Re: [R] Re moving levels of a factor
Dieter Menne wrote: Henrik Wahren wrote: How can one or more levels be removed from a factor of a data frame. There was a similar post on how to do this when a factor meets some criterion (e.g. = 2), but I can¹t seem to get that solution to work. Here, I simply want to drop some levels. Simply call factor again on the reduced set a = factor(c(a,b,b,b)) a[-1] factor(a[-1]) If a is in a data frame, replace a by by df$a. or a[-1, drop=TRUE] Uwe Ligges Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] prcomp() PCA vs fastICA() PCA?
Hi all,
I wonder what the difference is between the functions prcomp and the PCA
plotting method used in example 3 from the fastICA package. They give totally
different plots. The reason for asking is that I've earlier used prcomp, but
now I should do an ICA, and I guess I cannot compare the PCA plot from prcomp
with the ICA plot if the two PCA plots looks different?
Does anyone knows anything about this? Maybe there's a different approach
that's better?
if(require(MASS))
{
x - mvrnorm(n = 1000, mu = c(0, 0), Sigma = matrix(c(10, 3, 3, 1), 2, 2))
x1 - mvrnorm(n = 1000, mu = c(-1, 2), Sigma = matrix(c(10, 3, 3, 1), 2, 2))
X - rbind(x, x1)
a - fastICA(X, 2, alg.typ = deflation, fun = logcosh, alpha = 1,
method = R, row.norm = FALSE, maxit = 200,
tol = 0.0001, verbose = TRUE)
par(mfrow = c(1, 3))
plot(a$X, main = Pre-processed data)
plot(a$X%*%a$K, main = PCA components)
plot(a$S, main = ICA components)
}
PC=prcomp (X, center=T, scale=T)
hcl=hclust(dist(df))
plot(PC$x[,1],PC$x[,2], main=PCA components (prcomp))
Best regards,
Joel
_
Nya Windows 7 gör allt lite enklare. Hitta en dator som passar dig!
http://windows.microsoft.com/shop
[[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] gsub does not support \b?
Tan, Richard wrote: Hello, can someone help? How come gsub(\bINDS\b,INDUSTRIES,ADVANCED ENERGY INDS) [1] ADVANCED ENERGY INDS It does, but you need to escape it: gsub(\\bINDS\\b,INDUSTRIES,ADVANCED ENERGY INDS) Uwe Ligges not ADVANCED ENERGY INDUSTRIES Thanks. Richard [[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] What parts of 'Statistical Models in S' are not applicable to R?
Peng Yu wrote: According to Amazon review, 'Statistical Models in S' is a key reference for understanding the methods implemented in several of S-PLUS' high-end statistical functions, including 'lm()', predict()', 'design()', 'aov()', 'glm()', 'gam()', 'loess()', 'tree()', 'burl.tree()', 'nls()' and 'ms()'. But since it is for S, some part of the book may not be applicable to R. Some examples (e.g. interaction.plot()) discussed in this book are not available in R. Without, working examples, it is sometimes difficult for me to understand the materials in the book. Besides the functions mentioned in the Amazon review, could somebody give me a hint on what chapters (or sections) in this book are not appropriate to R? They all are appropriate, but nuances differ these days, as some nuances differ for recent S-PLUS versions, 17 years later. It should still be fine to learn some relevant concepts. Best wishes, 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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] please add this email to the maillist
Hi, You can ad it yourself: http://stat.ethz.ch/mailman/listinfo/r-help -Ista On Wed, Nov 11, 2009 at 12:57 AM, 杨江伟 jwyan...@gmail.com wrote: thanks a lot -- Pleasure should be subordinate to duty. [[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. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.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.
[R] Polynomial fitting
Dear R helpers Suppose I have a following data y - c(9.21, 9.51, 9.73, 9.88, 10.12. 10.21) t - c(0, 0.25, 1, 3, 6, 12) I want to find out the polynomial which fits y in terms of t i.e. y = f(t) some function of t. e.g. y = bo + b1*t + (b2 * t^2) + (b3 * t^3) + .. and so on. In Excel I have defined y as independent variable, then defined t, t^2 and t^3 and using regression I could arrive at the equation y = 9.505799 + (0.191092 * t) - (0.0225 * t^2) + (0.001245 * t^3) However I feel this is wrong as I am trying to use linear regression but here I am having polynomial in t. I am not that good in stats as well as in mathmatics. I request you to kindly help me as to how to express the 'y' in polynomial in terms of t. Thanking you in advance Julia Only a man of Worth sees Worth in other men [[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] [Rd] NetCDF output in R
Dear Robert,
Thanks for the information.
I tried this but it still did not work well.
I will try some other option.
Nana Browne
There is no key to happiness. The door is always open.
--- On Tue, 11/10/09, Robert J. Hijmans r.hijm...@gmail.com wrote:
From: Robert J. Hijmans r.hijm...@gmail.com
Subject: Re: [Rd] NetCDF output in R
To: nana amabr...@yahoo.com
Cc: r-help@r-project.org, Alice favre favre_al...@yahoo.fr
Date: Tuesday, November 10, 2009, 12:21 PM
Hi Nana,
This is not a r-devel question.
I suspect this should be something like:
write.netcdf.time(paste(path,'cam_fore.nc',sep=), fore[,,times]
,lon,lat,times)
Robert
On Tue, Nov 10, 2009 at 8:31 AM, nana amabr...@yahoo.com wrote:
Dear CSAG R users,
I will be glad if someone can point out what I am doing wrong or not doing at
all in this.
I am trying to write out netcdf file in R. I have 26 time step but only the
first time step is written.
For example:
library(ncdf)
path - '/home/work/'
forecast - open.ncdf(paste(path,'cam.1980.2005.nc',sep=))
fore - get.var.ncdf(forecast,'ppt')
lon - get.var.ncdf(forecast,'lon')
lat - get.var.ncdf(forecast,'lat')
dim(fore)[3]
26
times - 1:dim(fore)[3]
write.netcdf.time(paste(path,'cam_fore.nc',sep=), fore,lon,lat,times)
[1] put.var.ncdf: warning: you asked to write 1440 values, but the passed
data array has 37440 entries!
[[1]]
[1] 6
Warning message:
In 1:nt : numerical expression has 26 elements: only the first used
# function for writing out the netcdf file #
write.netcdf.time - function(filename='outputfile.nc',data,lons,lats,nt){
lon - dim.def.ncdf('lon','degrees_east',lons)
lat - dim.def.ncdf('lat','degrees_north',lats)
times - 1:nt
tdim - dim.def.ncdf('time','days since 1980-01-01', times, unlim=TRUE)
# levs - dim.def.ncdf('lev','pressure',levs)
var - var.def.ncdf('data','unitless',list(lon,lat,tdim),-999.9)
ncid - create.ncdf(filename,list(var))
put.var.ncdf(ncid, var, data)
close.ncdf(ncid)
}
##end of function###
Thank you.
Nana Browne
There is no key to happiness. The door is always open.
[[alternative HTML version deleted]]
__
r-de...@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-devel
[[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] Nested ANOVA
Hello, I have the following data and would like to get some hints on how to analyze tis data, nested analysis? Habitats: 2 (Seagrass meadows and sandy bottoms) Seasons: 4 (Winter, Spring, Summer and Autumn) Locations: 4 (2 locations for for each habitat and season) Replicates: 3 replicates for each location within each season and within each habitat Kindly regards, Stefanos Kalogirou -- View this message in context: http://old.nabble.com/Nested-ANOVA-tp26298891p26298891.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What parts of 'Statistical Models in S' are not applicable to R?
Le mercredi 11 novembre 2009 à 10:22 +0100, Uwe Ligges a écrit : Peng Yu wrote: According to Amazon review, 'Statistical Models in S' is a key reference for understanding the methods implemented in several of S-PLUS' high-end statistical functions, including 'lm()', predict()', 'design()', 'aov()', 'glm()', 'gam()', 'loess()', 'tree()', 'burl.tree()', 'nls()' and 'ms()'. But since it is for S, some part of the book may not be applicable to R. Some examples (e.g. interaction.plot()) discussed in this book are not available in R. Without, working examples, it is sometimes difficult for me to understand the materials in the book. Besides the functions mentioned in the Amazon review, could somebody give me a hint on what chapters (or sections) in this book are not appropriate to R? They all are appropriate, but nuances differ these days, as some nuances differ for recent S-PLUS versions, 17 years later. It should still be fine to learn some relevant concepts. You could also note that, at least in the 4th (last) edition of the book, the authors have marked passages with differences between R and S+ with a marginal R. Now this book has grow a bit out of date since its lst edition (2002 ?) : Newer R packages implements various things previously not implemented in R (e.g. multiple comparisons after ANOVA, previously available in S+ with the multicomp function, nd implemented (with a lot more generalizability) in the multcomp package). A 5th edition might be in order, but that would be a *daunting* task : The language R has grew (e. g. namespaces), the nature and extend of avilable tasks has grew *enormously*, and I don't think that producing a book that would be to 2009 R what VR4 was to 2002 R is doable by a two-person team, as talented, dedicated and productive as these two persons might be (modulo Oxford sarcasm :-). Furthermore, these two persons already give an enormous amount of time and effort to other R development (search for R-help activity of BV and BDR, or meditate on the recently published stats on R source activity...). Such a document would probably have to be something other than a book to stay up to date and accurate, and even coordinating such a task would need serious time... Even if it would exclude anything present in the vrious packages help files, and should limit to tutorial introductions, examples and discussions, the sheer volume (1700+ packages last time I looked) and the difficulty of coordination (how do you discuss 5 different packages, implementing various means to solve the same problem ?) would involve serious organizational difficulties. So I doubt such a document will get produced in the foreseeable future. Frequent R-help reading and note-taking is the second-best option... To come back to R-vs-S+ topic : unless I'm mistaken, R seems to be currently the dominant version of the S language, and most published S material will nowadays (implicitly) be aimed at R. This should reduce the importance of the problem. Sincerely, Emmanuel Charpentier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] What parts of 'Statistical Models in S' are not applicable to R?
Emmanuel Charpentier wrote: Le mercredi 11 novembre 2009 à 10:22 +0100, Uwe Ligges a écrit : Peng Yu wrote: According to Amazon review, 'Statistical Models in S' is a key reference for understanding the methods implemented in several of S-PLUS' high-end statistical functions, including 'lm()', predict()', 'design()', 'aov()', 'glm()', 'gam()', 'loess()', 'tree()', 'burl.tree()', 'nls()' and 'ms()'. But since it is for S, some part of the book may not be applicable to R. Some examples (e.g. interaction.plot()) discussed in this book are not available in R. Without, working examples, it is sometimes difficult for me to understand the materials in the book. Besides the functions mentioned in the Amazon review, could somebody give me a hint on what chapters (or sections) in this book are not appropriate to R? They all are appropriate, but nuances differ these days, as some nuances differ for recent S-PLUS versions, 17 years later. It should still be fine to learn some relevant concepts. You could also note that, at least in the 4th (last) edition of the book, the authors have marked passages with differences between R and S+ with a marginal R. I guess you are talking about a very different book . Uwe Ligges Now this book has grow a bit out of date since its lst edition (2002 ?) : Newer R packages implements various things previously not implemented in R (e.g. multiple comparisons after ANOVA, previously available in S+ with the multicomp function, nd implemented (with a lot more generalizability) in the multcomp package). A 5th edition might be in order, but that would be a *daunting* task : The language R has grew (e. g. namespaces), the nature and extend of avilable tasks has grew *enormously*, and I don't think that producing a book that would be to 2009 R what VR4 was to 2002 R is doable by a two-person team, as talented, dedicated and productive as these two persons might be (modulo Oxford sarcasm :-). Furthermore, these two persons already give an enormous amount of time and effort to other R development (search for R-help activity of BV and BDR, or meditate on the recently published stats on R source activity...). Such a document would probably have to be something other than a book to stay up to date and accurate, and even coordinating such a task would need serious time... Even if it would exclude anything present in the vrious packages help files, and should limit to tutorial introductions, examples and discussions, the sheer volume (1700+ packages last time I looked) and the difficulty of coordination (how do you discuss 5 different packages, implementing various means to solve the same problem ?) would involve serious organizational difficulties. So I doubt such a document will get produced in the foreseeable future. Frequent R-help reading and note-taking is the second-best option... To come back to R-vs-S+ topic : unless I'm mistaken, R seems to be currently the dominant version of the S language, and most published S material will nowadays (implicitly) be aimed at R. This should reduce the importance of the problem. Sincerely, Emmanuel Charpentier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Error in lm() function
Hi all, I wanted to have a seasonality study like whether a particular month has significant effect as compared to others. Here is my data : 0.10499 0 0 1 0 0 0 0 0 0 0 0 0.00259 0 0 0 1 0 0 0 0 0 0 0 -0.060150 0 0 0 1 0 0 0 0 0 0 0.10721 0 0 0 0 0 1 0 0 0 0 0 0.03597 0 0 0 0 0 0 1 0 0 0 0 0.10584 0 0 0 0 0 0 0 1 0 0 0 0.02063 0 0 0 0 0 0 0 0 1 0 0 -0.035090 0 0 0 0 0 0 0 0 1 0 -0.034850 0 0 0 0 0 0 0 0 0 1 0.01632 0 0 0 0 0 0 0 0 0 0 0 0.06844 1 0 0 0 0 0 0 0 0 0 0 -0.017660 1 0 0 0 0 0 0 0 0 0 0.00989 0 0 1 0 0 0 0 0 0 0 0 0.11673 0 0 0 1 0 0 0 0 0 0 0 0.01789 0 0 0 0 1 0 0 0 0 0 0 -0.003230 0 0 0 0 1 0 0 0 0 0 0.06811 0 0 0 0 0 0 1 0 0 0 0 -0.012920 0 0 0 0 0 0 1 0 0 0 -0.122440 0 0 0 0 0 0 0 1 0 0 -0.066450 0 0 0 0 0 0 0 0 1 0 -0.033550 0 0 0 0 0 0 0 0 0 1 0.02308 0 0 0 0 0 0 0 0 0 0 0 -0.117111 0 0 0 0 0 0 0 0 0 0 0.06116 0 1 0 0 0 0 0 0 0 0 0 0.02832 0 0 1 0 0 0 0 0 0 0 0 0.01441 0 0 0 1 0 0 0 0 0 0 0 -0.044120 0 0 0 1 0 0 0 0 0 0 0.05558 0 0 0 0 0 1 0 0 0 0 0 0.08363 0 0 0 0 0 0 1 0 0 0 0 -0.010630 0 0 0 0 0 0 1 0 0 0 i.e. all explanatory variables here are dichotomous. However once I run lm() i got following error : lm(dat[,1]~dat[,-1]) Error in model.frame.default(formula = dat[, 1] ~ dat[, -1], drop.unused.levels = TRUE) : invalid type (list) for variable 'dat[, -1]' Can anyone please tell me what to do? Best, -- View this message in context: http://old.nabble.com/Error-in-lm%28%29-function-tp26300324p26300324.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] Data transformation
Try this:
x - read.table(textConnection(idcode1code2 p
+ 148 0.1
+ 157 0.9
+ 218 0.4
+ 262 0.2
+ 243 0.6
+ 356 0.7
+ 375 0.9), header=TRUE)
closeAllConnections()
# create object like output from 'melt'
x.m - data.frame(id=c(x$id, x$id),
+var=paste('var', c(x$code1, x$code2), sep=''),
+variable=rep('p', 2*nrow(x)),
+value=c(x$p, x$p))
require(reshape) # use the reshape package
(x.n - cast(x.m, id ~ var, function(.dat){
+ if (length(.dat) == 0) return(0) # test for no data; return
zero if that is the case
+ mean(.dat)
+ }))
id var1 var2 var3 var4 var5 var6 var7 var8
1 1 0.0 0.0 0.0 0.1 0.9 0.0 0.9 0.1
2 2 0.4 0.2 0.6 0.6 0.0 0.2 0.0 0.4
3 3 0.0 0.0 0.0 0.0 0.8 0.7 0.9 0.0
On Tue, Nov 10, 2009 at 11:10 PM, legen lege...@gmail.com wrote:
Thank you for your kind help. Your script works very well. Would you please
show me how to change NaN to zero and column variables 1, 2, ..., 8 to var1,
var2, ..., var8? Thanks again.
Legen
jholtman wrote:
Is this what you want:
x - read.table(textConnection(id code1 code2 p
+ 1 4 8 0.1
+ 1 5 7 0.9
+ 2 1 8 0.4
+ 2 6 2 0.2
+ 2 4 3 0.6
+ 3 5 6 0.7
+ 3 7 5 0.9), header=TRUE)
closeAllConnections()
# create object like output from 'melt'
x.m - data.frame(id=c(x$id, x$id), var=c(x$code1, x$code2),
+ variable=rep('p', 2*nrow(x)), value=c(x$p, x$p))
require(reshape) # use the reshape package
cast(x.m, id ~ var, mean)
id 1 2 3 4 5 6 7 8
1 1 NaN NaN NaN 0.1 0.9 NaN 0.9 0.1
2 2 0.4 0.2 0.6 0.6 NaN 0.2 NaN 0.4
3 3 NaN NaN NaN NaN 0.8 0.7 0.9 NaN
On Tue, Nov 10, 2009 at 4:30 PM, legen lege...@gmail.com wrote:
Dear all,
I have a dataset as below:
id code1 code2 p
1 4 8 0.1
1 5 7 0.9
2 1 8 0.4
2 6 2 0.2
2 4 3 0.6
3 5 6 0.7
3 7 5 0.9
I just want to rewrite it as this (vertical to horizontal):
id var1 var2 var3 var4 var5 var6 var7 var8
1 0 0 0 0.1 0.9 0 0.9 0.1
2 0.4 0.2 0.6 0.6 0 0.2 0 0.4
3 0 0 0 0 0.8 0.7 0.9 0
For the third subject, there are two values being equal to 5 in code1 and
code2, but different values in p: 0.7 and 0.9, so I assigned their
average
0.8 in var5.
Does anybody can help me to handle this? Many thanks for your
consideration
and time.
Legen
--
View this message in context:
http://old.nabble.com/Data-transformation-tp26291568p26291568.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.
--
View this message in context:
http://old.nabble.com/Data-transformation-tp26291568p26295766.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.
[R] architecture of statistical programming language R
hi, I'm mustafa. I'm a master student Dokuz Eylül University in Izmir/Turkey. nbsp; I focus on Statistical programming language R. I learn architecture of R. But i don't find out any material in internet. Would you like tonbsp;help me this subject? If you have documentations, would you like to share with me? nbsp; If you help me, i have informationsnbsp;about architecture of R and i'mnbsp;happy. nbsp; More power to you! thanks. ___ Ücretsiz dinlemek için yüzbinlerce þarký Kavun'da! Týkla, dinle. [[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] Data transformation
Your script works very well. Thank you very much.
Legen
Henrique Dallazuanna wrote:
Try this also:
xtabs(rep(p, 2) ~ rep(id, 2) + sprintf(var%d, c(code1, code2)), data =
x)
On Wed, Nov 11, 2009 at 2:10 AM, legen lege...@gmail.com wrote:
Thank you for your kind help. Your script works very well. Would you
please
show me how to change NaN to zero and column variables 1, 2, ..., 8 to
var1,
var2, ..., var8? Thanks again.
Legen
jholtman wrote:
Is this what you want:
x - read.table(textConnection(id code1 code2 p
+ 1 4 8 0.1
+ 1 5 7 0.9
+ 2 1 8 0.4
+ 2 6 2 0.2
+ 2 4 3 0.6
+ 3 5 6 0.7
+ 3 7 5 0.9), header=TRUE)
closeAllConnections()
# create object like output from 'melt'
x.m - data.frame(id=c(x$id, x$id), var=c(x$code1, x$code2),
+ variable=rep('p', 2*nrow(x)), value=c(x$p, x$p))
require(reshape) # use the reshape package
cast(x.m, id ~ var, mean)
id 1 2 3 4 5 6 7 8
1 1 NaN NaN NaN 0.1 0.9 NaN 0.9 0.1
2 2 0.4 0.2 0.6 0.6 NaN 0.2 NaN 0.4
3 3 NaN NaN NaN NaN 0.8 0.7 0.9 NaN
On Tue, Nov 10, 2009 at 4:30 PM, legen lege...@gmail.com wrote:
Dear all,
I have a dataset as below:
id code1 code2 p
1 4 8 0.1
1 5 7 0.9
2 1 8 0.4
2 6 2 0.2
2 4 3 0.6
3 5 6 0.7
3 7 5 0.9
I just want to rewrite it as this (vertical to horizontal):
id var1 var2 var3 var4 var5 var6 var7 var8
1 0 0 0 0.1 0.9 0 0.9 0.1
2 0.4 0.2 0.6 0.6 0 0.2 0 0.4
3 0 0 0 0 0.8 0.7 0.9 0
For the third subject, there are two values being equal to 5 in code1
and
code2, but different values in p: 0.7 and 0.9, so I assigned their
average
0.8 in var5.
Does anybody can help me to handle this? Many thanks for your
consideration
and time.
Legen
--
View this message in context:
http://old.nabble.com/Data-transformation-tp26291568p26291568.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.
--
View this message in context:
http://old.nabble.com/Data-transformation-tp26291568p26295766.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.
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
View this message in context:
http://old.nabble.com/Data-transformation-tp26291568p26301029.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] LINEAR MIXED EFFECT
CAN ANYONE PLEASE HELP ME WITH THIS i HAVE TO DO A MIXED EFFECT LINEAR MODEL WITH MY DATA DUE TO THE FACT THAT I have pseudoreplication! Although after reading and trying it for several times can get around due to Error in na.fail.default(list(date = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, : missing values in object I uploaded my data file Thank you so much Kind regards AG http://old.nabble.com/file/p26300394/rawoctobercalciumexperiment2.txt rawoctobercalciumexperiment2.txt -- View this message in context: http://old.nabble.com/LINEAR-MIXED-EFFECT-tp26300394p26300394.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] Data transformation
That's what I want. Many thanks for your help.
Legen
jholtman wrote:
Try this:
x - read.table(textConnection(idcode1code2 p
+ 148 0.1
+ 157 0.9
+ 218 0.4
+ 262 0.2
+ 243 0.6
+ 356 0.7
+ 375 0.9), header=TRUE)
closeAllConnections()
# create object like output from 'melt'
x.m - data.frame(id=c(x$id, x$id),
+var=paste('var', c(x$code1, x$code2), sep=''),
+variable=rep('p', 2*nrow(x)),
+value=c(x$p, x$p))
require(reshape) # use the reshape package
(x.n - cast(x.m, id ~ var, function(.dat){
+ if (length(.dat) == 0) return(0) # test for no data; return
zero if that is the case
+ mean(.dat)
+ }))
id var1 var2 var3 var4 var5 var6 var7 var8
1 1 0.0 0.0 0.0 0.1 0.9 0.0 0.9 0.1
2 2 0.4 0.2 0.6 0.6 0.0 0.2 0.0 0.4
3 3 0.0 0.0 0.0 0.0 0.8 0.7 0.9 0.0
On Tue, Nov 10, 2009 at 11:10 PM, legen lege...@gmail.com wrote:
Thank you for your kind help. Your script works very well. Would you
please
show me how to change NaN to zero and column variables 1, 2, ..., 8 to
var1,
var2, ..., var8? Thanks again.
Legen
jholtman wrote:
Is this what you want:
x - read.table(textConnection(id code1 code2 p
+ 1 4 8 0.1
+ 1 5 7 0.9
+ 2 1 8 0.4
+ 2 6 2 0.2
+ 2 4 3 0.6
+ 3 5 6 0.7
+ 3 7 5 0.9), header=TRUE)
closeAllConnections()
# create object like output from 'melt'
x.m - data.frame(id=c(x$id, x$id), var=c(x$code1, x$code2),
+ variable=rep('p', 2*nrow(x)), value=c(x$p, x$p))
require(reshape) # use the reshape package
cast(x.m, id ~ var, mean)
id 1 2 3 4 5 6 7 8
1 1 NaN NaN NaN 0.1 0.9 NaN 0.9 0.1
2 2 0.4 0.2 0.6 0.6 NaN 0.2 NaN 0.4
3 3 NaN NaN NaN NaN 0.8 0.7 0.9 NaN
On Tue, Nov 10, 2009 at 4:30 PM, legen lege...@gmail.com wrote:
Dear all,
I have a dataset as below:
id code1 code2 p
1 4 8 0.1
1 5 7 0.9
2 1 8 0.4
2 6 2 0.2
2 4 3 0.6
3 5 6 0.7
3 7 5 0.9
I just want to rewrite it as this (vertical to horizontal):
id var1 var2 var3 var4 var5 var6 var7 var8
1 0 0 0 0.1 0.9 0 0.9 0.1
2 0.4 0.2 0.6 0.6 0 0.2 0 0.4
3 0 0 0 0 0.8 0.7 0.9 0
For the third subject, there are two values being equal to 5 in code1
and
code2, but different values in p: 0.7 and 0.9, so I assigned their
average
0.8 in var5.
Does anybody can help me to handle this? Many thanks for your
consideration
and time.
Legen
--
View this message in context:
http://old.nabble.com/Data-transformation-tp26291568p26291568.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.
--
View this message in context:
http://old.nabble.com/Data-transformation-tp26291568p26295766.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.
--
View this message in context:
http://old.nabble.com/Data-transformation-tp26291568p26300980.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org
Re: [R] Error in lm() function
Hi Bogaso, Try this vecnames-names(test[,2:11]) fmla - as.formula(paste(test[,1] ~ , paste(vecnames, collapse= +))) res-lm(fmla) Regards M Bogaso a écrit : Hi all, I wanted to have a seasonality study like whether a particular month has significant effect as compared to others. Here is my data : 0.10499 0 0 1 0 0 0 0 0 0 0 0 0.00259 0 0 0 1 0 0 0 0 0 0 0 -0.060150 0 0 0 1 0 0 0 0 0 0 0.10721 0 0 0 0 0 1 0 0 0 0 0 0.03597 0 0 0 0 0 0 1 0 0 0 0 0.10584 0 0 0 0 0 0 0 1 0 0 0 0.02063 0 0 0 0 0 0 0 0 1 0 0 -0.035090 0 0 0 0 0 0 0 0 1 0 -0.034850 0 0 0 0 0 0 0 0 0 1 0.01632 0 0 0 0 0 0 0 0 0 0 0 0.06844 1 0 0 0 0 0 0 0 0 0 0 -0.017660 1 0 0 0 0 0 0 0 0 0 0.00989 0 0 1 0 0 0 0 0 0 0 0 0.11673 0 0 0 1 0 0 0 0 0 0 0 0.01789 0 0 0 0 1 0 0 0 0 0 0 -0.003230 0 0 0 0 1 0 0 0 0 0 0.06811 0 0 0 0 0 0 1 0 0 0 0 -0.012920 0 0 0 0 0 0 1 0 0 0 -0.122440 0 0 0 0 0 0 0 1 0 0 -0.066450 0 0 0 0 0 0 0 0 1 0 -0.033550 0 0 0 0 0 0 0 0 0 1 0.02308 0 0 0 0 0 0 0 0 0 0 0 -0.117111 0 0 0 0 0 0 0 0 0 0 0.06116 0 1 0 0 0 0 0 0 0 0 0 0.02832 0 0 1 0 0 0 0 0 0 0 0 0.01441 0 0 0 1 0 0 0 0 0 0 0 -0.044120 0 0 0 1 0 0 0 0 0 0 0.05558 0 0 0 0 0 1 0 0 0 0 0 0.08363 0 0 0 0 0 0 1 0 0 0 0 -0.010630 0 0 0 0 0 0 1 0 0 0 i.e. all explanatory variables here are dichotomous. However once I run lm() i got following error : lm(dat[,1]~dat[,-1]) Error in model.frame.default(formula = dat[, 1] ~ dat[, -1], drop.unused.levels = TRUE) : invalid type (list) for variable 'dat[, -1]' Can anyone please tell me what to do? Best, -- Mohamed Lajnef INSERM Unité 955. 40 rue de Mesly. 94000 Créteil. Courriel : mohamed.laj...@inserm.fr tel.: 01 49 81 31 31 (poste 18470) Sec : 01 49 81 32 90 fax : 01 49 81 30 99 Portable:06 15 60 01 62 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Passing MULTIPLE arguments from php file to r scripts
Hi friends, Again i don't know how to pass multiple arguments from php file to r scripts. Please have a look at this link ; it gives a very simple explanation of passing variables from a PHP page to r scipts.Because i have done the same thing. http://www.math.ncu.edu.tw/~chenwc/R_note/index.php?item=phpsubitem=ex_1http://www.math.ncu.edu.tw/%7Echenwc/R_note/index.php?item=phpsubitem=ex_1 *CONTENTS OF MY PHP FILE: ex_1.php* ?php** $cmd = echo 'argv1 - \ex_1.R\; source(argv1)' | . /usr/bin/R --vanilla --slave ; $ret = system($cmd); ? Assigned the file name to argv1 and passed argv1 to source() function for running r scripts *CONTENTS OF MY R FILE: ex_1.r* print(hello) print(argv1) When I open the php file in the browser I get these results:--- [1] hello [1] ex_1.R See here the php file could invoked the desired r script as well as pass the argument argv1 from the php file and made it available to the r script Till passing of one argument everything was fine. *Then I changed the php file ex_1.php:---* ?php** $a=1; $cmd = echo 'argv1 - \ex_1.R\; source(argv1)' | . /usr/bin/R --vanilla --slave ; //print($cmd); $ret = system($cmd); ? *Then I changed the r script ex_1.r:---* print(hello) print(argv1) print(argv2) When I ran the php file neither the php file could invoke the r script nor the arguments got printed(none of the arghuments got printed,not even argv1 ) So with this change in $cmd :-- $cmd = echo 'argv1 - \ex_1.R\; source(argv1)' | .echo 'argv2 -.$a.; source(argv1)' | ./usr/bin/R --vanilla --slave; Nothing worked. Again when I changed $cmd to this :-- $cmd = echo 'argv1 - \ex_1.R\; source(argv1)' | .echo 'argv2 -.$a.; source(argv1)' | ./usr/bin/R --vanilla --slave--args; The r script could be invoked and bothe the arguments were available but with warnings:- WARNING: unknown option '--slave--args' R version 2.7.1 (2008-06-23) Copyright (C) 2008 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. argv2 -1; source(argv1) *How should I pass multiple arguments from a php page to invoke a r scripts as well as make the arguments passed available to the r script?* Thanks Moumita [[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] Error: cannot allocate vector of size...
For me with ff - on a 3 GB notebook - 3e6x100 works out of the box even without
compression: doubles consume 2.2 GB on disk, but the R process remains under
100MB, rest of RAM used by file-system-cache.
If you are under windows, you can create the ffdf files in a compressed folder.
For the random doubles this reduces size on disk to 230MB - which should even
work on a 1GB notebook.
BTW: the most compressed datatype (vmode) that can handle NAs is logical:
consumes 2bit per tri-bool. The nextmost compressed is byte covering c(NA,
-127:127) and consuming its name on disk and in fs-cache.
The code below should give an idea of how to do pairwise stats on columns where
each pair fits easily into RAM. In the real world, you would not create the
data but import it using read.csv.ffdf (expect that reading your file takes
longer than reading/writing the ffdf).
Regards
Jens Oehlschlägel
library(ff)
k - 100
n - 3e6
# creating a ffdf dataframe of the requires size
l - vector(list, k)
for (i in 1:k)
l[[i]] - ff(vmode=double, length=n, update=FALSE)
names(l) - paste(c, 1:k, sep=)
d - do.call(ffdf, l)
# writing 100 columns of 1e6 random data takes 90 sec
system.time(
for (i in 1:k){
cat(i, )
print(system.time(d[,i] - rnorm(n))[elapsed])
}
)[elapsed]
m - matrix(as.double(NA), k, k)
# pairwise correlating one column against all others takes ~ 17.5 sec
# pairwise correlating all combinations takes 15 min
system.time(
for (i in 2:k){
cat(i, )
print(system.time({
x - d[[i]][]
for (j in 1:(i-1)){
m[i,j] - cor(x, d[[j]][])
}
})[elapsed])
}
)[elapsed]
--
GRATIS für alle GMX-Mitglieder: Die maxdome Movie-FLAT!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: cannot allocate vector of size 3.4 Gb
Hi Peng,
the major problem about your specific case is that when creating the
final object, we need to set dimnames() appropriately. This triggers a
copy of the object and that's where you get the error you describe.
With the current release, unfortunately, there isn't much to do
(unless you're willing to add more memory). For the next release, I
have plans of addressing issues like this and reduce memory footprint
when processing data with the oligo package. It's nothing that can be
done right now, as it takes some time, but I expect everything to be
ready for the next release.
If you're trying to run RMA on your data, I can think of ways of
working around this problem.
Cheers,
b
On Nov 7, 2009, at 5:46 PM, Benilton Carvalho wrote:
ok, i'll take a look at this and get back to you during the week. b
On Nov 7, 2009, at 1:19 PM, Peng Yu wrote:
Most of the 8GB was available, when I run the code, because R was the
only computation session running.
On Sat, Nov 7, 2009 at 7:51 AM, Benilton Carvalho
bcarv...@jhsph.edu wrote:
you haven't answered how much resource you have available when you
try
reading in the data.
with the mouse exon chip, the math is the same i mentioned before.
having 8 GB, you should be able to read in 70 samples of this chip.
if you
can't, that's because you don't have enough resources when trying
to read.
best,
b
On Nov 7, 2009, at 10:12 AM, Peng Yu wrote:
On Fri, Nov 6, 2009 at 8:19 PM, Benilton Carvalho bcarv...@jhsph.edu
wrote:
this is converging to bioc.
let me know what your sessionInfo() is and what type of CEL files
you're
trying to read, additionally provide exactly how you reproduce the
problem.
Here is my sessionInfo(). pname is 'moex10stv1cdf'.
for (f in list.celfiles('.',full.names=T,recursive=T)) {
+ print(f)
+ pname=cleancdfname(whatcdf(f))
+ print(pname)
+ }
sessionInfo()
R version 2.9.2 (2009-08-24)
x86_64-unknown-linux-gnu
locale:
LC_CTYPE
=
en_US
.UTF
-8
;LC_NUMERIC
=
C
;LC_TIME
=
en_US
.UTF
-8
;LC_COLLATE
=
en_US
.UTF
-8
;LC_MONETARY
=
C
;LC_MESSAGES
=
en_US
.UTF
-8
;LC_PAPER
=
en_US
.UTF
-8
;LC_NAME
=
C
;LC_ADDRESS
=C;LC_TELEPHONE=C;LC_MEASUREMENT=en_US.UTF-8;LC_IDENTIFICATION=C
attached base packages:
[1] stats graphics grDevices utils datasets methods
base
other attached packages:
[1] pd.moex.1.0.st.v1_2.4.1 RSQLite_0.7-2 DBI_0.2-4
[4] oligo_1.8.3 preprocessCore_1.6.0
oligoClasses_1.6.0
[7] Biobase_2.4.1
loaded via a namespace (and not attached):
[1] affxparser_1.16.0 affyio_1.12.0 Biostrings_2.12.9
IRanges_1.2.3
[5] splines_2.9.2
it appears to me, i'm not sure, that you start a fresh session of
R and
then
tries to read in the data - how much resource do you have
available when
you
try reading in the data? having 8GB RAM does not mean that you
have 8GB
when
you tried the task.
b
On Nov 7, 2009, at 12:08 AM, Peng Yu wrote:
On Fri, Nov 6, 2009 at 5:00 PM, Marc Schwartz marc_schwa...@me.com
wrote:
On Nov 6, 2009, at 4:19 PM, Peng Yu wrote:
On Fri, Nov 6, 2009 at 3:39 PM, Charlie Sharpsteen
ch...@sharpsteen.net
wrote:
On Fri, Nov 6, 2009 at 1:30 PM, Peng Yu pengyu...@gmail.com
wrote:
I run R on a linux machine that has 8GB memory. But R gives
me an
error Error: cannot allocate vector of size 3.4 Gb. I'm
wondering
why it can not allocate 3.4 Gb on a 8GB memory machine. How
to fix
the
problem?
Is it 32-bit R or 64-bit R?
Are you running any other programs besides R?
How far into your data processing does the error occur?
The more statements you execute, the more fragmented R's
available
memory pool becomes. A 3.4 Gb chunk may no longer be
available.
I'm pretty sure it is 64-bit R. But I need to double check.
What
command I should use to check?
It seems that it didn't do anything but just read a lot of
files
before it showed up the above errors.
Check the output of:
.Machine$sizeof.pointer
If it is 4, R was built as 32 bit, if it is 8, R was built as
64 bit.
See
?.Machine for more information.
It is 8. The code that give the error is listed below. There are
70
celfiles. I'm wondering how to investigate what cause the
problem and
fix it.
library(oligo)
cel_files = list.celfiles('.', full.names=T,recursive=T)
data=read.celfiles(cel_files)
You can also check:
R.version$arch
and
.Platform$r_arch
which for 64 bit should show x86_64.
HTH,
Marc Schwartz
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible
code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Loadings and scores from fastICA?
Hi all,
Does anyone know how to get the independent components and loadings from an
Independent Component Analysis (ICA), as well as principal components and
loadings from a Pricipal Component analysis (PCA) using the fastICA package? Or
perhaps if there's another way to do ICAs in R?
Below is an example from the fastICA manual
(http://cran.r-project.org/web/packages/fastICA/fastICA.pdf)
if(require(MASS))
{
x - mvrnorm(n = 1000, mu = c(0, 0), Sigma = matrix(c(10, 3, 3, 1), 2, 2))
x1 - mvrnorm(n = 1000, mu = c(-1, 2), Sigma = matrix(c(10, 3, 3, 1), 2,
2))
X - rbind(x, x1)
a - fastICA(X, 2, alg.typ = deflation, fun = logcosh, alpha = 1,
method = R, row.norm = FALSE, maxit = 200, tol = 0.0001, verbose = TRUE)
par(mfrow = c(1, 3))
plot(a$X, main = Pre-processed data)
plot(a$X%*%a$K, main = PCA components)
plot(a$S, main = ICA components)
}
Best regards,
Joel
_
Hitta kärleken i vinter!
http://dejting.se.msn.com/channel/index.aspx?trackingid=1002952
[[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] Error in lm() function
On Nov 11, 2009, at 7:14 AM, Bogaso wrote: Hi all, I wanted to have a seasonality study like whether a particular month has significant effect as compared to others. Here is my data : 0.10499 0 0 1 0 0 0 0 0 0 0 0 0.00259 0 0 0 1 0 0 0 0 0 0 0 -0.060150 0 0 0 1 0 0 0 0 0 0 0.10721 0 0 0 0 0 1 0 0 0 0 0 0.03597 0 0 0 0 0 0 1 0 0 0 0 0.10584 0 0 0 0 0 0 0 1 0 0 0 0.02063 0 0 0 0 0 0 0 0 1 0 0 -0.035090 0 0 0 0 0 0 0 0 1 0 -0.034850 0 0 0 0 0 0 0 0 0 1 0.01632 0 0 0 0 0 0 0 0 0 0 0 0.06844 1 0 0 0 0 0 0 0 0 0 0 -0.017660 1 0 0 0 0 0 0 0 0 0 0.00989 0 0 1 0 0 0 0 0 0 0 0 0.11673 0 0 0 1 0 0 0 0 0 0 0 0.01789 0 0 0 0 1 0 0 0 0 0 0 -0.003230 0 0 0 0 1 0 0 0 0 0 0.06811 0 0 0 0 0 0 1 0 0 0 0 -0.012920 0 0 0 0 0 0 1 0 0 0 -0.122440 0 0 0 0 0 0 0 1 0 0 -0.066450 0 0 0 0 0 0 0 0 1 0 -0.033550 0 0 0 0 0 0 0 0 0 1 0.02308 0 0 0 0 0 0 0 0 0 0 0 -0.117111 0 0 0 0 0 0 0 0 0 0 0.06116 0 1 0 0 0 0 0 0 0 0 0 0.02832 0 0 1 0 0 0 0 0 0 0 0 0.01441 0 0 0 1 0 0 0 0 0 0 0 -0.044120 0 0 0 1 0 0 0 0 0 0 0.05558 0 0 0 0 0 1 0 0 0 0 0 0.08363 0 0 0 0 0 0 1 0 0 0 0 -0.010630 0 0 0 0 0 0 1 0 0 0 i.e. all explanatory variables here are dichotomous. However once I run lm() i got following error : lm( Error in model.frame.default(formula = dat[, 1] ~ dat[, -1], drop.unused.levels = TRUE) : invalid type (list) for variable 'dat[, -1]' Assuming that dat really is a data.frame, then the error message is saying you have attempted to pass most of it (since it is a list) to the rhs of the formula and the lm procedure was expecting a vector or vectors. What do you get with: lm( dat[,1]~. , data = dat) (It would be preferable for readability and possibly even for code correctness to refer to the column name for data[,1].) Can anyone please tell me what to do? Better to post the output of str(dat) , or even better, the output of dput(dat). Best, -- 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] dividing a matrix by positive sum or negative sum depending on the sign
Hi,
I have a matrix with positive numbers, negative numbers, and NAs. An
example of the matrix is as follows
-1 -1 2 NA
3 3 -2 -1
1 1 NA -2
I need to compute a scaled version of this matrix. The scaling method is
dividing each positive numbers in each row by the sum of positive numbers
in that row and dividing each negative numbers in each row by the sum of
absolute value of negative numbers in that row.
So the resulting matrix would be
-1/2 -1/2 2/2 NA
3/6 3/6 -2/3 -1/3
1/2 1/2 NA -2/2
Is there an efficient way to do that in R? One way I am using is
1. rowSums for positive numbers in the matrix
2. rowSums for negative numbers in the matrix
3. sweep(mat, 1, posSumVec, posDivFun)
4. sweep(mat, 1, negSumVec, negDivFun)
posDivFun = function(x,y) {
xPosId = x0 !is.na(x)
x[xPosId] = x[xPosId]/y[xPosId]
return(x)
}
negDivFun = function(x,y) {
xNegId = x0 !is.na(x)
x[xNegId] = -x[xNegId]/y[xNegId]
return(x)
}
It is not fast enough though. This scaling is to be applied to large data
sets repetitively. I would like to make it as fast as possible. Any
thoughts on improving it would be appreciated.
Thanks
Jeff
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] R installer can't find math.h, stdio.h
This is possibly the wrong list, but anyway ... Using Linux Debian-4.0 Etch (regularly upgraded). I set about installing Simon Woods' soap package -- see: http://www.maths.bath.ac.uk/~sw283/simon/software.html So I downloaded the tar archive soap_0.1-3.tar.gz and then ran (as root) R CMD INSTALL as follows (and with the following errors): # R CMD INSTALL /home/ted/Downloads/soap_0.1-3.tar.gz * installing to library ?/usr/local/lib/R/site-library? * installing *source* package ?soap? ... ** libs gcc -std=gnu99 -I/usr/share/R/include -fpic -g -O2 -c soap.c -o soap.o soap.c:6:18: error: math.h: No such file or directory soap.c:7:19: error: stdio.h: No such file or directory [the remaining output being assorted warnings] Now, I do have math.h and stdio.h: /usr/i486-linuxlibc1/include/math.h /usr/i486-linuxlibc1/include/stdio.h which may be a non-standard place (usually, in the past at any rate, in /usr/include). The soap bundle has one file to be compiled: soap/src/soap.c I have tried to pick my way around what happens when you call R CMD INSTALL, in order to find where the -I/usr/share/R/include comes from, so as to enable it to locate math.h and stdio.h, but I got lost! Any advice welcome! With thanks, Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 11-Nov-09 Time: 14:42:21 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] update.packages()
Dear all, W.XP I recently upgraded to R2.10.0 and did what I usually do: copied all special libraries from the old installation and then used update.packages() on the command line. Nothing happened. Then I clicked on the same command on the droplist which generated update.packages(ask='graphics'), but still with no effect. When I tried to load the gregmisc bundle I got error messages telling me to upgrade the individual packages (like gmodels and gplot). What has happened and why? I really liked the automatic scan of my installation followed by semi automatic installation that used to get triggered by update.packages() Thanks /CG -- CG Pettersson, Ph.D. Swedish University of Agricultural Sciences (SLU) Dept. of Crop Production Ecology. Box 7043. SE-750 07 Uppsala, Sweden cg.petters...@vpe.slu.se __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unexpected behavior for as.date()
The date library was written 20 or so years ago. It was a very good first effort, but the newer Date library has superior functionality in nearly every way. The date library is still available, for legacy projects such as yours, but I do not advise it for new work. To answer your specific questions: 1. What you have is a real bug. The underlying C routine that scans through the text returns 0 as a marker for any string it can't figure out, a year of 'abc' or month 'charlie' for example. The S function then turns these into NA. I never, ever thought about year 0. In our longer term studies at Mayo we have birth dates in the 1800s, it is rather surprising that a birth date of 1900 hadn't caught me sometime in the past. I'll fix this. 2. The date library predates the strptime function by over 10 years. It is not a huge surprise that I neglected to include support for it -- my oracular abilities are limited. For an inherited project such as this I would suggest reading the date() documentation as a first step; it is not very long since the package is simple. You want the date.mdy function, which is more straightforward than strptime (with less capabilities of course). Terry Therneau [[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] End of Month date capture
Dear R-users, I have the following zoo object: x1x2 x3 x4 x5 x6 1998-08-31 -0.0704375904 NA NA NA NA NA 1998-09-01 0.0379028122 NA NA NA NA 0.00609139 1998-09-02 -0.0038191639 NA NA NA NA NA 1998-09-03 -0.0083235389 NA NA NA NA NA 1998-09-04 -0.0085576782 NA NA 0.0028570541 NA NA 1998-09-07 0.00 NA NA NA NA NA 1998-09-08 0.0496459618 NA NA NA NA NA 1998-09-09 -0.0170081847 NA NA NA NA NA 1998-09-10 -0.0261897076 NA NA NA NA NA 1998-09-11 0.0290280530 NA NA NA NA NA 1998-09-14 0.0202677162 NA NA NA NA NA 1998-09-15 0.0077005314 NA NA NA NA NA 1998-09-16 0.0074886581 NA NA NA -0.002710978 NA 1998-09-17 -0.0257819401 NA NA NA NA NA 1998-09-18 0.0011966887 NA NA NA NA NA 1998-09-21 0.0037182403 NA NA NA NA NA 1998-09-22 0.0055904154 NA NA NA NA NA 1998-09-23 0.0347982355 NA NA NA NA NA 1998-09-24 -0.0221650663 NA NA NA NA NA 1998-09-25 0.0019449387 NA 0.007833611 NA NA NA 1998-09-28 0.0037641439 NA NA NA NA NA 1998-09-29 0.0003146288 NA NA NA NA NA 1998-09-30 -0.0309894451 NA NA NA NA NA 1998-10-01 -0.0305704149 NA NA NA NA 0. 1998-10-02 0.0163000909 -0.03409975 NA 0.0004991463 NA NA 1998-10-05 -0.0141025660 NA NA NA NA NA 1998-10-06 -0.0040240279 NA NA NA NA NA 1998-10-07 -0.0142284540 NA NA NA NA NA 1998-10-08 -0.0116470759 NA NA NA NA NA 1998-10-09 0.0256723791 NA NA NA NA NA 1998-10-12 0.0134404929 NA NA NA NA NA 1998-10-13 -0.0029209410 NA NA NA NA NA 1998-10-14 0.0107283327 NA NA NA NA NA 1998-10-15 0.0408820605 NA NA NA NA NA 1998-10-16 0.0084890073 NA NA NA 0.006675831 NA 1998-10-19 0.0056352536 NA NA NA NA NA 1998-10-20 0.0014485122 NA NA NA NA NA 1998-10-21 0.0056142800 NA NA NA NA NA 1998-10-22 0.0079687631 NA NA NA NA NA 1998-10-23 -0.0072680217 NA NA NA NA NA I need to create second one which contains the end of each month rows only. Is there an easy way of doing this? The data starts at 1998-08-31 and ends at 2009-11-06. Many thanks in advance, Costas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Introducing R to statisticians
Dear all, I will present R language and R software environment to the Statistical Society of Serbia. As I will doing it to professional statisticians it seems unneccesary to me to present them how R language works in details. I am more interested to present them with the latest facts regarding R (approximately number of users, number of add-on packages etc.) and in general why they should start using it. I would be grateful if anyone could point to me a place where I can find the information. Also, any comparisons with other stats packages and languages would be much appreciated. Thanks in advance. With kind regards DK _ Use Hotmail to send and receive mail from your different email accounts. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] dividing a matrix by positive sum or negative sum depending on the sign
one approach is the following:
mat - rbind(c(-1, -1, 2, NA), c(3, 3, -2, -1), c(1, 1, NA, -2))
mat / ave(abs(mat), row(mat), sign(mat), FUN = sum)
I hope it helps.
Best,
Dimitris
Hao Cen wrote:
Hi,
I have a matrix with positive numbers, negative numbers, and NAs. An
example of the matrix is as follows
-1 -1 2 NA
3 3 -2 -1
1 1 NA -2
I need to compute a scaled version of this matrix. The scaling method is
dividing each positive numbers in each row by the sum of positive numbers
in that row and dividing each negative numbers in each row by the sum of
absolute value of negative numbers in that row.
So the resulting matrix would be
-1/2 -1/2 2/2 NA
3/6 3/6 -2/3 -1/3
1/2 1/2 NA -2/2
Is there an efficient way to do that in R? One way I am using is
1. rowSums for positive numbers in the matrix
2. rowSums for negative numbers in the matrix
3. sweep(mat, 1, posSumVec, posDivFun)
4. sweep(mat, 1, negSumVec, negDivFun)
posDivFun = function(x,y) {
xPosId = x0 !is.na(x)
x[xPosId] = x[xPosId]/y[xPosId]
return(x)
}
negDivFun = function(x,y) {
xNegId = x0 !is.na(x)
x[xNegId] = -x[xNegId]/y[xNegId]
return(x)
}
It is not fast enough though. This scaling is to be applied to large data
sets repetitively. I would like to make it as fast as possible. Any
thoughts on improving it would be appreciated.
Thanks
Jeff
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] architecture of statistical programming language R
mustafa_binar wrote: hi, I'm mustafa. I'm a master student Dokuz Eylül University in Izmir/Turkey. nbsp; I focus on Statistical programming language R. I learn architecture of R. But i don't find out any material in internet. Would you like tonbsp;help me this subject? If you have documentations, would you like to share with me? nbsp; If you help me, i have informationsnbsp;about architecture of R and i'mnbsp;happy. We'd like to make you happy, but this is much too vague. There's quite a bit of documentation on the main R project site http://www.r-project.org, including all of the R manuals, lists of books, etc.. If you have more specific questions, someone on this list will probably try to help. -- View this message in context: http://old.nabble.com/architecture-of-statistical-programming-language-R-tp26301393p26302999.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] Introducing R to statisticians
On 11/11/2009 10:26 AM, Damjan Krstajic wrote: Dear all, I will present R language and R software environment to the Statistical Society of Serbia. As I will doing it to professional statisticians it seems unneccesary to me to present them how R language works in details. I am more interested to present them with the latest facts regarding R (approximately number of users, number of add-on packages etc.) and in general why they should start using it. Number of users: I don't know any reliable estimates. If you don't want to make something up, just do a Google search, and pick someone else's estimate. For example, this article (recently cited by David Smith on his blog) http://www.information-management.com/blogs/business_intelligence_bi_statistics-10016491-1.html says an estimated 2M worldwide users. Number of add-on packages: This is easier. The big repositories are CRAN and Bioconductor, with other smaller and private ones not making a big difference to the number of packages available. You can just go to those sites and count. Why should they use it? That depends on what's wrong with what they're using now. If they (and their clients) are satisfied with what they've got, then it's harder to argue that they should switch. If you know their applications you can show how well R does there, but if what they've got now is just as good and the price difference doesn't matter to them, they won't want to switch. Duncan Murdoch I would be grateful if anyone could point to me a place where I can find the information. Also, any comparisons with other stats packages and languages would be much appreciated. Thanks in advance. With kind regards DK _ Use Hotmail to send and receive mail from your different email accounts. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] dividing a matrix by positive sum or negative sum depending on the sign
On Nov 11, 2009, at 10:36 AM, Dimitris Rizopoulos wrote:
one approach is the following:
mat - rbind(c(-1, -1, 2, NA), c(3, 3, -2, -1), c(1, 1, NA, -2))
mat / ave(abs(mat), row(mat), sign(mat), FUN = sum)
Very elegant. My solution was a bit more pedestrian, but may have some
speed advantage:
t( apply(mat, 1, function(x) ifelse( x 0, -x/sum(x[x0], na.rm=T), x/
sum(x[x0], na.rm=T) ) ) )
system.time(replicate(1, t( apply(mat, 1, function(x) ifelse( x
0, -x/sum(x[x0], na.rm=T), x/sum(x[x0], na.rm=T) ) ) ) ) )
user system elapsed
5.958 0.027 5.977
system.time(replicate(1, mat / ave(abs(mat), row(mat),
sign(mat), FUN = sum) ) )
user system elapsed
12.886 0.064 12.886
--
David
I hope it helps.
Best,
Dimitris
Hao Cen wrote:
Hi,
I have a matrix with positive numbers, negative numbers, and NAs. An
example of the matrix is as follows
-1 -1 2 NA
3 3 -2 -1
1 1 NA -2
I need to compute a scaled version of this matrix. The scaling
method is
dividing each positive numbers in each row by the sum of positive
numbers
in that row and dividing each negative numbers in each row by the
sum of
absolute value of negative numbers in that row.
So the resulting matrix would be
-1/2 -1/2 2/2 NA
3/6 3/6 -2/3 -1/3
1/2 1/2 NA -2/2
Is there an efficient way to do that in R? One way I am using is
1. rowSums for positive numbers in the matrix
2. rowSums for negative numbers in the matrix
3. sweep(mat, 1, posSumVec, posDivFun)
4. sweep(mat, 1, negSumVec, negDivFun)
posDivFun = function(x,y) {
xPosId = x0 !is.na(x)
x[xPosId] = x[xPosId]/y[xPosId]
return(x)
}
negDivFun = function(x,y) {
xNegId = x0 !is.na(x)
x[xNegId] = -x[xNegId]/y[xNegId]
return(x)
}
It is not fast enough though. This scaling is to be applied to
large data
sets repetitively. I would like to make it as fast as possible. Any
thoughts on improving it would be appreciated.
Thanks
Jeff
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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.
[R] partial cumsum
Hello, I am searching for a function to calculate partial cumsums. For example it should calculate the cumulative sums until a NA appears, and restart the cumsum calculation after the NA. this: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) should become this: 1 3 6 NA 5 11 18 26 35 45 any ideas? thank you and best regards, stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 get the names of list elements when iterating over a list?
I need to get the names of the list elements when I iterate over a
list. I'm wondering how to do so?
alist=list(a=c(1,3),b=c(-1,3),c=c(-2,1))
sapply(alist,function(x){
#need to use the name of x for some subsequent process
})
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] problems in installing biomart
Dear Forum my machine runs Ubuntu 9.04. I am trying to install Biomart, what I realize is that something is missing and it can't install XML, RCurl but I don't know what more to do, I looked in previous posts but I did not find infoprmation that helped. Thanks Andreia biocLite(biomaRt) Running biocinstall version 2.3.14 with R version 2.8.1 Your version of R requires version 2.3 of Bioconductor. Warning in install.packages(pkgs = pkgs, repos = repos, dependencies = dependencies, : argument 'lib' is missing: using '/home/user/R/x86_64-pc-linux-gnu-library/2.8' also installing the dependencies XML, RCurl trying URL 'http://cran.fhcrc.org/src/contrib/XML_2.6-0.tar.gz' Content type 'application/x-gzip' length 680366 bytes (664 Kb) opened URL == downloaded 664 Kb trying URL 'http://cran.fhcrc.org/src/contrib/RCurl_1.3-0.tar.gz' Content type 'application/x-gzip' length 799689 bytes (780 Kb) opened URL == downloaded 780 Kb trying URL ' http://bioconductor.org/packages/2.3/bioc/src/contrib/biomaRt_1.16.0.tar.gz' Content type 'application/x-gzip' length 278561 bytes (272 Kb) opened URL == downloaded 272 Kb * Installing *source* package 'XML' ... checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ANSI C... none needed checking how to run the C preprocessor... gcc -E checking for sed... /bin/sed checking for pkg-config... /usr/bin/pkg-config checking for xml2-config... no Cannot find xml2-config ERROR: configuration failed for package 'XML' ** Removing '/home/user/R/x86_64-pc-linux-gnu-library/2.8/XML' * Installing *source* package 'RCurl' ... checking for curl-config... no Cannot find curl-config ERROR: configuration failed for package 'RCurl' ** Removing '/home/user/R/x86_64-pc-linux-gnu-library/2.8/RCurl' * Installing *source* package 'biomaRt' ... ** R ** inst ** preparing package for lazy loading Error in loadNamespace(i, c(lib.loc, .libPaths())) : there is no package called 'RCurl' Calls: Anonymous ... code2LazyLoadDB - loadNamespace - namespaceImport - loadNamespace Execution halted ERROR: lazy loading failed for package 'biomaRt' ** Removing '/home/user/R/x86_64-pc-linux-gnu-library/2.8/biomaRt' The downloaded packages are in /tmp/RtmpzdpweV/downloaded_packages Warning messages: 1: In install.packages(pkgs = pkgs, repos = repos, dependencies = dependencies, : installation of package 'XML' had non-zero exit status 2: In install.packages(pkgs = pkgs, repos = repos, dependencies = dependencies, : installation of package 'RCurl' had non-zero exit status 3: In install.packages(pkgs = pkgs, repos = repos, dependencies = dependencies, : installation of package 'biomaRt' had non-zero exit status [[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] problems in installing biomart
Hi Andreia -- Andreia Fonseca wrote: Dear Forum my machine runs Ubuntu 9.04. I am trying to install Biomart, what I realize is that something is missing and it can't install XML, RCurl but I don't know what more to do, I looked in previous posts but I did not find infoprmation that helped. Thanks Andreia biocLite(biomaRt) Running biocinstall version 2.3.14 with R version 2.8.1 Your version of R requires version 2.3 of Bioconductor. Warning in install.packages(pkgs = pkgs, repos = repos, dependencies = dependencies, : argument 'lib' is missing: using '/home/user/R/x86_64-pc-linux-gnu-library/2.8' also installing the dependencies ‘XML’, ‘RCurl’ trying URL 'http://cran.fhcrc.org/src/contrib/XML_2.6-0.tar.gz' Content type 'application/x-gzip' length 680366 bytes (664 Kb) opened URL == downloaded 664 Kb trying URL 'http://cran.fhcrc.org/src/contrib/RCurl_1.3-0.tar.gz' Content type 'application/x-gzip' length 799689 bytes (780 Kb) opened URL == downloaded 780 Kb trying URL ' http://bioconductor.org/packages/2.3/bioc/src/contrib/biomaRt_1.16.0.tar.gz' Content type 'application/x-gzip' length 278561 bytes (272 Kb) opened URL == downloaded 272 Kb * Installing *source* package 'XML' ... checking for gcc... gcc checking for C compiler default output file name... a.out checking whether the C compiler works... yes checking whether we are cross compiling... no checking for suffix of executables... checking for suffix of object files... o checking whether we are using the GNU C compiler... yes checking whether gcc accepts -g... yes checking for gcc option to accept ANSI C... none needed checking how to run the C preprocessor... gcc -E checking for sed... /bin/sed checking for pkg-config... /usr/bin/pkg-config checking for xml2-config... no Cannot find xml2-config ERROR: configuration failed for package 'XML' use your operating system package installation manager to install libxml2. This is a Bioconductor package, so consider asking on the Biocondcutor mailing list. Martin ** Removing '/home/user/R/x86_64-pc-linux-gnu-library/2.8/XML' * Installing *source* package 'RCurl' ... checking for curl-config... no Cannot find curl-config ERROR: configuration failed for package 'RCurl' ** Removing '/home/user/R/x86_64-pc-linux-gnu-library/2.8/RCurl' * Installing *source* package 'biomaRt' ... ** R ** inst ** preparing package for lazy loading Error in loadNamespace(i, c(lib.loc, .libPaths())) : there is no package called 'RCurl' Calls: Anonymous ... code2LazyLoadDB - loadNamespace - namespaceImport - loadNamespace Execution halted ERROR: lazy loading failed for package 'biomaRt' ** Removing '/home/user/R/x86_64-pc-linux-gnu-library/2.8/biomaRt' The downloaded packages are in /tmp/RtmpzdpweV/downloaded_packages Warning messages: 1: In install.packages(pkgs = pkgs, repos = repos, dependencies = dependencies, : installation of package 'XML' had non-zero exit status 2: In install.packages(pkgs = pkgs, repos = repos, dependencies = dependencies, : installation of package 'RCurl' had non-zero exit status 3: In install.packages(pkgs = pkgs, repos = repos, dependencies = dependencies, : installation of package 'biomaRt' had non-zero exit status [[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. -- Martin Morgan Computational Biology / Fred Hutchinson Cancer Research Center 1100 Fairview Ave. N. PO Box 19024 Seattle, WA 98109 Location: Arnold Building M1 B861 Phone: (206) 667-2793 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Introducing R to statisticians
On Wed, 11 Nov 2009 10:51:53 -0500 Duncan Murdoch murd...@stats.uwo.ca wrote: If you know their applications you can show how well R does there, And do mention the (increasing) number of books available. It's only a slight exaggeration to say that there are R books on almost any application you could think of. -- Karl Ove Hufthammer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Implementation of the Shuffled Complex Evolution (SCE-UA) Algorithm
Simon Seibert simon.seibert at mytum.de writes: Good evening list, I'm looking for an R implementation of the Shuffled Complex Evolutionâ (SCE-UA) algorithm after Duan et al. (1993). Does anybody know if there is an extension/ package existing that contains it? Thanks very much for your help! Cheers, Simon I am looking into stochastic global optimization routines, such as variants of genetic algorithms (GA), particle swarm optimization (PSO), and shuffled complex (SCE) or differential evolution (DE). At the moment I am testing a free Matlab implementation of SCE-UA. If that turns out to be interesting and effective, I will translate it into an R function, maybe within the next few weeks -- except, of course, someone else is going to do an implementation of a similar procedure. Right now I am not totally convinced. Maybe you try your problem with other approaches available in R (see e.g. the Optimization task view). Regards Hans Werner Duan QY, Gupta KV, Sorooshian S (1993) Shuffled Complex Evolution Approach for Effective and Efficient Global Minimization. In Jour. of optimization theorie and applications. Vol 76, 3, 501-521 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 get the names of list elements when iterating over alist?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Peng Yu
Sent: Wednesday, November 11, 2009 8:02 AM
To: r-h...@stat.math.ethz.ch
Subject: [R] How to get the names of list elements when
iterating over alist?
I need to get the names of the list elements when I iterate over a
list. I'm wondering how to do so?
alist=list(a=c(1,3),b=c(-1,3),c=c(-2,1))
sapply(alist,function(x){
#need to use the name of x for some subsequent process
})
sapply(X,FUN) and lapply(X,FUN) do the equivalent of
for(i in seq_along(X))
FUN(X[[i]])
X[[i]] represents i'th elemenent of the list X,
not a list containing the i'th element. Hence the name
from the original list X is not part of X[[i]]. If FUN
expects a named list of length 1 then you can do
sapply(seq_along(X), function(i)FUN(X[i]))
or write your own for loop
for(i in seq_along(X))
FUN(X[i])
If FUN wants the name as a separate argument you could do
sapply(seq_along(X), function(i)FUN(X[[i]], names(X)[i]))
or you could write the equivalent loop by hand.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.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] partial cumsum
Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of smu Sent: Wednesday, November 11, 2009 7:58 AM To: r-help@r-project.org Subject: [R] partial cumsum Hello, I am searching for a function to calculate partial cumsums. For example it should calculate the cumulative sums until a NA appears, and restart the cumsum calculation after the NA. this: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) should become this: 1 3 6 NA 5 11 18 26 35 45 Perhaps ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum) [1] 1 3 6 NA 5 11 18 26 35 45 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com any ideas? thank you and best regards, stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Sweave() within a function: objects not found
Dear list subscriber,
suppose, I do have a minimal Sweave file 'test.Rnw':
\documentclass{article}
\begin{document}
printx=
x
@
\end{document}
Within R, I define the following function:
f - function(x){
Sweave(test.Rnw)
}
The call:
f(x = 1:10)
results in the following error message:
f(x = 1:10)
Writing to file test.tex
Processing code chunks ...
1 : echo term verbatim (label=printx)
Error: chunk 1 (label=printx)
Error in eval(expr, envir, enclos) : object 'x' not found
In principle, I could assign x to the global environment and then the Sweave
file will be processed correctly:
f2 - function(x){
+ attach(list(x = x))
+ Sweave(test.Rnw)
+ }
f2(x = 1:10)
Writing to file test.tex
Processing code chunks ...
1 : echo term verbatim (label=printx)
You can now run LaTeX on 'test.tex'
Kind of a dum question, but how could it be achieved that Sweave recognizes the
objects within this function call?
Any pointers are most welcome,
Bernhard
sessionInfo()
R version 2.10.0 (2009-10-26)
i386-pc-mingw32
locale:
[1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252
[3] LC_MONETARY=German_Germany.1252 LC_NUMERIC=C
[5] LC_TIME=German_Germany.1252
attached base packages:
[1] stats graphics datasets utils grDevices methods base
other attached packages:
[1] fortunes_1.3-6
Dr. Bernhard Pfaff
Director
Global Quantitative Equity
Invesco Asset Management Deutschland GmbH
An der Welle 5
D-60322 Frankfurt am Main
Tel: +49 (0)69 29807 230
Fax: +49 (0)69 29807 178
www.institutional.invesco.com
Email: bernhard_pf...@fra.invesco.com
Geschäftsführer: Karl Georg Bayer, Bernhard Langer, Dr. Jens Langewand,
Alexander Lehmann, Christian Puschmann
Handelsregister: Frankfurt am Main, HRB 28469
Sitz der Gesellschaft: Frankfurt am Main
*
Confidentiality Note: The information contained in this ...{{dropped:10}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there a way to specify drop=FALSE as the global default?
On Mon, Oct 19, 2009 at 7:57 PM, Peng Yu pengyu...@gmail.com wrote: tmp - matrix(1:2) tmp tmp[,1,drop=FALSE] See the above example. Is there a way to make 'drop=FALSE' as global default, so that when I say 'tmp[,1]', R will treat it as 'tmp[,1,drop=FALSE]'? Is there a way to set drop=FALSE globally? I can't remember to put drop=FALSE to all the '[]'s. But this indeed causes some weird bugs for me to figure out. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] partial cumsum
On Wed, 11 Nov 2009 08:53:50 -0800 William Dunlap wdun...@tibco.com wrote: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) should become this: 1 3 6 NA 5 11 18 26 35 45 Perhaps ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum) [1] 1 3 6 NA 5 11 18 26 35 45 Clever way of putting the NA values at the end of each group. (I had to study it quite a bit to understand what was going on, and why! :-) ) I wish cumsum took a 'na.rm' argument, though. It would make stuff like this much easier. -- Karl Ove Hufthammer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] dividing a matrix by positive sum or negative sum depending on the sign
On Nov 11, 2009, at 10:57 AM, David Winsemius wrote:
On Nov 11, 2009, at 10:36 AM, Dimitris Rizopoulos wrote:
one approach is the following:
mat - rbind(c(-1, -1, 2, NA), c(3, 3, -2, -1), c(1, 1, NA, -2))
mat / ave(abs(mat), row(mat), sign(mat), FUN = sum)
Very elegant. My solution was a bit more pedestrian, but may have
some speed advantage:
I am wondering if there might be further performance improvements if
sums were pre-calculated before the ifelse scaling step.
Perhaps:
mat - matrix(sample(-4:4, 100, replace=T), ncol=10)
system.time(replicate(1, t(apply(mat, 1, function(x) {negs -
sum(x[x0], na.rm=T); poss - sum(x[x0], na.rm=T); ifelse( x 0, -x/
negs, x/poss)} ) ) ) )
user system elapsed
9.420 0.103 9.619
system.time(replicate(1, t( apply(mat, 1, function(x) ifelse( x
0, -x/sum(x[x0], na.rm=T), x/sum(x[x0], na.rm=T) ) ) ) ) )
user system elapsed
8.206 0.035 8.231
That was only a 15% improvement but I got a 50% improvement by
replacing the ifelse() with its Boolean algebra equivalent:
t( apply(mat, 1, function(x) -x*(x 0)/sum(x[x0], na.rm=T) +
x*(x0)/sum(x[x0], na.rm=T) ) )
[,1] [,2] [,3] [,4]
[1,] -0.5 -0.5 1.000 NA
[2,] 0.5 0.5 -0.667 -0.333
[3,] 0.5 0.5 NA -1.000
system.time(replicate(1, t( apply(mat, 1, function(x) -x*(x
0)/sum(x[x0], na.rm=T) + x*(x0)/sum(x[x0], na.rm=T) ) ) ))
user system elapsed
4.805 0.041 4.839
I could not figure out the Jeff's method of applying the two functions
he presented, so I am unable to compare any of these methods to his
strategy.
--
David.
system.time(replicate(1, t( apply(mat, 1, function(x)
ifelse( x 0, -x/sum(x[x0], na.rm=T), x/sum(x[x0],
na.rm=T) ) ) ) ) )
user system elapsed
5.958 0.027 5.977
system.time(replicate(1, mat / ave(abs(mat), row(mat),
sign(mat), FUN = sum) ) )
user system elapsed
12.886 0.064 12.886
--
David
I hope it helps.
Best,
Dimitris
Hao Cen wrote:
Hi,
I have a matrix with positive numbers, negative numbers, and NAs. An
example of the matrix is as follows
-1 -1 2 NA
3 3 -2 -1
1 1 NA -2
I need to compute a scaled version of this matrix. The scaling
method is
dividing each positive numbers in each row by the sum of positive
numbers
in that row and dividing each negative numbers in each row by the
sum of
absolute value of negative numbers in that row.
So the resulting matrix would be
-1/2 -1/2 2/2 NA
3/6 3/6 -2/3 -1/3
1/2 1/2 NA -2/2
Is there an efficient way to do that in R? One way I am using is
1. rowSums for positive numbers in the matrix
2. rowSums for negative numbers in the matrix
3. sweep(mat, 1, posSumVec, posDivFun)
4. sweep(mat, 1, negSumVec, negDivFun)
posDivFun = function(x,y) {
xPosId = x0 !is.na(x)
x[xPosId] = x[xPosId]/y[xPosId]
return(x)
}
negDivFun = function(x,y) {
xNegId = x0 !is.na(x)
x[xNegId] = -x[xNegId]/y[xNegId]
return(x)
}
It is not fast enough though. This scaling is to be applied to
large data
sets repetitively. I would like to make it as fast as possible. Any
thoughts on improving it would be appreciated.
Thanks
Jeff
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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.
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] how to use # in a rd doc in url address
I am writing a rd doc, and need to use # in a url adress. This would make:
\url{http://www..org/myfolder/#myanchor}
Of course, I suppose this will not work because # is a special character
starting a comment line in the rd dialect. I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know how
to get the thing right ?
Patrick
[[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] partial cumsum
On Wed, 11 Nov 2009 18:11:51 +0100 Karl Ove Hufthammer k...@huftis.org wrote: I wish cumsum took a 'na.rm' argument, though. It would make stuff like this much easier. Or, more specifically, a 'na.value', which could perhaps default to 'NA' to get the current behaviour, but which one could set 0 for 'cumsum', 1 for 'cumprod', Inf for 'cummin' and -Inf for 'cummax'. -- Karl Ove Hufthammer __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 use # in a rd doc in url address
x=\url{http://www..org/myfolder/#myanchor};
print(x,quote=F)
Does this work for you?
Daniel
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Patrick Giraudoux
Gesendet: Wednesday, November 11, 2009 12:15 PM
An: r-help@r-project.org
Betreff: [R] how to use # in a rd doc in url address
I am writing a rd doc, and need to use # in a url adress. This would make:
\url{http://www..org/myfolder/#myanchor}
Of course, I suppose this will not work because # is a special character
starting a comment line in the rd dialect. I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know how to
get the thing right ?
Patrick
[[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] R.oo and S4?
I'm very familiar with C++. In this sense, it is easier for me to learn R.oo according to your advice. On the other hand, S3 and S4 are the most used. I'm wondering what would be the best choice for me. Do you have any recommendation considering the pros and cons of both ways? Is it true that packages based on R.oo can not be deposited to CRAN? On Mon, Oct 26, 2009 at 1:28 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: S3 and S4 are part of the core of R so they would presumably be the most used. S4 is an extension to S3 but adds strong typing and a number of other features. R.oo and proto are packages on CRAN which give access to different models of object oriented programming than S3 and S4. R.oo uses a more conventional model of OO than S3 or S4 that is probably closer to what you are used to if you are coming form another language while proto uses the prototype model (or pure object model). proto tends to apply in user interface applications and there is some info on which other packages make use of proto on the proto home page at: http://r-proto.googlecode.com On Mon, Oct 26, 2009 at 2:47 PM, Peng Yu pengyu...@gmail.com wrote: There are different way to make R classes. I know R.oo and S4. I'm wondering which one is the current popular one. Which one is current recommended when make new R packages? Thank you! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] partial cumsum
On Wed, Nov 11, 2009 at 08:53:50AM -0800, William Dunlap wrote: Perhaps ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum) [1] 1 3 6 NA 5 11 18 26 35 45 it takes some time to understand how it works, but it's perfect. thank you, stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Sweave() within a function: objects not found
On 11/11/2009 12:09 PM, Pfaff, Bernhard Dr. wrote:
Dear list subscriber,
suppose, I do have a minimal Sweave file 'test.Rnw':
\documentclass{article}
\begin{document}
printx=
x
@
\end{document}
Within R, I define the following function:
f - function(x){
Sweave(test.Rnw)
}
The call:
f(x = 1:10)
results in the following error message:
f(x = 1:10)
Writing to file test.tex
Processing code chunks ...
1 : echo term verbatim (label=printx)
Error: chunk 1 (label=printx)
Error in eval(expr, envir, enclos) : object 'x' not found
In principle, I could assign x to the global environment and then the Sweave
file will be processed correctly:
f2 - function(x){
+ attach(list(x = x))
+ Sweave(test.Rnw)
+ }
f2(x = 1:10)
Writing to file test.tex
Processing code chunks ...
1 : echo term verbatim (label=printx)
You can now run LaTeX on 'test.tex'
Kind of a dum question, but how could it be achieved that Sweave recognizes the
objects within this function call?
The way you did it is close. You can attach all the local variables by
using
attach(environment())
though global variables will take precedence, because attach puts the
environment 2nd in the search list. And you'd better remember to detach
them.
I'd say it's better to make Sweave files self-contained, so that you can
run R CMD Sweave outside of R, and get the right results. But if you
really want to do this, then you can write your own Sweave driver and
replace the default RweaveEvalWithOpt with a function that looks
elsewhere for variables.
Duncan Murdoch
Any pointers are most welcome,
Bernhard
sessionInfo()
R version 2.10.0 (2009-10-26)
i386-pc-mingw32
locale:
[1] LC_COLLATE=German_Germany.1252 LC_CTYPE=German_Germany.1252
[3] LC_MONETARY=German_Germany.1252 LC_NUMERIC=C
[5] LC_TIME=German_Germany.1252
attached base packages:
[1] stats graphics datasets utils grDevices methods base
other attached packages:
[1] fortunes_1.3-6
Dr. Bernhard Pfaff
Director
Global Quantitative Equity
Invesco Asset Management Deutschland GmbH
An der Welle 5
D-60322 Frankfurt am Main
Tel: +49 (0)69 29807 230
Fax: +49 (0)69 29807 178
www.institutional.invesco.com
Email: bernhard_pf...@fra.invesco.com
Geschäftsführer: Karl Georg Bayer, Bernhard Langer, Dr. Jens Langewand,
Alexander Lehmann, Christian Puschmann
Handelsregister: Frankfurt am Main, HRB 28469
Sitz der Gesellschaft: Frankfurt am Main
*
Confidentiality Note: The information contained in this ...{{dropped:10}}
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Introducing R to statisticians
To me, R is the language of choice for a rapidly increasing number of people involved in new statistical algorithm development. If they are happy with the tools they currently use, learning R may be a lot of pain for little gain. However, if they want to stay current with the latest developments in almost any area of statistics, I know of no better way than to subscribe to some of the R mailing lists (or related lists like Bioconductor). To me, reading those is like attending a professional meeting a few minutes per day. Hope this helps. Spencer Graves Ove Hufthammer wrote: On Wed, 11 Nov 2009 10:51:53 -0500 Duncan Murdoch murd...@stats.uwo.ca wrote: If you know their applications you can show how well R does there, And do mention the (increasing) number of books available. It's only a slight exaggeration to say that there are R books on almost any application you could think of. -- Spencer Graves, PE, PhD President and Chief Operating Officer Structure Inspection and Monitoring, Inc. 751 Emerson Ct. San José, CA 95126 ph: 408-655-4567 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] partial cumsum
-Original Message- From: smu [mailto:m...@z107.de] Sent: Wednesday, November 11, 2009 9:26 AM To: William Dunlap Cc: r-help@r-project.org Subject: Re: [R] partial cumsum On Wed, Nov 11, 2009 at 08:53:50AM -0800, William Dunlap wrote: Perhaps ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum) [1] 1 3 6 NA 5 11 18 26 35 45 it takes some time to understand how it works, but it's perfect. Note that the 2nd argument assigns a group number based on the number of NA's prior to the current position in the vector. The odd repeated calls to rev() are there to put the NA's at the ends of the groups, instead of at the beginnings: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) rev(cumsum(rev(is.na(x [1] 1 1 1 1 0 0 0 0 0 0 A more natural way to do this is cumsum(is.na(c(NA,x[-length(x)]))) [1] 1 1 1 1 2 2 2 2 2 2 Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com thank you, stefan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 use # in a rd doc in url address
Daniel Malter a écrit :
x=\url{http://www..org/myfolder/#myanchor};
print(x,quote=F)
Does this work for you?
Daniel
I am not working on consol mode (which would make your suggestion
straight applicable), but writing a rd documentationn (the documentation
that comes out with the command ?myfunction). The rd file has a Latex
style syntax and I just want to insert the url within this
documentation. Eg.
\details{
You may want to connect to \url{http://www..org/myfolder/#myanchor}
}
I am not sure one can define a variable and print it in such context...
Best
Patrick
-
cuncta stricte discussurus
-
-Ursprüngliche Nachricht-
Von: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] Im
Auftrag von Patrick Giraudoux
Gesendet: Wednesday, November 11, 2009 12:15 PM
An: r-help@r-project.org
Betreff: [R] how to use # in a rd doc in url address
I am writing a rd doc, and need to use # in a url adress. This would make:
\url{http://www..org/myfolder/#myanchor}
Of course, I suppose this will not work because # is a special character
starting a comment line in the rd dialect. I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know how to
get the thing right ?
Patrick
[[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] Comparison of vectors in a matrix
Yes, thanks for this, this is exactly what I want to do. However, I have a remaining problem which is how to get R to understand that each entry in my matrix is a vector of names. I have been trying to import my text file with the names in each vector of names enclosed in quotes and separated by commas, or separated by spaces, or without quotes, etc, with no luck. Everytime, R seems to consider the vector of names as just one long name. In my first colum, for example, I have henry, in the second, mary, ruth, and in the third mary, joseph, and I have no idea how to get R to see that mary, ruth, for example, is composed of two strings of text, rather than just one. Thanks for any further help! http://old.nabble.com/file/p26305756/ffoexample.txt ffoexample.txt Tony Plate wrote: Nice problem! If I understand you correctly, here's how to do it (with list-based matrices): set.seed(1) (x - matrix(lapply(rpois(10,2)+1, function(k) sample(letters[1:10], size=k)), ncol=2, dimnames=list(1:5,c(A,B A B 1 Character,2 Character,5 2 Character,2 Character,5 3 Character,3 Character,3 4 Character,5 Character,3 5 Character,2 i x[1,1] [[1]] [1] c b x[1,2] [[1]] [1] c d a j f (y - cbind(x, A-B=apply(x, 1, function(ab) setdiff(ab[[1]], ab[[2]] A B A-B 1 Character,2 Character,5 b 2 Character,2 Character,5 g 3 Character,3 Character,3 Character,3 4 Character,5 Character,3 Character,2 5 Character,2 i Character,2 y[1,3] [[1]] [1] b -- Tony Plate esterhazy wrote: Hi, I have a matrix with two columns, and the elements of the matrix are vectors. So for example, in line 3 of column 1 I have a vector v31=(marc, robert, marie). What I need to do is to compare all vectors in column 1 and 2, so as to get, for example setdiff(v31,v32) into a new column. Is there a way to do this in R? Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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://old.nabble.com/Comparison-of-vectors-in-a-matrix-tp26284855p26305756.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] lme4 glmer how to extract the z values?
Ben , Thanks! This worked after I updated my version of R (from 2.8.1 to 2.10) . Best, Spencer On Tue, Nov 10, 2009 at 7:04 PM, Ben Bolker bol...@ufl.edu wrote: sj ssj1364 at gmail.com writes: I am using glmer() from lmer(lme4) to run generalized linear mixed models. I can't figure out how to extract the z values for the fixed effects that are reported using the summary function . Any help would be appreciated. library(lme4) example(glmer) coef(summary(gm1)) coef(summary(gm1))[,z value] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] how to use # in a rd doc in url address
On 11/11/2009 12:15 PM, Patrick Giraudoux wrote:
I am writing a rd doc, and need to use # in a url adress. This would make:
\url{http://www..org/myfolder/#myanchor}
That should work.
Of course, I suppose this will not work because # is a special character
starting a comment line in the rd dialect.
That's not correct. # is only special in R code, and with \url{} the
text is considered as verbatim text, i.e. only \, %, { and } are special.
I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know how
to get the thing right ?
I don't understand this question.
Duncan Murdoch
Patrick
[[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] Comparison of vectors in a matrix
On Nov 11, 2009, at 1:02 PM, esterhazy wrote: Yes, thanks for this, this is exactly what I want to do. However, I have a remaining problem which is how to get R to understand that each entry in my matrix is a vector of names. I have been trying to import my text file with the names in each vector of names enclosed in quotes and separated by commas, or separated by spaces, or without quotes, etc, with no luck. Those are all subsumed under the R whitespace separators that scan and the read.* functions use by default. Everytime, R seems to consider the vector of names as just one long name. In my first colum, for example, I have henry, in the second, mary, ruth, and in the third mary, joseph, and I have no idea how to get R to see that mary, ruth, for example, is composed of two strings of text, rather than just one. Thanks for any further help! http://old.nabble.com/file/p26305756/ffoexample.txt ffoexample.txt I copied and pasted into a textConnection: lines - readLines(textConnection(usernamefriends friendof + 26376 paradisacorbasi zannechaos filmstarusa jelliclecat status jennilee_rose obiwaynekenobi shaycaron jillthepill 26376 witchy1 jumbach oscarwinner slai jenniebennie tirwen estall bront zarchasmpgmr pink_kimono reidzilla miz_anneliese ambelies jrianne henry3730 thufer_hawat mom2bunky venaeli grbenda ikkin56 paradisacorbasi filmstarusa jelliclecat jennilee_rose shaycaron jillthepill 26376 witchy1 jumbach oscarwinner tirwen estall bront zarchasmpgmr pink_kimono reidzilla miz_anneliese ambelies jrianne henry3730 ahahaha_ha thufer_hawat mom2bunky venaeli grbenda donna_stewart gjanyn + aaandy ma monotony porcelian abrokenstarr seraphimsigrist ballena cbaqir _cassyandra darkspree erishkigal pinkdevildances lightning_geek a_life_verbatim frozen_wishes lijago tindernight crashing_angel sabrina_g giggly_teapot smarties_2087 theinimitable_l african_sunset sirenlunaris eagan_bryhtm robomonkeyninja maid_ov_metal aubloomiel countdownish queennola pearl007 princess_macaw wolfie_sara gypsy_jack may_cash morningchorus vanityflair lyingpeacefully squashedfrogs d_e_r_v_i_s_h the_koira lakshmichithra muffinbits amaya_aneko jenikaandzhaodi blood_gypsy aura_oneill icemodeled wannabesnorlax2 ma frida monotony porcelian abrokenstarr seraphimsigrist ballena cbaqir _cassyandra shepardshadows darkspree erishkigal pinkdevildances lightning_geek a_life_verbatim frozen_wishes lijago tindernight crashing_angel sabrina_g giggly_teapot laudanum_tea smarties_2087 theinimitable_l african_sunset sirenlunaris eagan_bryhtm robomonkeyninja aaskie maid_ov_metal aubloomiel countdownish queennola pearl007 princess_macaw wolfie_sara gypsy_jack may_cash morningchorus vanityflair lyingpeacefully squashedfrogs d_e_r_v_i_s_h the_koira lakshmichithra muffinbits amaya_aneko jenikaandzhaodi blood_gypsy aura_oneill icemodeled sixtycents wannabesnorlax2)) #realLines will bring in everything up to an EOL. # and then used scan() on the individual lines: col1 - scan(textConnection(lines[1]), what=character) Read 3 items col1 [1] username friends friendof col2 - scan(textConnection(lines[2]), what=character) Read 58 items col3 - scan(textConnection(lines[3]), what=character) Read 102 items I have not figured out what you and Tony are doing with these collections of character vectors, but this should help with you doing the basic data entry. Tony Plate wrote: Nice problem! If I understand you correctly, here's how to do it (with list-based matrices): set.seed(1) (x - matrix(lapply(rpois(10,2)+1, function(k) sample(letters[1:10], size=k)), ncol=2, dimnames=list(1:5,c(A,B A B 1 Character,2 Character,5 2 Character,2 Character,5 3 Character,3 Character,3 4 Character,5 Character,3 5 Character,2 i x[1,1] [[1]] [1] c b x[1,2] [[1]] [1] c d a j f (y - cbind(x, A-B=apply(x, 1, function(ab) setdiff(ab[[1]], ab[[2]] A B A-B 1 Character,2 Character,5 b 2 Character,2 Character,5 g 3 Character,3 Character,3 Character,3 4 Character,5 Character,3 Character,2 5 Character,2 i Character,2 y[1,3] [[1]] [1] b -- Tony Plate esterhazy wrote: Hi, I have a matrix with two columns, and the elements of the matrix are vectors. So for example, in line 3 of column 1 I have a vector v31=(marc, robert, marie). What I need to do is to compare all vectors in column 1 and 2, so as to get, for example setdiff(v31,v32) into a new column. Is there a way to do this in R? Thanks! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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:
Re: [R] LINEAR MIXED EFFECT
Dear Ana Golveia, It is completelly impossible someone realise what kind or help you need or what is happening. I suggest you give a look on the posting guide, mainly that part about a minimum reproducible code with self explaining information, etc. Cheers milton On Wed, Nov 11, 2009 at 7:22 AM, ANARPCG a.gouvei...@imperial.ac.uk wrote: CAN ANYONE PLEASE HELP ME WITH THIS i HAVE TO DO A MIXED EFFECT LINEAR MODEL WITH MY DATA DUE TO THE FACT THAT I have pseudoreplication! Although after reading and trying it for several times can get around due to Error in na.fail.default(list(date = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, : missing values in object I uploaded my data file Thank you so much Kind regards AG http://old.nabble.com/file/p26300394/rawoctobercalciumexperiment2.txt rawoctobercalciumexperiment2.txt -- View this message in context: http://old.nabble.com/LINEAR-MIXED-EFFECT-tp26300394p26300394.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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to use # in a rd doc in url address
Duncan Murdoch a écrit :
On 11/11/2009 12:15 PM, Patrick Giraudoux wrote:
I am writing a rd doc, and need to use # in a url adress. This
would make:
\url{http://www..org/myfolder/#myanchor}
That should work.
Of course, I suppose this will not work because # is a special
character starting a comment line in the rd dialect.
That's not correct. # is only special in R code, and with \url{} the
text is considered as verbatim text, i.e. only \, %, { and } are special.
I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know
how to get the thing right ?
I don't understand this question.
You answered it above... There is no reason for using special
quotation considering your reminder: with \url{} the text is
considered as verbatim text
Thanks for the focus,
Best,
Patrick
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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 use # in a rd doc in url address
Patrick Giraudoux a écrit :
Duncan Murdoch a écrit :
On 11/11/2009 12:15 PM, Patrick Giraudoux wrote:
I am writing a rd doc, and need to use # in a url adress. This
would make:
\url{http://www..org/myfolder/#myanchor}
That should work.
Of course, I suppose this will not work because # is a special
character starting a comment line in the rd dialect.
That's not correct. # is only special in R code, and with \url{} the
text is considered as verbatim text, i.e. only \, %, { and } are
special.
I did not found a similar
example in Writing R exentions. I am not sure bout using \dQuote{a
quotation}), and use \sQuote and \dQuote correctly. Does anyone know
how to get the thing right ?
I don't understand this question.
You answered it above... There is no reason for using special
quotation considering your reminder: with \url{} the text is
considered as verbatim text
Thanks for the focus,
Best,
Patrick
Yes, can confirmed it works perfect without any complication... Good
lesson. Being used to prepare oneself to the worst, one over-anticipates
it, but occasionally it does not happen
Cheers,
Patrick
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] source() vs attach()0
Duncan Murdoch wrote:
Just declaring it there is the only reasonable way, i.e.
test-function(foo) {
subtest - function() {
foo - foo+1
}
subtest()
return(foo)
}
The reason you can't somehow assign it within an existing test is that
subtest is a different closure every time. Its environment will be
the local frame of the call to test, so the foo on the right hand
side of the assignment will be the value that was passed to test.
An unreasonable way to get what you want is to mess with the
environment, e.g.
subtest - function() {
foo - foo+1
}
test - function(foo) {
mysubtest - subtest # make a local copy
environment(mysubtest) - environment() # attach the local frame
mysubtest()
return(foo)
}
This is ugly programming, likely to bite you at some future date.
Duncan Murdoch
Duncan,
Thank you a lot for this clarification.
Unfortunately, I have to get through this at the moment: I have
subfunctions that are to be used by in the environments of several
'main' functions: and I prefer not to declare them separately each time.
Therefore I will have to mess with the environment (for now) :-(
In this case, would it be wiser to shift the 'environment messing' to
the subfunction?
I.e. tell it to manipulate its respective parent environment, as in the
example below?
subtest - function() {
eval(expression(foo - foo+1),parent.frame())
}
test - function(foo) {
subtest()
return(foo)
}
Stefan
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] as.Date from data.frame
#Hello, #I loaded data using read.table - I needed to convert a row in the data frame to date class: data V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 2 38.16999817 30.7008 36.8661 35.18999863 3 37.4754 29.9576 36.4508 35.3661 4 37.4754 30.1462 36.47000122 35.3661 5 37.84999847 30.5669 36.8415 35.74000168 6 38.3839 31.1462 37.3415 36.2746 7 39.1161 31.9085 38.02999878 36.9754 8 39.81000137 32.65000153 38.6831 37.6339 9 40.47000122 33.34999847 39.27999878 38.2746 data[1,] V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 dates1-as.Date(data[1,]) do not know how to convert 'dates[1,]' to class Date # However, I can individually convert them all: dates1.1-as.Date(data[1,1]) dates1.1 [1] 2008-05-19 # How can I change the class of the date over the entire row (my original file contains more than one hundred rows)? # Thank you very much. -- View this message in context: http://old.nabble.com/as.Date-from-data.frame-tp26305830p26305830.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] Comparison of vectors in a matrix
This is a tricky data entry problem. The right technique will depend on the fine details of the
data, and it's not clear what those are. E.g., when you say In my first column, for example,
I have henry , it's unclear to me whether or not the double quotes are part of
the data or not - which is why it's nice to provide reproducible examples.
But, if you do have quoted strings in your data fields as they exist in an R
matrix, you can do something like the following:
# each element of the matrix x contains one or more quoted strings, separated
by commas
x - matrix(c('a, b', 'c', 'b', 'd'), ncol=2, dimnames=list(c(row1, row2),
c(X,Y)))
x
X Y
row1 \a\, \b\ \b\
row2 \c\\d\
# use R's parsing and evaluation to turn 'a, b' into c(a, b), and turn
that
# into a matrix containing character vectors of various lengths.
matrix(lapply(parse(text=paste(c(, x, ))), eval), ncol=ncol(x),
dimnames=dimnames(x))
X Y
row1 Character,2 b
row2 c d
- Tony Plate
esterhazy wrote:
Yes, thanks for this, this is exactly what I want to do.
However, I have a remaining problem which is how to get R to understand that
each entry in my matrix is a vector of names.
I have been trying to import my text file with the names in each vector of
names enclosed in quotes and separated by commas, or separated by spaces, or
without quotes, etc, with no luck.
Everytime, R seems to consider the vector of names as just one long name.
In my first colum, for example, I have henry, in the second, mary,
ruth, and in the third mary, joseph, and I have no idea how to get R
to see that mary, ruth, for example, is composed of two strings of text,
rather than just one.
Thanks for any further help!
http://old.nabble.com/file/p26305756/ffoexample.txt ffoexample.txt
Tony Plate wrote:
Nice problem!
If I understand you correctly, here's how to do it (with list-based
matrices):
set.seed(1)
(x - matrix(lapply(rpois(10,2)+1, function(k) sample(letters[1:10],
size=k)), ncol=2, dimnames=list(1:5,c(A,B
A B
1 Character,2 Character,5
2 Character,2 Character,5
3 Character,3 Character,3
4 Character,5 Character,3
5 Character,2 i
x[1,1]
[[1]]
[1] c b
x[1,2]
[[1]]
[1] c d a j f
(y - cbind(x, A-B=apply(x, 1, function(ab) setdiff(ab[[1]],
ab[[2]]
A B A-B
1 Character,2 Character,5 b
2 Character,2 Character,5 g
3 Character,3 Character,3 Character,3
4 Character,5 Character,3 Character,2
5 Character,2 i Character,2
y[1,3]
[[1]]
[1] b
-- Tony Plate
esterhazy wrote:
Hi,
I have a matrix with two columns, and the elements of the matrix are
vectors.
So for example, in line 3 of column 1 I have a vector v31=(marc,
robert,
marie).
What I need to do is to compare all vectors in column 1 and 2, so as to
get,
for example setdiff(v31,v32) into a new column.
Is there a way to do this in R?
Thanks!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Joint counts and spatial autocorrelation - binary data
Hi All, I am new to this form and new to R, having just initiated the analysis of my first project using R. I have been working on a logistic model of land use change and am concerned about 1) measuring spatial autocorrelation and 2) including an autocovariate in my model. Here is what I think I need to do. My dependent variable takes on the value of either 1 or 0 depending on whether a particular pixel from a random sample of pixels transitioned from undeveloped to developed during the study’s time period. I’d like to measure spatial autocorrelation by first creating a distance matrix which includes the distances from each of my sample points to every other sample point using latitude and longitude coordinates. I would then like to divide these distances into bins (e.g. 0 to 500m, 501 to 1000m etc.). I have done this using both the dist() function in geoR and the earth.dist() function in fossil. Both seem to work. I would then like to calculate the number of similar states (i.e. joint counts) between pairs of sample points at various distances (i.e. bin size). I can then graph this as proportion of disconcordance in land use change versus distance (i.e. disconcordance is when one point of the pair transitions and the other does not) very similar to that done by McDonald and Urban (2006). I’d expect that as distance increases, the proportion disconcordance eventually reaches that of the entire sample or more correctly, that of a totally random spatial process. How can I assess the joint counts most easily? Are there other approaches folks would recommend? Remember, I am somewhat of a newbie. The last wrinkle is that I am using multi-level models, which with varying intercepts and/or slopes seems to account for some spatial heterogeneity. I am not sure how to think about the pairing of multilevel and autologistic approaches. Any insights? Sample data below LongLat Trans -87.942447.464950 -88.045147.464 0 -82.752442.894771 -86.690545.5972 0 -87.731646.579880 -82.476943.056741 -83.431342.428280 -86.259844.370780 -86.255944.668410 -86.246744.679790 -- View this message in context: http://old.nabble.com/Joint-counts-and-spatial-autocorrelation---binary-data-tp26306008p26306008.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] as.Date from data.frame
Hi, On Nov 11, 2009, at 1:06 PM, separent wrote: #Hello, #I loaded data using read.table - I needed to convert a row in the data frame to date class: data V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 2 38.16999817 30.7008 36.8661 35.18999863 3 37.4754 29.9576 36.4508 35.3661 4 37.4754 30.1462 36.47000122 35.3661 5 37.84999847 30.5669 36.8415 35.74000168 6 38.3839 31.1462 37.3415 36.2746 7 39.1161 31.9085 38.02999878 36.9754 8 39.81000137 32.65000153 38.6831 37.6339 9 40.47000122 33.34999847 39.27999878 38.2746 data[1,] V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 dates1-as.Date(data[1,]) I think this should work if data[1,] returns a vector of characters: R d - c(2008-05-19, 2008-05-30) R dd - as.Date(d) R is(dd) [1] Date oldClass Is your first row being treated as factors? Maybe if you convert to character first it should work? R dates1-as.Date(as.character(data[1,])) But the fact that it works for as.Date(data[1,1]) make it a bit weird ... -steve do not know how to convert 'dates[1,]' to class Date # However, I can individually convert them all: dates1.1-as.Date(data[1,1]) dates1.1 [1] 2008-05-19 # How can I change the class of the date over the entire row (my original file contains more than one hundred rows)? # Thank you very much. -- View this message in context: http://old.nabble.com/as.Date-from-data.frame-tp26305830p26305830.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. -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] as.Date from data.frame
On Nov 11, 2009, at 12:06 PM, separent wrote: #Hello, #I loaded data using read.table - I needed to convert a row in the data frame to date class: data V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 2 38.16999817 30.7008 36.8661 35.18999863 3 37.4754 29.9576 36.4508 35.3661 4 37.4754 30.1462 36.47000122 35.3661 5 37.84999847 30.5669 36.8415 35.74000168 6 38.3839 31.1462 37.3415 36.2746 7 39.1161 31.9085 38.02999878 36.9754 8 39.81000137 32.65000153 38.6831 37.6339 9 40.47000122 33.34999847 39.27999878 38.2746 data[1,] V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 dates1-as.Date(data[1,]) do not know how to convert 'dates[1,]' to class Date # However, I can individually convert them all: dates1.1-as.Date(data[1,1]) dates1.1 [1] 2008-05-19 # How can I change the class of the date over the entire row (my original file contains more than one hundred rows)? # Thank you very much. In a data frame, you cannot change the data type for a row, without changing the data type for the entire column. That being said, it looks to me like each column contains continuous data for a chronological series. If that is correct, then you want the dates to be the column names and not the first row., Thus, when you used read.table(), you should have included the argument 'header = TRUE', which would do just that. With read.table(), header is FALSE by default. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] as.Date from data.frame
On Nov 11, 2009, at 1:06 PM, separent wrote: #Hello, #I loaded data using read.table - I needed to convert a row in the data frame to date class: You _should_ have given the actual code (as the Posting Guide requests), but I would guess that you should have added either as.is=TRUE or stringsAsFactors=FALSE. As it is, you probably have all factors. data V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 2 38.16999817 30.7008 36.8661 35.18999863 3 37.4754 29.9576 36.4508 35.3661 4 37.4754 30.1462 36.47000122 35.3661 5 37.84999847 30.5669 36.8415 35.74000168 6 38.3839 31.1462 37.3415 36.2746 7 39.1161 31.9085 38.02999878 36.9754 8 39.81000137 32.65000153 38.6831 37.6339 9 40.47000122 33.34999847 39.27999878 38.2746 data[1,] V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 dates1-as.Date(data[1,]) do not know how to convert 'dates[1,]' to class Date You should have given the actual error message which was probably more informative than what your brain interpreted it to mean. # However, I can individually convert them all: dates1.1-as.Date(data[1,1]) dates1.1 [1] 2008-05-19 # How can I change the class of the date over the entire row (my original file contains more than one hundred rows)? Off hand I would say you want to work on a transpose of this data. You cannot mix datatypes within columns and it seems fairly clear that the rows below number 1 are numeric. Back to the drawing board. -- 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] Comparison of vectors in a matrix
That is wonderful, now I think I am all set! Thanks again!
Tony Plate wrote:
This is a tricky data entry problem. The right technique will depend on
the fine details of the data, and it's not clear what those are. E.g.,
when you say In my first column, for example, I have henry , it's
unclear to me whether or not the double quotes are part of the data or not
- which is why it's nice to provide reproducible examples.
But, if you do have quoted strings in your data fields as they exist in an
R matrix, you can do something like the following:
# each element of the matrix x contains one or more quoted strings,
separated by commas
x - matrix(c('a, b', 'c', 'b', 'd'), ncol=2,
dimnames=list(c(row1, row2), c(X,Y)))
x
X Y
row1 \a\, \b\ \b\
row2 \c\\d\
# use R's parsing and evaluation to turn 'a, b' into c(a, b), and
turn that
# into a matrix containing character vectors of various lengths.
matrix(lapply(parse(text=paste(c(, x, ))), eval), ncol=ncol(x),
dimnames=dimnames(x))
X Y
row1 Character,2 b
row2 c d
- Tony Plate
esterhazy wrote:
Yes, thanks for this, this is exactly what I want to do.
However, I have a remaining problem which is how to get R to understand
that
each entry in my matrix is a vector of names.
I have been trying to import my text file with the names in each vector
of
names enclosed in quotes and separated by commas, or separated by spaces,
or
without quotes, etc, with no luck.
Everytime, R seems to consider the vector of names as just one long name.
In my first colum, for example, I have henry, in the second, mary,
ruth, and in the third mary, joseph, and I have no idea how to get
R
to see that mary, ruth, for example, is composed of two strings of
text,
rather than just one.
Thanks for any further help!
http://old.nabble.com/file/p26305756/ffoexample.txt ffoexample.txt
Tony Plate wrote:
Nice problem!
If I understand you correctly, here's how to do it (with list-based
matrices):
set.seed(1)
(x - matrix(lapply(rpois(10,2)+1, function(k) sample(letters[1:10],
size=k)), ncol=2, dimnames=list(1:5,c(A,B
A B
1 Character,2 Character,5
2 Character,2 Character,5
3 Character,3 Character,3
4 Character,5 Character,3
5 Character,2 i
x[1,1]
[[1]]
[1] c b
x[1,2]
[[1]]
[1] c d a j f
(y - cbind(x, A-B=apply(x, 1, function(ab) setdiff(ab[[1]],
ab[[2]]
A B A-B
1 Character,2 Character,5 b
2 Character,2 Character,5 g
3 Character,3 Character,3 Character,3
4 Character,5 Character,3 Character,2
5 Character,2 i Character,2
y[1,3]
[[1]]
[1] b
-- Tony Plate
esterhazy wrote:
Hi,
I have a matrix with two columns, and the elements of the matrix are
vectors.
So for example, in line 3 of column 1 I have a vector v31=(marc,
robert,
marie).
What I need to do is to compare all vectors in column 1 and 2, so as to
get,
for example setdiff(v31,v32) into a new column.
Is there a way to do this in R?
Thanks!
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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.
--
View this message in context:
http://old.nabble.com/Comparison-of-vectors-in-a-matrix-tp26284855p26306896.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] partial cumsum
William Dunlap wrote:
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of smu
Sent: Wednesday, November 11, 2009 7:58 AM
To: r-help@r-project.org
Subject: [R] partial cumsum
Hello,
I am searching for a function to calculate partial cumsums.
For example it should calculate the cumulative sums until a
NA appears,
and restart the cumsum calculation after the NA.
this:
x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10)
should become this:
1 3 6 NA 5 11 18 26 35 45
Perhaps
ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum)
[1] 1 3 6 NA 5 11 18 26 35 45
Nice simple function! Here's a different approach I use that's faster for long
vectors with many NA values. Note however, that this approach can suffer from
catastrophic round-off error because it does a cumsum over the whole vector
(after replacing NA's with zeros) and then subtracting out the cumsum at the
most recent NA values. Most of the body of this function is devoted to
allowing (an unreasonable degree of) flexibility in specification of where to
reset.
cumsum.reset - function(x, reset.at=which(is.na(x)), na.rm=F) {
# compute the cumsum of x, resetting the cumsum to 0 at each element indexed
by reset.at
if (is.logical(reset.at)) {
if (length(reset.at)length(x)) {
if ((length(reset.at) %% length(x))!=0)
stop(length of reset.at must be a multiple of length of x)
x - rep(x, len=length(reset.at))
} else if (length(reset.at)length(x)) {
if ((length(x) %% length(reset.at))!=0)
stop(length of x must be a multiple of length of reset.at)
reset.at - rep(reset.at, len=length(x))
}
reset.at - which(reset.at)
} else if (!is.numeric(reset.at)) {
stop(reset.at must be logical or numeric)
}
if (length(i - which(reset.at=1)))
reset.at - reset.at[-i]
if (any(i - is.na(x[reset.at])))
x[reset.at[i]] - 0
if (na.rm any(i - which(is.na(x
x[i] - 0
y - cumsum(x)
d - diff(c(0, y[reset.at-1]))
r - rep(0, length(x))
r[reset.at] - d
return(y - cumsum(r))
}
x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10)
cumsum.reset(x)
[1] 1 3 6 0 5 11 18 26 35 45
ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum)
[1] 1 3 6 NA 5 11 18 26 35 45
The speedup from not breaking the input vector into smaller vectors is (to me)
surprisingly small -- only a factor of 3:
x - replace(rnorm(1e6), sample(1e6, 1), NA)
all.equal(replace(ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum), is.na(x), 0),
cumsum.reset(x))
[1] TRUE
system.time(cumsum.reset(x))
user system elapsed
0.310.030.35
system.time(ave(x, rev(cumsum(rev(is.na(x, FUN=cumsum))
user system elapsed
0.990.051.15
So, I'd go with the ave() approach unless this type of cumsum is the core of a
long computationally intensive job. And if that's the case, it would make
sense to code it in C.
-- Tony Plate
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
any ideas?
thank you and best regards,
stefan
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] fitting a glm with matrix of responses
hi all - quick question: i have a matrix 'y' of response values, with two explanatory variables 'x1' and 'x2'. tested values of 'x1' and 'x2' are sitting in two vectors 'x1' and 'x2'. i want to learn model parameters without unrolling the matrix of response values. example below: # some fake data for the example x1 - 1:5 x2 - 1:10 y - matrix(runif(50), nrow = 5) # current method: z - vector() for(i in x1) for(j in x2) z - c(z, i, j, y[i, j]) z - data.frame(matrix(z, ncol = 3, byrow = TRUE)) colnames(z) - c(x1, x2, y) m - glm(y ~ x1 + x2 + x1:x2, family = binomial, data = z) # what i'd like to do, kind of: m - glm(y ~ x1 + x2 + x1:x2) basically, i have to unfold the matrix 'y' to a data frame 'z' then solve. this is somewhat tedious. anyone know of a way i can do this more generally, especially if working in even higher dimensions than 2 (i.e. with an arbitrary- dimension array of response values)? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Automating Plot Commands using a Loop
Hello, Thank you for responding! The data is in the same format as the example I showed earlier. The data is different from matrix to matrix, but the general format is: Time1Time 2 Time3 Time4 Species1 134 5 Species2 345 6 Species33 4 5 6 The problem I had was that I wanted to graph the species values (which are different from matrix to matrix) through time - i.e. values y-axis and time x-axis, so I transposed the matrix such that the species would be the columns and time would be the rows so that this would be easier to code. If I code it without using a formula: plot(Data$Species1,type=o,axes=F,ann=F,ylim=c(0,1)) It works perfectly fine. Thanks, Andrew Z. Why did you transpose the dataframe (TData - t(Data))? Is your data in the same structure as is expected. Provide either the data file you are plotting, or at least an 'str' of the object. -- View this message in context: http://old.nabble.com/Automating-Plot-Commands-using-a-Loop-tp26273268p26306647.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] fisher.test negative value error
Hi all, I am new to python, R and rpy2. I having few errors when I am using 'fisher.test' function where I am struck now. Please help to fix these bugs. My code for fisher.test goes this way *from rpy2 import * import rpy2.robjects as robjects def fisherExact(a,b,c,d): v = [a,b,c,d] m = robjects.r['matrix'](v,2,2) s = robjects.r['fisher.test'](m) return s * Here the value of a=0,b=1,c=0 and d=1. so the matrix m should be like this [,1] [,2] [1,] 00 [2,] 11 But I am getting an error like this *File /usr/lib64/python2.6/site-packages/rpy2/robjects/__init__.py, line 422, in __call__ res = super(RFunction, self).__call__(*new_args, **new_kwargs) rinterface.RRuntimeError: Error in function (x, y = NULL, workspace = 2e+05, hybrid = FALSE, control = list(), : all entries of 'x' must be nonnegative and finite* clearly every element of matrix m is +ve but I don't know why I am getting this error. I am using python 2.6,R-core-2.9.2-1 and rpy2. Please help me in this regard. Thanks, Bhanu. [[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] fitting a glm with matrix of responses
On Nov 11, 2009, at 2:24 PM, Murat Tasan wrote: hi all - quick question: i have a matrix 'y' of response values, with two explanatory variables 'x1' and 'x2'. tested values of 'x1' and 'x2' are sitting in two vectors 'x1' and 'x2'. i want to learn model parameters without unrolling the matrix of response values. example below: # some fake data for the example x1 - 1:5 x2 - 1:10 y - matrix(runif(50), nrow = 5) # current method: z - vector() for(i in x1) for(j in x2) z - c(z, i, j, y[i, j]) z - data.frame(matrix(z, ncol = 3, byrow = TRUE)) colnames(z) - c(x1, x2, y) m - glm(y ~ x1 + x2 + x1:x2, family = binomial, data = z) # what i'd like to do, kind of: m - glm(as.vector(y) ~ expand.grid(x1 , x2)^2) Perhaps: zdf - expand.grid(x1,x2) zdf$y - as.vector(y) m - glm(as.vector(y) ~ (Var1 + Var2)^2, data=zdf) m Call: glm(formula = as.vector(y) ~ (Var1 + Var2)^2, data = zdf) Coefficients: (Intercept) Var1 Var2Var1:Var2 0.425943 0.066960-0.001198-0.006480 Degrees of Freedom: 49 Total (i.e. Null); 46 Residual Null Deviance: 4.067 Residual Deviance: 3.759AIC: 22.5 basically, i have to unfold the matrix 'y' to a data frame 'z' then solve. as.vector would do that, but I don't know how well it works within a formula. this is somewhat tedious. anyone know of a way i can do this more generally, especially if working in even higher dimensions than 2 (i.e. with an arbitrary- dimension array of response values)? ?formula You can get all two way interactions with ( )^2 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] fisher.test negative value error
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Bhanu Mangipudi Sent: Wednesday, November 11, 2009 11:16 AM To: r-help@r-project.org Subject: [R] fisher.test negative value error Hi all, I am new to python, R and rpy2. I having few errors when I am using 'fisher.test' function where I am struck now. Please help to fix these bugs. My code for fisher.test goes this way *from rpy2 import * import rpy2.robjects as robjects def fisherExact(a,b,c,d): v = [a,b,c,d] m = robjects.r['matrix'](v,2,2) s = robjects.r['fisher.test'](m) return s * Here the value of a=0,b=1,c=0 and d=1. so the matrix m should be like this [,1] [,2] [1,] 00 [2,] 11 But I am getting an error like this *File /usr/lib64/python2.6/site-packages/rpy2/robjects/__init__.py, line 422, in __call__ res = super(RFunction, self).__call__(*new_args, **new_kwargs) rinterface.RRuntimeError: Error in function (x, y = NULL, workspace = 2e+05, hybrid = FALSE, control = list(), : all entries of 'x' must be nonnegative and finite* I think you are passing a list matrix to fisher.test, not a numeric matrix. E.g., fisher.test(matrix(list(0,1,0,1),2,2)) Error in fisher.test(matrix(list(0, 1, 0, 1), 2, 2)) : all entries of 'x' must be nonnegative and finite The python v should be a numeric array instead of a list. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com clearly every element of matrix m is +ve but I don't know why I am getting this error. I am using python 2.6,R-core-2.9.2-1 and rpy2. Please help me in this regard. Thanks, Bhanu. [[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] as.Date from data.frame
Hello Gentlemen, All of your answers were helpfull. Indeed, 'as.is=TRUE' or 'stringsAsFactors=FALSE' avoid importing as factors, as it did when I imported the table without specifying anything. However, data frames will not allow different datatypes within a single column, so, for time series, I should work with a transpose of the table. However, when the time series is the label of a profile, I should read the table with 'header = TRUE'. In other words: * Plotting values of a column (let's say againts observation number) for a selected time (where time is the name of the series): import data with 'header = TRUE' * Plotting the evolution of the values for a single observation number against time: 'header = TRUE' transpose 'as.is=TRUE' Kind Regards, Serge-Étienne Parent Golder Associés Canada separent wrote: #Hello, #I loaded data using read.table - I needed to convert a row in the data frame to date class: data V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 2 38.16999817 30.7008 36.8661 35.18999863 3 37.4754 29.9576 36.4508 35.3661 4 37.4754 30.1462 36.47000122 35.3661 5 37.84999847 30.5669 36.8415 35.74000168 6 38.3839 31.1462 37.3415 36.2746 7 39.1161 31.9085 38.02999878 36.9754 8 39.81000137 32.65000153 38.6831 37.6339 9 40.47000122 33.34999847 39.27999878 38.2746 data[1,] V1 V2 V3 V4 1 2008-05-19 2008-04-19 2008-03-21 2008-02-22 dates1-as.Date(data[1,]) do not know how to convert 'dates[1,]' to class Date # However, I can individually convert them all: dates1.1-as.Date(data[1,1]) dates1.1 [1] 2008-05-19 # How can I change the class of the date over the entire row (my original file contains more than one hundred rows)? # Thank you very much. -- View this message in context: http://old.nabble.com/as.Date-from-data.frame-tp26305830p26307735.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] LINEAR MIXED EFFECT
Milton's point is dead-on, and I would highly encourage you to give the posting guide a look. That said... you might try na.action = na.omit in your call to... actually, we don't know what function you are using (see first point). Regardless, sounds like you have missing data and na.action is set to na.fail (ie, fail if any missing data). cheers, Dave milton ruser wrote: Dear Ana Golveia, It is completelly impossible someone realise what kind or help you need or what is happening. I suggest you give a look on the posting guide, mainly that part about a minimum reproducible code with self explaining information, etc. Cheers milton On Wed, Nov 11, 2009 at 7:22 AM, ANARPCG a.gouvei...@imperial.ac.uk wrote: CAN ANYONE PLEASE HELP ME WITH THIS i HAVE TO DO A MIXED EFFECT LINEAR MODEL WITH MY DATA DUE TO THE FACT THAT I have pseudoreplication! Although after reading and trying it for several times can get around due to Error in na.fail.default(list(date = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, : missing values in object I uploaded my data file Thank you so much Kind regards AG http://old.nabble.com/file/p26300394/rawoctobercalciumexperiment2.txt rawoctobercalciumexperiment2.txt -- View this message in context: http://old.nabble.com/LINEAR-MIXED-EFFECT-tp26300394p26300394.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.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://old.nabble.com/LINEAR-MIXED-EFFECT-tp26300394p26308114.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] Is there a way to specify drop=FALSE as the global default?
On Wed, 11 Nov 2009, Peng Yu wrote: On Mon, Oct 19, 2009 at 7:57 PM, Peng Yu pengyu...@gmail.com wrote: tmp - matrix(1:2) tmp tmp[,1,drop=FALSE] See the above example. Is there a way to make 'drop=FALSE' as global default, so that when I say 'tmp[,1]', R will treat it as 'tmp[,1,drop=FALSE]'? Is there a way to set drop=FALSE globally? No. I can't remember to put drop=FALSE to all the '[]'s. But this indeed causes some weird bugs for me to figure out. It would probably cause a lot more weird bugs if the behavior of [ depended on a global option. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error: cannot allocate vector of size...
On Wed, 11 Nov 2009, Larry Hotchkiss wrote: Hi, I'm responding to the question about storage error, trying to read a 300 x 100 dataset into a data.frame. I wonder whether you can read the data as strings. If the numbers are all one digit, each cell would require just 1 byte instead of 8. Um, no. a-rep(1,10) object.size(a) 400056 bytes object.size(a)/length(a) 4.00056 bytes They are character strings, not individual characters, so some overhead is unavoidable. Even if all the strings are identical you need four bytes per string, the same as if the data were read as integers. -thomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Partial correlations and p-values
I'm trying to write code to calculate partial correlations (along with p-values). I'm new to R, and I don't know how to do this. I have searched and come across different functions, but I haven't been able to get any of them to work (for example, pcor and pcor.test from the ggm package). In the following example, I am trying to compute the correlation between x and y, while controlling for z (partial correlation): x - c(1,20,14,7,9) y - c(5,6,7,9,10) z - c(13,27,16,5,4) What function can I append to this to find this partial correlation? Many thanks! -- View this message in context: http://old.nabble.com/Partial-correlations-and-p-values-tp26308463p26308463.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] R.oo and S4?
I think it would be best to learn S3 first since that is a fundamental part of R and S4 is an extension of it and also its very simple so there is not much to learn. After that you can branch out. On Mon, Oct 26, 2009 at 2:28 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: S3 and S4 are part of the core of R so they would presumably be the most used. S4 is an extension to S3 but adds strong typing and a number of other features. R.oo and proto are packages on CRAN which give access to different models of object oriented programming than S3 and S4. R.oo uses a more conventional model of OO than S3 or S4 that is probably closer to what you are used to if you are coming form another language while proto uses the prototype model (or pure object model). proto tends to apply in user interface applications and there is some info on which other packages make use of proto on the proto home page at: http://r-proto.googlecode.com On Mon, Oct 26, 2009 at 2:47 PM, Peng Yu pengyu...@gmail.com wrote: There are different way to make R classes. I know R.oo and S4. I'm wondering which one is the current popular one. Which one is current recommended when make new R packages? Thank you! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loadings and scores from fastICA?
The help for fastICA says:
The data matrix X is considered to be a linear combination of
non-Gaussian (independent) components i.e. X = SA where columns of
S contain the independent components and A is a linear mixing
matrix.
The value of fastICA is a list with components S (the estimated source matrix) and
A (the estimated mixing matrix). Are these what you want?
-- Tony Plate
Joel Fürstenberg-Hägg wrote:
Hi all,
Does anyone know how to get the independent components and loadings from an
Independent Component Analysis (ICA), as well as principal components and
loadings from a Pricipal Component analysis (PCA) using the fastICA package? Or
perhaps if there's another way to do ICAs in R?
Below is an example from the fastICA manual
(http://cran.r-project.org/web/packages/fastICA/fastICA.pdf)
if(require(MASS))
{
x - mvrnorm(n = 1000, mu = c(0, 0), Sigma = matrix(c(10, 3, 3, 1), 2, 2))
x1 - mvrnorm(n = 1000, mu = c(-1, 2), Sigma = matrix(c(10, 3, 3, 1), 2,
2))
X - rbind(x, x1)
a - fastICA(X, 2, alg.typ = deflation, fun = logcosh, alpha = 1, method =
R, row.norm = FALSE, maxit = 200, tol = 0.0001, verbose = TRUE)
par(mfrow = c(1, 3))
plot(a$X, main = Pre-processed data)
plot(a$X%*%a$K, main = PCA components)
plot(a$S, main = ICA components)
}
Best regards,
Joel
_
Hitta kärleken i vinter!
http://dejting.se.msn.com/channel/index.aspx?trackingid=1002952
[[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] thousands separator in console output?
Hello, I'm beginning with R and I'm wondering if there's any way to display large values (e.g. 161651654167) using my system's thousands separator or scientific notation. 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] plotting time series
Hi, I'm new to R. I've two sort of time series, one sampled every second and other every 30 minutes stored in two data.frames zy1 and zy2. I've joined the two series using date-merge(zy1,zy2, all.x=T,all.y=T) when I want to make a plot, using the code below, it doesn't work. It gives me the following error produced NA error plot( a~ time,data=z, type=l, ylim=range(30,80)) lines( b~ time,data=z, type=l) Can you help me? Thanks Morgi -- View this message in context: http://old.nabble.com/plotting-time-series-tp26308476p26308476.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] eacf
Dear Michel, The only thing you need to do is download the TSA package, and read its document. In this document you're gonna find the command eacf, so it's as simple as using the source TSA, and in ur Tinn-R just call the instruction with the options you want. I expect this information to be useful for you. Michel Helcias wrote: I am Brazilian and I don't know how to speak English, for that I apologize for my writing. I'd like of informations about the extended (sample) autocorrelation function (*EACF*) for the time series. where to acquire some related command. thak you. [[alternative HTML version deleted]] __ r-h...@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- View this message in context: http://old.nabble.com/eacf-tp4938311p26308603.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] Suppressing final spaces in data.frame printouts
When printing data.frames, R aligns columns by padding with spaces.
For example,
print(data.frame(x=c('a','bb','ccc')),right=FALSE)
x
1 a |-- vertical bar shows end of line
2 bb |-- vertical bar shows end of line
3 ccc|-- vertical bar shows end of line
Is there some way to suppress the padding for the final column? I
often have data frames which contain a handful of long strings in the
final column which, when printed out, cause wraparound on all the
rows, even those not containing long strings, something like this:
print(data.frame(q=1:3,x=c('a','bb','this is a very long string')),right=FALSE)
q x |
|
1 1 a |
|
2 2 bb |
|
3 3 this is a very l|
ong string|
where I'd rather have
print(data.frame(q=1:3,x=c('a','bb','this is a very long string')),right=FALSE)
q x|
1 1 a|
2 2 bb|
3 3 this is a very l|
ong string|
I could of course write my own print function for this, but was
wondering if there was a standard way of doing it. If not in R,
perhaps there is some way to have ESS delete the final spaces?
Thanks,
-s
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] thousands separator in console output?
On Nov 11, 2009, at 3:52 PM, Patricio Cuarón wrote: Hello, I'm beginning with R and I'm wondering if there's any way to display large values (e.g. 161651654167) using my system's thousands separator or scientific notation. Thanks! To get scientific notation as a default for given large values, see ? options and note 'scipen', which by default is 0. 161651654167 [1] 161651654167 options(scipen = -5) 161651654167 [1] 1.616517e+11 That is about the only way, by default, to adjust the output of numeric values, for which R uses ?print.default. You can play with the value of scipen to get the behavior you wish for some definition of 'large' numbers. If you want to format large numbers with commas as the thousand separator, that will not happen by default, but you can use format() to do this when outputting numbers as you may need in a function and/ or for pretty printing in tables, etc.: format(161651654167, big.mark = ,) [1] 161,651,654,167 Note that the result is a character vector and not numeric. See ?format and perhaps ?formatC for more information. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R.oo and S4?
Please what I already wrote in my previous message of this thread. Also, everything in R.oo is based on S3 and it uses standard R constructs and data types to achieve what it does. You can submit packages based on R.oo, and there are several such packages on CRAN, see 'Reverse dependencies' on http://cran.r-project.org/web/packages/R.oo/. Who told you it is not possible? /Henik On Wed, Nov 11, 2009 at 6:26 PM, Peng Yu pengyu...@gmail.com wrote: I'm very familiar with C++. In this sense, it is easier for me to learn R.oo according to your advice. On the other hand, S3 and S4 are the most used. I'm wondering what would be the best choice for me. Do you have any recommendation considering the pros and cons of both ways? Is it true that packages based on R.oo can not be deposited to CRAN? On Mon, Oct 26, 2009 at 1:28 PM, Gabor Grothendieck ggrothendi...@gmail.com wrote: S3 and S4 are part of the core of R so they would presumably be the most used. S4 is an extension to S3 but adds strong typing and a number of other features. R.oo and proto are packages on CRAN which give access to different models of object oriented programming than S3 and S4. R.oo uses a more conventional model of OO than S3 or S4 that is probably closer to what you are used to if you are coming form another language while proto uses the prototype model (or pure object model). proto tends to apply in user interface applications and there is some info on which other packages make use of proto on the proto home page at: http://r-proto.googlecode.com On Mon, Oct 26, 2009 at 2:47 PM, Peng Yu pengyu...@gmail.com wrote: There are different way to make R classes. I know R.oo and S4. I'm wondering which one is the current popular one. Which one is current recommended when make new R packages? Thank you! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Suppressing final spaces in data.frame printouts
Stavros Macrakis wrote: I could of course write my own print function for this, but was wondering if there was a standard way of doing it. If not in R, perhaps there is some way to have ESS delete the final spaces? ESS, or more precisely emacs, can handle that. Use the M-x toggle-truncate-lines command: Toggle whether to fold or truncate long lines for the current buffer. Rich __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] partial cumsum
quote: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) rev(cumsum(rev(is.na(x [1] 1 1 1 1 0 0 0 0 0 0 A more natural way to do this is cumsum(is.na(c(NA,x[-length(x)]))) [1] 1 1 1 1 2 2 2 2 2 2 endquote Both of which suggest the original problem could also be dealt with by using rle(). Something like xna-is.na(x) rle(xna) and then apply cumsum to sections of x based onthe lengths returned by rle Carl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there a way to specify drop=FALSE as the global default?
See the above example. Is there a way to make 'drop=FALSE' as global
default, so that when I say 'tmp[,1]', R will treat it as
'tmp[,1,drop=FALSE]'?
The following code won't change the defaults, but it would at least
let you know when you're making the mistake:
trace_all - function(fs, tracer) {
lapply(fs, trace, exit = tracer, print=FALSE)
invisible()
}
functions_with_arg - function(arg, pos) {
fs - ls(pos=pos)
present - unlist(lapply(fs, function(x)
is.function(get(x)) !is.null(formals(x)[[arg]])))
fs[present]
}
trace_all(
functions_with_arg(drop, package:base),
quote(if (drop) warning(drop = TRUE, call. = F))
)
mtcars[1, 2]
[1] 6
Warning message:
drop = TRUE
Unfortunately it doesn't pick up on the generic [ because it is a primitive.
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] partial cumsum
On Nov 11, 2009, at 4:45 PM, Carl Witthoft wrote: By Bill Dunlap: quote: x - c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) rev(cumsum(rev(is.na(x [1] 1 1 1 1 0 0 0 0 0 0 A more natural way to do this is cumsum(is.na(c(NA,x[-length(x)]))) [1] 1 1 1 1 2 2 2 2 2 2 endquote Both of which suggest the original problem could also be dealt with by using rle(). Something like xna-is.na(x) rle(xna) and then apply cumsum to sections of x based onthe lengths returned by rle Carl Which would be something along the lines of the following: rle.x - rle(!is.na(x))$lengths rle.x [1] 3 1 6 as.numeric(unlist(sapply(split(x, rep(seq(along = rle.x), rle.x)), cumsum))) [1] 1 3 6 NA 5 11 18 26 35 45 Which is what I had been working on until I saw Bill's more elegant one-liner solution... :-) The interim use of split() gets you 'x' split by where the NA's occur: rep(seq(along = rle.x), rle.x) [1] 1 1 1 2 3 3 3 3 3 3 split(x, rep(seq(along = rle.x), rle.x)) $`1` [1] 1 2 3 $`2` [1] NA $`3` [1] 5 6 7 8 9 10 and then you use cumsum() in sapply() (or lapply()) on each list element and coerce the result back to an un-named numeric vector. Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Is there a way to specify drop=FALSE as the global default?
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of hadley wickham
Sent: Wednesday, November 11, 2009 3:03 PM
To: Peng Yu
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] Is there a way to specify drop=FALSE as the
global default?
See the above example. Is there a way to make 'drop=FALSE' as global
default, so that when I say 'tmp[,1]', R will treat it as
'tmp[,1,drop=FALSE]'?
The following code won't change the defaults, but it would at least
let you know when you're making the mistake:
Or you could write wrapper functions for [ and [- that
add the drop=FALSE argument. E.g.,
SS - function(x, ..., drop=FALSE) x[..., drop=drop] # 'Safe
Subscripting'
`SS-` - function(x, ..., value) { x[...] - value ; x }
Use them as
SS(x, 1, i) # instead of x[1, i]
Use grep (or codetools::walkCode) to make sure you don't
have any other calls to [ in your code. (The ss- is
there only so calls to [ on the left side of the assignment
don't attract grep's attention.)
S+ has a subscript2d() that does this and also lets
you use 2 subscripts on vectors (the 2nd must be 1
in that case, but you don't need to use if statements).
subscript2d doesn't even have the drop= argument - if
you want to make a vector from a one column matrix use
as.vector.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
trace_all - function(fs, tracer) {
lapply(fs, trace, exit = tracer, print=FALSE)
invisible()
}
functions_with_arg - function(arg, pos) {
fs - ls(pos=pos)
present - unlist(lapply(fs, function(x)
is.function(get(x)) !is.null(formals(x)[[arg]])))
fs[present]
}
trace_all(
functions_with_arg(drop, package:base),
quote(if (drop) warning(drop = TRUE, call. = F))
)
mtcars[1, 2]
[1] 6
Warning message:
drop = TRUE
Unfortunately it doesn't pick up on the generic [ because it
is a primitive.
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.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] Convert to time from epoch
Hello R users, Is anyone familiar with an R function that converts a time expression ( POSIx for example ) to time (seconds/minutes) from epoch? I was unable to find any Best, Alon [[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] Convert to time from epoch
try this: x - Sys.time() str(x) POSIXct[1:1], format: 2009-11-11 18:40:10 x [1] 2009-11-11 18:40:10 EST as.numeric(x) # secs from 1/1/1970 [1] 1257982810 On Wed, Nov 11, 2009 at 6:29 PM, Alon Ben-Ari alon.ben...@gmail.com wrote: Hello R users, Is anyone familiar with an R function that converts a time expression ( POSIx for example ) to time (seconds/minutes) from epoch? I was unable to find any Best, Alon [[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] Implementation of the Shuffled Complex Evolution (SCE-UA) Algorithm
http://code.google.com/p/ihacres/source/browse/trunk/man/SCEoptim.Rd http://code.google.com/p/ihacres/source/browse/trunk/R/sce.R source(http://ihacres.googlecode.com/svn/trunk/R/sce.R;) 2009/11/12 Hans W Borchers hwborch...@googlemail.com: Simon Seibert simon.seibert at mytum.de writes: Good evening list, I'm looking for an R implementation of the Shuffled Complex Evolution†(SCE-UA) algorithm after Duan et al. (1993). Does anybody know if there is an extension/ package existing that contains it? Thanks very much for your help! Cheers, Simon I am looking into stochastic global optimization routines, such as variants of genetic algorithms (GA), particle swarm optimization (PSO), and shuffled complex (SCE) or differential evolution (DE). At the moment I am testing a free Matlab implementation of SCE-UA. If that turns out to be interesting and effective, I will translate it into an R function, maybe within the next few weeks -- except, of course, someone else is going to do an implementation of a similar procedure. Right now I am not totally convinced. Maybe you try your problem with other approaches available in R (see e.g. the Optimization task view). Regards Hans Werner Duan QY, Gupta KV, Sorooshian S (1993) Shuffled Complex Evolution Approach for Effective and Efficient Global Minimization. In Jour. of optimization theorie and applications. Vol 76, 3, 501-521 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Felix Andrews / 安福立 Postdoctoral Fellow Integrated Catchment Assessment and Management (iCAM) Centre Fenner School of Environment and Society [Bldg 48a] The Australian National University Canberra ACT 0200 Australia M: +61 410 400 963 T: + 61 2 6125 4670 E: felix.andr...@anu.edu.au CRICOS Provider No. 00120C -- http://www.neurofractal.org/felix/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] loop through variable names
Often I perform the same task on a series of variables in a dataframe, by
looping through a character vector that holds the names and using paste(),
eval(), and parse() inside the loop.
For instance:
thesevars-names(environmental)
environmental$ToyOutcome-rnorm(nrow(environmental))
tableOfResults-data.frame(var=thesevars)
tableOfResults$Beta- NA
rownames(tableOfResults)-thesevars
for( thisvar in thesevars) {
thiscommand- paste(thislm - lm( ToyOutcome ~ , thisvar, ,
data=environmental))
eval(parse(text=thiscommand))
tableOfResults[thisvar, Beta] - coef(thislm)[thisvar]
}
print(tableOfResults)
Note that it's not always as simple a task as in this example. Within the loop,
I might first figure out whether the variable is continuous or categorical,
then perform an operation depending on its type--maybe lm() for continuous but
wilcox.test() for dichotomous.
But the use of paste(), eval(), and parse() seems awkward. Is there a more
elegant way to approach this?
Thanks
Jacob A. Wegelin
Department of Biostatistics
Virginia Commonwealth University
Richmond VA 23298-0032
U.S.A.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] loop through variable names
Try this:
lm(environmental[c(ToyOutcome, thisvar)])
or
lm(ToyOutcome ~., environmental[c(ToyOutcome, thisvar)])
On Wed, Nov 11, 2009 at 6:49 PM, Jacob Wegelin jacobwege...@fastmail.fm wrote:
Often I perform the same task on a series of variables in a dataframe, by
looping through a character vector that holds the names and using paste(),
eval(), and parse() inside the loop.
For instance:
thesevars-names(environmental)
environmental$ToyOutcome-rnorm(nrow(environmental))
tableOfResults-data.frame(var=thesevars)
tableOfResults$Beta- NA
rownames(tableOfResults)-thesevars
for( thisvar in thesevars) {
thiscommand- paste(thislm - lm( ToyOutcome ~ , thisvar, ,
data=environmental))
eval(parse(text=thiscommand))
tableOfResults[thisvar, Beta] - coef(thislm)[thisvar]
}
print(tableOfResults)
Note that it's not always as simple a task as in this example. Within the
loop, I might first figure out whether the variable is continuous or
categorical, then perform an operation depending on its type--maybe lm() for
continuous but wilcox.test() for dichotomous.
But the use of paste(), eval(), and parse() seems awkward. Is there a more
elegant way to approach this?
Thanks
Jacob A. Wegelin
Department of Biostatistics
Virginia Commonwealth University
Richmond VA 23298-0032
U.S.A.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] naive collinear weighted linear regression
Hi there Sorry for what may be a naive or dumb question. I have the following data: x - c(1,2,3,4) # predictor vector y - c(2,4,6,8) # response vector. Notice that it is an exact, perfect straight line through the origin and slope equal to 2 error - c(0.3,0.3,0.3,0.3) # I have (equal) ``errors'', for instance, in the measured responses Of course the best fit coefficients should be 0 for the intercept and 2 for the slope. Furthermore, it seems completely plausible (or not?) that, since the y_i have associated non-vanishing ``errors'' (dispersions), there should be corresponding non-vanishing ``errors'' associated to the best fit coefficients, right? When I try: fit_mod - lm(y~x,weights=1/error^2) I get Warning message: In lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : extra arguments weigths are just disregarded. Keeping on, despite the warning message, which I did not quite understand, when I type: summary(fit_mod) I get Call: lm(formula = y ~ x, weigths = 1/error^2) Residuals: 1 2 3 4 -5.067e-17 8.445e-17 -1.689e-17 -1.689e-17 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 0.000e+00 8.776e-17 0.000e+001 x 2.000e+00 3.205e-17 6.241e+16 2e-16 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 7.166e-17 on 2 degrees of freedom Multiple R-squared: 1, Adjusted R-squared: 1 F-statistic: 3.895e+33 on 1 and 2 DF, p-value: 2.2e-16 Naively, should not the column Std. Error be different from zero?? What I have in mind, and sure is not what Std. Error means, is that if I carried out a large simulation, assuming each response y_i a Gaussian random variable with mean y_i and standard deviation 2*error=0.6, and then making an ordinary least squares fitting of the slope and intercept, I would end up with a mean for these simulated coefficients which should be 2 and 0, respectively, and, that's the point, a non-vanishing standard deviation for these fitted coefficients, right?? This somehow is what I expected should be an estimate or, at least, a good indicator, of the degree of uncertainty which I should assign to the fitted coefficients; it seems to me these deviations, thus calculated as a result of the simulation, will certainly not be zero (or 3e-17, for that matter). So this Std. Error does not provide what I, naively, think should be given as a measure of the uncertainties or errors in the fitted coefficients... What am I not getting right?? Thanks and sorry for the naive and non-expert question! -- ### Prof. Mauricio Ortiz Calvao Federal University of Rio de Janeiro Institute of Physics, P O Box 68528 CEP 21941-972 Rio de Janeiro, RJ Brazil Email: o...@if.ufrj.br Phone: (55)(21)25627483 Homepage: http://www.if.ufrj.br/~orca ### __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Suppressing final spaces in data.frame printouts
On Wed, Nov 11, 2009 at 8:12 PM, Stavros Macrakis macra...@alum.mit.edu wrote: Thanks for the suggestion. I'mm familiar with the truncate-lines variable, but that's not quite what I was looking for. I don't want the padding spaces displayed, but I do want to see long strings at the end of the line. Then we can use a different emacs trick. delete-trailing-whitespaceM-x ... RET Command: Delete all the trailing whitespace across the current buffer. ess-nuke-trailing-whitespace M-x ... RET Command: Nuke all trailing whitespace in the buffer. whitespace-toggle-trailing-check M-x ... RET Command: Toggle the check for trailing space in the local buffer. Rich __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Suppressing final spaces in data.frame printouts
Thanks for the suggestion. I'mm familiar with the truncate-lines variable, but that's not quite what I was looking for. I don't want the padding spaces displayed, but I do want to see long strings at the end of the line. Thanks anyway, -s On Wed, Nov 11, 2009 at 5:40 PM, Richard M. Heiberger r...@temple.eduwrote: Stavros Macrakis wrote: I could of course write my own print function for this, but was wondering if there was a standard way of doing it. If not in R, perhaps there is some way to have ESS delete the final spaces? ESS, or more precisely emacs, can handle that. Use the M-x toggle-truncate-lines command: Toggle whether to fold or truncate long lines for the current buffer. Rich [[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.
