function(d) -1.81 * d / (0.79 + d)
sigmoid - function(d) -1.70 * d^5 / (4^5 + d^5)
xyplot(flat(d) + linear(d) + logistic(d) + umbrella(d) + emax(d) +
sigmoid(d) ~ d,
data = list(d = seq(0, 8, length = 101)),
ylab='Expected change from baseline in VAS at Week 6', xlab='Dose',
type = l, outer = TRUE, )
Many thanks
On Tue, Jul 22, 2008 at 4:31 PM, Deepayan Sarkar
[EMAIL PROTECTED]wrote:
On 7/22/08, John Smith [EMAIL PROTECTED] wrote:
Thanks, Deepayan,
I actually want to have a picture like the Figure 5.9 in Mixed-Effects
Models in S and S-PLUS, which have a separate panel for each function
and
has the function name been printed in strip.
xyplot(linear(d) + logistic(d) + umbrella(d) ~ d,
data = list(d = seq(-5, 5, length = 101)),
type = l, outer = TRUE)
There are several ways to make the strip labels just the names; the
simplest is to create a data frame with all the necessary columns
beforehand.
-Deepayan
Thanks
On Tue, Jul 22, 2008 at 1:08 PM, Deepayan Sarkar
[EMAIL PROTECTED]
wrote:
On 7/21/08, John Smith [EMAIL PROTECTED] wrote:
Dear R Users:
I have a list function as:
Flat: y = 0
Linear: y = -(1.65/8)d
Logistic: y = 0.015 - 1.73/{1+exp[1.2(4-d)]}
Umbrella: y= -(1.65/3)d + (1.65/36)d^2
Emax: y = -1.81d/(0.79+d)
Sigmoid Emax: y = -1.70d^5/(4^5+d^5)
And want draw the figure as attached (those material are extracted
from
a
paper). Could anyone give me a sample code to do this?
The attachment didn't come through, but here's one approach (with
three of your functions):
linear - function(d) -(1.65/8) * d
logistic - function(d) 0.015 - 1.73 / (1 + exp(1.2 * (4-d)))
umbrella - function(d) -(1.65/3) * d + (1.65 / 36) * d^2
xyplot(linear(d) + logistic(d) + umbrella(d) ~ d,
data = list(d = seq(-5, 5, length = 101)),
type = l,
auto.key = list(lines = TRUE, points = FALSE))
-Deepayan
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and provide commented, minimal, self-contained, reproducible code.
[R] simulate data based on partial correlation matrix
Given four known and fixed vectors, x1,x2,x3,x4, I am trying to generate a fifth vector,z, with specified known and fixed partial correlations. How can I do this? In the past I have used the following (thanks to Greg Snow) to generate a fifth vector based on zero order correlations---however I'd like to modify it so that it can generate a fifth vector with specific partial correlations rather than zero order correlations: # create x1-x4 x1 - rnorm(100, 50, 3) x2 - rnorm(100) + x1/5 x3 - rnorm(100) + x2/5 x4 - rnorm(100) + x3/5 # find current correlations cor1 - cor( cbind(x1,x2,x3,x4) ) cor1 # create 1st version of z z - rnorm(100) # combine in a matrix m1 - cbind( x1, x2, x3, x4, z ) # center and scale m2 - scale(m1) # find cholesky decomp c1 - chol(var(m2)) # force to be independent m3 - m2 %*% solve(c1) # create new correlation matrix: cor2 - cbind( rbind( cor1, z=c(.5,.3,.1,.05) ), z=c(.5,.3,.1,.05,1) ) # create new matrix m4 - m3 %*% chol(cor2) # uncenter and unscale m5 - sweep( m4, 2, attr(m2, 'scaled:scale'), '*') m5 - sweep( m5, 2, attr(m2, 'scaled:center'), '+') ##Check they are equal zapsmall(cor(m5))==zapsmall(cor2) Thanks, ben __ 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] Long Range Dependence: Hurst exponent estimation
There is the 'fdim' package that computes the fractal dimension D. Between D and the Hurst exponent H there should be a relation D = 2 - H I wonder if this is true when computing D and H with different approaches Regards, Hans Werner Borchers ABB Corporate Research tolga.i.uzuner at jpmorgan.com writes: Thanks Gary. That package is a bit weird. When one installs and loads it up, you don't actually get any of those functions. [...] It appears to be the case that the fSeries package currently does not really have that functionality. Perhaps they meant to put it in there but decided to exclude it in the current release for some reason. I have also cc'd the RMetrics group here to see if they can cast a light on this issue. Thanks, Tolga __ 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] Multivariate Regression with Weights
Hi all, I'd like to fit a multivariate regression with the variance of the error term porportional to the predictors, like the WLS in the univariate case. y_1~x_1+x_2 y_2~x_1+x_2 var(y_1)=x_1*sigma_1^2 var(y_2)=x_2*sigma_2^2 cov(y_1,y_2)=sqrt(x_1*x_2)*sigma_12^2 How can I specify this in R? Is there a corresponding function to the univariate specification lm(y~x,weights=x)?? Thanks. Sincerely, Yanwei Zhang Department of Actuarial Research and Modeling Munich Re America Tel: 609-275-2176 Email: [EMAIL PROTECTED]mailto:[EMAIL PROTECTED] [[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] about the 95%CI around the median...
On 5/08/2008, at 1:31 AM, Frank E Harrell Jr wrote:
snip
I wonder why we don't just use the exact nonparametric confidence
interval for the median, which is just as easy to compute. Also,
it will be asymmetric if the data are skewed, as it should be.
snip
The boxplot.stats() function gives the same ``confidence interval''
for the median as the
boxplot() function. This isn't however *really* a confidence interval.
The help on boxplot() (see the ``notch'' argument) indicates that
this ``confidence interval''
actually provides a rough *hypothesis test* of the equality of *two*
medians. The null hypothesis
is rejected if the notches (confidence intervals) don't overlap.
cheers,
Rolf
##
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Re: [R] Multivariate Regression with Weights
Thanks. Sincerely, Yanwei Zhang Department of Actuarial Research and Modeling Munich Re America Tel: 609-275-2176 Email: [EMAIL PROTECTED] -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Sent: Monday, August 04, 2008 5:15 PM To: Zhang Yanwei - Princeton-MRAm Subject: RE: [R] Multivariate Regression with Weights the systemfit package can do that and the documentation is quite nice also. On Mon, Aug 4, 2008 at 4:39 PM, Zhang Yanwei - Princeton-MRAm wrote: Hi all, I'd like to fit a multivariate regression with the variance of the error term porportional to the predictors, like the WLS in the univariate case. y_1~x_1+x_2 y_2~x_1+x_2 var(y_1)=x_1*sigma_1^2 var(y_2)=x_2*sigma_2^2 cov(y_1,y_2)=sqrt(x_1*x_2)*sigma_12^2 How can I specify this in R? Is there a corresponding function to the univariate specification lm(y~x,weights=x)?? Thanks. Sincerely, Yanwei Zhang Department of Actuarial Research and Modeling Munich Re America Tel: 609-275-2176 Email: [EMAIL PROTECTED]mailto:[EMAIL PROTECTED] [[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] Are there any guis out there, which will allow editing of the graph?
Duncan Murdoch murdoch at stats.uwo.ca writes: On 04/08/2008 12:50 PM, Arthur Roberts wrote: Hi, all, I would like to know if there is any gui interface out there (academic or commercial) that allows one to edit R-language generated graphs (e.g positioning x axis labels.) It would be nice to have something like the user interface of Igor or Origin. I have already used JGR and R-gui. These are good, but they don't allow one to easily edit graphs. I have also tried locator() and the package iplots. Your input is greatly appreciated. I imagine there are a number of commercial packages available that would allow you to edit PDF or PS files, but I haven't used any of them, so I can't recommend one. There is this fantastic open source editor for vector graphics, *Inkscape* (http://www.inkscape.org/). Inkscape can read PDF files generated by R as PDFs. By 'ungrouping' graphs, Inkscape is able to modify any single piece of the diagram. Inkscape will save modified diagrams in PDF, EPS, SVG or any other reasonable graphics format. Regards, Hans Werner Borchers ABB Corporate Research Duncan Murdoch __ 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] log and Log Histogram
Hi All, I have a X,Y and Z text file. I wish modify the X value in a logarithmic value and to plot all log(Z) with Histogram. I try different code but I have a problem to find the solution. Thank you very much Ale [[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] log and Log Histogram
On Mon, 4 Aug 2008, Alessandro wrote: Hi All, I have a X,Y and Z text file. I wish modify the X value in a logarithmic value and to plot all log(Z) with Histogram. I try different code but I have a problem to find the solution. My package HyperbolicDist has a log histogram function. David Scott _ David Scott Department of Statistics, Tamaki Campus The University of Auckland, PB 92019 Auckland 1142,NEW ZEALAND Phone: +64 9 373 7599 ext 86830 Fax: +64 9 373 7000 Email: [EMAIL PROTECTED] Graduate Officer, Department of Statistics Director of Consulting, Department of Statistics __ 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] R init file and source()
In the context of calling R from another program (namely gretl, http://gretl.sourceforge.net ) I'm trying to understand the interactions of the R init file (corresponding to the environment variable RPROFILE) and the source() function. I'll illustrate my problem with the following simplified contrast implemented in the bash shell (with R 2.7.1). 1. Works fine: [EMAIL PROTECTED]:~/Rfoo$ cat fooProfile1 # nothing here [EMAIL PROTECTED]:~/Rfoo$ cat doit.1 export RPROFILE=fooProfile1 R --no-save --no-init-file --no-restore-data --slave fooSrc [EMAIL PROTECTED]:~/Rfoo$ cat fooSrc library(graphics) postscript(myfile.ps) [EMAIL PROTECTED]:~/Rfoo$ sh doit.1 (silently produces a well-formed but empty myfile.ps) 2. Doesn't work: [EMAIL PROTECTED]:~/Rfoo$ cat fooProfile2 source(fooSrc, verbose=TRUE, echo=TRUE) [EMAIL PROTECTED]:~/Rfoo$ cat doit.2 export RPROFILE=fooProfile2 R --no-save --no-init-file --no-restore-data --slave [EMAIL PROTECTED]:~/Rfoo$ cat fooSrc library(graphics) postscript(myfile.ps) [EMAIL PROTECTED]:~/Rfoo$ sh doit.2 'envir' chosen:environment: R_GlobalEnv encoding = native.enc chosen -- parsed 2 expressions; now eval(.)ing them: eval(expression_nr. 1 ) = library(graphics) curr.fun: symbol library .. after `expression(library(graphics))' eval(expression_nr. 2 ) = postscript(myfile.ps) Error in eval.with.vis(expr, envir, enclos) : could not find function postscript == end of output == I'd like to understand why, after loading the graphics library, the function postscript() is not available in the second case while it is in the first. (The file fooSrc has the same content in the two cases: in the first case it's loaded via the shell, while in the second it's loaded using R's source() function.) Thanks for any help. -- Allin Cottrell Department of Economics Wake Forest University, NC __ 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] Constrained Optimization
Hello, I am trying to run a constrained optimization in R. constrOptim is really useful and has helped me a lot, but unfortunately, it doesn't provide the hessian. Is there a solution to this problem? I've tried optim with penalty-functions and L-BFGS-B, but it doesn't help. Alan. -- http://games.entertainment.gmx.net/de/entertainment/games/free/puzzle/6169196 __ 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] R: log and Log Histogram
Hi david, I tried HyperbolicDist but I have this problem (I wish transform Z value) Library (HyperbolicDist) Data (testground) # X,Y and Z value in txt file changetestground - testground[-length(testground)]/testground[-1] hist(change) Error in hist.default(change) : 'x' deve essere numeric logHist(change) Error in hist.default(x, plot = FALSE) : 'x' must be numeric THANK Ale -Messaggio originale- Da: David Scott [mailto:[EMAIL PROTECTED] Inviato: lunedì 4 agosto 2008 16.30 A: Alessandro Cc: r-help@r-project.org Oggetto: Re: [R] log and Log Histogram On Mon, 4 Aug 2008, Alessandro wrote: Hi All, I have a X,Y and Z text file. I wish modify the X value in a logarithmic value and to plot all log(Z) with Histogram. I try different code but I have a problem to find the solution. My package HyperbolicDist has a log histogram function. David Scott _ David Scott Department of Statistics, Tamaki Campus The University of Auckland, PB 92019 Auckland 1142,NEW ZEALAND Phone: +64 9 373 7599 ext 86830 Fax: +64 9 373 7000 Email: [EMAIL PROTECTED] Graduate Officer, Department of Statistics Director of Consulting, Department of Statistics __ 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] Lattice: How to draw curves from given formulae
On Mon, Aug 4, 2008 at 1:39 PM, John Smith [EMAIL PROTECTED] wrote: I have another questions. How can I type specific names into strips of the resulting plot? For instance, in the resulting figure from the attached code, instead of 'umbrella(d)', I want have 'UMBRELLA' in the strip. library(lattice) flat - function(d) 0 * d linear - function(d) -(1.65/8) * d logistic - function(d) 0.015 - 1.73 / (1 + exp(1.2 * (4-d))) umbrella - function(d) -(1.65/3) * d + (1.65 / 36) * d^2 emax - function(d) -1.81 * d / (0.79 + d) sigmoid - function(d) -1.70 * d^5 / (4^5 + d^5) xyplot(flat(d) + linear(d) + logistic(d) + umbrella(d) + emax(d) + sigmoid(d) ~ d, data = list(d = seq(0, 8, length = 101)), ylab='Expected change from baseline in VAS at Week 6', xlab='Dose', type = l, outer = TRUE, ) Here's one solution. xyplot(flat(d) + linear(d) + logistic(d) + umbrella(d) + emax(d) + sigmoid(d) ~ d, data = list(d = seq(0, 8, length = 101)), ylab='Expected change from baseline in VAS at Week 6', xlab='Dose', type = l, outer = TRUE, strip = strip.custom(factor.levels = c(FLAT, LINEAR, LOGISTIC, UMBRELLA, EMAX, SIGMOID))) -Deepayan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] problem with nested loop for regression
Exactly what problem are you having? There is nothing wrong with
nested for loops, so what is leading you to believe you have a
problem? I ran your program and it seems to terminate. Most of the
time seems to have been spent in the following statement:
if(!all(is.na(sel_col))) {Preg[,k]-coef(lm(tt~sel_col))}
Given that you appear to be executing this statement about 9000 times,
it is not surprising. So can you provide more details?
0 51.2 root
1. 44.5 coef
2. . 44.1 lm
3. . . 15.7 eval
4. . . . 15.6 eval
5. . . . | 15.6 model.frame
6. . . . | . 15.3 model.frame.default
7. . . . | . .6.0 sapply
8. . . . | . . .3.8 lapply
9. . . . | . . . .3.0 FUN
10. . . . | . . . . |1.1 %in%
11. . . . | . . . . | .1.0 match
12. . . . | . . . . | . .0.8 is.factor
13. . . . | . . . . | . . .0.7 inherits
10. . . . | . . . . |1.0 .deparseOpts
11. . . . | . . . . | .0.3 pmatch
9. . . . | . . . .0.5 as.list
8. . . . | . . .1.7 unique
9. . . . | . . . .0.7 unique.default
9. . . . | . . . .0.6 unlist
10. . . . | . . . . |0.4 lapply
8. . . . | . . .0.3 unlist
7. . . . | . .4.0 na.omit
8. . . . | . . .3.8 na.omit.data.frame
9. . . . | . . . .3.0 [
10. . . . | . . . . |2.9 [.data.frame
11. . . . | . . . . | .0.6 duplicated
12. . . . | . . . . | . .0.3 duplicated.default
11. . . . | . . . . | .0.3 [
12. . . . | . . . . | . .0.2 [.ts
11. . . . | . . . . | .0.3 [[
On Mon, Aug 4, 2008 at 12:58 PM, rcoder [EMAIL PROTECTED] wrote:
Hi,
I guess my question is really more about the nested for loop construct and
whether it is doing what I intend it to do in my code in the previous post.
I would be grateful if anyone who has used nested loops could let me know if
I am doing something wrong.
Thanks,
rcoder
rcoder wrote:
Hi everyone,
I'm experiencing difficulty getting the results I want when I use a nested
for loop. I have a data set to which I perform some calculations, and then
try to apply a regression over a rolling window. The code runs, but the
regression results I am getting (intercept and slope) are simply the same,
repeated again and again in the results matrix. The regression does not
seem to be obeying the instructions of the nested loop, which intend it to
calculate regression coefficients over a data one row at a time. I have
been struggling with the code for many days now, testing various parts,
and I cannot seem to get the nested loop to work as I want it to. I would
be very grateful for any suggestions. Below is a brief version of my code:
#Code start
library(zoo)
seed.set(1)
Pmat-matrix(rnorm(1000), nrow=100, ncol=100)
maxcol-ncol(Pmat)
maxrow-nrow(Pmat)
startrow-10
period-10
wind-2 #roll window
subdiv-period/wind
rollstart-11 #start roll at period+1
#converting Pmat into ts for rollapply() to work...
PmatTS-ts(Pmat)
Preg-matrix(NA,ncol=maxcol,nrow=2*maxrow)
PmatWt-matrix(NA, nrow=subdiv,ncol=maxcol)
mult_col-matrix(1:5, nrow=5, ncol=1)
#rolling calculations...
for (i in (rollstart):maxrow)
{
#extract the relevant timeframe...
dslice-PmatTS[(i-period):i,]
dslicets-ts(dslice)
#operating on sliced data...
Pmin-rollapply(dslicets, wind, by=wind, min, na.rm=F)
Pmax-rollapply(dslicets, wind, by=wind, max, na.rm=F)
Pmult-Pmin*Pmax#calculating product
tt-time(Pmult)
for (j in 1:5)#1st nested loop
{
PmatWt[j,]-Pmult[j,]*mult_col[j,]
}
#rolling regression analysis...
for (k in 1:maxcol) #2nd nested loop
{
sel_col-PmatWt[,k]
if(!all(is.na(sel_col))) {Preg[,k]-coef(lm(tt~sel_col))}
}
}
#Code End
Thanks,
rcoder
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--
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Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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[R] Create data frame from header only
Hi all, Given a string list, paste(A,1:5,sep=) [1] A1 A2 A3 A4 A5 I would like to create an empty data frame using that list as the header, so I can access my data frame column using, df [ list [ i ] ] Anyone ? Thanks, Etienne [[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] Sweave and ggplot2
Dear Thierry,
Your guess was correct. I was not aware that I needed print( ). It was not
needed with plot( ). I have yet to try xyplot( ) but I suspect that the
print( ) might be needed here as well. Thank you very much for your help.
Cheers,
Sorn
ONKELINX, Thierry [EMAIL PROTECTED]
04/08/2008 06:02 PM
To
[EMAIL PROTECTED], r-help@r-project.org
cc
Subject
RE: [R] Sweave and ggplot2
Dear Sorn,
It's hard to guess what your problem is, as you don't provide any sample
code. My guess is that the graphics are empty. Did you use
print(qplot(...)) or just qplot(). The latter won't work. You need
print(qplot(...))
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature
and Forest
Cel biometrie, methodologie en kwaliteitszorg / Section biometrics,
methodology and quality assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium
tel. + 32 54/436 185
[EMAIL PROTECTED]
www.inbo.be
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data.
~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
Namens [EMAIL PROTECTED]
Verzonden: maandag 4 augustus 2008 6:55
Aan: r-help@r-project.org
Onderwerp: [R] Sweave and ggplot2
Hi all,
I've been trying to run Sweave with R code embedded - using the ggplot2
package and in particular the qplot command. There appears to be a
problem
in Sweave not picking up that qplot is a function. Has anybody else
tried
to use qplot in Sweave and have you been successful? Any help would be
very much appreciated.
Kind Regards,
Sorn
Notice:
This email and any attachments may contain information\ ...{{dropped:26}}
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Re: [R] Create data frame from header only
Hi Etienne It is not so elegant, bu I think that work. colname.list-paste(A,1:5,sep=) df-data.frame(matrix(matrix(rep(1,length(colname.list)),1),1)) df colnames(df)-colname.list df df-df[-1,] df Cheers, Miltinho Astronauta Brazil On 8/4/08, Etienne Bellemare Racine [EMAIL PROTECTED] wrote: Hi all, Given a string list, paste(A,1:5,sep=) [1] A1 A2 A3 A4 A5 I would like to create an empty data frame using that list as the header, so I can access my data frame column using, df [ list [ i ] ] Anyone ? Thanks, Etienne [[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] problem with nested loop for regression
Actually now that I read it closer, I see what your problem is. what
did you think the statement:
Preg[,k]-coef(lm(tt~sel_col))
was going to do? Preg is a 200x100 matrix and you are only storing
two values (the coefficients) so they will be repeated 100 times in
the column. So there is nothing wrong with your nested for loops;
it is in the algorithm that you are using. You might want to use the
browser and you would see something like this:
Browse[1] str(sel_col)
num [1:5] -0.115 -2.666 -0.811 1.440 -0.879
Browse[1] str(tt)
Time-Series [1:5] from 2 to 10: 2 4 6 8 10
Browse[1] n
debug: if (!all(is.na(sel_col))) {
Preg[, k] - coef(lm(tt ~ sel_col))
}
Browse[1] n
debug: Preg[, k] - coef(lm(tt ~ sel_col))
Browse[1] k
[1] 1
Browse[1] n
debug: k
Browse[1] n
debug: sel_col - PmatWt[, k]
Browse[1] k
[1] 2
Browse[1] str(sel_col)
num [1:5] -0.115 -2.666 -0.811 1.440 -0.879
Browse[1] str(Preg)
num [1:200, 1:100] 6.356 0.587 6.356 0.587 6.356 ...
Browse[1] str(PmatWt)
num [1:5, 1:100] -0.115 -2.666 -0.811 1.440 -0.879 ...
Browse[1] coef(lm(tt~sel_col))
(Intercept) sel_col
6.3557600.586699
Browse[1] str(Preg)
num [1:200, 1:100] 6.356 0.587 6.356 0.587 6.356 ...
Browse[1]
This would help you to understand what is happening.
On Mon, Aug 4, 2008 at 8:35 PM, jim holtman [EMAIL PROTECTED] wrote:
Exactly what problem are you having? There is nothing wrong with
nested for loops, so what is leading you to believe you have a
problem? I ran your program and it seems to terminate. Most of the
time seems to have been spent in the following statement:
if(!all(is.na(sel_col))) {Preg[,k]-coef(lm(tt~sel_col))}
Given that you appear to be executing this statement about 9000 times,
it is not surprising. So can you provide more details?
0 51.2 root
1. 44.5 coef
2. . 44.1 lm
3. . . 15.7 eval
4. . . . 15.6 eval
5. . . . | 15.6 model.frame
6. . . . | . 15.3 model.frame.default
7. . . . | . .6.0 sapply
8. . . . | . . .3.8 lapply
9. . . . | . . . .3.0 FUN
10. . . . | . . . . |1.1 %in%
11. . . . | . . . . | .1.0 match
12. . . . | . . . . | . .0.8 is.factor
13. . . . | . . . . | . . .0.7 inherits
10. . . . | . . . . |1.0 .deparseOpts
11. . . . | . . . . | .0.3 pmatch
9. . . . | . . . .0.5 as.list
8. . . . | . . .1.7 unique
9. . . . | . . . .0.7 unique.default
9. . . . | . . . .0.6 unlist
10. . . . | . . . . |0.4 lapply
8. . . . | . . .0.3 unlist
7. . . . | . .4.0 na.omit
8. . . . | . . .3.8 na.omit.data.frame
9. . . . | . . . .3.0 [
10. . . . | . . . . |2.9 [.data.frame
11. . . . | . . . . | .0.6 duplicated
12. . . . | . . . . | . .0.3 duplicated.default
11. . . . | . . . . | .0.3 [
12. . . . | . . . . | . .0.2 [.ts
11. . . . | . . . . | .0.3 [[
On Mon, Aug 4, 2008 at 12:58 PM, rcoder [EMAIL PROTECTED] wrote:
Hi,
I guess my question is really more about the nested for loop construct and
whether it is doing what I intend it to do in my code in the previous post.
I would be grateful if anyone who has used nested loops could let me know if
I am doing something wrong.
Thanks,
rcoder
rcoder wrote:
Hi everyone,
I'm experiencing difficulty getting the results I want when I use a nested
for loop. I have a data set to which I perform some calculations, and then
try to apply a regression over a rolling window. The code runs, but the
regression results I am getting (intercept and slope) are simply the same,
repeated again and again in the results matrix. The regression does not
seem to be obeying the instructions of the nested loop, which intend it to
calculate regression coefficients over a data one row at a time. I have
been struggling with the code for many days now, testing various parts,
and I cannot seem to get the nested loop to work as I want it to. I would
be very grateful for any suggestions. Below is a brief version of my code:
#Code start
library(zoo)
seed.set(1)
Pmat-matrix(rnorm(1000), nrow=100, ncol=100)
maxcol-ncol(Pmat)
maxrow-nrow(Pmat)
startrow-10
period-10
wind-2 #roll window
subdiv-period/wind
rollstart-11 #start roll at period+1
#converting Pmat into ts for rollapply() to work...
PmatTS-ts(Pmat)
Preg-matrix(NA,ncol=maxcol,nrow=2*maxrow)
PmatWt-matrix(NA, nrow=subdiv,ncol=maxcol)
mult_col-matrix(1:5, nrow=5, ncol=1)
#rolling calculations...
for (i in (rollstart):maxrow)
{
#extract the relevant timeframe...
dslice-PmatTS[(i-period):i,]
dslicets-ts(dslice)
#operating on sliced data...
Pmin-rollapply(dslicets, wind, by=wind, min, na.rm=F)
Pmax-rollapply(dslicets, wind, by=wind, max, na.rm=F)
Pmult-Pmin*Pmax#calculating product
tt-time(Pmult)
for (j in 1:5)#1st nested loop
{
PmatWt[j,]-Pmult[j,]*mult_col[j,]
}
Re: [R] Create data frame from header only
on 08/04/2008 07:40 PM Etienne Bellemare Racine wrote: Hi all, Given a string list, paste(A,1:5,sep=) [1] A1 A2 A3 A4 A5 I would like to create an empty data frame using that list as the header, so I can access my data frame column using, df [ list [ i ] ] Anyone ? Thanks, Etienne You don't indicate how you might want the column classes to be defined, but: Cols - paste(A, 1:5, sep=) Cols [1] A1 A2 A3 A4 A5 DF - read.table(textConnection(), col.names = Cols) DF [1] A1 A2 A3 A4 A5 0 rows (or 0-length row.names) str(DF) 'data.frame': 0 obs. of 5 variables: $ A1: logi $ A2: logi $ A3: logi $ A4: logi $ A5: logi Or, as an example of using 'colClasses': DF - read.table(textConnection(), col.names = Cols, colClasses = character) str(DF) 'data.frame': 0 obs. of 5 variables: $ A1: chr $ A2: chr $ A3: chr $ A4: chr $ A5: chr Note from the Value section of ?read.table: Empty input is an error unless col.names is specified, when a 0-row data frame is returned: similarly giving just a header line if header = TRUE results in a 0-row data frame. Note that in either case tthe columns will logical unless colClasses was supplied. In this case, I am supplying an empty value using textConnection() in place of the typical source file name. HTH, Marc Schwartz __ 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] Create data frame from header only
Even if it's not so elegant, I will bind it in a function, so I don't have to
bother with that anymore. I think there should be a simple function in R to
initialize an empty data frame. From what I've read, it is a recurrent
question on that list.
#Create an empty data frame from a header list
empty.df- function(header){
df-data.frame(matrix(matrix(rep(1,length(header)),1),1))
colnames(df)-header
return(df[NULL,])
}
# Usage
new.df - empty.df(paste(A,1:5,sep=))
new.df
[1] A1 A2 A3 A4 A5
0 rows (or 0-length row.names)
Thanks again,
Etienne
milton ruser wrote:
Hi Etienne
It is not so elegant, bu I think that work.
colname.list-paste(A,1:5,sep=)
df-data.frame(matrix(matrix(rep(1,length(colname.list)),1),1))
df
colnames(df)-colname.list
df
df-df[-1,]
df
Cheers,
Miltinho Astronauta
Brazil
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Re: [R] Create data frame from header only
See dataFrame() in R.utils. It was design for the purpose of
allocating empty data frames in an efficient way.
Example:
library(R.utils);
df - dataFrame(colClasses=c(a=integer, b=double), nrow=10);
str(df)
'data.frame': 10 obs. of 2 variables:
$ a: int 0 0 0 0 0 0 0 0 0 0
$ b: num 0 0 0 0 0 0 0 0 0 0
/Henrik
On Mon, Aug 4, 2008 at 6:42 PM, Etienne B. Racine [EMAIL PROTECTED] wrote:
Even if it's not so elegant, I will bind it in a function, so I don't have to
bother with that anymore. I think there should be a simple function in R to
initialize an empty data frame. From what I've read, it is a recurrent
question on that list.
#Create an empty data frame from a header list
empty.df- function(header){
df-data.frame(matrix(matrix(rep(1,length(header)),1),1))
colnames(df)-header
return(df[NULL,])
}
# Usage
new.df - empty.df(paste(A,1:5,sep=))
new.df
[1] A1 A2 A3 A4 A5
0 rows (or 0-length row.names)
Thanks again,
Etienne
milton ruser wrote:
Hi Etienne
It is not so elegant, bu I think that work.
colname.list-paste(A,1:5,sep=)
df-data.frame(matrix(matrix(rep(1,length(colname.list)),1),1))
df
colnames(df)-colname.list
df
df-df[-1,]
df
Cheers,
Miltinho Astronauta
Brazil
--
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http://www.nabble.com/Create-data-frame-from-header-only-tp18822575p18823082.html
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[R] normalize a data-set
Dear All, I have a dataset (X,Y and Z) with no-normalize distribution (I saw with qqnorm) but with Gaussian shape. I am a not good statistic and I wish to know a method to normalize Z values. Thanks for help. It's very useful for my PhD ALe [[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] optimize simultaneously two binomials inequalities using nlm( ) or optim( )
Dear R users,
I´m trying to optimize simultaneously two binomials inequalities (used in
acceptance sampling) which are nonlinear solution, so there is no simple
direct solution. Please, let me explain shortly the the problem and the
question as following.
The objective is to obtain the smallest value of 'n' (sample size)
satisfying both inequalities:
(1-alpha) = pbinom(c, n, p1) pbinom(c, n, p2) = beta
Where p1 (AQL) and p2 (LTPD) are probabilities parameters (Consumer and
Producer), the alpha and beta are conumer and producer risks, and finally,
the 'n' represents the sample size and the 'c' an acceptance number or
maximum number of defects (nonconforming) in sample size.
Considering that the 'n' and 'c' values are integer variables, it is
commonly not possible to derive an OC curve including the both (p1,1-alpha)
and (p2,beta) points. Some adjacency compromise is commonly required,
achieved by searching a more precise OC curve with respect to one of the
points.
I´m using Mathematica 6 but it is a Trial, so I would like use R intead (or
better, I need it)!
To exemplify, In Mathematica I call the function using NMinimize passing
the restriction and parameters:
/* function name findOpt and parameters... */
restriction = (1 - alpha) = CDF[BinomialDistribution[sample_n, p1], c]
betha = CDF[BinomialDistribution[sample_n, p2], c]
0 alpha alphamax 0 betha bethamax
1 sample_n = lot_Size 0 = c lot_size
p1 p2 p2max ;
fcost = sample_n/lot_Size;
result = NMinimize[{fcost, restriction}, {sample_n, c, alpha, betha, p2max},
Method - NelderMead, AccuracyGoal - 10];
/* Calling the function findOpt */
findOpt[p1=0.005, lot_size=1000, alphamax=0.05, bethamax =0.05, p2max =
0.04]
/* and I got the return of values of; minimal n, c, alpha, betha and
the p2 or (LTPD) were computed */ {0.514573, {alpha$74 - 0.0218683,
sample_n$74 - 155.231, betha$74 - 0.05,
c$74 - 2, p2$74 - 0.04}}
Now, using R, I would define the pbinom(c, n, prob) given only the minimum
and maximum values to n and c and limits to p1 and p2 probabilities
(Consumer and Producer), similar on the example above.
I found some examples to optimize equations in R and some tips, but I not be
able to define the sintaxe to use with that functions. Among the functions
that could be used to resolve the problem presented, I found the function
optim() that it is used for unconstrained optimization and the nlm() which
is used for solving nonlinear unconstrained minimization problems.
May I wrong, but the nlm() function would be appropriate to solve this
problem, is it right?
Can I get a pointer to solve this problem using the nlm() function or where
could I get some tips/example to help me, please?
// (1-alpha) = pbinom(c, n, p1) pbinom(c, n, p2) = beta
It was used betha parameter name to avoid the 'beta' function used in
Mathematica...
findS - function(p1='numeric', lot_size='numeric', alphamax='numeric',
bethamax ='numeric', p2max ='numeric')
{
(1 - alpha) = pbinom(c, sample_n, p1) betha = pbinom(c,
sample_n, p2)
0 alpha alphamax 0 betha bethamax
1 sample_n = lot_Size 0 = c lot_size
p1 p2 p2max ;
}
Parameters:
p1=0.005,
lot_size=1000,
alphamax=0.05,
bethamax =0.05,
p2max = 0.04
Minimize results should return/printing the following values:
sample_n, (minimal sample size)
c , (critical level of defectives)
alpha , (producer's risk)
betha , (consumer's risk)
p2max (consumer's probability p2)
Could one help me understand how can desing the optimize nonlinear function
using R for two binomials or point me some tips?
Thanks in advance.
EToktar
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[R] I have an array of plots and I would like to put a title on the whole array.
Hi, all, Does anyone know of a way to get a title on an array of plots? I have tried to use title, mtext, and text to get the title on the array, but to no avail. I would like something like the following. title(Big Array) par(mfrow=c(2,4)) plot(x1,y1) plot(x2,y2) .. .. .. Much appreciated, Art __ 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] Create data frame from header only
On Mon, 4 Aug 2008, Etienne B. Racine wrote:
Even if it's not so elegant, I will bind it in a function, so I don't have to
bother with that anymore. I think there should be a simple function in R to
initialize an empty data frame. From what I've read, it is a recurrent
question on that list.
Not from what I have seen. But it is easy with data.frame
colname.list - paste(A,1:5,sep=) # note this is not a list.
cols - lapply(colname.list, function(x) numeric(0))
names(cols) - colname.list
do.call(data.frame, cols)
#Create an empty data frame from a header list
empty.df- function(header){
df-data.frame(matrix(matrix(rep(1,length(header)),1),1))
colnames(df)-header
return(df[NULL,])
}
# Usage
new.df - empty.df(paste(A,1:5,sep=))
new.df
[1] A1 A2 A3 A4 A5
0 rows (or 0-length row.names)
Thanks again,
Etienne
milton ruser wrote:
Hi Etienne
It is not so elegant, bu I think that work.
colname.list-paste(A,1:5,sep=)
df-data.frame(matrix(matrix(rep(1,length(colname.list)),1),1))
df
colnames(df)-colname.list
df
df-df[-1,]
df
Cheers,
Miltinho Astronauta
Brazil
--
View this message in context:
http://www.nabble.com/Create-data-frame-from-header-only-tp18822575p18823082.html
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and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] I have an array of plots and I would like to put a title on the whole array.
par(mfrow=c(2,4), oma=c(0,0,1,0)) plots ... title(Big Array, outer=TRUE) You need to set up an outer margin. See 'An Introduction to R' for that topic. You probably want to reset the margins (mar=) for such a large array of plots. On Mon, 4 Aug 2008, Arthur Roberts wrote: Hi, all, Does anyone know of a way to get a title on an array of plots? I have tried to use title, mtext, and text to get the title on the array, but to no avail. I would like something like the following. title(Big Array) par(mfrow=c(2,4)) plot(x1,y1) plot(x2,y2) .. .. .. Much appreciated, Art __ 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. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] 95% CI bands on a Lowess smoother
Hi there, I'm plotting some glass RI values just by plotting plot(x) then I put on my lowess smoother lines(lowess(x)) now I want to put on some 95% Confidence Interval bands of the lowess smoother, but don't know how?? Thanks -- Gareth Campbell PhD Candidate The University of Auckland P +649 815 3670 M +6421 256 3511 E [EMAIL PROTECTED] [EMAIL PROTECTED] [[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] Constrained Optimization
alan.ng at gmx.net writes: Hello, I am trying to run a constrained optimization in R. constrOptim is really useful and has helped me a lot, but unfortunately, it doesn't provide the hessian. Is there a solution to this problem? You didn't provide an example to understand why 'optim' doesn't solve your problem. Perhaps you can find an optimal point with 'constrOptim' and then compute the hessian at that point with some other R function, for instance 'numDeriv::hessian'. // Hans Werner Borchers I've tried optim with penalty-functions and L-BFGS-B, but it doesn't help. Alan. __ 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] 95% CI bands on a Lowess smoother
On Tue, 5 Aug 2008, Gareth Campbell wrote: Hi there, I'm plotting some glass RI values just by plotting plot(x) then I put on my lowess smoother lines(lowess(x)) now I want to put on some 95% Confidence Interval bands of the lowess smoother, but don't know how?? You can have pointwise confidence intervals or a simultaneous confidence band, so which did you mean? I don't know of any theory for the latter, whereas for the former lowess is a locally linear fit and hence there will be asymptotic results not very relevant for local fits. Looks like a case for simulation-based inference, e.g. via a bootstrap. Thanks -- Gareth Campbell PhD Candidate The University of Auckland P +649 815 3670 M +6421 256 3511 E [EMAIL PROTECTED] [EMAIL PROTECTED] [[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. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] Including exogenous X-variables in ARFIMA models
Does anybody know if there is a way to include exogenous X-variables in an ARFIMA model? It appears the ARFIMA function in fracdiff does not support this. Supposing there is nothing already written, and I wanted to modify the existing arfima function, how would I go about this? Would I need to modify the fracdiff() binaries and compile them on my machine? I'm running WindowsI think I'm screwed = ( -- Scotty _ Get more from your digital life. Find out how. [[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.
