Re: [R] bar chart with means - using ggplot
I would recommend not representing means by bars at all. Bars are for counts, stacking from zero. Means are point estimates. ggplot has a lot of routines for displaying means with errors bars: geom_linerange, geom_pointrange. To hone your ggplot skills I recommend looking into these geoms. On Sep 11, 2009, at 4:32 PM, Felipe Carrillo wrote: Like this? # example using qplot library(ggplot2) meanprice - tapply(diamonds$price, diamonds$cut, mean);meanprice cut - factor(levels(diamonds$cut), levels = levels(diamonds$cut)) qplot(cut, meanprice, geom=bar, stat=identity, fill = I(grey50)) dev.new() # create a new graph to compare with qplot # Example using ggplot ggdata - data.frame(meanprice,cut);ggdata ggplot(ggdata,aes(y=meanprice,x=cut)) + geom_bar(fill=grey50,stat='identity') Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA --- On Fri, 9/11/09, Andreas Christoffersen achristoffer...@gmail.com wrote: From: Andreas Christoffersen achristoffer...@gmail.com Subject: bar chart with means - using ggplot To: ggplot2 ggpl...@googlegroups.com Date: Friday, September 11, 2009, 1:50 PM In the help pages hadley provides the following example showing how to achieve bar charts with y being means of a variable instead of counts. The example uses qplot however - and not ggplot. Since i would like to understand ggplot better I would really like to see how this could be done in ggplot. # example using qplot library(ggplot2) meanprice - tapply(diamonds$price, diamonds$cut, mean) cut - factor(levels(diamonds$cut), levels = levels(diamonds$cut)) qplot(cut, meanprice, geom=bar, stat=identity, fill = I(grey50)) --~--~-~--~~~---~--~~ You received this message because you are subscribed to the ggplot2 mailing list. To post to this group, send email to ggpl...@googlegroups.com To unsubscribe from this group, send email to ggplot2+unsubscr...@googlegroups.com For more options, visit this group at http://groups.google.com/group/ggplot2 -~--~~~~--~~--~--~--- --- Dianne Cook dic...@iastate.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] mclustBIC version 3.3.1
Hi there, I started getting a new error with the latest mclust package version 3.3.1. My only solution was to install the older package 3.2.1 or even 3.1-10. , (i think older ones will still work since i ve been using it for a while) the sentence giving trouble is: BIC - mclustBIC(as.vector(data), G = 1:3, modelNames=c(E)) Quite often i get missing groups suggesting the data is not appropriate for that clustering into 3 groups. I could not cast the results with NULL values in order to continue my process. Does anybody know anything about the new version of this function and how its effectiveness got reduced so drastically? Its functionalities seem to behave expanded handling noise etc etc as per changelog. Any ideas or workaround? Thanks a lot for the contribution in advance. Max -- View this message in context: http://www.nabble.com/mclustBIC-version-3.3.1-tp25411550p25411550.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] State Space models in R
Hi, Rob Hyndman's forecast package does exponential smoothing forecasting based on state space models (and lots of other stuff, ARIMA et al.). It's not exactly the companion package to his book, but it comes close. The book's (Forecasting with Exponential Smoothing - The State Space Approach) webpage is here: http://www.exponentialsmoothing.net/ HTH, Stephan Giovanni Petris schrieb: Hello everybody, I am writing a review paper about State Space models in R, and I would like to cover as many packages as I reasonably can. So far I am familiar with the following tools to deal with SS models: * StructTS, Kalman* (in stats) * packages dse[1-2] * package sspir * package dlm I would like to have some input from users who work with SS models: are there any other packages for SS models that I am missing?, which package do you use and why?, what do you think are advantages/ disadvantages of the package you use? Of course I do have my own preferences (biased, of course) and opinions about the different packages, but I would also like to summarize in my paper the feedback I get from the R community. Thank you in advance. Best, Giovanni Petris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Completion for proto objects
On Fri, 11 Sep 2009 18:56:36 +0200, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
in the devel version. If that does not help let me know offline
and I will try to help you.
Thanks Gabor,
I solved the problem.
Here is the code in case somebody else wants to have full completions for
proto objects:
# slightly modified utils:::specialCompletions
specialCompletions1 - function (text, spl)
{
wm - which.max(spl)
op - names(spl)[wm]
opStart - spl[wm]
opEnd - opStart + nchar(op)
if (opStart 1)
return(character(0L))
prefix - substr(text, 1L, opStart - 1L)
suffix - substr(text, opEnd, 100L)
if (op == ?)
return(helpCompletions(prefix, suffix))
if (opStart = 1)
return(character(0L))
tryToEval - function(s) {
try(eval(parse(text = s), envir = .GlobalEnv), silent = TRUE)
}
comps - switch(op, `$` = {
if (.CompletionEnv$settings[[ops]]) {
object - tryToEval(prefix)
if (inherits(object, try-error))
suffix
else {
if (!inherits(object, 'proto') is.environment(object))
{ ## this line is modified!!
ls(object, all.names = TRUE, pattern = sprintf(^%s,
makeRegexpSafe(suffix)))
}
else {
grep(sprintf(^%s, makeRegexpSafe(suffix)),
names(object), value = TRUE)
}
}
}
else suffix
}, `...@` = {
if (.CompletionEnv$settings[[ops]]) {
object - tryToEval(prefix)
if (inherits(object, try-error))
suffix
else {
grep(sprintf(^%s, makeRegexpSafe(suffix)),
methods::slotNames(object), value = TRUE)
}
}
else suffix
}, `::` = {
if (.CompletionEnv$settings[[ns]]) {
nse - try(getNamespaceExports(prefix), silent = TRUE)
if (inherits(nse, try-error))
suffix
else {
grep(sprintf(^%s, makeRegexpSafe(suffix)),
nse, value = TRUE)
}
}
else suffix
}, `:::` = {
if (.CompletionEnv$settings[[ns]]) {
ns - try(getNamespace(prefix), silent = TRUE)
if (inherits(ns, try-error))
suffix
else {
ls(ns, all.names = TRUE, pattern = sprintf(^%s,
makeRegexpSafe(suffix)))
}
}
else suffix
}, `[` = , `[[` = {
comps - normalCompletions(suffix)
if (length(comps))
comps
else suffix
})
if (length(comps) == 0L)
comps -
sprintf(%s%s%s, prefix, op, comps)
}
environment(specialCompletions1) - asNamespace(utils)
assignInNamespace(specialCompletions, specialCompletions1, utils)
## names for proto objects
names.proto - function(x){
.local - function(x){
if(inherits(x[[.super]], proto)) c(ls(x),
Recall(x[[.super]]))
else ls(x)
}
unlist(.local(x))
}
## checks
tl - list(abc=afdas)
p1 - proto(p1_abc=123)
p2 - proto(.=p1, p2_abc=23423)
names(p2)
#[1] p2_abc p1_abc
names(p1)
#[1] p1_abc
Completion works fine.
Vitalie.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] How to visualize paired t-test results?
I would like to know if you have any suggestions how to visualize the results from a paired t-test (see the example data below). I tried to produce plots that show the mean and CI's from the original data and the estimate of the difference between means and the confidence intervals (see below) from the t-test. I really don't know what would be the best way to graphically display the results. Thanks in advance. data ## sam1 - c(51.7, 54.2, 53.3, 57, 56.4, 61.5, 57.2, 56.2, 58.4, 55.8) sam2 - c(62.5, 65.2, 67.6, 69.9, 69.4, 70.1, 67.8, 67, 68.5, 62.4) DF - stack(data.frame(sam1 = sam1, sam2 = sam2)) paired-t-test ## res - t.test(values ~ ind, data=DF, paired = TRUE) plots # library(gregmisc) opar - par(mfrow=c(1,2)) tapply(DF$values, list(DF$ind), ci) plotmeans(values ~ ind, data=DF) d - sam1 - sam2 stripchart(d, vertical=T, pch=16) points(1, res$estimate, col=red, pch=16) arrows(1, res$estimate, 1, res$conf.int[1], col=red, lwd=2, angle=90, length=0.1) arrows(1, res$estimate, 1, res$conf.int[2], col=red, lwd=2, angle=90, length=0.1) [[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] could not find function Varcov after upgrade of R?
After upgrading R to 2.9.2, I can't use the anova() fuction. It says could not find function Varcov . What's wrong with my computer? Help needed, thanks! Yao Zhu Department of Urology Fudan University Shanghai Cancer Center No. 270 Dongan Road, Shanghai, China [[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] consistent results with heatmap.2
Hi, I am trying to create a heatmap with some specific requirements. Specifically, I need to be able to center the color-scale around 0, and I need to truncate the data so that a few extreme values do not cause the rest of the heatmap to appear black (on a red/green scale). After reading through and experimenting with heatmap, heatmap.2, heatmap_plus, and heatmap_2, I believe heatmap.2 will provide the best solution - it appears to be the only option that will center around 0 without me manually adjusting the zlim. In addition, the color key and distributional information is useful (at least for the non-truncated data). Although I have not used truncated data at this point, I have so far been unable to produce consistent dendrograms when I'm not producing the dendrogram as part of the heatmap. What am I missing? ### R version 2.9.0 (2009-04-17) Copyright (C) 2009 The R Foundation for Statistical Computing ISBN 3-900051-07-0 # The correct results hv - heatmap.2(x, col=col.palette, Colv=FALSE, key=TRUE, Rowv=cc1$order, dendrogram=row, scale=none, tracecol=white, cexCol=.5) hv$rowInd [1] 26 23 5 19 15 4 2 16 21 13 22 12 10 20 7 3 25 1 8 11 24 9 18 14 17 [26] 6 # Attempt 1 cc = as.dendrogram(hclust(dist(x))) hv1 - heatmap.2(x, col=col.palette, Rowv=cc, Colv=FALSE, key=TRUE, dendrogram=row, scale=none, tracecol=white, cexCol=.5) hv1$rowInd [1] 5 23 26 1 25 8 11 9 24 18 14 6 17 2 4 16 21 13 22 12 10 20 3 7 15 [26] 19 # Attempt 2 cutting - cutree(hclust(dist(x)),k=8) # I counted about 8 divisions in correct clustering hv2 - heatmap.2(x, col=col.palette, Rowv=cutting, Colv=FALSE, key=TRUE, dendrogram=row, scale=none, tracecol=white, cexCol=.5) hv2$rowInd [1] 5 23 26 1 25 8 11 9 24 18 14 6 17 15 19 2 4 16 21 13 22 3 7 12 10 [26] 20 # Attempt 3 correct.order - hv$rowInd hv3 - heatmap.2(x, col=col.palette, Rowv=correct.order, Colv=FALSE, key=TRUE, dendrogram=row, scale=none, tracecol=white, cexCol=.5) hv3$rowInd [1] 5 26 23 15 19 22 13 7 3 12 20 10 16 21 4 2 8 11 25 1 24 9 18 14 6 [26] 17 My understanding is that my first attempt is the same call heatmap.2 makes - so there is no reason for the different results. Note - if there is another way to create the heatmaps meeting my requirements using other packages, that works too - I just need something to produce consistent results. Thanks! Melissa Key Graduate Student Department of Statistics Purdue 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.
Re: [R] could not find function Varcov after upgrade of R?
I uses the Design library.
take this example:
library(Design)
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('Male','Female'), n,
rep=TRUE, prob=c(.6, .4)))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='Female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
dd - datadist(age, sex)
options(datadist='dd')
Srv - Surv(dt,e)
f - cph(Srv ~ rcs(age,4) + sex, x=TRUE, y=TRUE)
cox.zph(f, rank) # tests of PH
anova(f)
# Error in anova.Design(f) : could not find function Varcov
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
2009/9/12 Ronggui Huang ronggui.hu...@gmail.com
I cannot reproduce the problem you mentioned.
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
anova(lm.D9 - lm(weight ~ group))
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=Chinese (Simplified)_People's Republic of
China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of
China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of
China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic
of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
2009/9/12 zhu yao mailzhu...@gmail.com:
After upgrading R to 2.9.2, I can't use the anova() fuction.
It says could not find function Varcov .
What's wrong with my computer? Help needed, thanks!
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
[[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.
--
HUANG Ronggui, Wincent
Doctoral Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
[[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] could not find function Varcov after upgrade of R?
On Sep 12, 2009, at 5:24 AM, zhu yao wrote: After upgrading R to 2.9.2, I can't use the anova() fuction. It says could not find function Varcov . What's wrong with my computer? Help needed, thanks! You have not given us very much information but from the naming conventions I would guess that you need to reinstall the Design/Hmisc combination of packages after having updated your R from an an earlier version of R that was different in the second digit. The libraries for the 2.9 series of R versions is stored in a different directory than the 2.8 or 2.7 versions. There is a simple command to the update.packages function that will reinstall updated versions of the packages you had previously installed. My practice, unnecessarily incremental, has been to reinstall each of the packages from scratch. The automatic method would be: update.packages() # which will ask you to confirm eachnpackage # or update.packages(ask=FALSE) # which will not ask -- David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Save data in txt
Hi everyone I have a little problem with R. I built a lineal regression equation using stepAIC function in both directions. Once I get this formula (lineal regression), I would like to save in a txt file data refered to p-value, r-squared, coefficients,...from the choosen model previously using stepAIC and p-value,r-squared, coefficients,...from each predictor. I try to use function Write to create a txt and save that data but I can't. So, Does anyone know a function to save specific data in a txt? I would like to save that specific data in separated columns, I mean, first column first predictor, second column p-value, third column r-squared, etc... Refered to data from each predictor (p-value and r-squared), is there an order in R to obtain this data or I have to calculte them by myself? I know I can use r.squared but it gives me a value from the choosen model not from an specific predictor. Cheers, Lucas _ [[elided Hotmail spam]] [[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] could not find function Varcov after upgrade of R?
I've had the same problem with predict.Design, and have sent an email to the
maintainer of the Design package at Vanderbilt University. I wasn't even
able to run the examples given on the help page of predict.Design - I
received the same error about Varcov that you did.
I *think* it's a problem with the package, rather than R 2.9.2, and I hope
the problem will soon be fixed. I was able to use predict.Design with 2.9.2
until I updated the Design package a few days ago.
david freedman
zhu yao wrote:
I uses the Design library.
take this example:
library(Design)
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('Male','Female'), n,
rep=TRUE, prob=c(.6, .4)))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='Female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
dd - datadist(age, sex)
options(datadist='dd')
Srv - Surv(dt,e)
f - cph(Srv ~ rcs(age,4) + sex, x=TRUE, y=TRUE)
cox.zph(f, rank) # tests of PH
anova(f)
# Error in anova.Design(f) : could not find function Varcov
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
2009/9/12 Ronggui Huang ronggui.hu...@gmail.com
I cannot reproduce the problem you mentioned.
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
anova(lm.D9 - lm(weight ~ group))
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=Chinese (Simplified)_People's Republic of
China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of
China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of
China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic
of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
2009/9/12 zhu yao mailzhu...@gmail.com:
After upgrading R to 2.9.2, I can't use the anova() fuction.
It says could not find function Varcov .
What's wrong with my computer? Help needed, thanks!
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
[[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.
--
HUANG Ronggui, Wincent
Doctoral Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
[[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.
--
View this message in context:
http://www.nabble.com/could-not-find-function-%22Varcov%22-after-upgrade-of-R--tp25412881p25414017.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] could not find function Varcov after upgrade of R?
Did you type library(Hmisc,T) before loading Design?
Carlos
--
From: David Freedman 3.14da...@gmail.com
Sent: Saturday, September 12, 2009 8:26 AM
To: r-help@r-project.org
Subject: Re: [R] could not find function Varcov after upgrade of R?
I've had the same problem with predict.Design, and have sent an email to
the
maintainer of the Design package at Vanderbilt University. I wasn't even
able to run the examples given on the help page of predict.Design - I
received the same error about Varcov that you did.
I *think* it's a problem with the package, rather than R 2.9.2, and I hope
the problem will soon be fixed. I was able to use predict.Design with
2.9.2
until I updated the Design package a few days ago.
david freedman
zhu yao wrote:
I uses the Design library.
take this example:
library(Design)
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('Male','Female'), n,
rep=TRUE, prob=c(.6, .4)))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='Female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
dd - datadist(age, sex)
options(datadist='dd')
Srv - Surv(dt,e)
f - cph(Srv ~ rcs(age,4) + sex, x=TRUE, y=TRUE)
cox.zph(f, rank) # tests of PH
anova(f)
# Error in anova.Design(f) : could not find function Varcov
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
2009/9/12 Ronggui Huang ronggui.hu...@gmail.com
I cannot reproduce the problem you mentioned.
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
anova(lm.D9 - lm(weight ~ group))
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=Chinese (Simplified)_People's Republic of
China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of
China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of
China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic
of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
2009/9/12 zhu yao mailzhu...@gmail.com:
After upgrading R to 2.9.2, I can't use the anova() fuction.
It says could not find function Varcov .
What's wrong with my computer? Help needed, thanks!
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
[[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.
--
HUANG Ronggui, Wincent
Doctoral Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
[[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.
--
View this message in context:
http://www.nabble.com/could-not-find-function-%22Varcov%22-after-upgrade-of-R--tp25412881p25414017.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.
__
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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] could not find function Varcov after upgrade of R?
I am unable to reproduce that problem, either. Have both of you
updated both the Design and Hmisc packages? Varcov is in Hmisc.
--
David Winsemius
On Sep 12, 2009, at 8:26 AM, David Freedman wrote:
I've had the same problem with predict.Design, and have sent an
email to the
maintainer of the Design package at Vanderbilt University. I wasn't
even
able to run the examples given on the help page of predict.Design - I
received the same error about Varcov that you did.
I *think* it's a problem with the package, rather than R 2.9.2, and
I hope
the problem will soon be fixed. I was able to use predict.Design
with 2.9.2
until I updated the Design package a few days ago.
david freedman
zhu yao wrote:
I uses the Design library.
take this example:
library(Design)
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('Male','Female'), n,
rep=TRUE, prob=c(.6, .4)))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='Female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
dd - datadist(age, sex)
options(datadist='dd')
Srv - Surv(dt,e)
f - cph(Srv ~ rcs(age,4) + sex, x=TRUE, y=TRUE)
cox.zph(f, rank) # tests of PH
anova(f)
# Error in anova.Design(f) : could not find function Varcov
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
2009/9/12 Ronggui Huang ronggui.hu...@gmail.com
I cannot reproduce the problem you mentioned.
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
anova(lm.D9 - lm(weight ~ group))
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=Chinese (Simplified)_People's Republic of
China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of
China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of
China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's
Republic
of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
2009/9/12 zhu yao mailzhu...@gmail.com:
After upgrading R to 2.9.2, I can't use the anova() fuction.
It says could not find function Varcov .
What's wrong with my computer? Help needed, thanks!
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
--
HUANG Ronggui, Wincent
Doctoral Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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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] could not find function Varcov after upgrade of R?
I apologize for this change in the Hmisc package which breaks the
current version of Design in CRAN. Design should be updated by early
next week. In the meantime there are two solutions:
1. Install the rms package which is a replacement for Design with better
graphics, and note the few changes you would need to make in your code
as described in http://biostat.mc.vanderbilt.edu/Rrms (the main change
is for plot(fit object)), or
2. Run source('http://biostat.mc.vanderbilt.edu/tmp/Varcov.r') to define
the Varcov functions until Design is updated
Frank
David Freedman wrote:
I've had the same problem with predict.Design, and have sent an email to the
maintainer of the Design package at Vanderbilt University. I wasn't even
able to run the examples given on the help page of predict.Design - I
received the same error about Varcov that you did.
I *think* it's a problem with the package, rather than R 2.9.2, and I hope
the problem will soon be fixed. I was able to use predict.Design with 2.9.2
until I updated the Design package a few days ago.
david freedman
zhu yao wrote:
I uses the Design library.
take this example:
library(Design)
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('Male','Female'), n,
rep=TRUE, prob=c(.6, .4)))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='Female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
dd - datadist(age, sex)
options(datadist='dd')
Srv - Surv(dt,e)
f - cph(Srv ~ rcs(age,4) + sex, x=TRUE, y=TRUE)
cox.zph(f, rank) # tests of PH
anova(f)
# Error in anova.Design(f) : could not find function Varcov
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
2009/9/12 Ronggui Huang ronggui.hu...@gmail.com
I cannot reproduce the problem you mentioned.
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
anova(lm.D9 - lm(weight ~ group))
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=Chinese (Simplified)_People's Republic of
China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of
China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of
China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic
of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
2009/9/12 zhu yao mailzhu...@gmail.com:
After upgrading R to 2.9.2, I can't use the anova() fuction.
It says could not find function Varcov .
What's wrong with my computer? Help needed, thanks!
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
HUANG Ronggui, Wincent
Doctoral Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
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Re: [R] Save data in txt
On Sep 12, 2009, at 7:56 AM, Lucas Sevilla García wrote: Hi everyone I have a little problem with R. I built a lineal regression equation using stepAIC function in both directions. Once I get this formula (lineal regression), I would like to save in a txt file data refered to p-value, r-squared, coefficients,...from the choosen model previously using stepAIC and p-value,r-squared, coefficients,...from each predictor. I try to use function Write to create a txt and save that data but I can't. As I expected I get an error trying ?Write. Are you aware that complete case dependence for function names is enforced by R? Or are you using a package that has such a function but failed to tell us about that fact? So, Does anyone know a function to save specific data in a txt? ?cat ?capture.output I would like to save that specific data in separated columns, I mean, first column first predictor, second column p-value, third column r-squared, etc... Refered to data from each predictor (p-value and r-squared), is there an order in R to obtain this data or I have to calculte them by myself? A specific example would help here. You almost certainly do not need to calculate such information. They are probably available in either the model object itself or in the results of summary or aov. I know I can use r.squared but it gives me a value from the choosen model not from an specific predictor. If you can be more specific about what you are attempting to derive, it would be helpful. I suspect some sort of partial correlation coefficient but you are extremely vague in what you need. Cheers, Lucas _ [[elided Hotmail spam]] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] could not find function Varcov after upgrade of R?
Hi Zhu,
could not find function Varcov after upgrade of R?
Frank Harrell (author of Design) has noted in another thread that Hmisc has
changed... The problem is that functions like anova.Design call a function
in the _old_ Hmisc package called Varcov.default. In the new version of
Hmisc this is called something else (vcov.default).
The best way to fix this is to install the new (i.e. current) version of
Hmisc and Frank's replacement for his Design package, which is called rms
(Regression Modeling Strategies).
Regards, Mark.
zhu yao wrote:
I uses the Design library.
take this example:
library(Design)
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('Male','Female'), n,
rep=TRUE, prob=c(.6, .4)))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='Female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
dd - datadist(age, sex)
options(datadist='dd')
Srv - Surv(dt,e)
f - cph(Srv ~ rcs(age,4) + sex, x=TRUE, y=TRUE)
cox.zph(f, rank) # tests of PH
anova(f)
# Error in anova.Design(f) : could not find function Varcov
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
2009/9/12 Ronggui Huang ronggui.hu...@gmail.com
I cannot reproduce the problem you mentioned.
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
anova(lm.D9 - lm(weight ~ group))
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=Chinese (Simplified)_People's Republic of
China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of
China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of
China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic
of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
2009/9/12 zhu yao mailzhu...@gmail.com:
After upgrading R to 2.9.2, I can't use the anova() fuction.
It says could not find function Varcov .
What's wrong with my computer? Help needed, thanks!
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
[[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.
--
HUANG Ronggui, Wincent
Doctoral Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
[[alternative HTML version deleted]]
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View this message in context:
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Re: [R] Working with large matrix
On Fri, Sep 11, 2009 at 12:15 PM, A Ezhil ezhi...@yahoo.com wrote: Dear All, I have large matrix (46000 x 11250). I would like to do the linear regression for each row. I wrote a simple function that has lm() and used apply(mat,1,func). The issue is that it takes ages to load the file and also to finish the lm. I am using LINUX 64 bit with 32G mem. Is there an elegant and fast way of completing this task? I'm not quite sure what you mean by do the linear regression for each row but you may find it convenient to use a matrix as the response in the call to lm. In that case lm fits the same model, described by the right hand side of the formula, to each column of the matrix on the left hand side. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SPSS Statistics-R Integration Plug-In
Hi Michael, I've used the R plug-in and really like it. You can read my instructions on how to install and use it by going to Amazon.com, searching for the book, R for SAS and SPSS Users and then search inside the book for the section, Running R from SPSS. I've only got about 3 pages on it (28 bottom through 31), and Amazon will let you read them all. You do need to choose a specific version of R, but the old versions are kept on CRAN, so they're easy to get. Version 18 of SPSS actually ships with the version of R it needs on the SPSS DVD. SPSS makes it really easy to transfer data and results back and forth, and it formats the results nicely as SPSS pivot tables. I also have an example of how to make an R program appear on the SPSS menus in R you taking advantage of R?, an SPSS Directions 2009 talk. It's at: http://RforSASandSPSSusers.com, right-hand side. Cheers, Bob -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Michael Chajewski Sent: Tuesday, September 08, 2009 9:43 AM To: r-help@r-project.org Subject: [R] SPSS Statistics-R Integration Plug-In Dear All, Has anyone tried to use this plug-in? Since I am running R-2.9.1 it will not even let me install it. Further, since I am running Windows I cannot use the R provided R-2.7.0 Linux installation file from the archive (tried to install it through cygwin and it was a mess). Suggestions? Ideas? Has anybody used this plug-in? Michael -- Michael Chajewski, M.A. Department of Psychology Fordham University Dealy Hall Room 239 441 East Fordham Road Bronx, NY 10458 (718) 817-0654 http://www.chajewski.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Working with large matrix
Another possible interpretation is that each row contains several vectors, possibly laid end to end. For example, if the row J were length 30 (which it isn't), the first 10 might be Y, the next 10 X1, and the next N2. The function might be lm(J[1:10] ~ J[11:20] + J[21:30]). If this is what is going on, then one possible approach is to transform this matrix into the long form, with one observation (value of Y) and two predictors per row instead of (here) 10 values of Y, then use lmList() in the lme4 package. I suspect that this is faster than using apply() with lm(), but I'm not sure. I don't think this will help load the file. Jon On 09/12/09 08:32, Douglas Bates wrote: On Fri, Sep 11, 2009 at 12:15 PM, A Ezhil ezhi...@yahoo.com wrote: Dear All, I have large matrix (46000 x 11250). I would like to do the linear regression for each row. I wrote a simple function that has lm() and used apply(mat,1,func). The issue is that it takes ages to load the file and also to finish the lm. I am using LINUX 64 bit with 32G mem. Is there an elegant and fast way of completing this task? I'm not quite sure what you mean by do the linear regression for each row but you may find it convenient to use a matrix as the response in the call to lm. In that case lm fits the same model, described by the right hand side of the formula, to each column of the matrix on the left hand side. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jonathan Baron, Professor of Psychology, University of Pennsylvania Home page: http://www.sas.upenn.edu/~baron Editor: Judgment and Decision Making (http://journal.sjdm.org) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] mclustBIC version 3.3.1
madmax1425 wrote: Hi there, I started getting a new error with the latest mclust package version 3.3.1. My only solution was to install the older package 3.2.1 or even 3.1-10. , (i think older ones will still work since i ve been using it for a while) the sentence giving trouble is: BIC - mclustBIC(as.vector(data), G = 1:3, modelNames=c(E)) Quite often i get missing groups suggesting the data is not appropriate for that clustering into 3 groups. I could not cast the results with NULL values in order to continue my process. Does anybody know anything about the new version of this function and how its effectiveness got reduced so drastically? Its functionalities seem to behave expanded handling noise etc etc as per changelog. Any ideas or workaround? Thanks a lot for the contribution in advance. Max If you think there is a bug please mail to the package maintainer and report your findings. Uwe Ligges __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to find maximum for multivariable data
Dear all,
I have a data set as follows:
ID cycle.number cycle.result
1 2525 1 38
2 2525 2 38
3 2525 3 25
4 2525 4 25
5 2525 5 25
6 2525 6 25
7 2531 1 38
8 2531 2 38
9 2078 1 38
102078 2 38
I want to find out the maximum cycle.number for each ID, and later find the
corresponding cycle.result for that cycle.
I have already managed to pull out the maximum cycle by using a for loop:
max.cycle - vector()
patients - (levels(factor(ID)))
for (i in 1:length(patients)) {
max.cycle[i] - max(cycle.number[(ID %in% patients[i] )])
}
But i would like to know if there is a better or more elegant way of pulling
out the maximum cycle.number for each ID? Perhaps without the need for using
a for loop?
Many thanks,
Caroline
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2::qplot() -- arbitary transformations of coordinate system?
Is there a place to find a list of the legal values for the coord_trans parameters. I spent a bunch of time searching the ggplot2 docs and r-help for same without success. I also made an attempt at looking at the code in R which also failed. In the book, or with apropos(^Trans, ignore = F) Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to find maximum for multivariable data
Try this:
If cycle.number is ordered:
do.call(rbind, lapply(split(x, x$ID), tail, 1))
On Sat, Sep 12, 2009 at 12:36 PM, caroline choong carolinevcho...@gmail.com
wrote:
Dear all,
I have a data set as follows:
ID cycle.number cycle.result
1 2525 1 38
2 2525 2 38
3 2525 3 25
4 2525 4 25
5 2525 5 25
6 2525 6 25
7 2531 1 38
8 2531 2 38
9 2078 1 38
102078 2 38
I want to find out the maximum cycle.number for each ID, and later find the
corresponding cycle.result for that cycle.
I have already managed to pull out the maximum cycle by using a for loop:
max.cycle - vector()
patients - (levels(factor(ID)))
for (i in 1:length(patients)) {
max.cycle[i] - max(cycle.number[(ID %in% patients[i] )])
}
But i would like to know if there is a better or more elegant way of
pulling
out the maximum cycle.number for each ID? Perhaps without the need for
using
a for loop?
Many thanks,
Caroline
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to find maximum for multivariable data
On Sep 12, 2009, at 11:36 AM, caroline choong wrote:
Dear all,
I have a data set as follows:
ID cycle.number cycle.result
1 2525 1 38
2 2525 2 38
3 2525 3 25
4 2525 4 25
5 2525 5 25
6 2525 6 25
7 2531 1 38
8 2531 2 38
9 2078 1 38
102078 2 38
I want to find out the maximum cycle.number for each ID, and later
find the
corresponding cycle.result for that cycle.
I have already managed to pull out the maximum cycle by using a for
loop:
max.cycle - vector()
patients - (levels(factor(ID)))
for (i in 1:length(patients)) {
max.cycle[i] - max(cycle.number[(ID %in% patients[i] )])
}
But i would like to know if there is a better or more elegant way of
pulling
out the maximum cycle.number for each ID?
There is:
?tapply
patients - read.table(textConnection(ID
cycle.number cycle.result
+ 1 2525 1 38
+ 2 2525 2 38
+ 3 2525 3 25
+ 4 2525 4 25
+ 5 2525 5 25
+ 6 2525 6 25
+ 7 2531 1 38
+ 8 2531 2 38
+ 9 2078 1 38
+ 102078 2 38), header=TRUE)
tapply(patients$cycle.number, patients$ID, max)
2078 2525 2531
262
Perhaps without the need for using
a for loop?
Many thanks,
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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[R] [R-pkgs] New package sos for searching help pages of contributed packages
Searching help pages of contributed packages just got easier with the release of the new sos package. This is a replacement for and substantial enhancement of the existing RSiteSearch package. To learn more about it, try vignette(sos). We hope you find this as useful as we have. Spencer Graves, Sundar Dorai-Raj, Romain Francois ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Survreg function + deviation
I am trying to estimate by MLE weibull coefficients and deviation. But how can i estimate deviation? I tried to find at the list and at the internet this but couldn't make it. Anyone could help me on this? 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] median() function on zoo objects
Hello R-help community: I have what I think is a simple question that I hope someone can answer. When using the median() function on any zoo object (in particular, mine is an irregular time series), I get the following error, which is thrown from .gt(): Error in if (xi == xj) 0L else if (xi xj) 1L else -1L : argument is of length zero median() of course works fine on the coredata() of my zoo object, but not on the object itself. Is the median.default function coded incorrectly for zoo objects? If not, then I don't understand why functions like mean() and sd() work correctly on the same zoo object, but not median(). Am I asking the right questions? Thanks for any help you can offer. Regards, Marc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] could not find function Varcov after upgrade of R?
Thanks for the reminder - actually I think I've become sloppy about the 'T'
(and the order of loading the Hmisc and Design packages), but that doesn't
seem to be the problem. Also, all of my packages have been updated
david
Carlos Alzola wrote:
Did you type library(Hmisc,T) before loading Design?
Carlos
--
From: David Freedman 3.14da...@gmail.com
Sent: Saturday, September 12, 2009 8:26 AM
To: r-help@r-project.org
Subject: Re: [R] could not find function Varcov after upgrade of R?
I've had the same problem with predict.Design, and have sent an email to
the
maintainer of the Design package at Vanderbilt University. I wasn't even
able to run the examples given on the help page of predict.Design - I
received the same error about Varcov that you did.
I *think* it's a problem with the package, rather than R 2.9.2, and I
hope
the problem will soon be fixed. I was able to use predict.Design with
2.9.2
until I updated the Design package a few days ago.
david freedman
zhu yao wrote:
I uses the Design library.
take this example:
library(Design)
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('Male','Female'), n,
rep=TRUE, prob=c(.6, .4)))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='Female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
dd - datadist(age, sex)
options(datadist='dd')
Srv - Surv(dt,e)
f - cph(Srv ~ rcs(age,4) + sex, x=TRUE, y=TRUE)
cox.zph(f, rank) # tests of PH
anova(f)
# Error in anova.Design(f) : could not find function Varcov
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
2009/9/12 Ronggui Huang ronggui.hu...@gmail.com
I cannot reproduce the problem you mentioned.
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
anova(lm.D9 - lm(weight ~ group))
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=Chinese (Simplified)_People's Republic of
China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of
China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of
China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic
of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
2009/9/12 zhu yao mailzhu...@gmail.com:
After upgrading R to 2.9.2, I can't use the anova() fuction.
It says could not find function Varcov .
What's wrong with my computer? Help needed, thanks!
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
[[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.
--
HUANG Ronggui, Wincent
Doctoral Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
[[alternative HTML version deleted]]
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to determine if a variable is already set?
Thanks for your replies. I use the following script: if(!exists(i)) stop (set the variable i, call. = FALSE) but before the stop expression, Error gets displayed: Error: set the variable i Is there another function that stops the execution, prints an expression without printing Error or any other expression except the expression parameter? Best, --- On Fri, 9/11/09, Marc Schwartz marc_schwa...@me.com wrote: From: Marc Schwartz marc_schwa...@me.com Subject: Re: [R] how to determine if a variable is already set? To: carol white wht_...@yahoo.com Cc: r-h...@stat.math.ethz.ch Date: Friday, September 11, 2009, 10:21 AM On Sep 11, 2009, at 12:15 PM, carol white wrote: Hi, It might be a primitive question but how it is possible to determine if a variable is initialized in an environment? Suppose that we start a R session and wants to run a script which use the variable i. Which function could evaluate if i is already initialized or not and if not, then ask interactively the user to set it? This is to avoid the error message: object i is not found. Regards, Carol See ?exists Note that this will tell you if the object exists, not if it contains a specifically desired initial value. You would have to check for the latter after determining that the object does indeed exist. HTH, Marc Schwartz [[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] median() function on zoo objects
Can you provide a reproducible example. See last line to every message on r-help. When I try it it works: median(zoo(1:3)) 2 2 On Sat, Sep 12, 2009 at 1:14 PM, Marc Chiarini marc.chiar...@tufts.edu wrote: Hello R-help community: I have what I think is a simple question that I hope someone can answer. When using the median() function on any zoo object (in particular, mine is an irregular time series), I get the following error, which is thrown from .gt(): Error in if (xi == xj) 0L else if (xi xj) 1L else -1L : argument is of length zero median() of course works fine on the coredata() of my zoo object, but not on the object itself. Is the median.default function coded incorrectly for zoo objects? If not, then I don't understand why functions like mean() and sd() work correctly on the same zoo object, but not median(). Am I asking the right questions? Thanks for any help you can offer. Regards, Marc __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Triangular distribution for kernel regression
Hello, I am trying to get fitted/estimated values using kernel regression and a triangular kernel. I have found packages that easily fit values from a kernel regression (e.g. ksmooth) but do not have a triangular distribution option, and density estimators that have triangular distribution options that I can't seem to use to produce estimated values (e.g. density). Any help is appreciated. Bryan [[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] Triangular distribution for kernel regression
Try: RSiteSearch(kernel triangular) On Sat, Sep 12, 2009 at 1:51 PM, Bryan thespamho...@gmail.com wrote: Hello, I am trying to get fitted/estimated values using kernel regression and a triangular kernel. I have found packages that easily fit values from a kernel regression (e.g. ksmooth) but do not have a triangular distribution option, and density estimators that have triangular distribution options that I can't seem to use to produce estimated values (e.g. density). Any help is appreciated. Bryan [[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] stack overflow when loading workspace
Dear all, I can't load my workspace of 25Mo on R version 2.9.2, because of a stack overflow. But I saved it normally (save.image()), and I didn't get any notification... Does anyone know what that can be due to? Is there any limitation of number of objects (+/-63000)? Thanks Edwige Polus. [[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] could not find function Varcov after upgrade of R?
David Freedman wrote:
Thanks for the reminder - actually I think I've become sloppy about the 'T'
(and the order of loading the Hmisc and Design packages), but that doesn't
seem to be the problem. Also, all of my packages have been updated
Note that Frank Harrell already answered that there is a Hmisc - Design
issue which will be fixed early next week.
You had the unfortune to be too up to date and got the bug that was
introduced very recently (and most people who answered had not so recent
packages).
Uwe Ligges
david
Carlos Alzola wrote:
Did you type library(Hmisc,T) before loading Design?
Carlos
--
From: David Freedman 3.14da...@gmail.com
Sent: Saturday, September 12, 2009 8:26 AM
To: r-help@r-project.org
Subject: Re: [R] could not find function Varcov after upgrade of R?
I've had the same problem with predict.Design, and have sent an email to
the
maintainer of the Design package at Vanderbilt University. I wasn't even
able to run the examples given on the help page of predict.Design - I
received the same error about Varcov that you did.
I *think* it's a problem with the package, rather than R 2.9.2, and I
hope
the problem will soon be fixed. I was able to use predict.Design with
2.9.2
until I updated the Design package a few days ago.
david freedman
zhu yao wrote:
I uses the Design library.
take this example:
library(Design)
n - 1000
set.seed(731)
age - 50 + 12*rnorm(n)
label(age) - Age
sex - factor(sample(c('Male','Female'), n,
rep=TRUE, prob=c(.6, .4)))
cens - 15*runif(n)
h - .02*exp(.04*(age-50)+.8*(sex=='Female'))
dt - -log(runif(n))/h
label(dt) - 'Follow-up Time'
e - ifelse(dt = cens,1,0)
dt - pmin(dt, cens)
units(dt) - Year
dd - datadist(age, sex)
options(datadist='dd')
Srv - Surv(dt,e)
f - cph(Srv ~ rcs(age,4) + sex, x=TRUE, y=TRUE)
cox.zph(f, rank) # tests of PH
anova(f)
# Error in anova.Design(f) : could not find function Varcov
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
2009/9/12 Ronggui Huang ronggui.hu...@gmail.com
I cannot reproduce the problem you mentioned.
ctl - c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14)
trt - c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69)
group - gl(2,10,20, labels=c(Ctl,Trt))
weight - c(ctl, trt)
anova(lm.D9 - lm(weight ~ group))
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=Chinese (Simplified)_People's Republic of
China.936;LC_CTYPE=Chinese (Simplified)_People's Republic of
China.936;LC_MONETARY=Chinese (Simplified)_People's Republic of
China.936;LC_NUMERIC=C;LC_TIME=Chinese (Simplified)_People's Republic
of China.936
attached base packages:
[1] stats graphics grDevices utils datasets methods base
2009/9/12 zhu yao mailzhu...@gmail.com:
After upgrading R to 2.9.2, I can't use the anova() fuction.
It says could not find function Varcov .
What's wrong with my computer? Help needed, thanks!
Yao Zhu
Department of Urology
Fudan University Shanghai Cancer Center
No. 270 Dongan Road, Shanghai, China
[[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.
--
HUANG Ronggui, Wincent
Doctoral Candidate
Dept of Public and Social Administration
City University of Hong Kong
Home page: http://asrr.r-forge.r-project.org/rghuang.html
[[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.
--
View this message in context:
http://www.nabble.com/could-not-find-function-%22Varcov%22-after-upgrade-of-R--tp25412881p25414017.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] stack overflow when loading workspace
sebed1110-div...@yahoo.fr wrote: Dear all, I can't load my workspace of 25Mo on R version 2.9.2, because of a stack overflow. But I saved it normally (save.image()), and I didn't get any notification... Does anyone know what that can be due to? Is there any limitation of number of objects (+/-63000)? At least, the error message would be helpful. Uwe Ligges Thanks Edwige Polus. [[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] median() function on zoo objects
My apologies. I don't know how it happened, but somehow the library must have been corrupted. I re-installed the zoo package, and all was well again. Thanks! Regards, Marc Gabor Grothendieck wrote: Can you provide a reproducible example. See last line to every message on r-help. When I try it it works: median(zoo(1:3)) 2 2 On Sat, Sep 12, 2009 at 1:14 PM, Marc Chiarini marc.chiar...@tufts.edu wrote: Hello R-help community: I have what I think is a simple question that I hope someone can answer. When using the median() function on any zoo object (in particular, mine is an irregular time series), I get the following error, which is thrown from .gt(): Error in if (xi == xj) 0L else if (xi xj) 1L else -1L : argument is of length zero median() of course works fine on the coredata() of my zoo object, but not on the object itself. Is the median.default function coded incorrectly for zoo objects? If not, then I don't understand why functions like mean() and sd() work correctly on the same zoo object, but not median(). Am I asking the right questions? Thanks for any help you can offer. Regards, Marc [[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] ggplot2: deterministic position_jitter geom_line with position_jitter
Dear guRus, I am starting to work with the ggplot2 package and have two very dumb questions: 1) deterministic position_jitter - the jittering is stochastic; is there any way to get a deterministic jittering? For instance: example.data - data.frame(group=c(foo,bar,foo,bar,foo,bar),x=c(1,1,2,2,3,3),y=c(1,1,0,2,1,1)) set.seed(2009) qplot(x,y,data=example.data,shape=group,position=position_jitter(w=0.1,h=0)) For x=1, the foo point is to the left of the bar point, and for x=3 the other way around. I would like to have all foo points at seq(1,3)-epsilon and all bar points at seq(1,3)+epsilon. Do I need to manually modify example.data$x groupwise for this? 2) geom_line with position_jitter - when I call multiple geoms with position_jitter, each geom gets its own jittering. For example (continuing with example.data above): set.seed(2009) qplot(x,y,data=example.data,geom=c(point,line),shape=group,position=position_jitter(w=0.1,h=0)) The lines do not connect the points - is there any way to have the geom_line connect all the foo points on the one hand and all the bar points on the other hand? -- What I've done: searched through HW's book, googled, searched RSeek. For point 2) above, I tried using multiple layers and resetting the seed in between, to wit: pp - ggplot(example.data,aes(x,y,shape=group)) set.seed(2009) pp - pp+layer(geom=point,position=position_jitter(w=0.1,h=0)) set.seed(2009) pp - pp+layer(geom=line,position=position_jitter(w=0.1,h=0)) print(pp) This doesn't do what I want, either... Why I'm doing this: example.data actually contains group means across a covariate x, and they need to be plotted as dots plus error bars (psychologists' convention), so use boxplots instead is a perfectly correct reply to my questions which unfortunately does not help me. Thanks for your time! Stephan -- sessionInfo(): R version 2.9.2 (2009-08-24) i386-pc-mingw32 locale: LC_COLLATE=German_Germany.1252;LC_CTYPE=German_Germany.1252;LC_MONETARY=German_Germany.1252;LC_NUMERIC=C;LC_TIME=German_Germany.1252 attached base packages: [1] grid grDevices datasets splines graphics stats tcltk utils methods base other attached packages: [1] ggplot2_0.8.3 reshape_0.8.3 plyr_0.1.9 proto_0.3-8 svSocket_0.9-43 svMisc_0.9-48 TinnR_1.0.3 R2HTML_1.59-1 Hmisc_3.7-0 [10] survival_2.35-4 loaded via a namespace (and not attached): [1] cluster_1.12.0 lattice_0.17-25 tools_2.9.2 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ggplot2: deterministic position_jitter geom_line with position_jitter
On Sat, Sep 12, 2009 at 1:34 PM, Stephan Kolassa stephan.kola...@gmx.de wrote: Dear guRus, I am starting to work with the ggplot2 package and have two very dumb questions: 1) deterministic position_jitter - the jittering is stochastic; is there any way to get a deterministic jittering? For instance: example.data - data.frame(group=c(foo,bar,foo,bar,foo,bar),x=c(1,1,2,2,3,3),y=c(1,1,0,2,1,1)) set.seed(2009) qplot(x,y,data=example.data,shape=group,position=position_jitter(w=0.1,h=0)) For x=1, the foo point is to the left of the bar point, and for x=3 the other way around. I would like to have all foo points at seq(1,3)-epsilon and all bar points at seq(1,3)+epsilon. Do I need to manually modify example.data$x groupwise for this? Yes. The plyr package generally makes this sort of manipulation pain free. 2) geom_line with position_jitter - when I call multiple geoms with position_jitter, each geom gets its own jittering. For example (continuing with example.data above): set.seed(2009) qplot(x,y,data=example.data,geom=c(point,line),shape=group,position=position_jitter(w=0.1,h=0)) The lines do not connect the points - is there any way to have the geom_line connect all the foo points on the one hand and all the bar points on the other hand? Again, you'll have to adding the jittering yourself. -- What I've done: searched through HW's book, googled, searched RSeek. For point 2) above, I tried using multiple layers and resetting the seed in between, to wit: pp - ggplot(example.data,aes(x,y,shape=group)) set.seed(2009) pp - pp+layer(geom=point,position=position_jitter(w=0.1,h=0)) set.seed(2009) pp - pp+layer(geom=line,position=position_jitter(w=0.1,h=0)) print(pp) This doesn't do what I want, either... The jittering isn't evaluated until render time, so this won't help. You'll notice if you draw the plot again, you'll get a different rendering. Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to visualize paired t-test results?
Hello johannes, You're example looks nice and I am hoping to see more ideas from other members. Just one tiny idea: In - stripchart Also use ylim to include the ZERO line (and then actually add the line) like this: plot.ylim - c(min(d, 0), max(d, 0)) abline(h = 0, lty = 2, col = blue, lwd =2) stripchart(d, vertical=T, pch=16, ylim = plot.ylim ) This way you get to show how far you got the difference to be from the zero (otherwise, all the graph gives is the variance, without any useful knowledge of the distance of the differences from the zero) Also, consider using this function for CI plotting: http://finzi.psych.upenn.edu/R/library/plotrix/html/plotCI.html Best, Tal Galili On Sat, Sep 12, 2009 at 12:00 PM, johannes rara johannesr...@gmail.comwrote: I would like to know if you have any suggestions how to visualize the results from a paired t-test (see the example data below). I tried to produce plots that show the mean and CI's from the original data and the estimate of the difference between means and the confidence intervals (see below) from the t-test. I really don't know what would be the best way to graphically display the results. Thanks in advance. data ## sam1 - c(51.7, 54.2, 53.3, 57, 56.4, 61.5, 57.2, 56.2, 58.4, 55.8) sam2 - c(62.5, 65.2, 67.6, 69.9, 69.4, 70.1, 67.8, 67, 68.5, 62.4) DF - stack(data.frame(sam1 = sam1, sam2 = sam2)) paired-t-test ## res - t.test(values ~ ind, data=DF, paired = TRUE) plots # library(gregmisc) opar - par(mfrow=c(1,2)) tapply(DF$values, list(DF$ind), ci) plotmeans(values ~ ind, data=DF) d - sam1 - sam2 stripchart(d, vertical=T, pch=16) points(1, res$estimate, col=red, pch=16) arrows(1, res$estimate, 1, res$conf.int[1], col=red, lwd=2, angle=90, length=0.1) arrows(1, res$estimate, 1, res$conf.int[2], col=red, lwd=2, angle=90, length=0.1) [[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. -- -- My contact information: Tal Galili Phone number: 972-50-3373767 FaceBook: Tal Galili My Blogs: http://www.r-statistics.com/ http://www.talgalili.com http://www.biostatistics.co.il [[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] Triangular distribution for kernel regression
Gabor, Thanks for your quick reply (on a weekend even!) I've looked through the results of the search you recommended, and several related searches, and don't see anything exceptionally helpful. Kernel regression is a relatively new analysis for me; I apologize for needing a little more direction. I've understand that it is connected to local polynomial regression but I can't seem to have any success from that direction either. At this point the only package that is giving smoothed estimates as I would expect is ksmooth - which doesn't include the appropriate distribution. Best, Bryan On Sat, Sep 12, 2009 at 1:55 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Try: RSiteSearch(kernel triangular) On Sat, Sep 12, 2009 at 1:51 PM, Bryan thespamho...@gmail.com wrote: Hello, I am trying to get fitted/estimated values using kernel regression and a triangular kernel. I have found packages that easily fit values from a kernel regression (e.g. ksmooth) but do not have a triangular distribution option, and density estimators that have triangular distribution options that I can't seem to use to produce estimated values (e.g. density). Any help is appreciated. Bryan [[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] R on Multi Core
Also have a look at the foreach package: http://finzi.psych.upenn.edu/R/library/foreach/html/foreach-package.html On Fri, Sep 11, 2009 at 11:05 PM, Noah Silverman n...@smartmediacorp.comwrote: Hi, Our discussions about 64 bit R has led me to another thought. I have a nice dual core 3.0 chip inside my Linux Box (Running Fedora 11.) Is there a version of R that would take advantage of BOTH cores?? (Watching my system performance meter now is interesting, Running R will hold a single core at 100% perfectly, but the other core sites idle.) Thanks! -- Noah __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- -- My contact information: Tal Galili Phone number: 972-50-3373767 FaceBook: Tal Galili My Blogs: http://www.r-statistics.com/ http://www.talgalili.com http://www.biostatistics.co.il [[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] Triangular distribution for kernel regression
What about kknn -- that was listed as having the triangular distribution? On Sat, Sep 12, 2009 at 3:42 PM, Bryan thespamho...@gmail.com wrote: Gabor, Thanks for your quick reply (on a weekend even!) I've looked through the results of the search you recommended, and several related searches, and don't see anything exceptionally helpful. Kernel regression is a relatively new analysis for me; I apologize for needing a little more direction. I've understand that it is connected to local polynomial regression but I can't seem to have any success from that direction either. At this point the only package that is giving smoothed estimates as I would expect is ksmooth - which doesn't include the appropriate distribution. Best, Bryan On Sat, Sep 12, 2009 at 1:55 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Try: RSiteSearch(kernel triangular) On Sat, Sep 12, 2009 at 1:51 PM, Bryan thespamho...@gmail.com wrote: Hello, I am trying to get fitted/estimated values using kernel regression and a triangular kernel. I have found packages that easily fit values from a kernel regression (e.g. ksmooth) but do not have a triangular distribution option, and density estimators that have triangular distribution options that I can't seem to use to produce estimated values (e.g. density). Any help is appreciated. Bryan [[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] medcouple-based outlier detection in R
I need to detect outliers in a large data set which is highly right-skewed. I plan to use medcouple-based outlier detection. Is there any support for medcouple-based outlier detection in R? Are there any other routines in R to perform outlier detection in highly right-skewed data? Manuj Sharma See the Web#39;s breaking stories, chosen by people like you. Check out Yahoo! Buzz. http://in.buzz.yahoo.com/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] medcouple-based outlier detection in R
On Sep 12, 2009, at 3:54 PM, Manuj Sharma wrote: I need to detect outliers in a large data set which is highly right- skewed. I plan to use medcouple-based outlier detection. Is there any support for medcouple-based outlier detection in R? library(sos) # a package that links to Baron's r-search ???medcouple 7 hits in 2 different packages (robustbase and LambertW) So I would guess that the answer is yes. Are there any other routines in R to perform outlier detection in highly right-skewed data? Yes. Suggest you look over the Robust Statistics Task View. Manuj Sharma See the Web#39;s breaking stories, chosen by people like you. Check out Yahoo! Buzz. http://in.buzz.yahoo.com/ [[alternative HTML version deleted]] David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Triangular distribution for kernel regression
Hi Brian, I am trying to get fitted/estimated values using kernel regression and a triangular kernel. Look at Loader's locfit package. You are likely to be pleasantly surprised. Regards, Mark. Bryan-65 wrote: Hello, I am trying to get fitted/estimated values using kernel regression and a triangular kernel. I have found packages that easily fit values from a kernel regression (e.g. ksmooth) but do not have a triangular distribution option, and density estimators that have triangular distribution options that I can't seem to use to produce estimated values (e.g. density). Any help is appreciated. Bryan [[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. -- View this message in context: http://www.nabble.com/Triangular-distribution-for-kernel-regression-tp25416706p25417878.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] Re : stack overflow when loading workspace
Hi, the error message is Error: protect(): protection stack overflow Edwige. Dear all, I can't load my workspace of 25Mo on R version 2.9.2, because of But I saved it normally (save.image()), and I didn't get any notification... Does anyone know what that can be due to? Is there any limitation of number of objects (+/-63000)? At least, the error message would be helpful. Uwe Ligges Thanks Edwige Polus. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Best Subset Selection - Leaps Package
Dear Experts, I'm a new R user and I'll appreciate your help regarding the following. I'm trying to generate an exhaustive search of all candidate models in a simple linear regression and select the one with the lowest CV-error (or alternatively the lowest Error on a Test set -- if I have lots of data). The leaps package can generate this exhaustive search but all models are evaluated on the train data (without cross-validation). How can I implement what I'm trying to achieve? Any guidance will help... library(ElemStatLearn) #Follow the example of Page 58 in Elements of Stat Learning Book train - subset(prostate, train==TRUE )[,1:9] test - subset(prostate, train=FALSE )[,1:9] #Best subset selection library(leaps) prostate.leaps - regsubsets( lpsa ~ . , data=train, nbest=100, nvmax=8, method=exhaustive, really.big=T) prostate.leaps.sum - summary(prostate.leaps) prostate.models - prostate.leaps.sum$which prostate.models prostate.models.rss - prostate.leaps.sum$rss prostate.models.rss prostate.models.size - as.numeric(attr(prostate.models, dimnames)[[1]]) prostate.models.best.rss -tapply(prostate.models.rss, prostate.models.size, min) prostate.models.best.rss Thanks a lot! Lars. [[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] Triangular distribution for kernel regression
I originally looked over kknn because I need to be able to specify a bandwidth parameter. I am trying to replicate some previous non-R work in R, so I can't stray to far from the procedure used there. In reading the paper referenced in the docs, I see that kknn can reduce to the NadarayaWatson estimator, which is where I need to be, but I'm not sure how to manipulate the bandwidth, as would be possible in other methods. Can you clarify this at all? Bryan On Sat, Sep 12, 2009 at 3:46 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: What about kknn -- that was listed as having the triangular distribution? On Sat, Sep 12, 2009 at 3:42 PM, Bryan thespamho...@gmail.com wrote: Gabor, Thanks for your quick reply (on a weekend even!) I've looked through the results of the search you recommended, and several related searches, and don't see anything exceptionally helpful. Kernel regression is a relatively new analysis for me; I apologize for needing a little more direction. I've understand that it is connected to local polynomial regression but I can't seem to have any success from that direction either. At this point the only package that is giving smoothed estimates as I would expect is ksmooth - which doesn't include the appropriate distribution. Best, Bryan On Sat, Sep 12, 2009 at 1:55 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Try: RSiteSearch(kernel triangular) On Sat, Sep 12, 2009 at 1:51 PM, Bryan thespamho...@gmail.com wrote: Hello, I am trying to get fitted/estimated values using kernel regression and a triangular kernel. I have found packages that easily fit values from a kernel regression (e.g. ksmooth) but do not have a triangular distribution option, and density estimators that have triangular distribution options that I can't seem to use to produce estimated values (e.g. density). Any help is appreciated. Bryan [[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] Triangular distribution for kernel regression
You can specify the number of nearest neighbors. On Sat, Sep 12, 2009 at 5:16 PM, Bryan thespamho...@gmail.com wrote: I originally looked over kknn because I need to be able to specify a bandwidth parameter. I am trying to replicate some previous non-R work in R, so I can't stray to far from the procedure used there. In reading the paper referenced in the docs, I see that kknn can reduce to the Nadaraya–Watson estimator, which is where I need to be, but I'm not sure how to manipulate the bandwidth, as would be possible in other methods. Can you clarify this at all? Bryan On Sat, Sep 12, 2009 at 3:46 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: What about kknn -- that was listed as having the triangular distribution? On Sat, Sep 12, 2009 at 3:42 PM, Bryan thespamho...@gmail.com wrote: Gabor, Thanks for your quick reply (on a weekend even!) I've looked through the results of the search you recommended, and several related searches, and don't see anything exceptionally helpful. Kernel regression is a relatively new analysis for me; I apologize for needing a little more direction. I've understand that it is connected to local polynomial regression but I can't seem to have any success from that direction either. At this point the only package that is giving smoothed estimates as I would expect is ksmooth - which doesn't include the appropriate distribution. Best, Bryan On Sat, Sep 12, 2009 at 1:55 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: Try: RSiteSearch(kernel triangular) On Sat, Sep 12, 2009 at 1:51 PM, Bryan thespamho...@gmail.com wrote: Hello, I am trying to get fitted/estimated values using kernel regression and a triangular kernel. I have found packages that easily fit values from a kernel regression (e.g. ksmooth) but do not have a triangular distribution option, and density estimators that have triangular distribution options that I can't seem to use to produce estimated values (e.g. density). Any help is appreciated. Bryan [[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] Re : stack overflow when loading workspace
On 12/09/2009 5:04 PM, sebed1110-div...@yahoo.fr wrote: Hi, the error message is Error: protect(): protection stack overflow 63000 objects is far more than most people would use, so you may have hit an internal limit. The protection stack is used internally by R to temporarily mark objects as in use even if they haven't been assigned anywhere; overflows usually signal a bug in some code. By default, it is limited to 5 entries, and it's very rare that any code would generate more than that number of temporary objects. However, if you have 63000 objects, it's conceivable that they are all protected at some point during loading. The easiest way to fix this may be not to produce so many objects. I suspect that they aren't all meaningful to you, you probably produced them automatically. So why not store them within a list, or some other structure? That's the normal way to program in R. If you can't do that, you may be able to read your file by recompiling R with a larger stack limit. The line to change is in src/include/Defn.h around line 214, currently #define R_PPSSIZE 5L Change that to a larger number and you might get lucky. Duncan Murdoch Edwige. Dear all, I can't load my workspace of 25Mo on R version 2.9.2, because of But I saved it normally (save.image()), and I didn't get any notification... Does anyone know what that can be due to? Is there any limitation of number of objects (+/-63000)? At least, the error message would be helpful. Uwe Ligges Thanks Edwige Polus. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] (no subject)
Dear All I hope you can help me with this small problem. I want to draw a normal distribution line to this data: p-rnorm(100, mean=5, sd=3000) hist(p) Kabeli [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to determine if a variable is already set?
Do watch out, however, for *where* i exits.
That is, if you type search() you will see a list of environments in
which i might be found. You're probably assuming that i, if
exists(i) is true, is in .GlobalEnv, but it might be in one of the
other environments, in which case exists('i') will be TRUE, but it
won't be the i you are looking for.
I believe you need to quote then name also
use:exits('i')
not:exists(i)
See ?exists (again)
-Don
At 10:19 AM -0700 9/12/09, carol white wrote:
Thanks for your replies.
I use the following script:
if(!exists(i)) stop (set the variable i, call. = FALSE)
but before the stop expression, Error gets displayed:
Error: set the variable i
Is there another function that stops the execution, prints an
expression without printing Error or any other expression except the
expression parameter?
Best,
--- On Fri, 9/11/09, Marc Schwartz marc_schwa...@me.com wrote:
From: Marc Schwartz marc_schwa...@me.com
Subject: Re: [R] how to determine if a variable is already set?
To: carol white wht_...@yahoo.com
Cc: r-h...@stat.math.ethz.ch
Date: Friday, September 11, 2009, 10:21 AM
On Sep 11, 2009, at 12:15 PM, carol white wrote:
Hi,
It might be a primitive question but how it is possible to
determine if a variable is initialized in an environment? Suppose
that we start a R session and wants to run a script which use the
variable i. Which function could evaluate if i is already
initialized or not and if not, then ask interactively the user to
set it? This is to avoid the error message: object i is not found.
Regards,
Carol
See ?exists
Note that this will tell you if the object exists, not if it
contains a specifically desired initial value. You would have to
check for the latter after determining that the object does indeed
exist.
HTH,
Marc Schwartz
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--
--
Don MacQueen
Environmental Protection Department
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Livermore, CA, USA
925-423-1062
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Re: [R] Normal distribution
Dear All let me go one step further by asking you if you could help me show that the distribution of this data in normal. have a little idea (by trial and error) but i seem to not fully understand how its done. H-rnorm(100, mean=5, sd=3000) par(las=1) hist(H, breaks=seq(4, 6, 1000), freq=F) f- function(x) exp(-(x-5000)^2/1800)/sqrt(1800*pi) x- seq(4, 6, 1000) lines(x,f(x)) The code works when its like this: H-rnorm(100, mean=5000, sd=1000) par(las=1) hist(H, breaks=seq(2000, 8000, 100), freq=F) f- function(x) exp(-(x-5000)^2/200)/sqrt(200*pi) x- seq(2000, 8000, 50) lines(x,f(x)) Please tell me or advice of an easier or understandable 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] (no subject)
A better subject for your question might have been helpful. There are many options for hist and truehist in the MASS package, but this might help: x=rnorm(100, mean=5, sd=3000) hist(x, prob=T) x2=density(x) lines(x2$x,x2$y) KABELI MEFANE wrote: Dear All I hope you can help me with this small problem. I want to draw a normal distribution line to this data: p-rnorm(100, mean=5, sd=3000) hist(p) Kabeli [[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. -- View this message in context: http://www.nabble.com/%28no-subject%29-tp25418417p25418513.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] (no subject)
Thank you Sir I have been trying to come up with the code for 6 hours thats why i ended up forgetting the title. Kabeli --- On Sat, 12/9/09, David Freedman 3.14da...@gmail.com wrote: From: David Freedman 3.14da...@gmail.com Subject: Re: [R] (no subject) To: r-help@r-project.org Date: Saturday, 12 September, 2009, 10:49 PM A better subject for your question might have been helpful. There are many options for hist and truehist in the MASS package, but this might help: x=rnorm(100, mean=5, sd=3000) hist(x, prob=T) x2=density(x) lines(x2$x,x2$y) KABELI MEFANE wrote: Dear All I hope you can help me with this small problem. I want to draw a normal distribution line to this data: p-rnorm(100, mean=5, sd=3000) hist(p) Kabeli [[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. -- View this message in context: http://www.nabble.com/%28no-subject%29-tp25418417p25418513.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Normal distribution
Why don't you just do this: H-rnorm(100, mean=5, sd=3000) par(las=1) hist(H, breaks=seq(4, 6, 1000), freq=F) f- function(x) exp(-(x-5000)^2/1800)/sqrt(1800*pi) x- seq(4, 6, 100) lines(x, dnorm(x, 5, 3000)) On Sat, Sep 12, 2009 at 5:47 PM, KABELI MEFANEkabelimef...@yahoo.co.uk wrote: Dear All let me go one step further by asking you if you could help me show that the distribution of this data in normal. have a little idea (by trial and error) but i seem to not fully understand how its done. H-rnorm(100, mean=5, sd=3000) par(las=1) hist(H, breaks=seq(4, 6, 1000), freq=F) f- function(x) exp(-(x-5000)^2/1800)/sqrt(1800*pi) x- seq(4, 6, 1000) lines(x,f(x)) The code works when its like this: H-rnorm(100, mean=5000, sd=1000) par(las=1) hist(H, breaks=seq(2000, 8000, 100), freq=F) f- function(x) exp(-(x-5000)^2/200)/sqrt(200*pi) x- seq(2000, 8000, 50) lines(x,f(x)) Please tell me or advice of an easier or understandable 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. -- 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] Normal distribution
Thank You It really works fine and i do understand it. Now i can sleep, its 0200 hours here. Why don't you just do this: H-rnorm(100, mean=5, sd=3000) par(las=1) hist(H, breaks=seq(4, 6, 1000), freq=F) f- function(x) exp(-(x-5000)^2/1800)/sqrt(1800*pi) x- seq(4, 6, 100) lines(x, dnorm(x, 5, 3000)) On Sat, Sep 12, 2009 at 5:47 PM, KABELI MEFANEkabelimef...@yahoo.co.uk wrote: Dear All let me go one step further by asking you if you could help me show that the distribution of this data in normal. have a little idea (by trial and error) but i seem to not fully understand how its done. H-rnorm(100, mean=5, sd=3000) par(las=1) hist(H, breaks=seq(4, 6, 1000), freq=F) f- function(x) exp(-(x-5000)^2/1800)/sqrt(1800*pi) x- seq(4, 6, 1000) lines(x,f(x)) The code works when its like this: H-rnorm(100, mean=5000, sd=1000) par(las=1) hist(H, breaks=seq(2000, 8000, 100), freq=F) f- function(x) exp(-(x-5000)^2/200)/sqrt(200*pi) x- seq(2000, 8000, 50) lines(x,f(x)) Please tell me or advice of an easier or understandable 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. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[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] the integer of a given location
I have a huge number such as 78923698701 z-78923698701 I want to find the integer of a given location, for example here, what is the 2nd number? 8. Thanks in advance! -- View this message in context: http://www.nabble.com/the-integer-of-a-given-location-tp25418729p25418729.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] the integer of a given location
Is this what you want: z-78923698701 substring(as.character(z), 2, 2) [1] 8 On Sat, Sep 12, 2009 at 6:16 PM, darkhorn darkh...@gmail.com wrote: I have a huge number such as 78923698701 z-78923698701 I want to find the integer of a given location, for example here, what is the 2nd number? 8. Thanks in advance! -- View this message in context: http://www.nabble.com/the-integer-of-a-given-location-tp25418729p25418729.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. -- 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] the integer of a given location
Hi darkhorn, Here are two suggestions: # Suggestion 1 z - 78923698701 foo - function(x, pos) strsplit(as.character(x),)[[1]][pos] foo(z,2) # [1] 8 # Suggestion 2 foo2 - function(x, pos) substr(x,pos,pos) foo2(z,2) # [1] 8 See ?strsplit and ?substr for more details. HTH, Jorge On Sat, Sep 12, 2009 at 6:16 PM, darkhorn darkh...@gmail.com wrote: I have a huge number such as 78923698701 z-78923698701 I want to find the integer of a given location, for example here, what is the 2nd number? 8. Thanks in advance! -- View this message in context: http://www.nabble.com/the-integer-of-a-given-location-tp25418729p25418729.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] the integer of a given location
Thank you very much. :-) as.integer(substring(as.character(z), 2, 2)) [1] 8 as.integer(foo2(z,2)) [1] 8 -- View this message in context: http://www.nabble.com/the-integer-of-a-given-location-tp25418729p25419555.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] Barplot+Table
Marc Schwartz-3 wrote: Using the data that is in the online plot rather than the above, here is a first go. Note that I am not drawing the background grid in the barplot or the lines for table below it. These could be added if you really need them. snip HTH, Marc Schwartz Wow. That looks absolutely amazing. I really can't thank you enough. This not only gives me the layout for what I am trying to do, it gives me a much better understanding of how the graphcs systems in R fit together. -- View this message in context: http://www.nabble.com/Barplot%2BTable-tp25406046p25419738.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] Manage an unknown and variable number of data frames
Hi, In the code below I create a small data.frame (dat) and then cut it into different groups using CutList. The lists in CutList allow to me choose whatever columns I want from dat and allow me to cut it into any number of groups by changing the lists. It seems to work OK but when I'm done I have a variable number of data frames what I need to do further operations on and I don't know how to manage them as a collection. How do experience R coders handle keeping all this straight so that if I add another column from dat and more groups in the cuts it all stays straight? I need to send each dataf rame to another function to add columns of specific data calcuations to each of them. Best for me (I think) would be to enumerate each data frame using the row.name number from CutTable if possible, but that's just my thought. If each data frame became an element of CutTable then I'd always know where they are. Really I'm needing to get a handle on keeping a variable and unknown number of these things straight. Thanks, Mark dat = data.frame( a=round(runif(100,-20,30),2), b=round(runif(100,-40,50),2) ) # Give each cut list a name matching the column in dat that you # want to use as criteria for making the cut. # Create any number of cuts in each row. CutList = list( a=c(-Inf,-10,10,Inf), b=c(-Inf,0,20,Inf) ) CutResults = mapply(cut,x=dat[,names(CutList)],CutList,SIMPLIFY=FALSE) CutTable = as.data.frame(table(CutResults)) CutResultsDF = as.data.frame(CutResults) head(CutResultsDF, n=15) dat$aRange = CutResultsDF$a dat$bRange = CutResultsDF$b head(dat, 15) # I don't want to do the following as it doesn't # get managed automatically. Subset1 = subset(subset(dat, aRange==CutTable$a[1]), bRange==CutTable$b[1])[1:2] Subset2 = subset(subset(dat, aRange==CutTable$a[2]), bRange==CutTable$b[2])[1:2] Subset3 = subset(subset(dat, aRange==CutTable$a[3]), bRange==CutTable$b[3])[1:2] Subset4 = subset(subset(dat, aRange==CutTable$a[4]), bRange==CutTable$b[4])[1:2] Subset1 Subset2 Subset3 Subset4 CutTable __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] zoo plot: yearly marks on X-Axis
On Sun, 2009-07-26 at 13:17 +0200, Poersching wrote: stvienna wiener schrieb: Hi all, I am plotting a financial time series, but I need a more detailed X-Axis. Example: x - zoo(rnorm(1:6000), as.Date(1992-11-11)+c(1:6000)) plot(x) The X-Axis is labeled 1995, 2000 and 2005. I would need either 1995, 1997, etc. or maybe yearly I used google first, then look at ?plot.zoo but could't get it working. Regards, Steve __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Hey, try something like the following: plot(x, y, xaxt=n) axis.Date(1, at=seq(as.Date(1960-01-01), max(as.Date(x)), years), labels = FALSE, tcl = -0.2) axis.Date(1, at=seq(as.Date(1960-01-01), max(as.Date(x)), 5 years), labels = TRUE, las=3, tcl = -0.2) Regards, Christian Hello, I'm actually struggling with the similar problem. I applied Your script into mine like this: - library(Rdbi) library(RdbiPgSQL) conn - dbConnect(PgSQL(), host=localhost, dbname=BVS, user=postgres, password = **) query_duj_kal - dbSendQuery(conn, select zdroj as well, cas as date, fe2, fe3, mn, nh4, no2, no3, o2, teplota as temperature from analystats where zdroj like 'Dunaj Kalinkovo' order by date) watchem_duj_kal - dbGetResult(query_duj_kal) date - (watchem_duj_kal$date) date_p - strptime(date, %m-%d-%Y) no3 - (watchem_duj_kal$no3) nh4 - (watchem_duj_kal$nh4) par(mfrow=c(3,1), ylog = TRUE, yaxp = c(0.01, 100, 3)) maxy - 100 miny - 0.005 plot(date_p, no3,log = y, type = l, ylim = c(miny, maxy), col = darkred, main = Dunaj Kalinkovo, xlab = time, ylab = log(NO3-,NH4 +), xaxt=n) axis.Date(1, at=seq(as.Date(1972-01-01),max(as.Date(date_p))), years), labels = FALSE, tcl = -0.2) axis.Date(1, at=seq(as.Date(1970-01-01),max(as.Date(date_p))), 5 years), labels = TRUE, las=3, tcl = -0.2) - Everything goes well, except that I cannot get no tickles nor labels at the moment and I cannot find what I did wrong. Many thanks for any advice Tomas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] zoo plot: yearly marks on X-Axis
Try this: library(zoo) x - zoo(rnorm(1:6000), as.Date(1992-11-11)+c(1:6000)) plot(x, xaxt = n) # cut time into years, remove duplicate years and convert back to Date yy - as.Date(unique(cut(time(x), year))) axis(1, yy, format(yy, %y)) There are a number of examples of custom X axes in the examples section of ?plot.zoo On Sat, Sep 12, 2009 at 10:33 PM, lanc...@fns.uniba.sk wrote: On Sun, 2009-07-26 at 13:17 +0200, Poersching wrote: stvienna wiener schrieb: Hi all, I am plotting a financial time series, but I need a more detailed X-Axis. Example: x - zoo(rnorm(1:6000), as.Date(1992-11-11)+c(1:6000)) plot(x) The X-Axis is labeled 1995, 2000 and 2005. I would need either 1995, 1997, etc. or maybe yearly I used google first, then look at ?plot.zoo but could't get it working. Regards, Steve __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Hey, try something like the following: plot(x, y, xaxt=n) axis.Date(1, at=seq(as.Date(1960-01-01), max(as.Date(x)), years), labels = FALSE, tcl = -0.2) axis.Date(1, at=seq(as.Date(1960-01-01), max(as.Date(x)), 5 years), labels = TRUE, las=3, tcl = -0.2) Regards, Christian Hello, I'm actually struggling with the similar problem. I applied Your script into mine like this: - library(Rdbi) library(RdbiPgSQL) conn - dbConnect(PgSQL(), host=localhost, dbname=BVS, user=postgres, password = **) query_duj_kal - dbSendQuery(conn, select zdroj as well, cas as date, fe2, fe3, mn, nh4, no2, no3, o2, teplota as temperature from analystats where zdroj like 'Dunaj Kalinkovo' order by date) watchem_duj_kal - dbGetResult(query_duj_kal) date - (watchem_duj_kal$date) date_p - strptime(date, %m-%d-%Y) no3 - (watchem_duj_kal$no3) nh4 - (watchem_duj_kal$nh4) par(mfrow=c(3,1), ylog = TRUE, yaxp = c(0.01, 100, 3)) maxy - 100 miny - 0.005 plot(date_p, no3,log = y, type = l, ylim = c(miny, maxy), col = darkred, main = Dunaj Kalinkovo, xlab = time, ylab = log(NO3-,NH4 +), xaxt=n) axis.Date(1, at=seq(as.Date(1972-01-01),max(as.Date(date_p))), years), labels = FALSE, tcl = -0.2) axis.Date(1, at=seq(as.Date(1970-01-01),max(as.Date(date_p))), 5 years), labels = TRUE, las=3, tcl = -0.2) - Everything goes well, except that I cannot get no tickles nor labels at the moment and I cannot find what I did wrong. Many thanks for any advice Tomas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Manage an unknown and variable number of data frames
On Sep 12, 2009, at 10:13 PM, Mark Knecht wrote: Hi, In the code below I create a small data.frame (dat) and then cut it into different groups using CutList. The lists in CutList allow to me choose whatever columns I want from dat and allow me to cut it into any number of groups by changing the lists. It seems to work OK but when I'm done I have a variable number of data frames what I need to do further operations on and I don't know how to manage them as a collection. List processing. How do experience R coders handle keeping all this straight so that if I add another column from dat and more groups in the cuts it all stays straight? I need to send each dataf rame to another function to add columns of specific data calcuations to each of them. Best for me (I think) would be to enumerate each data frame using the row.name number from CutTable if possible, but that's just my thought. If each data frame became an element of CutTable then I'd always know where they are. Really I'm needing to get a handle on keeping a variable and unknown number of these things straight. Thanks, Mark dat = data.frame( a=round(runif(100,-20,30),2), b=round(runif(100,-40,50),2) ) # Give each cut list a name matching the column in dat that you # want to use as criteria for making the cut. # Create any number of cuts in each row. CutList = list( a=c(-Inf,-10,10,Inf), b=c(-Inf,0,20,Inf) ) CutResults = mapply(cut,x=dat[,names(CutList)],CutList,SIMPLIFY=FALSE) CutTable = as.data.frame(table(CutResults)) CutResultsDF = as.data.frame(CutResults) head(CutResultsDF, n=15) dat$aRange = CutResultsDF$a dat$bRange = CutResultsDF$b head(dat, 15) You could have gotten the same labeling of columns into categories with a combination of ave and cut. dat$arng2 - ave(dat$a, FUN=function(x) cut(x, breaks=CutList$a) ) dat a b aRangebRange arng2 1 -10.45 43.30 (-Inf,-10] (20, Inf] 1 2 9.09 -33.66 (-10,10] (-Inf,0] 2 329.27 18.34 (10, Inf](0,20] 3 428.92 46.55 (10, Inf] (20, Inf] 3 5 2.07 -8.23 (-10,10] (-Inf,0] 2 618.28 -35.13 (10, Inf] (-Inf,0] 3 7 -16.26 40.59 (-Inf,-10] (20, Inf] 1 snip # I don't want to do the following as it doesn't # get managed automatically. It is possibly unclear what you are hoping to accomplish with that subset(subset(.)) construction. Are you trying to accomplish what a logical conjunction for subset= , coupled with a select= parameter would do inside a single subset? subset(dat, aRange==CutTable$a[1] bRange==CutTable$b[1], select=c(a,b) ) a b 26 -17.50 -18.46 28 -15.48 -34.37 31 -10.04 -21.55 38 -11.73 -29.40 46 -18.28 -17.42 95 -11.62 -22.94 96 -12.16 -1.57 97 -15.44 -19.89 Subset1 = subset(subset(dat, , Subset2 = subset(subset(dat, aRange==CutTable$a[2]), bRange==CutTable $b[2])[1:2] Subset3 = subset(subset(dat, aRange==CutTable$a[3]), bRange==CutTable $b[3])[1:2] Subset4 = subset(subset(dat, aRange==CutTable$a[4]), bRange==CutTable $b[4])[1:2] You could automate that with work.list - lapply(1:4, function(x) subset(dat, aRange==CutTable $a[x] bRange==CutTable$b[x], select=c(a,b) ) ) work.list[[1]] # first element of a 4 element list a b 26 -17.50 -18.46 28 -15.48 -34.37 31 -10.04 -21.55 38 -11.73 -29.40 46 -18.28 -17.42 95 -11.62 -22.94 96 -12.16 -1.57 97 -15.44 -19.89 Subset1 Subset2 Subset3 Subset4 CutTable David Winsemius, MD Heritage Laboratories West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Re : stack overflow when loading workspace
There is a command-line flag for this: tystie% R --help ... --max-ppsize=NSet max size of protect stack to N so you do not need to re-compile. It is not uncommon to need more space to load objects than to save them, and the need to raise this limit to re-load a worskpace has come up before on R-help (it is why there is a command-line flag). The PP stack is not the only stack that can overflow; it is not even the commonest one. So we could not have guessed without the error message, and if you raise this limit you may hit others. I don't think 63000 objects is large, but 63000 named objects would be. Remember that each character string etc is an object (see memory.profile() for a real count). On Sat, 12 Sep 2009, Duncan Murdoch wrote: On 12/09/2009 5:04 PM, sebed1110-div...@yahoo.fr wrote: Hi, the error message is Error: protect(): protection stack overflow 63000 objects is far more than most people would use, so you may have hit an internal limit. The protection stack is used internally by R to temporarily mark objects as in use even if they haven't been assigned anywhere; overflows usually signal a bug in some code. By default, it is limited to 5 entries, and it's very rare that any code would generate more than that number of temporary objects. However, if you have 63000 objects, it's conceivable that they are all protected at some point during loading. The easiest way to fix this may be not to produce so many objects. I suspect that they aren't all meaningful to you, you probably produced them automatically. So why not store them within a list, or some other structure? That's the normal way to program in R. If you can't do that, you may be able to read your file by recompiling R with a larger stack limit. The line to change is in src/include/Defn.h around line 214, currently #define R_PPSSIZE 5L Change that to a larger number and you might get lucky. Duncan Murdoch Edwige. Dear all, I can't load my workspace of 25Mo on R version 2.9.2, because of But I saved it normally (save.image()), and I didn't get any notification... Does anyone know what that can be due to? Is there any limitation of number of objects (+/-63000)? At least, the error message would be helpful. Uwe Ligges Thanks Edwige Polus. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
