[R] Help in installing and loading the BradleyTerry add on package in R
How do I install and load the BradleyTerry add on package in R 2.5.1 in MSWindowsXP environment? Kalyan Roy Indian Market Research Bureau (IMRB) International New Delhi, India [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in installing and loading the BradleyTerry add on package in R
However you may also need to install the package brlr, since the BradleyTerry package depends on this. For Windows users, it's usually easiest to install packages using the Packages menu in the RGui - any dependencies are then automatically installed. Heather -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Jim Lemon Sent: 10 September 2007 12:07 To: [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Help in installing and loading the BradleyTerry add on package in R Kalyan Roy (DEL/MSG) wrote: How do I install and load the BradleyTerry add on package in R 2.5.1 in MSWindowsXP environment? Hi Kalyan, If R CMD INSTALL doesn't work, you can use WinZip or Zip Reader to unzip the package to: C:\Program Files\R-2.5.1\library or whatever your path to the library directory is, and then hand edit the packages.html file in: C:\Program Files\R-2.5.1\doc\html to include the new package in your HTML listing. This will allow you to access the help files and use the package. Jim __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in installing and loading the BradleyTerry add on package in R
Kalyan Roy (DEL/MSG) wrote: How do I install and load the BradleyTerry add on package in R 2.5.1 in MSWindowsXP environment? Hi Kalyan, If R CMD INSTALL doesn't work, you can use WinZip or Zip Reader to unzip the package to: C:\Program Files\R-2.5.1\library or whatever your path to the library directory is, and then hand edit the packages.html file in: C:\Program Files\R-2.5.1\doc\html to include the new package in your HTML listing. This will allow you to access the help files and use the package. Jim __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with color coded bar graph
Luis Naver wrote: I have a list of observations that are -1, 1 or 0. I would like to represent them in a horizontal bar color coded based on value like a stacked bar graph. I can achieve this in the form of a png with the following code: A = floor(runif(10)*3) - 1 png(width=100, height=10) par(mar=c(0,0,0,0)) image(matrix(A), col=grey(c(0.1, 0.5, 0.9))) dev.off() However I would like to do this with one of the standard plotting tools (i.e. barplot) to take advantage of labels and multiple series. Any help would be appreciated. Hi Luis, I understood your request as wanting a single horizontal bar with 10 segments, each colored according to the value of A. If this is correct, you might want: library(plotrix) plot(1,xlim=c(-1,1),ylim=c(-1,1),xlab=,ylab=,type=n,axes=FALSE) gradient.rect(-1,-0.1,1,0.1,col=grey(c(0.1,0.5,0.9))[A+2]) Jim __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help on replacing values
Your columns are factors, not character strings. Use as.is = TRUE as an argument to read.table. Also its a bit dangerous to use T although not wrong. Its safer to use TRUE. On 9/7/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Dear List, I have a newbie question. I have read in a data.frame as follows: data = read.table(table.txt, header = T) data X1 X2 X3 X4 A AB AC AB AC B AB AC AA AB C AA AB AA AB D AA AB AB AC E AB AA AA AB F AB AA AB AC B AB AC AB AA I would like to replace AA values by BB in column X2. I have tried using replace() with no success, although I am not sure this is the right function. This is the code I have used: data$X2 - replace(data$X2, data$X2 ==AA,BB) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = BB) What is wrong with the code? How can I get this done? how about changing AA values by BB in all 4 columns simultaneously? Actually this is a small example dataframe, the real one would have about 1000 columns. Extendind this, I found a similar thread dated July 2006 that used replace() on iris dataset, but I have tried reproducing it obtaining same warning message iris$Species - replace(iris$Species, iris$Species == setosa,NewName) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = NewName) Thanks in advance your help, David __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help on replacing values
Dear List, I have a newbie question. I have read in a data.frame as follows: data = read.table(table.txt, header = T) data X1 X2 X3 X4 A AB AC AB AC B AB AC AA AB C AA AB AA AB D AA AB AB AC E AB AA AA AB F AB AA AB AC B AB AC AB AA I would like to replace AA values by BB in column X2. I have tried using replace() with no success, although I am not sure this is the right function. This is the code I have used: data$X2 - replace(data$X2, data$X2 ==AA,BB) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = BB) What is wrong with the code? How can I get this done? how about changing AA values by BB in all 4 columns simultaneously? Actually this is a small example dataframe, the real one would have about 1000 columns. Extendind this, I found a similar thread dated July 2006 that used replace() on iris dataset, but I have tried reproducing it obtaining same warning message iris$Species - replace(iris$Species, iris$Species == setosa,NewName) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = NewName) Thanks in advance your help, David __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help on replacing values
Thanks a lot Gabor, that was very helpful. All sorted now! Best David Your columns are factors, not character strings. Use as.is = TRUE as an argument to read.table. Also its a bit dangerous to use T although not wrong. Its safer to use TRUE. On 9/7/07, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote: Dear List, I have a newbie question. I have read in a data.frame as follows: data = read.table(table.txt, header = T) data X1 X2 X3 X4 A AB AC AB AC B AB AC AA AB C AA AB AA AB D AA AB AB AC E AB AA AA AB F AB AA AB AC B AB AC AB AA I would like to replace AA values by BB in column X2. I have tried using replace() with no success, although I am not sure this is the right function. This is the code I have used: data$X2 - replace(data$X2, data$X2 ==AA,BB) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = BB) What is wrong with the code? How can I get this done? how about changing AA values by BB in all 4 columns simultaneously? Actually this is a small example dataframe, the real one would have about 1000 columns. Extendind this, I found a similar thread dated July 2006 that used replace() on iris dataset, but I have tried reproducing it obtaining same warning message iris$Species - replace(iris$Species, iris$Species == setosa,NewName) Warning message: invalid factor level, NAs generated in: `[-.factor`(`*tmp*`, list, value = NewName) Thanks in advance your help, David __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help with color coded bar graph
I have a list of observations that are -1, 1 or 0. I would like to represent them in a horizontal bar color coded based on value like a stacked bar graph. I can achieve this in the form of a png with the following code: A = floor(runif(10)*3) - 1 png(width=100, height=10) par(mar=c(0,0,0,0)) image(matrix(A), col=grey(c(0.1, 0.5, 0.9))) dev.off() However I would like to do this with one of the standard plotting tools (i.e. barplot) to take advantage of labels and multiple series. Any help would be appreciated. - Luis Naver __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with color coded bar graph
On Fri, 2007-09-07 at 12:45 -0700, Luis Naver wrote: I have a list of observations that are -1, 1 or 0. I would like to represent them in a horizontal bar color coded based on value like a stacked bar graph. I can achieve this in the form of a png with the following code: A = floor(runif(10)*3) - 1 png(width=100, height=10) par(mar=c(0,0,0,0)) image(matrix(A), col=grey(c(0.1, 0.5, 0.9))) dev.off() However I would like to do this with one of the standard plotting tools (i.e. barplot) to take advantage of labels and multiple series. Any help would be appreciated. - Luis Naver How about this: barplot(rep(1, length(A)), col = black, space = 0, border = 0) barplot(A, col = grey(0.9), space = 0, border = 0, add = TRUE) The first call sets the plot region to black, ensuring that the x and y axes are consistent with the second call. Alternatively, you can use barplot2() in the gplots CRAN package to do this in a single call, as it has an argument to color the plot region. HTH, Marc Schwartz __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with color coded bar graph
On Fri, 7 Sep 2007, Luis Naver wrote: I have a list of observations that are -1, 1 or 0. I would like to represent them in a horizontal bar color coded based on value like a stacked bar graph. I can achieve this in the form of a png with the following code: A = floor(runif(10)*3) - 1 png(width=100, height=10) par(mar=c(0,0,0,0)) image(matrix(A), col=grey(c(0.1, 0.5, 0.9))) dev.off() If I understand you correctly, you want a sequence of bars with equal height and colors coded by A (treated like a factor). So Maybe something like cA - grey.colors(3)[factor(A)] barplot(rep(1, length(A)), col = cA, border = cA) or barplot(rep(1, length(A)), col = cA, border = cA, space = 0, xaxs = i, axes = FALSE) ? hth, Z However I would like to do this with one of the standard plotting tools (i.e. barplot) to take advantage of labels and multiple series. Any help would be appreciated. - Luis Naver __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with color coded bar graph
On Fri, 2007-09-07 at 15:07 -0500, Marc Schwartz wrote: On Fri, 2007-09-07 at 12:45 -0700, Luis Naver wrote: I have a list of observations that are -1, 1 or 0. I would like to represent them in a horizontal bar color coded based on value like a stacked bar graph. I can achieve this in the form of a png with the following code: A = floor(runif(10)*3) - 1 png(width=100, height=10) par(mar=c(0,0,0,0)) image(matrix(A), col=grey(c(0.1, 0.5, 0.9))) dev.off() However I would like to do this with one of the standard plotting tools (i.e. barplot) to take advantage of labels and multiple series. Any help would be appreciated. - Luis Naver How about this: barplot(rep(1, length(A)), col = black, space = 0, border = 0) barplot(A, col = grey(0.9), space = 0, border = 0, add = TRUE) The first call sets the plot region to black, ensuring that the x and y axes are consistent with the second call. Alternatively, you can use barplot2() in the gplots CRAN package to do this in a single call, as it has an argument to color the plot region. Actually, here is an easier way: barplot(rep(1, length(A)), col = ifelse(A == 0, black, grey(0.9)), space = 0, border = 0) Just set 'col' based upon the value in 'A'. HTH, Marc __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with color coded bar graph
Thanks to all who replied (and very quickly). Unfortunatly I was not clear enough as to my intentions. My goal is to replicate a graph I saw in the work by Perry, Miller and Enright in A comparison of methods for the statistical analysis of spatial point patterns in plant ecology (http://www.springerlink.com/content/ 013275pp7376v0hx). For those without the article here is a copy of the graph in question (replicated without permission) http:// img511.imageshack.us/img511/8720/barexamplejl8.png. As you can see in the example, there are several hoizontal bars, colored by the values in an array (one for each bar). I've been thinking of following your examples but setting it to stack, such that all the elements would be placed one on top another. While this may work it seems particularly ungraceful. Again, thanks for the help. -Luis Naver On Sep 7, 2007, at 1:21 PM, Marc Schwartz wrote: On Fri, 2007-09-07 at 15:07 -0500, Marc Schwartz wrote: On Fri, 2007-09-07 at 12:45 -0700, Luis Naver wrote: I have a list of observations that are -1, 1 or 0. I would like to represent them in a horizontal bar color coded based on value like a stacked bar graph. I can achieve this in the form of a png with the following code: A = floor(runif(10)*3) - 1 png(width=100, height=10) par(mar=c(0,0,0,0)) image(matrix(A), col=grey(c(0.1, 0.5, 0.9))) dev.off() However I would like to do this with one of the standard plotting tools (i.e. barplot) to take advantage of labels and multiple series. Any help would be appreciated. - Luis Naver How about this: barplot(rep(1, length(A)), col = black, space = 0, border = 0) barplot(A, col = grey(0.9), space = 0, border = 0, add = TRUE) The first call sets the plot region to black, ensuring that the x and y axes are consistent with the second call. Alternatively, you can use barplot2() in the gplots CRAN package to do this in a single call, as it has an argument to color the plot region. Actually, here is an easier way: barplot(rep(1, length(A)), col = ifelse(A == 0, black, grey(0.9)), space = 0, border = 0) Just set 'col' based upon the value in 'A'. HTH, Marc __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help: how can i build a constrained non-linear model?
Dear all I try to run the code as follows, test.model-nls(y~exp(A)*(x-PMA)^4+exp(B)*(x-PMA)^2+Const, data=test, start=list(A=8,B=5,Const=10,PMA=0), control=nls.control(maxiter = 50,minFactor=1/1048), trace=TRUE) But how can i build a selfSart, since i have much data ? Thanks for your help first! Vina From: [EMAIL PROTECTED]: [EMAIL PROTECTED]: Help: how can i build a constrained non-linear model?Date: Tue, 4 Sep 2007 07:53:41 + DearI have a data.frame, and want to fit a constrained non-linear model:data: x y -0.08 20.815 -0.065 19.8128 -0.05 19.1824 -0.03 18.7346 -0.015 18.3129 0.015 18.0269 0.03 18.4715 0.05 18.9517 0.065 19.4184 0.08 20.146 0 18.2947model:y~exp(a)*(x-m)^4+exp(b)*(x-m)^2+const I try to use nls() and set start=list(a=1,b=1,c=1,m=1), but which always give me a error message that Error in qr.solv(QR.B,cc): singular matrix 'a' in solve. How can i build a selfStart? or any suggestion! ThanksRegards,Vina Invite your mail contacts to join your friends list with Windows Live Spaces. It's easy! Try it! _ [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help: how can i build a constrained non-linear model?
Dear I have a data.frame, and want to fit a constrained non-linear model: data: x y -0.08 20.815 -0.065 19.8128 -0.05 19.1824 -0.03 18.7346 -0.015 18.3129 0.015 18.0269 0.03 18.4715 0.05 18.9517 0.065 19.4184 0.08 20.146 0 18.2947 model: y~exp(a)*(x-m)^4+exp(b)*(x-m)^2+const I try to use nls() and set start=list(a=1,b=1,c=1,m=1), but which always give me a error message that Error in qr.solv(QR.B,cc): singular matrix 'a' in solve. How can i build a selfStart? or any suggestion! ThanksRegards, Vina _ [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help on inverse distribution
Dear All, I need to use the inverse of some distributions in R for simulation, but I could not find it, can anyone tell me which package I should install? thanks [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on inverse distribution
To make it specific, I need to simulation Y with inverse beta distribution, that is, Y~inverseF(X), where F is the CDF of beta distribution. THANKS On 9/4/07, Lisa Hu [EMAIL PROTECTED] wrote: Dear All, I need to use the inverse of some distributions in R for simulation, but I could not find it, can anyone tell me which package I should install? thanks [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on inverse distribution
On 5/09/2007, at 11:16 AM, Lisa Hu wrote:
To make it specific, I need to simulation Y with inverse beta
distribution,
that is, Y~inverseF(X), where F is the CDF of beta distribution.
THANKS
On 9/4/07, Lisa Hu [EMAIL PROTECTED] wrote:
Dear All,
I need to use the inverse of some distributions in R for
simulation, but I
could not find it, can anyone tell me which package I should install?
thanks
help.search(distribution) ?Beta
##
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Re: [R] R-help Digest, Vol 54, Issue 30
Ron Crump wrote: Hi, I have a dataframe that contains pedigree information; that is individual, sire and dam identities as separate columns. It also has date of birth. These identifiers are not numeric, or not sequential. Obviously, an identifier can appear in one or two columns, depending on whether it was a parent or not. These should be consistent. Not all identifiers appear in the individual column - it is possible for a parent not to have its own record if its parents were not known. Missing parental (sire and/or dam) identifiers can occur. I need to export the data for use in another program that requires the pedigree to be coded as integers, increasing with date of birth (therefore sire and dam always have lower identifiers than their offspring) and with missing values coded as 0. How would I go about doing this? You might look at http://www.qimr.edu.au/davidD/sib-pair.R, specifically the read.pedigree() and wrlink() functions. The former is not very impressive speedwise -- I usually perform these tasks in the my Sib-pair (Fortran) program, which is on the same webpage. It will order the pedigree by generational position, so a DOB is not required to do the sort. Terry Therneau's kinship package does that ordering, but doesn't include output routines for the Linkage format. David Duffy. | David Duffy (MBBS PhD) ,-_|\ | email: [EMAIL PROTECTED] ph: INT+61+7+3362-0217 fax: -0101 / * | Epidemiology Unit, Queensland Institute of Medical Research \_,-._/ | 300 Herston Rd, Brisbane, Queensland 4029, Australia GPG 4D0B994A v __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] R Help
I got the Warning message below when I tried to load Locfit. What is wrong?
Regards
Ola Asteman
--
R version 2.4.0 (2006-10-03)
Copyright (C) 2006 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.
library(foreign)
library(mgcv)
This is mgcv 1.3-19
library(locfit)
Loading required package: akima
Error: package 'akima' could not be loaded
In addition: Warning message:
there is no package called 'akima' in: library(pkg, character.only = TRUE,
logical = TRUE, lib.loc = lib.loc)
--
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Re: [R] outdated R and package dependencies; was: R Help
Ola Asteman wrote:
I got the Warning message below when I tried to load Locfit. What is wrong?
Please read the posting guide which suggests to use a sensible subject line.
Regards
Ola Asteman
--
R version 2.4.0 (2006-10-03)
It makes sense to upgrade to a recent version of R.
Copyright (C) 2006 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.
library(foreign)
library(mgcv)
This is mgcv 1.3-19
library(locfit)
How was locfit installed?
install.packages(locfit, dependencies = TRUE)
should also install the package's dependencies - and you have not all
installed, among them akima.
Uwe Ligges
Loading required package: akima
Error: package 'akima' could not be loaded
In addition: Warning message:
there is no package called 'akima' in: library(pkg, character.only = TRUE,
logical = TRUE, lib.loc = lib.loc)
--
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Re: [R] R Help
You don't have installed the akima pakage.
install.packages(akima, dep=T)
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 28/08/07, Ola Asteman [EMAIL PROTECTED] wrote:
I got the Warning message below when I tried to load Locfit. What is
wrong?
Regards
Ola Asteman
--
R version 2.4.0 (2006-10-03)
Copyright (C) 2006 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.
library(foreign)
library(mgcv)
This is mgcv 1.3-19
library(locfit)
Loading required package: akima
Error: package 'akima' could not be loaded
In addition: Warning message:
there is no package called 'akima' in: library(pkg, character.only = TRUE,
logical = TRUE, lib.loc = lib.loc)
--
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Re: [R] R Help
[EMAIL PROTECTED] napsal dne 28.08.2007 13:33:13:
You don't have installed the akima pakage.
install.packages(akima, dep=T)
And wait about two months and update your R version to 2.6.0. Or update
now to 2.5.1
Regards
Petr
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40 S 49° 16' 22 O
On 28/08/07, Ola Asteman [EMAIL PROTECTED] wrote:
I got the Warning message below when I tried to load Locfit. What is
wrong?
Regards
Ola Asteman
--
R version 2.4.0 (2006-10-03)
Copyright (C) 2006 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.
library(foreign)
library(mgcv)
This is mgcv 1.3-19
library(locfit)
Loading required package: akima
Error: package 'akima' could not be loaded
In addition: Warning message:
there is no package called 'akima' in: library(pkg, character.only =
TRUE,
logical = TRUE, lib.loc = lib.loc)
--
This e-mail and any attachment may be confidential and m...{{dropped}}
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[R] help with aggregate(): tables of means for terms in an mlm
I'm trying to extend some work in the car and heplots packages
that requires getting a table of multivariate means for one
(or later, more) terms in an mlm object. I can do this for
concrete examples, using aggregate(), but can't figure out how to
generalize it. I want to return a result that has the factor-level
combinations as rownames, and the means as the body of the table
(aggregate returns the factors as initial columns).
# Examples: m1 m2 are desired results
library(car)
soils.mod - lm(cbind(pH,N,Dens,P,Ca,Mg,K,Na,Conduc) ~ Block +
Contour*Depth, data=Soils)
term.names(soils.mod)
[1] (Intercept) Block Contour Depth
[5] Contour:Depth
# response variables
resp- model.response(model.frame(soils.mod))
# 1-factor means: term=Contour
m1-aggregate(resp, list(Soils$Contour), mean)
rownames(m1) - m1[,1]
( m1 - m1[,-1] )
pH N Dens PCaMg KNa Conduc
Depression 4.692 0.08731 1.343 188.2 7.101 8.986 0.3781 5.823 6.946
Slope 4.746 0.10594 1.333 159.4 8.109 8.320 0.4156 6.103 6.964
Top4.570 0.11256 1.272 150.9 8.877 8.088 0.6050 4.872 5.856
# 2-factor means: term=Contour:Depth
m2-aggregate(resp, list(Soils$Contour, Soils$Depth), mean)
rownames(m2) - paste(m2[,1], m2[,2],sep=:)
( m2 - m2[,-(1:2)] )
pH N Dens P CaMg K Na
Conduc
Depression:0-10 5.353 0.17825 0.9775 333.0 10.685 7.235 0.6250 1.5125
1.473
Slope:0-10 5.508 0.21900 1.0500 258.0 12.248 7.232 0.6350 1.9900
2.050
Top:0-10 5.332 0.19550 1.0025 242.8 13.385 6.590 0.8000 0.9225
1.373
Depression:10-30 4.880 0.08025 1.3575 187.5 7.548 9.635 0.4500 4.6400
5.480
Slope:10-30 5.283 0.10100 1.3475 160.2 9.515 8.980 0.4800 4.9350
4.910
Top:10-304.850 0.11750 1.3325 147.5 10.238 8.090 0.6500 2.9800
3.583
Depression:30-60 4.362 0.05050 1.5350 124.2 5.402 9.918 0.2400 7.5875
9.393
Slope:30-60 4.268 0.06075 1.5100 114.5 5.877 8.968 0.3000 7.6300
8.925
Top:30-604.205 0.07950 1.3225 116.2 6.620 8.742 0.5450 6.2975
7.440
Depression:60-90 4.173 0.04025 1.5025 108.0 4.770 9.157 0.1975 9.5525
11.438
Slope:60-90 3.927 0.04300 1.4225 105.0 4.798 8.100 0.2475 9.8575
11.970
Top:60-903.893 0.05775 1.4300 97.0 5.268 8.928 0.4250 9.2900
11.030
Here is the current version of a function that doesn't work, because I
can't supply the factor names to aggregate in the proper way.
Can someone help me make it work?
termMeans.mlm - function( object, term ) {
resp- model.response(model.frame(object))
terms - term.names(soils.mod)
terms - terms[terms != (Intercept)]
factors - strsplit(term, :)
# browser()
means - aggregate(resp, factors, mean)
# rownames(means) - ...
# means - means[, -(1:length(factors)]
}
termMeans.mlm(soils.mod, Contour)
Error in FUN(X[[1L]], ...) : arguments must have same length
thanks,
-Michael
--
Michael Friendly Email: friendly AT yorku DOT ca
Professor, Psychology Dept.
York University Voice: 416 736-5115 x66249 Fax: 416 736-5814
4700 Keele Streethttp://www.math.yorku.ca/SCS/friendly.html
Toronto, ONT M3J 1P3 CANADA
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[R] Help on coxph
I am new to R. I just installed R2.5.1 in my computer and tried to use coxph, but it gives me this message: No documentation for 'coxph' in specified packages and libraries: you could try 'help.search(coxph)' can anyone tell me how to install this package? thanks a lot [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on coxph
In the R-surface menu, go to packages - install packages. Select a server and install the package survival. Then, on the prompt, type: library(survival) Then it works. Look up the numerous manual on using R if you need further help. If that's not the issue, give us more detailed information on the problem. Daniel PhD Program Strategy Dept. of Management and Organization Robert H. Smith School of Business University of Maryland Van Munching Hall College Park, MD 20742 www.rhsmith.umd.edu www.umd.edu mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] -Ursprüngliche Nachricht- Von: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] Im Auftrag von Wei Hu Gesendet: Sunday, August 26, 2007 9:49 PM An: r-help@stat.math.ethz.ch Betreff: [R] Help on coxph I am new to R. I just installed R2.5.1 in my computer and tried to use coxph, but it gives me this message: No documentation for 'coxph' in specified packages and libraries: you could try 'help.search(coxph)' can anyone tell me how to install this package? thanks a lot [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with vector gymnastics
try 5*which(tf)[cumsum(tf)] Gladwin, Philip schrieb: Hello, What is the best way of solving this problem? answer - ifelse(tf=TRUE, i * 5, previous answer) where as an initial condition tf[1] - TRUE For example if, tf - c(T,F,F,F,T,T,F) over i = 1 to 7 then the output of the function will be answer = 5 5 5 5 25 30 30 Thank you. Phil, __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 22046 Hamburg T ++49/40/42803-8243 F ++49/40/42803-7790 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help with vector gymnastics
Hello, What is the best way of solving this problem? answer - ifelse(tf=TRUE, i * 5, previous answer) where as an initial condition tf[1] - TRUE For example if, tf - c(T,F,F,F,T,T,F) over i = 1 to 7 then the output of the function will be answer = 5 5 5 5 25 30 30 Thank you. Phil, __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with vector gymnastics
library(zoo) tf - c(T,F,F,F,T,T,F) i - seq(7) answer - ifelse(tf, i*5, NA) answer - na.locf(answer) On 8/23/07, Gladwin, Philip [EMAIL PROTECTED] wrote: Hello, What is the best way of solving this problem? answer - ifelse(tf=TRUE, i * 5, previous answer) where as an initial condition tf[1] - TRUE For example if, tf - c(T,F,F,F,T,T,F) over i = 1 to 7 then the output of the function will be answer = 5 5 5 5 25 30 30 Thank you. Phil, __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Felix Andrews / 安福立 PhD candidate Integrated Catchment Assessment and Management Centre The Fenner School of Environment and Society The Australian National University (Building 48A), ACT 0200 Beijing Bag, Locked Bag 40, Kingston ACT 2604 http://www.neurofractal.org/felix/ voice:+86_1051404394 (in China) mobile:+86_13522529265 (in China) mobile:+61_410400963 (in Australia) xmpp:[EMAIL PROTECTED] 3358 543D AAC6 22C2 D336 80D9 360B 72DD 3E4C F5D8 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with vector gymnastics
Philip - I don't know if this is the best way, but it gives you the output you want. Using your tf, vals - rle(ifelse(tf, 5*which(tf), 0)) vals$values[vals$values == 0] - vals$values[which(vals$values==0) - 1] inverse.rle(vals) [1] 5 5 5 5 25 30 30 Gladwin, Philip wrote: Hello, What is the best way of solving this problem? answer - ifelse(tf=TRUE, i * 5, previous answer) where as an initial condition tf[1] - TRUE For example if, tf - c(T,F,F,F,T,T,F) over i = 1 to 7 then the output of the function will be answer = 5 5 5 5 25 30 30 Thank you. Phil, __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help with optimization using GENOUD
Dear Friends,
I have been trying to learn how to use the derivative free optimization
algorithms implemented in the package RGENOUD by Mebane and Sekhon. However, it
does not seem to work for reasons best described as my total ignorance. If
anybody has experience using this package, it would be really helpful if you
can point out where I'm making a mistake.
Thanks in advance
Anup
Sample code attached
library(rgenoud)
nobs - 5000
t.beta - c(0,1,-1)
X - as.matrix(cbind(rep(1, nobs), runif(nobs), runif(nobs))) # Creating the
design matrix
prodterm - (X%*%t.beta)+rnorm(nrow(X))
Y - as.matrix(ifelse(prodterm0, 0, 1))
# Defining the likelihood function
log.like - function(beta, Y, X)
{
term1 - pnorm(X%*%beta)
term2 - 1-term1
loglik - (sum(Y*log(term1))+sum((1-Y)*log(term2))) # Likelihood function to be
maximized
}
stval - c(0,0,0)
opt.output - optim(stval,log.like,Y=Y[,1], X=X[,1:3],
hessian=T, method=BFGS, control=c(fnscale=-1,trace=1))
opt.output
### Now using GENOUD gives me errors
genoud.output - genoud(log.like,beta=stval,X=X[,1:3], Y=Y[,1], nvars=3,
pop.size=3000, max=TRUE)
-
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[R] help and Firefox
My configuration is Windows XP, R-2.5.1patched. My standard browser in Windows is Firefox 2.0.6, and I am using htmlhelp. I have problems with starting the browser for displaying help. help(lm) works as it should when Firefox is already running. When I do help(lm) and the browser is not yet started, I get Error in shell.exec(url) : 'C:\PROGRA~2\R\R-25~1.1\library\stats\html\lm.html' not found but nevertheless the browser starts and the html file is displayed some seconds later. If I try to use help from within a function and the browser is not open, the browser will not start and therefore help will not be displayed. Has anybody else experienced the same problem? Is there a solution? -- Erich Neuwirth, University of Vienna Faculty of Computer Science Computer Supported Didactics Working Group Visit our SunSITE at http://sunsite.univie.ac.at Phone: +43-1-4277-39464 Fax: +43-1-4277-39459 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help and Firefox
I found a solution to my problem. It is describe here. http://kb.mozillazine.org/Windows_error_opening_Internet_shortcut_or_local_HTML_file_-_Firefox Essentially, it involves switching off DDE for some file associations in Windows Explorer. It really is a Firefox bug. Erich Neuwirth wrote: My configuration is Windows XP, R-2.5.1patched. My standard browser in Windows is Firefox 2.0.6, and I am using htmlhelp. I have problems with starting the browser for displaying help. help(lm) works as it should when Firefox is already running. When I do help(lm) and the browser is not yet started, I get Error in shell.exec(url) : 'C:\PROGRA~2\R\R-25~1.1\library\stats\html\lm.html' not found but nevertheless the browser starts and the html file is displayed some seconds later. If I try to use help from within a function and the browser is not open, the browser will not start and therefore help will not be displayed. Has anybody else experienced the same problem? Is there a solution? -- Erich Neuwirth, University of Vienna Faculty of Computer Science Computer Supported Didactics Working Group Visit our SunSITE at http://sunsite.univie.ac.at Phone: +43-1-4277-39464 Fax: +43-1-4277-39459 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with npmc
I cant't seem to get npmc to make a comparison to a control level summary(npmc(brain), type=BF, control=1) $`Data-structure` group.index class.level nobs c 1 c 30 l 2 l 30 r 3 r 30 $`Results of the multiple Behrens-Fisher-Test` cmpeffect lower.cl upper.cl p.value.1s p.value.2s 1 1-2 0.643 0.4610459 0.8256208 0.08595894 0.14750647 2 1-3 0.444 0.2576352 0.6312537 0.99636221 0.75376639 3 2-3 0.328 0.1602449 0.4964218 1. 0.04476692 What elementary error am I making. Thanks, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help with scatterplot3d
Hello, I am having a bit of trouble with scatterplot3d(). I was able to plot a 3d cloud of points using the following code: my.3dplot-scatterplot3d(my.coords, pch=19, zlim=c(0,1), scale.y=0.5, angle=30, box=FALSE) where my.coords is a data frame that contains x, y, and z coordinates for grid points whose elevation we sampled. The problem occurs when I try to add points using points3d. I tried to follow the code in the examples of the package pdf. First, I tried the following code to add all of the points that I wanted: my.3dplot$points3d(seq(400,600,0.19), seq(600,400,0.295), seq(800,500,0.24), seq(1000,1400,0.22), seq(1200,600,0.24), seq(1200,1500,0.28), seq(1300,1400,0.205), seq(1700,500,0.26), seq(1700,600,0.21), seq(1900,1400,0.255), seq(2300,1400,0.275), seq(2600,1300,0.225), seq(2700,400,0.235), seq(2700,1300,0.265), seq(3100,1000,0.135), col=blue, type=h, pch=16) and got the following error: Error in seq.default(600, 400, 0.295) : wrong sign in 'by' argument So I just tried to add the first point using the following code: my.3dplot$points3d(seq(400,600,0.19), col=blue, type=h, pch=16) and got the following error message: Error in xyz.coords(x, y, z) : 'x', 'y' and 'z' lengths differ. Does anyone know how I can use this function in the scatterplot3d package? Thanks so much! Ryan ~~ Ryan D. Briscoe Runquist Population Biology Graduate Group University of California, Davis [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with scatterplot3d
On 8/13/2007 3:03 PM, Ryan Briscoe Runquist wrote: Hello, I am having a bit of trouble with scatterplot3d(). I was able to plot a 3d cloud of points using the following code: my.3dplot-scatterplot3d(my.coords, pch=19, zlim=c(0,1), scale.y=0.5, angle=30, box=FALSE) where my.coords is a data frame that contains x, y, and z coordinates for grid points whose elevation we sampled. The problem occurs when I try to add points using points3d. I tried to follow the code in the examples of the package pdf. First, I tried the following code to add all of the points that I wanted: my.3dplot$points3d(seq(400,600,0.19), seq(600,400,0.295), seq(800,500,0.24), seq(1000,1400,0.22), seq(1200,600,0.24), seq(1200,1500,0.28), seq(1300,1400,0.205), seq(1700,500,0.26), seq(1700,600,0.21), seq(1900,1400,0.255), seq(2300,1400,0.275), seq(2600,1300,0.225), seq(2700,400,0.235), seq(2700,1300,0.265), seq(3100,1000,0.135), col=blue, type=h, pch=16) The header to the function (which you can see by evaluating my.3dplot$points3d) is function (x, y = NULL, z = NULL, type = p, ...) so you are setting x to seq(400,600,0.19), y to seq(600,400,0.295), etc. I think you probably want my.3dplot$points3d(x=c(400, 600, ...), y=c(600, 400, ...), z=c(0.19, 0.295, ...), ...) (where the ... is to be filled in by you.) and got the following error: Error in seq.default(600, 400, 0.295) : wrong sign in 'by' argument So I just tried to add the first point using the following code: my.3dplot$points3d(seq(400,600,0.19), col=blue, type=h, pch=16) and got the following error message: Error in xyz.coords(x, y, z) : 'x', 'y' and 'z' lengths differ. Does anyone know how I can use this function in the scatterplot3d package? try my.3dplot$points3d(x=400, y=600, z=0.19, col=blue, type=h, pch=16) Duncan Murdoch __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with scatterplot3d
Ryan Briscoe Runquist rdbriscoe at ucdavis.edu writes: Hello, I am having a bit of trouble with scatterplot3d(). I was able to plot a 3d cloud of points using the following code: my.3dplot-scatterplot3d(my.coords, pch=19, zlim=c(0,1), scale.y=0.5, angle=30, box=FALSE) where my.coords is a data frame that contains x, y, and z coordinates for grid points whose elevation we sampled. The problem occurs when I try to add points using points3d. I tried to follow the code in the examples of the package pdf. First, I tried the following code to add all of the points that I wanted: my.3dplot$points3d(seq(400,600,0.19), seq(600,400,0.295), seq(800,500,0.24), seq(1000,1400,0.22), seq(1200,600,0.24), seq(1200,1500,0.28), seq(1300,1400,0.205), seq(1700,500,0.26), seq(1700,600,0.21), seq(1900,1400,0.255), seq(2300,1400,0.275), seq(2600,1300,0.225), seq(2700,400,0.235), seq(2700,1300,0.265), seq(3100,1000,0.135), col=blue, type=h, pch=16) I think you probably want: xvals - c(400,600,800,1000,1200,1200,1300,1700,1700,1900, 2300,2600,2700,2700,3100) yvals - c(600,400,500,1400,600,1500,1400,500,600, 1400,1400,1300,400,1300,1000) zvals - c(0.19,0.295,0.24,0.22,0.24,0.28,0.205,0.26, 0.21,0.255,0.275,0.225,0.235,0.265,0.135) my.3dplot$points3d(x=xvals,y=yvals,z=zvals,col=blue,type=h,pch=16) Look carefully at ?seq and you may understand what you've done wrong ... Ben Bolker __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help using gPath
Here's a partial extract from a sample session after running your code (NOTE this is using the development version of R; grid.ls() does not exist in R 2.5.1 or earlier): Inspect the grob tree with grid.ls() (similar to Hadley's current.grobTree(), but with different formatting) ... (I'll probably remove current.grobTree as soon as grid.ls makes it to a released version of R) grid.ls() plot-surrounds GRID.cellGrob.118 background GRID.cellGrob.119 plot.gTree.113 background guide.gTree.90 background.rect.80 minor-horizontal.segments.82 minor-vertical.segments.84 # OUTPUT TRUNCATED The format is much nicer than mine! ... It is not necessarily obvious which grob is which, but a little trial and error (e.g., grid.edit() to change the colour of a grob) shows that the border on the first panel is 'guide.rect.92', which is a child of 'plot.gTree.113' (NOTE the numbers come from a fresh R session). I will try and rename these grobs so that they are more easily accessible (and reproducible across multiple calls). That should make things easier in the future. Use grid.get() to grab that gTree and inspect that further using grid.ls(), this time also showing the viewports involved ... What do all the upViewports represent? Could the downViewports be incorporating into the same place as the original definition? (The remaining code should work for you in your version of R; it is just grid.ls() that is new.) Remove the original border rect, ... grid.remove(guide.rect.92, global=TRUE) ... (need global=TRUE because the border appears twice as a child of 'plot.gTree.113' [not sure why that is]) then add some lines that only draw the top, right, and bottom borders ... grid.add(plot.gTree.113, linesGrob(c(0, 1, 1, 0), c(1, 1, 0, 0), gp=gpar(col=green), vp=vpPath(layout, panel_1_1))) ... (I drew the new lines green so that they are easy to see). NOTE that in order to put the new lines in the same place as the original border, the new lines are added as children of the gTree 'plot.gTree.113' and they have a vpPath to make sure they get drawn in the right viewport within that gTree. Do you think it would be worth drawing all these rectangles as lines to make them easier to edit? What would probably be ideal would be a graphical interface to the grid.ls()-type information (something like an object explorer) that would make it easier to see which object is which and also make it easier to add and remove objects. A nice student project perhaps :) That would be great! Hadley -- http://had.co.nz/ __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help using gPath
# OUTPUT TRUNCATED ... which shows that 'guide.rect.92' is drawn within the viewport 'panel_1_1' (which is within the viewport called 'layout'). (The remaining code should work for you in your version of R; it is just grid.ls() that is new.) Remove the original border rect, ... grid.remove(guide.rect.92, global=TRUE) ... (need global=TRUE because the border appears twice as a child of 'plot.gTree.113' [not sure why that is]) then add some lines that only draw the top, right, and bottom borders ... grid.add(plot.gTree.113, linesGrob(c(0, 1, 1, 0), c(1, 1, 0, 0), gp=gpar(col=green), vp=vpPath(layout, panel_1_1))) ... (I drew the new lines green so that they are easy to see). NOTE that in order to put the new lines in the same place as the original border, the new lines are added as children of the gTree 'plot.gTree.113' and they have a vpPath to make sure they get drawn in the right viewport within that gTree. What would probably be ideal would be a graphical interface to the grid.ls()-type information (something like an object explorer) that would make it easier to see which object is which and also make it easier to add and remove objects. A nice student project perhaps :) Paul -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help with cumsum function
Hello! I have a question regarding the cumsum function that I do not know how to solve. Would appreciate help from someone. I have imported data from a txtfile with 2 columns. I am interested in the seconds column, which contains numbers from i=0 to 40. I would like to count the number of instances of the numbers 0 to 40 in the seconds column. My problem is that if there are no observations of one number i in the second column cumsum skips this in the output. Below is an example of the output of the argument Y=cumsum(table(x[,2])) 0 1 2 3 5 6 789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 3 5 6 10 13 14 16 18 21 25 27 30 32 35 38 40 42 54 105 233 306 341 383 417 2526 27 28 2930 31 32 33 34 35 36 37 38 39 40 441 468 487 502 518 532 542 546 552 564 566 574 578 584 591 594 As you can see from the output there are no observations of i=4 in this column. Are there any way I could return the following result instead 0 1 2 3 45 6 78 and so on 3 5 6 10 10 13 14 16 18 When i=4 cumsum is the same as when i=3 (10) I would really appreciate if anyone could help me with this one:) Jan Moberg __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with cumsum function
Hectorman Hectorman wrote: Hello! I have a question regarding the cumsum function that I do not know how to solve. Would appreciate help from someone. I have imported data from a txtfile with 2 columns. I am interested in the seconds column, which contains numbers from i=0 to 40. I would like to count the number of instances of the numbers 0 to 40 in the seconds column. My problem is that if there are no observations of one number i in the second column cumsum skips this in the output. Below is an example of the output of the argument Y=cumsum(table(x[,2])) 0 1 2 3 5 6 789 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 3 5 6 10 13 14 16 18 21 25 27 30 32 35 38 40 42 54 105 233 306 341 383 417 2526 27 28 2930 31 32 33 34 35 36 37 38 39 40 441 468 487 502 518 532 542 546 552 564 566 574 578 584 591 594 As you can see from the output there are no observations of i=4 in this column. Are there any way I could return the following result instead 0 1 2 3 45 6 78 and so on 3 5 6 10 10 13 14 16 18 When i=4 cumsum is the same as when i=3 (10) Y - cumsum(table(factor(x[,2], levels=0:40))) Uwe Ligges I would really appreciate if anyone could help me with this one:) Jan Moberg __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help using gPath
Hi Emilio Gagliardi wrote: Hi Paul, I'm sorry for not posting code, I wasn't sure if it would be helpful without the data...should I post the code and a sample of the data? I will remember to do that next time! It's important not only to post code, but also to make sure that other people can run it (i.e., include real data or have the code generate data or use one of R's predefined data sets). Also, isn't this next time ? :) grid.gedit(gPath(ylabel.text.382), gp=gpar(fontsize=16)) OK, I think my confusion comes from the notation that current.grobTree() produces and what strings are required in order to make changes to the underlying grobs. But, from what you've provided, it looks like I can access each grob with its unique name, regardless of which parent it is nested in...that helps Yes. By default, grid will search the tree of all grobs to find the name you provide. You can even just provide part of the name and it will find partial matches (depending on argument settings). On the other hand, by specifying a path that specified parent and child grobs, you can make sure you get exactly the grob you want. like to remove the left border on the first panel. I'd like to adjust the I'd guess you'd have to remove the grob background.rect.345 and then draw in just the sides you want, which would require getting to the right viewport, for which you'll need to study the viewport tree (see current.vpTree()) I did some digging into this and it seems pretty complicated, is there an example anywhere that makes sense to the beginner? The whole viewport grob relationship is not clear to me. So, accessing viewports and removing objects and drawing new ones is beyond me at this point. I can get my mind around your example below because I can see the object I want to modify in the viewer, and the code changes a property of that object, click enter, and bang the object changes. When you start talking external pointers and finding viewports and pushing and popping grobs I just get lost. I found the viewports for the grobTree, it looks like this: There's a book that provides a full explanation and the (basic) grid chapter is online (see http://www.stat.auckland.ac.nz/~paul/RGraphics/rgraphics.html) viewport[ROOT]-(viewport[layout]-(viewport[axis_h_1_1]-(viewport[bottom_axis]-(viewport[labels], viewport[ticks])), viewport[axis_h_1_2]-(viewport[bottom_axis]-(viewport[labels], viewport[ticks])), viewport[axis_v_1_1]-(viewport[left_axis]-(viewport[labels], viewport[ticks])), viewport[panel_1_1], viewport[panel_1_2], viewport[strip_h_1_1], viewport[strip_h_1_2], viewport[strip_v_1_1])) at that point I was like, ok, I'm done. :S Yep, the facilities for investigating the viewport and grob tree are basically inadequate. Based on some work Hadley did for ggplot, the development version of R has a slightly better tool called grid.ls() that can show how the grob tree and the viewport tree intertwine. That would allow you to see which viewport each grob was drawn in, which would help you, for example, to know which viewport you had to go to to replace a rectangle you want to remove. Something like ... grid.gedit(geom_bar.rect, gp=gpar(col=green)) Again, it would really help to have some code to run. My apologies, I thought the grobTree was sufficient in this case. Thanks very much for your help. Sorry to harp on about it, but if I had your code I could show you an example of how grid.ls() might help. Paul -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with counting how many times each value occur in each column
[Gabor Grothendieck] table(col(mat), mat) Clever, simple, and elegant! :-) -- François Pinard http://pinard.progiciels-bpi.ca __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help wit matrices
An even simpler solution is: mat2 - 1 * (mat1 0.25) Ravi. --- Ravi Varadhan, Ph.D. Assistant Professor, The Center on Aging and Health Division of Geriatric Medicine and Gerontology Johns Hopkins University Ph: (410) 502-2619 Fax: (410) 614-9625 Email: [EMAIL PROTECTED] Webpage: http://www.jhsph.edu/agingandhealth/People/Faculty/Varadhan.html -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Lanre Okusanya Sent: Friday, August 10, 2007 2:20 PM To: jim holtman Cc: r-help@stat.math.ethz.ch Subject: Re: [R] Help wit matrices that was ridiculously simple. duh. THanks Lanre On 8/10/07, jim holtman [EMAIL PROTECTED] wrote: Is this what you want: x - matrix(runif(100), 10) round(x, 3) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 0.268 0.961 0.262 0.347 0.306 0.762 0.524 0.062 0.028 0.226 [2,] 0.219 0.100 0.165 0.131 0.578 0.933 0.317 0.109 0.527 0.131 [3,] 0.517 0.763 0.322 0.374 0.910 0.471 0.278 0.382 0.880 0.982 [4,] 0.269 0.948 0.510 0.631 0.143 0.604 0.788 0.169 0.373 0.327 [5,] 0.181 0.819 0.924 0.390 0.415 0.485 0.702 0.299 0.048 0.507 [6,] 0.519 0.308 0.511 0.690 0.211 0.109 0.165 0.192 0.139 0.681 [7,] 0.563 0.650 0.258 0.689 0.429 0.248 0.064 0.257 0.321 0.099 [8,] 0.129 0.953 0.046 0.555 0.133 0.499 0.755 0.181 0.155 0.119 [9,] 0.256 0.954 0.418 0.430 0.460 0.373 0.620 0.477 0.132 0.050 [10,] 0.718 0.340 0.854 0.453 0.943 0.935 0.170 0.771 0.221 0.929 ifelse(x .5, 1, 0) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,]010001100 0 [2,]000011001 0 [3,]110010001 1 [4,]011101100 0 [5,]011000100 1 [6,]101100000 1 [7,]110100000 0 [8,]010100100 0 [9,]010000100 0 [10,]101011010 1 On 8/10/07, Lanre Okusanya [EMAIL PROTECTED] wrote: Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2-as.matrix(as.numeric(mat10.25)) but that returns a 1:10 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help wit matrices
Will something like this help? mm - matrix(rnorm(100),nrow=10) mm nn - mm .5 nn --- Lanre Okusanya [EMAIL PROTECTED] wrote: Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2-as.matrix(as.numeric(mat10.25)) but that returns a 1:10 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help wit matrices
I hope you don't mind that I offer also two solutions. No.1 is really
bad. No.2 should be on par with the other ones.
Best,
Roland
mydata - matrix(rnorm(10*10), ncol=10)
threshold.value - 1.5
mydata2 - matrix(0, nrow=nrow(mydata), ncol=ncol(mydata))
mydata3 - matrix(0, nrow=nrow(mydata), ncol=ncol(mydata))
### not really the way to go:
for (i in 1:nrow(mydata)) {
for (j in 1:ncol(mydata)) {
if (mydata[i,j]threshold.value) {
mydata2[i,j] - 1
}
}
}
### a better way...
mydata3[mydata threshold.value] - 1
mydata2
mydata3
Lanre Okusanya wrote:
Hello all,
I am working with a 1000x1000 matrix, and I would like to return a
1000x1000 matrix that tells me which value in the matrix is greater
than a theshold value (1 or 0 indicator).
i have tried
mat2-as.matrix(as.numeric(mat10.25))
but that returns a 1:10 matrix.
I have also tried for loops, but they are grossly inefficient.
THanks for all your help in advance.
Lanre
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help using gPath
haha Paul, It's important not only to post code, but also to make sure that other people can run it (i.e., include real data or have the code generate data or use one of R's predefined data sets). Oh, I hadn't thought of using the predefined datasets, thats a good idea! Also, isn't this next time ? :) By next time I meant, when I ask a question in the future, I didn't think you'd respond! So here is some code! library(reshape) library(ggplot2) theme_t - list(grid.fill=white,grid.colour=lightgrey,background.colour= black,axis.colour=dimgrey) ggtheme(theme_t) grp - c(2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3) time - c(1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2) cc - c(0.7271,0.7563,0.6979,0.8208,0.7521,0.7875,0.7563,0.7771,0.8208, 0.7938,0.8083,0.7188,0.7521,0.7854,0.7979,0.7583,0.7646,0.6938,0.6813,0.7708 ,0.7375,0.8104,0.8104,0.7792,0.7833,0.8083,0.8021,0.7313,0.7958,0.7021, 0.8167,0.8167,0.7583,0.7167,0.6563,0.6896,0.7333,0.8208,0.7396,0.8063,0.7083 ,0.6708,0.7292,0.7646,0.7667,0.775,0.8021,0.8125,0.7646,0.6917,0.7458,0.7833 ,0.7396,0.7229,0.7708,0.7729,0.8083,0.7771,0.6854,0.8417,0.7667,0.7063,0.75, 0.7813,0.8271,0.7896,0.7979,0.625,0.7938,0.7583,0.7396,0.7583,0.7938,0.7333, 0.7875,0.8146) data - as.data.frame(cbind(time,grp,cc)) data$grp - factor(data$grp,labels=c(Group A,Group B)) data$time - factor(data$time,labels=c(Pre-test,Post-test)) boxplot - qplot(grp, cc, data=data, geom=boxplot, orientation=horizontal, ylim=c(0.5,1), main=Hello World!, xlab=Label X, ylab=Label Y, facets=.~time, colour=red, size=2) boxplot + geom_jitter(aes(colour=steelblue)) + scale_colour_identity() + scale_size_identity() grid.gedit(ylabel, gp=gpar(fontsize=16)) There's a book that provides a full explanation and the (basic) grid chapter is online (see http://www.stat.auckland.ac.nz/~paul/RGraphics/rgraphics.html) Awesome, I'll check that out. Yep, the facilities for investigating the viewport and grob tree are basically inadequate. Based on some work Hadley did for ggplot, the development version of R has a slightly better tool called grid.ls() that can show how the grob tree and the viewport tree intertwine. That would allow you to see which viewport each grob was drawn in, which would help you, for example, to know which viewport you had to go to to replace a rectangle you want to remove. okie dokie, I'm ready to be amazed! hehe. emilio [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help wit matrices
Is this what you want: x - matrix(runif(100), 10) round(x, 3) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 0.268 0.961 0.262 0.347 0.306 0.762 0.524 0.062 0.028 0.226 [2,] 0.219 0.100 0.165 0.131 0.578 0.933 0.317 0.109 0.527 0.131 [3,] 0.517 0.763 0.322 0.374 0.910 0.471 0.278 0.382 0.880 0.982 [4,] 0.269 0.948 0.510 0.631 0.143 0.604 0.788 0.169 0.373 0.327 [5,] 0.181 0.819 0.924 0.390 0.415 0.485 0.702 0.299 0.048 0.507 [6,] 0.519 0.308 0.511 0.690 0.211 0.109 0.165 0.192 0.139 0.681 [7,] 0.563 0.650 0.258 0.689 0.429 0.248 0.064 0.257 0.321 0.099 [8,] 0.129 0.953 0.046 0.555 0.133 0.499 0.755 0.181 0.155 0.119 [9,] 0.256 0.954 0.418 0.430 0.460 0.373 0.620 0.477 0.132 0.050 [10,] 0.718 0.340 0.854 0.453 0.943 0.935 0.170 0.771 0.221 0.929 ifelse(x .5, 1, 0) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,]010001100 0 [2,]000011001 0 [3,]110010001 1 [4,]011101100 0 [5,]011000100 1 [6,]101100000 1 [7,]110100000 0 [8,]010100100 0 [9,]010000100 0 [10,]101011010 1 On 8/10/07, Lanre Okusanya [EMAIL PROTECTED] wrote: Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2-as.matrix(as.numeric(mat10.25)) but that returns a 1:10 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with counting how many times each value occur in each column
Try this where we have constructed the example to illustrate that it does handle the case where not all values are in each column: mat - matrix(rep(1:6, each = 4), 6) table(col(mat), mat) On 8/10/07, Tom Cohen [EMAIL PROTECTED] wrote: Dear list, I have the following dataset and want to know how many times each value occur in each column. data [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -100 -100 -100000000 -100 [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -50 [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [10,] -100 -100 -100 -50 -100 -100 -100 -100 -100 -100 [11,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [12,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [13,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [14,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [15,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [16,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [17,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [18,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [19,] -100 -100 -100000000 -100 [20,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 The result matrix should look like -100 0 -50 [1] 20 [2] 20 [3] 20 [4] 17 [5] 18 [6] 18 [7] 18 and so on [8] [9] [10] How can I do this in R ? Thanks alot for your help, Tom - Jämför pris på flygbiljetter och hotellrum: http://shopping.yahoo.se/c-169901-resor-biljetter.html [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help wit matrices
mat2-matrix(as.numeric(mat10.25), ncol=1000) -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 10/08/07, Lanre Okusanya [EMAIL PROTECTED] wrote: Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2-as.matrix(as.numeric(mat10.25)) but that returns a 1:10 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre __ R-help@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 and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with counting how many times each value occur in each column
[Tom Cohen] I have the following dataset and want to know how many times each value occur in each column. data [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -100 -100 -100000000 -100 [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -50 [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [10,] -100 -100 -100 -50 -100 -100 -100 -100 -100 -100 [11,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [12,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [13,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [14,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [15,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [16,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [17,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [18,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [19,] -100 -100 -100000000 -100 [20,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 The result matrix should look like -100 0 -50 [1] 20 [2] 20 [3] 20 [4] 17 [5] 18 [6] 18 [7] 18 and so on [8] [9] [10] Presuming that data is a matrix, one could try a sequence like this: dataf - factor(data) dim(dataf) - dim(data) result - t(apply(dataf, 2, tabulate, nlevels(dataf))) colnames(result) - levels(dataf) result If you want the columns sorted, you might decide the order of the levels on the factor() call, or explicitly reorder columns afterwards. -- François Pinard http://pinard.progiciels-bpi.ca __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help with counting how many times each value occur in each column
Dear list, I have the following dataset and want to know how many times each value occur in each column. data [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -100 -100 -100000000 -100 [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -50 [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [10,] -100 -100 -100 -50 -100 -100 -100 -100 -100 -100 [11,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [12,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [13,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [14,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [15,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [16,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [17,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [18,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [19,] -100 -100 -100000000 -100 [20,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 The result matrix should look like -100 0 -50 [1] 20 [2] 20 [3] 20 [4] 17 [5] 18 [6] 18 [7] 18 and so on [8] [9] [10] How can I do this in R ? Thanks alot for your help, Tom - Jämför pris på flygbiljetter och hotellrum: http://shopping.yahoo.se/c-169901-resor-biljetter.html [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help wit matrices
that was ridiculously simple. duh. THanks Lanre On 8/10/07, jim holtman [EMAIL PROTECTED] wrote: Is this what you want: x - matrix(runif(100), 10) round(x, 3) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] 0.268 0.961 0.262 0.347 0.306 0.762 0.524 0.062 0.028 0.226 [2,] 0.219 0.100 0.165 0.131 0.578 0.933 0.317 0.109 0.527 0.131 [3,] 0.517 0.763 0.322 0.374 0.910 0.471 0.278 0.382 0.880 0.982 [4,] 0.269 0.948 0.510 0.631 0.143 0.604 0.788 0.169 0.373 0.327 [5,] 0.181 0.819 0.924 0.390 0.415 0.485 0.702 0.299 0.048 0.507 [6,] 0.519 0.308 0.511 0.690 0.211 0.109 0.165 0.192 0.139 0.681 [7,] 0.563 0.650 0.258 0.689 0.429 0.248 0.064 0.257 0.321 0.099 [8,] 0.129 0.953 0.046 0.555 0.133 0.499 0.755 0.181 0.155 0.119 [9,] 0.256 0.954 0.418 0.430 0.460 0.373 0.620 0.477 0.132 0.050 [10,] 0.718 0.340 0.854 0.453 0.943 0.935 0.170 0.771 0.221 0.929 ifelse(x .5, 1, 0) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,]010001100 0 [2,]000011001 0 [3,]110010001 1 [4,]011101100 0 [5,]011000100 1 [6,]101100000 1 [7,]110100000 0 [8,]010100100 0 [9,]010000100 0 [10,]101011010 1 On 8/10/07, Lanre Okusanya [EMAIL PROTECTED] wrote: Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2-as.matrix(as.numeric(mat10.25)) but that returns a 1:10 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help wit matrices
Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2-as.matrix(as.numeric(mat10.25)) but that returns a 1:10 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help wit matrices
On 10-Aug-07 18:05:50, Lanre Okusanya wrote: Hello all, I am working with a 1000x1000 matrix, and I would like to return a 1000x1000 matrix that tells me which value in the matrix is greater than a theshold value (1 or 0 indicator). i have tried mat2-as.matrix(as.numeric(mat10.25)) but that returns a 1:10 matrix. I have also tried for loops, but they are grossly inefficient. THanks for all your help in advance. Lanre Simple-minded, but: S-matrix(rnorm(25),nrow=5) S [,1][,2] [,3] [,4] [,5] [1,] -0.9283624 -0.44418487 1.1174555 1.9040999 -0.4675796 [2,] 0.2658770 -0.28492642 -1.2271013 -0.5713291 1.8036235 [3,] 0.7010885 -0.42972262 0.7576021 0.3407972 -1.0628487 [4,] -0.2003087 0.87006841 0.6233792 -0.9974902 -0.9104270 [5,] 0.2729014 0.09781886 -1.0004486 1.5987385 -0.4747125 T-0*S T[S0.25] - 1+0*S[S0.25] T [,1] [,2] [,3] [,4] [,5] [1,]00110 [2,]10001 [3,]10110 [4,]01100 [5,]10010 Does this work OK for your big matrix? HTH Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 10-Aug-07 Time: 19:50:37 -- XFMail -- __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with counting how many times each value occur in eachcolumn
Tom, If all values (-100,0,-50) would be in every column then simple apply(data,2,table) would work. Even if there aren0t all values in every column you could correct that and insert additional lines with all values for all columns like data - cbind(data,matrix(ncol=10,nrow=3,rep(c(-100,0,-50),10))) and then do apply(data,2,table)-1 to get correct results. But someone on a list can probably make much more elegant solution. Bye, Gasper Cankar, PhD Researcher National Examinations Centre Slovenia -Original Message- From: Tom Cohen [mailto:[EMAIL PROTECTED] Sent: Friday, August 10, 2007 2:02 PM To: r-help@stat.math.ethz.ch Subject: [R] help with counting how many times each value occur in eachcolumn Dear list, I have the following dataset and want to know how many times each value occur in each column. data [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [1,] -100 -100 -100000000 -100 [2,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [3,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [4,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [5,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -50 [6,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [7,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [8,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [9,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [10,] -100 -100 -100 -50 -100 -100 -100 -100 -100 -100 [11,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [12,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [13,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [14,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [15,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [16,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [17,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [18,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 [19,] -100 -100 -100000000 -100 [20,] -100 -100 -100 -100 -100 -100 -100 -100 -100 -100 The result matrix should look like -100 0 -50 [1] 20 [2] 20 [3] 20 [4] 17 [5] 18 [6] 18 [7] 18 and so on [8] [9] [10] How can I do this in R ? Thanks alot for your help, Tom - Jämför pris på flygbiljetter och hotellrum: http://shopping.yahoo.se/c-169901-resor-biljetter.html [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help on R performance using aov function
Hi,
Im trying to replace some SAS statistical functions by R (batch calling).
But Ive seen that calling R in a batch mode (under Unix) takes about 2or 3
times more than SAS software. So its a great problem of performance for me.
Here is an extract of the calculation:
stoutput-file(res_oneWayAnova.dat,w);
cat(Param|F|Prob,file=stoutput,\n);
for (i in 1:n) {
p-list_param[[i]]
aov_-aov(A[,p]~ A[,wafer],data=A);
anova_-summary(aov_);
if (!is.na(anova_[[1]][1,5]) anova_[[1]][1,5]=0.0001)
res_aov-cbind(p,anova_[[1]][1,4],0.0001) else
res_aov-cbind(p,anova_[[1]][1,4],anova_[[1]][1,5]);
cat(res_aov, file=stoutput, append = TRUE,sep = |,\n);
};
close(stoutput);
A is a data.frame of about (400 lines and 1800 parameters).
Im a new user of R and I dont know if its a problem in my code or if
there are some tips that I can use to optimise my treatment.
Thanks a lot for your help.
Françoise Pfiffelmann
Engineering Data Analysis Group
--
Crolles2 Alliance
860 rue Jean Monnet
38920 Crolles, France
Tel: +33 438 92 29 84
Email: [EMAIL PROTECTED]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help on R performance using aov function
aov() will handle multiple responses and that would be considerably more
efficient than running separate fits as you seem to be doing.
Your code is nigh unreadable: please use your spacebar and remove the
redundant semicolons: `Writing R Extensions' shows you how to tidy up
your code to make it presentable. But I think anova_[[1]] is really
coef(summary(aov_)) which is a lot more intelligible.
On Thu, 9 Aug 2007, Francoise PFIFFELMANN wrote:
Hi,
Im trying to replace some SAS statistical functions by R (batch calling).
But Ive seen that calling R in a batch mode (under Unix) takes about 2or 3
times more than SAS software. So its a great problem of performance for me.
Here is an extract of the calculation:
stoutput-file(res_oneWayAnova.dat,w);
cat(Param|F|Prob,file=stoutput,\n);
for (i in 1:n) {
p-list_param[[i]]
aov_-aov(A[,p]~ A[,wafer],data=A);
anova_-summary(aov_);
if (!is.na(anova_[[1]][1,5]) anova_[[1]][1,5]=0.0001)
res_aov-cbind(p,anova_[[1]][1,4],0.0001) else
res_aov-cbind(p,anova_[[1]][1,4],anova_[[1]][1,5]);
cat(res_aov, file=stoutput, append = TRUE,sep = |,\n);
};
close(stoutput);
A is a data.frame of about (400 lines and 1800 parameters).
Im a new user of R and I dont know if its a problem in my code or if
there are some tips that I can use to optimise my treatment.
Thanks a lot for your help.
Françoise Pfiffelmann
Engineering Data Analysis Group
--
Crolles2 Alliance
860 rue Jean Monnet
38920 Crolles, France
Tel: +33 438 92 29 84
Email: [EMAIL PROTECTED]
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595__
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Re: [R] Help using gPath
Hi Emilio Gagliardi wrote: Hi everyone,I'm trying to figure out how to use gPath and the documentation is not very helpful :( I have the following plot object: plot-surrounds:: background plot.gTree.378:: background guide.gTree.355:: (background.rect.345, minor-horizontal.segments.347, minor-vertical.segments.349, major-horizontal.segments.351, major-vertical.segments.353) guide.gTree.356:: (background.rect.345, minor-horizontal.segments.347, minor-vertical.segments.349, major-horizontal.segments.351, major-vertical.segments.353) yaxis.gTree.338:: ticks.segments.321 labels.gTree.335:: (label.text.324, label.text.326, label.text.328, label.text.330, label.text.332, label.text.334) xaxis.gTree.339:: ticks.segments.309 labels.gTree.315:: (label.text.312, label.text.314) xaxis.gTree.340:: ticks.segments.309 labels.gTree.315:: (label.text.312, label.text.314) strip.gTree.364:: (background.rect.361, label.text.363) strip.gTree.370:: (background.rect.367, label.text.369) guide.rect.357 guide.rect.358 boxplots.gTree.283:: geom_boxplot.gTree.273:: (GRID.segments.267, GRID.segments.268, geom_bar.rect.270, geom_bar.rect.272) geom_boxplot.gTree.281:: (GRID.segments.275, GRID.segments.276, geom_bar.rect.278, geom_bar.rect.280) boxplots.gTree.301:: geom_boxplot.gTree.291:: (GRID.segments.285, GRID.segments.286, geom_bar.rect.288, geom_bar.rect.290) geom_boxplot.gTree.299:: (GRID.segments.293, GRID.segments.294, geom_bar.rect.296, geom_bar.rect.298) geom_jitter.points.303 geom_jitter.points.305 guide.rect.357 guide.rect.358 ylabel.text.382 xlabel.text.380 title It would be easier to help if we also had the code used to produce this plot, but in the meantime ... Could someone be so kind and create the proper call to grid.gedit() to access a couple of different aspects of this graph? I tried: grid.gedit(gPath(ylabel.text.382,labels), gp=gpar(fontsize=16)) # error That is looking for a grob called labels that is the child of a grob called ylabel.text.382. I can see a grob called ylabel.text.382, but it has no children. Try just ... grid.gedit(gPath(ylabel.text.382), gp=gpar(fontsize=16)) I'd like to change the margins on the label for the yaxis (not the tick marks) to put more space between the label and the tick marks. I'd also Margins may be tricky because it likely depends on a layout generated by ggplot; Hadley Wickham may have to help us out with a ggplot argument here ... (?) like to remove the left border on the first panel. I'd like to adjust the I'd guess you'd have to remove the grob background.rect.345 and then draw in just the sides you want, which would require getting to the right viewport, for which you'll need to study the viewport tree (see current.vpTree()) size of the font for the axis labels independently of the tick marks. I'd That's the one we've already done, right? like to change the color of the lines that make up the boxplots. Plus, I'd Something like ... grid.gedit(geom_bar.rect, gp=gpar(col=green)) ...? Again, it would really help to have some code to run. Paul like to change the margins of the strip labels. If you could show me a couple of examples I'm sure I cold get the rest working. Thanks so much, emilio [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Dr Paul Murrell Department of Statistics The University of Auckland Private Bag 92019 Auckland New Zealand 64 9 3737599 x85392 [EMAIL PROTECTED] http://www.stat.auckland.ac.nz/~paul/ __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help with Filtering (interest rates related)
Dear, r-help, Long time reader, first time poster, I'm working on a paper regarding a term structure estimation using the Kalman Filter Algorithm. The model in question is the Generalized Vasicek, and since there are coupon-bonds being estimated, I'm supposed to make some changes on the Kalman Filter. Does anyone has already used R for these purposes? Any tips? Does anyone has a Kalman Filter code I could use as a starting point for an Extended Kalman Filter Approach? Thanks a lot for the patience and time, Bernardo Ribeiro [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help using gPath
Hi Paul, I'm sorry for not posting code, I wasn't sure if it would be helpful without the data...should I post the code and a sample of the data? I will remember to do that next time! grid.gedit(gPath(ylabel.text.382), gp=gpar(fontsize=16)) OK, I think my confusion comes from the notation that current.grobTree() produces and what strings are required in order to make changes to the underlying grobs. But, from what you've provided, it looks like I can access each grob with its unique name, regardless of which parent it is nested in...that helps like to remove the left border on the first panel. I'd like to adjust the I'd guess you'd have to remove the grob background.rect.345 and then draw in just the sides you want, which would require getting to the right viewport, for which you'll need to study the viewport tree (see current.vpTree()) I did some digging into this and it seems pretty complicated, is there an example anywhere that makes sense to the beginner? The whole viewport grob relationship is not clear to me. So, accessing viewports and removing objects and drawing new ones is beyond me at this point. I can get my mind around your example below because I can see the object I want to modify in the viewer, and the code changes a property of that object, click enter, and bang the object changes. When you start talking external pointers and finding viewports and pushing and popping grobs I just get lost. I found the viewports for the grobTree, it looks like this: viewport[ROOT]-(viewport[layout]-(viewport[axis_h_1_1]-(viewport[bottom_axis]-(viewport[labels], viewport[ticks])), viewport[axis_h_1_2]-(viewport[bottom_axis]-(viewport[labels], viewport[ticks])), viewport[axis_v_1_1]-(viewport[left_axis]-(viewport[labels], viewport[ticks])), viewport[panel_1_1], viewport[panel_1_2], viewport[strip_h_1_1], viewport[strip_h_1_2], viewport[strip_v_1_1])) at that point I was like, ok, I'm done. :S Something like ... grid.gedit(geom_bar.rect, gp=gpar(col=green)) Again, it would really help to have some code to run. My apologies, I thought the grobTree was sufficient in this case. Thanks very much for your help. emilio [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help using gPath
Hi everyone,I'm trying to figure out how to use gPath and the documentation is not very helpful :( I have the following plot object: plot-surrounds:: background plot.gTree.378:: background guide.gTree.355:: (background.rect.345, minor-horizontal.segments.347, minor-vertical.segments.349, major-horizontal.segments.351, major-vertical.segments.353) guide.gTree.356:: (background.rect.345, minor-horizontal.segments.347, minor-vertical.segments.349, major-horizontal.segments.351, major-vertical.segments.353) yaxis.gTree.338:: ticks.segments.321 labels.gTree.335:: (label.text.324, label.text.326, label.text.328, label.text.330, label.text.332, label.text.334) xaxis.gTree.339:: ticks.segments.309 labels.gTree.315:: (label.text.312, label.text.314) xaxis.gTree.340:: ticks.segments.309 labels.gTree.315:: (label.text.312, label.text.314) strip.gTree.364:: (background.rect.361, label.text.363) strip.gTree.370:: (background.rect.367, label.text.369) guide.rect.357 guide.rect.358 boxplots.gTree.283:: geom_boxplot.gTree.273:: (GRID.segments.267, GRID.segments.268, geom_bar.rect.270, geom_bar.rect.272) geom_boxplot.gTree.281:: (GRID.segments.275, GRID.segments.276, geom_bar.rect.278, geom_bar.rect.280) boxplots.gTree.301:: geom_boxplot.gTree.291:: (GRID.segments.285, GRID.segments.286, geom_bar.rect.288, geom_bar.rect.290) geom_boxplot.gTree.299:: (GRID.segments.293, GRID.segments.294, geom_bar.rect.296, geom_bar.rect.298) geom_jitter.points.303 geom_jitter.points.305 guide.rect.357 guide.rect.358 ylabel.text.382 xlabel.text.380 title Could someone be so kind and create the proper call to grid.gedit() to access a couple of different aspects of this graph? I tried: grid.gedit(gPath(ylabel.text.382,labels), gp=gpar(fontsize=16)) # error I'd like to change the margins on the label for the yaxis (not the tick marks) to put more space between the label and the tick marks. I'd also like to remove the left border on the first panel. I'd like to adjust the size of the font for the axis labels independently of the tick marks. I'd like to change the color of the lines that make up the boxplots. Plus, I'd like to change the margins of the strip labels. If you could show me a couple of examples I'm sure I cold get the rest working. Thanks so much, emilio [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help on glmmML
Hello! I am using glmmML for a logitic regression with random effect. I use the posterior.mode as an estimate for the random effects. These can be very different from the estimates obtained using SAS , NLMIXED in the random with out= option. (all the fixed and standard error of random effect estimators are almost identical) Can someone explain to me why is that. The codes I use: R: glmm1-glmmML(mort30 ~ x , data=dat2,cluster=hospital,family=binomial) print(sort(glmm1$posterior.mode)) SAS: * proc* *nlmixed* data*=*dat*;* eta = b0 + b1*x+ u; expeta = exp(eta); p = expeta/(*1*+expeta); model mort30 ~ binomial(*1*,p); random u ~ normal(*0*,s2) subject=hospital out=blue; *run*; * proc* *sort* data=blue;by estimate; * proc* *print* data=blue;*run*; ** *THANKS FOR THE HELP* *RON* [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help on getting weekday
Dear All, I'd like to know which weekday it is for any given date, such as, what is the weekday for 2006-06-01? Any help will be highly appreciated. Best, Baoqiang [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help on getting weekday
Baoqiang Cao wrote: Dear All, I'd like to know which weekday it is for any given date, such as, what is the weekday for 2006-06-01? Any help will be highly appreciated. See ?weekdays Uwe Ligges Best, Baoqiang [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with ROC curve
Hi Ritesh,
may be i can help, but yeah i will try ? you can reach help to ROCR
package by
help.search(ROCR)
What is the structure of your data ? can you give some sample i.e. few
lines of your dataset ?
To build ROC curve using only PSA(variable) alone of the original cohort
against the ROC of the Model of the original cohort.
what do you intend to do, please clarify more ? it sounds like you have
been given tutorial, or you are working this for corporate ?
:) cheers
Rithesh M. Mohan [EMAIL PROTECTED]
Sent by: [EMAIL PROTECTED]
07/30/2007 01:06 PM
To
r-help@stat.math.ethz.ch
cc
Subject
[R] help with ROC curve
Hi
I'm new to stats and R, so can you please help me or guide me building
ROC curve in an elaborate way with codes
I loaded ROCR package, but I'm not sure how to use it.
Requirement
To build ROC curve using only PSA(variable) alone of the original cohort
against the ROC of the Model of the original cohort.
It would be really great if you could help me with this.
Thanks
Rithesh M Mohan
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[R] help with ROC curve
Hi I'm new to stats and R, so can you please help me or guide me building ROC curve in an elaborate way with codes I loaded ROCR package, but I'm not sure how to use it. Requirement To build ROC curve using only PSA(variable) alone of the original cohort against the ROC of the Model of the original cohort. It would be really great if you could help me with this. Thanks Rithesh M Mohan [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with Dates
Are you using the latest version of fame? 1.05 and earlier had a bug in tisFromCsv that was fixed in 1.08. Below I show what I get with fame version 1.08. There is still a problem in that the frequency-figuring logic appears to think the frequency is bwsunday (biweekly with weeks ending on Sunday) rather than semimonthly, which would appear to be a better fit. That's why the 19860330 observation is getting filled in with NA's. Jeff Lines - Date Price Open.Int. Comm.Long Comm.Short net.comm 15-Jan-86 673.25175645 65910 2842537485 31-Jan-86 677.00167350 54060 2712026940 14-Feb-86 680.25157985 37955 2542512530 28-Feb-86 691.75162775 49760 1603033730 14-Mar-86 706.50163495 54120 2799526125 31-Mar-86 709.75164120 54715 3039024325 + + + + + + boink - tisFromCsv(textConnection(Lines), dateFormat = %d-%b-%y, dateCol = Date, sep = ) boink $Price [,1] 19860119 673.25 19860202 677.00 19860216 680.25 19860302 691.75 19860316 706.50 19860330 NA 19860413 709.75 class: tis $Open.Int. [,1] 19860119 175645 19860202 167350 19860216 157985 19860302 162775 19860316 163495 19860330 NA 19860413 164120 class: tis $Comm.Long [,1] 19860119 65910 19860202 54060 19860216 37955 19860302 49760 19860316 54120 19860330NA 19860413 54715 class: tis $Comm.Short [,1] 19860119 28425 19860202 27120 19860216 25425 19860302 16030 19860316 27995 19860330NA 19860413 30390 class: tis $net.comm [,1] 19860119 37485 19860202 26940 19860216 12530 19860302 33730 19860316 26125 19860330NA 19860413 24325 class: tis Gabor Grothendieck [EMAIL PROTECTED] writes: On 26 Jul 2007 09:59:31 -0400, Jeffrey J. Hallman [EMAIL PROTECTED] wrote: zoo is nice. 'tisFromCsv()' in the fame package is nicer. Jeff 1. What am I doing wrong here? I only get one data column. 2. I assume the regularized dates which do not exactly match the input ones are intended so as to make this a regularly spaced series. Is that right? 3. What is the cause of the warning message? 4. Why is a list returned with a single component containing the output? Thanks. library(fame) Lines - Date Price Open.Int. Comm.Long Comm.Short net.comm + 15-Jan-86 673.25175645 65910 2842537485 + 31-Jan-86 677.00167350 54060 2712026940 + 14-Feb-86 680.25157985 37955 2542512530 + 28-Feb-86 691.75162775 49760 1603033730 + 14-Mar-86 706.50163495 54120 2799526125 + 31-Mar-86 709.75164120 54715 3039024325 + tisFromCsv(textConnection(Lines), dateFormat = %d-%b-%y, dateCol = Date, sep = ) [[1]] [,1] 19860119 673.25 19860202 677.00 19860216 680.25 19860302 691.75 19860316 706.50 19860330 709.75 class: tis Warning message: number of items to replace is not a multiple of replacement length in: x[i] - value __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Jeff __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with Dates
Yes, I was using 1.05. I get the same result as you with 1.08. On 26 Jul 2007 11:39:41 -0400, Jeffrey J. Hallman [EMAIL PROTECTED] wrote: Are you using the latest version of fame? 1.05 and earlier had a bug in tisFromCsv that was fixed in 1.08. Below I show what I get with fame version 1.08. There is still a problem in that the frequency-figuring logic appears to think the frequency is bwsunday (biweekly with weeks ending on Sunday) rather than semimonthly, which would appear to be a better fit. That's why the 19860330 observation is getting filled in with NA's. Jeff Lines - Date Price Open.Int. Comm.Long Comm.Short net.comm 15-Jan-86 673.25175645 65910 2842537485 31-Jan-86 677.00167350 54060 2712026940 14-Feb-86 680.25157985 37955 2542512530 28-Feb-86 691.75162775 49760 1603033730 14-Mar-86 706.50163495 54120 2799526125 31-Mar-86 709.75164120 54715 3039024325 + + + + + + boink - tisFromCsv(textConnection(Lines), dateFormat = %d-%b-%y, dateCol = Date, sep = ) boink $Price [,1] 19860119 673.25 19860202 677.00 19860216 680.25 19860302 691.75 19860316 706.50 19860330 NA 19860413 709.75 class: tis $Open.Int. [,1] 19860119 175645 19860202 167350 19860216 157985 19860302 162775 19860316 163495 19860330 NA 19860413 164120 class: tis $Comm.Long [,1] 19860119 65910 19860202 54060 19860216 37955 19860302 49760 19860316 54120 19860330NA 19860413 54715 class: tis $Comm.Short [,1] 19860119 28425 19860202 27120 19860216 25425 19860302 16030 19860316 27995 19860330NA 19860413 30390 class: tis $net.comm [,1] 19860119 37485 19860202 26940 19860216 12530 19860302 33730 19860316 26125 19860330NA 19860413 24325 class: tis Gabor Grothendieck [EMAIL PROTECTED] writes: On 26 Jul 2007 09:59:31 -0400, Jeffrey J. Hallman [EMAIL PROTECTED] wrote: zoo is nice. 'tisFromCsv()' in the fame package is nicer. Jeff 1. What am I doing wrong here? I only get one data column. 2. I assume the regularized dates which do not exactly match the input ones are intended so as to make this a regularly spaced series. Is that right? 3. What is the cause of the warning message? 4. Why is a list returned with a single component containing the output? Thanks. library(fame) Lines - Date Price Open.Int. Comm.Long Comm.Short net.comm + 15-Jan-86 673.25175645 65910 2842537485 + 31-Jan-86 677.00167350 54060 2712026940 + 14-Feb-86 680.25157985 37955 2542512530 + 28-Feb-86 691.75162775 49760 1603033730 + 14-Mar-86 706.50163495 54120 2799526125 + 31-Mar-86 709.75164120 54715 3039024325 + tisFromCsv(textConnection(Lines), dateFormat = %d-%b-%y, dateCol = Date, sep = ) [[1]] [,1] 19860119 673.25 19860202 677.00 19860216 680.25 19860302 691.75 19860316 706.50 19860330 709.75 class: tis Warning message: number of items to replace is not a multiple of replacement length in: x[i] - value __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Jeff __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with Dates
On 26 Jul 2007 09:59:31 -0400, Jeffrey J. Hallman [EMAIL PROTECTED] wrote: zoo is nice. 'tisFromCsv()' in the fame package is nicer. Jeff 1. What am I doing wrong here? I only get one data column. 2. I assume the regularized dates which do not exactly match the input ones are intended so as to make this a regularly spaced series. Is that right? 3. What is the cause of the warning message? 4. Why is a list returned with a single component containing the output? Thanks. library(fame) Lines - Date Price Open.Int. Comm.Long Comm.Short net.comm + 15-Jan-86 673.25175645 65910 2842537485 + 31-Jan-86 677.00167350 54060 2712026940 + 14-Feb-86 680.25157985 37955 2542512530 + 28-Feb-86 691.75162775 49760 1603033730 + 14-Mar-86 706.50163495 54120 2799526125 + 31-Mar-86 709.75164120 54715 3039024325 + tisFromCsv(textConnection(Lines), dateFormat = %d-%b-%y, dateCol = Date, sep = ) [[1]] [,1] 19860119 673.25 19860202 677.00 19860216 680.25 19860302 691.75 19860316 706.50 19860330 709.75 class: tis Warning message: number of items to replace is not a multiple of replacement length in: x[i] - value __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with Dates
zoo is nice. 'tisFromCsv()' in the fame package is nicer. Jeff __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help in HMM
Hi all, I have a time series data.. Data consist of only the counts. I want to model it using a two or three state HMM. I am using J K Lindsey's repeated package. I am not very clear with the arguments like mu, cmu.. etc.. How can I substitute the value for these arguments.Howthe initial estimates are calculate... Thanking You.. With Regards, Regina -- [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help on looping problem needed!
I am wondering if someone could help me out with following problem:
I have written a for loop which generates a random normal distribution
let us say 1000 times.
When the restriction is met (mean0.01), the loop stops, prints
the mean value and plots a histogram.
for(i in 1:1000) {
a-rnorm(1000,0,.2)
b-abs(mean(a))
if(b.01) next else {print(b);hist(a);break}}
How to reshape the loop when I want to find at least 5 distibutions
that meet my restriction and save them (assign) under
names R1R5.
Could you help me please?
Michael
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Re: [R] Help on looping problem needed!
try this:
tol - 0.01
mat - matrix(as.numeric(NA), 1000, 5)
k - 1
while(any(is.na(mat))){
x - rnorm(1000, sd = 0.02)
if (abs(mean(x)) tol) {
mat[, k] - x
k - k + 1
}
}
abs(colMeans(mat))
par(mfrow = c(2, 3))
apply(mat, 2, hist)
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Ing. Michal Kneifl, Ph.D. [EMAIL PROTECTED]
To: Rhelp r-help@stat.math.ethz.ch
Sent: Monday, July 23, 2007 4:40 PM
Subject: [R] Help on looping problem needed!
I am wondering if someone could help me out with following problem:
I have written a for loop which generates a random normal
distribution
let us say 1000 times.
When the restriction is met (mean0.01), the loop stops, prints
the mean value and plots a histogram.
for(i in 1:1000) {
a-rnorm(1000,0,.2)
b-abs(mean(a))
if(b.01) next else {print(b);hist(a);break}}
How to reshape the loop when I want to find at least 5 distibutions
that meet my restriction and save them (assign) under
names R1R5.
Could you help me please?
Michael
__
R-help@stat.math.ethz.ch mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Help with Dates
I am taking an excel dataset and reading it into R using read.table. (actually I am dumping the data into a .txt file first and then reading data in to R). If you are on *windows* you could also try my xlsReadWrite package which contains some datetime functions. Exceldates (e.g. formatted as dd-mmm-yy) can be read as COleDateTime (floating point) values or as character strings. The first one is preferable imo as it avoids a typecast and it is the type commonly used in OLE automation. But how can I subset for multiple periods e.g 00- 05? Floating point numbers dates can be converted to year-strings with dateTimeToStr( value, yy ) and then subset as shown in a previous post. The (paid) pro version contains many more date functions, e.g. yearOf. Details see http://treetron.googlepages.com/xls.oledatetimeex.html. -- Regards, Hans-Peter __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with Dates
Alex: I am taking an excel dataset and reading it into R using read.table. This sets up a data.frame object. The data you have are probably more conveniently represented as a time series, storing the date in an appropriate format, e.g., in class Date. (actually I am dumping the data into a .txt file first and then reading data in to R). Then you can do both steps (calling read.table() and transformation to a time series) in one go using the function read.zoo() from package zoo. If your text file looks like Date Price Open.Int. Comm.Long Comm.Short net.comm 15-Jan-86 673.25175645 65910 2842537485 31-Jan-86 677.00167350 54060 2712026940 14-Feb-86 680.25157985 37955 2542512530 28-Feb-86 691.75162775 49760 1603033730 14-Mar-86 706.50163495 54120 2799526125 31-Mar-86 709.75164120 54715 3039024325 then you can read it in via z - read.zoo(mydata.txt, format = %d-%b-%y, header = TRUE) Then you can do all sorts of standard things for time series, such as plot(z) or... The dataset runs from 1986 to 2007. I want to be able to take subsets of my data based on date e.g. data between 2000 - 2005. ...subsetting z2 - window(z, start = as.Date(2000-01-01), end = as.Date(2005-12-31)) etc. Look at the zoo package vignettes for more information vignette(zoo-quickref, package = zoo) vignette(zoo, package = zoo) hth, Z As it stands, I can't work with the dates as they are not in correct format. I tried successfully converting the dates to just the year using: transform(data, Yr = format(as.Date(as.character(Date),format = '%d-%b-%y'), %y))) This gives the following format: Date Price Open.Int. Comm.Long Comm.Short net.comm Yr 1 15-Jan-86 673.25175645 65910 2842537485 86 2 31-Jan-86 677.00167350 54060 2712026940 86 3 14-Feb-86 680.25157985 37955 2542512530 86 4 28-Feb-86 691.75162775 49760 1603033730 86 5 14-Mar-86 706.50163495 54120 2799526125 86 6 31-Mar-86 709.75164120 54715 3039024325 86 I can subset for a single year e.g: head(subset(df, Yr ==00) But how can I subset for multiple periods e.g 00- 05? The following won't work: head(subset(df, Yr ==00 Yr==01) or head(subset(df, Yr = c(00,01,02,03) I can't help but feeling that I am missing something and there is a simpler route. I leafed through R newletter 4.1 which deals with dates and times but it seemed that strptime and POSIXct / POSIXlt are not what I need either. Can anybody help me? Regards Alex __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with heatmap - how to remove annoying X before numeric values?
Sorry, I just realized I didn't send this to the list! (See below)
Thanks for all the help! All is working fine now.
If anyone knows of a more straightforward way to change the Value string
for the Key, please let me know (just to satisfy my curiosity). I got it to
work by modifying the source code (specifically, heatmap.2.R in the
gplots package).
However, before, I didn't make the call
source(heatmap.2.R)
before I called
source(mysillyheatmap.R)
Making the additional call to heatmap.2.R fixed everything.
Thanks again for all your help!
On 7/19/07, Suzanne Matthews [EMAIL PROTECTED] wrote:
Thank you all for your prompt replies! The check.names=FALSE parameter
fixed things entirely.
One more question:
Is there a straightforward way to modify the the Values string that
labels the key to a user-defined value? For example, let's say I want to
change Values to Silly Values. So far, what I have found that I need to
do is actually go and change a static string in the source code. Is there a
more direct way?
Also, after I make the source code change (in gplots package, file:
heatmap.2.R), how do I have R build from that? If I remember correctly, if
I put the new heatmap.2.R in my directory, R is supposed to check for
functions and such there before it goes and builds it from the main source
code base (located at /usr/bin/R). I am a touch confused on which directory
is my directory, where R will check first. I tried putting the modified
heatmap.2.R file in the directory that my script is, and where I initially
run R. But that didn't work!
Is there anything that I should add to my R script that will force it to
read from that from my directory? If not, which directory should I place
this in?
My OS is OS X, so I think Unix-based instructions will work.
Thank you once again for your time and patience!
Sincerely,
Suzanne
On 7/18/07, Moshe Olshansky [EMAIL PROTECTED] wrote:
I was right saying that my solution was not the best
possible!
--- Prof Brian Ripley [EMAIL PROTECTED] wrote:
read.table('temp.txt', check.names = FALSE)
would be easier (and more general, since make.names
can do more than
prepend an 'X').
On Wed, 18 Jul 2007, Moshe Olshansky wrote:
Hi Suzanne,
My solution (which I am sure is not the best)
would
be:
heat - read.table('temp.txt')
heat
X1905 X1910 X1950 X1992 X2011 X2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat0.40 0.54 0.52 0.60 0.60 0.63
Cat0.16 0.27 0.29 0.39 0.37 0.41
Dog0.43 0.54 0.52 0.61 0.60 0.62
Lynx 0.50 0.57 0.54 0.59 0.50 0.59
a-names(heat)
b-strsplit(a,split=X)
w-unlist(b)
w
[1] 1905 1910 1950
1992 2011 2020
z - w[seq(2,length(w),by=2)]
z
[1] 1905 1910 1950 1992 2011 2020
names(heat) - z
heat
1905 1910 1950 1992 2011 2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat 0.40 0.54 0.52 0.60 0.60 0.63
Cat 0.16 0.27 0.29 0.39 0.37 0.41
Dog 0.43 0.54 0.52 0.61 0.60 0.62
Lynx 0.50 0.57 0.54 0.59 0.50 0.59
Regards,
Moshe.
--- Suzanne Matthews
[EMAIL PROTECTED]
wrote:
Hello All,
I have a simple question based on how things are
labeled on my heat map;
particularly, there is this annoying X that
appears before the numeric
value of all the labels of my columns.
Let's say I have the following silly data, stored
in
temp.txt
19051910195019922011
2020
Gnat0.080.290.290.370.39
0.43
Snake 0.160.340.320.400.41
0.53
Bat 0.400.540.520.60 0.60
0.63
Cat 0.160.270.290.390.37
0.41
Dog 0.430.540.520.610.60
0.62
Lynx0.500.570.540.59 0.5
0.59
I use the following commands to generate my
heatmap:
heat - read.table('temp.txt')
x - as.matrix(heat)
heatmap.2(x, keysize=1.2, dendrogram=none,
trace=none, Colv = FALSE,
main = Silly Data, labCol=
NULL, margin=c(7,8))
This generates a very nice heatmap, but there is
one
thing I have an issue
with: How do I get rid of the 'X' that seems to
come
automatically before my
numeric column values? I just want those columns
to
be labeled 1905, 1910,
1950, and so on. I cannot find anything in the
heatmap.2 documentation that
suggests how I should do this.
Thank you very much for your time, and patience
in
reading this!
Sincerely,
Suzanne
[[alternative HTML version deleted]]
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read
Re: [R] help with heatmap - how to remove annoying X before numeric values?
- read.table('temp.txt')
x - as.matrix(heat)
heatmap.2(x, keysize=1.2, dendrogram=none,
trace=none, Colv = FALSE,
main = Silly Data, labCol=
NULL, margin=c(7,8))
This generates a very nice heatmap, but there is
one
thing I have an issue
with: How do I get rid of the 'X' that seems to
come
automatically before my
numeric column values? I just want those columns
to
be labeled 1905, 1910,
1950, and so on. I cannot find anything in the
heatmap.2 documentation that
suggests how I should do this.
Thank you very much for your time, and patience
in
reading this!
Sincerely,
Suzanne
[[alternative HTML version deleted]]
__
R-help@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
and provide commented, minimal, self-contained,
reproducible code.
__
R-help@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
and provide commented, minimal, self-contained,
reproducible code.
--
Brian D. Ripley,
[EMAIL PROTECTED]
Professor of Applied Statistics,
http://www.stats.ox.ac.uk/~ripley/http://www.stats.ox.ac.uk/%7Eripley/
University of Oxford, Tel: +44 1865
272861 (self)
1 South Parks Road, +44 1865
272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865
272595
[[alternative HTML version deleted]]
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
[R] Help with Dates
R I am taking an excel dataset and reading it into R using read.table. (actually I am dumping the data into a .txt file first and then reading data in to R). Here is snippet: head(data); Date Price Open.Int. Comm.Long Comm.Short net.comm 1 15-Jan-86 673.25175645 65910 2842537485 2 31-Jan-86 677.00167350 54060 2712026940 3 14-Feb-86 680.25157985 37955 2542512530 4 28-Feb-86 691.75162775 49760 1603033730 5 14-Mar-86 706.50163495 54120 2799526125 6 31-Mar-86 709.75164120 54715 3039024325 The dataset runs from 1986 to 2007. I want to be able to take subsets of my data based on date e.g. data between 2000 - 2005. As it stands, I can't work with the dates as they are not in correct format. I tried successfully converting the dates to just the year using: transform(data, Yr = format(as.Date(as.character(Date),format = '%d-%b-%y'), %y))) This gives the following format: Date Price Open.Int. Comm.Long Comm.Short net.comm Yr 1 15-Jan-86 673.25175645 65910 2842537485 86 2 31-Jan-86 677.00167350 54060 2712026940 86 3 14-Feb-86 680.25157985 37955 2542512530 86 4 28-Feb-86 691.75162775 49760 1603033730 86 5 14-Mar-86 706.50163495 54120 2799526125 86 6 31-Mar-86 709.75164120 54715 3039024325 86 I can subset for a single year e.g: head(subset(df, Yr ==00) But how can I subset for multiple periods e.g 00- 05? The following won't work: head(subset(df, Yr ==00 Yr==01) or head(subset(df, Yr = c(00,01,02,03) I can't help but feeling that I am missing something and there is a simpler route. I leafed through R newletter 4.1 which deals with dates and times but it seemed that strptime and POSIXct / POSIXlt are not what I need either. Can anybody help me? Regards Alex __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with Dates
Try some of the following: head(subset(df, Yr %in% c(00,01,02,03))) subset(df, (Yr = '00') (Yr = '03')) # same as above subset(df, (Yr == '00') | (Yr == '01') | (Yr == '02') |(Yr == '03')) # same On 7/19/07, Alex Park [EMAIL PROTECTED] wrote: R I am taking an excel dataset and reading it into R using read.table. (actually I am dumping the data into a .txt file first and then reading data in to R). Here is snippet: head(data); Date Price Open.Int. Comm.Long Comm.Short net.comm 1 15-Jan-86 673.25175645 65910 2842537485 2 31-Jan-86 677.00167350 54060 2712026940 3 14-Feb-86 680.25157985 37955 2542512530 4 28-Feb-86 691.75162775 49760 1603033730 5 14-Mar-86 706.50163495 54120 2799526125 6 31-Mar-86 709.75164120 54715 3039024325 The dataset runs from 1986 to 2007. I want to be able to take subsets of my data based on date e.g. data between 2000 - 2005. As it stands, I can't work with the dates as they are not in correct format. I tried successfully converting the dates to just the year using: transform(data, Yr = format(as.Date(as.character(Date),format = '%d-%b-%y'), %y))) This gives the following format: Date Price Open.Int. Comm.Long Comm.Short net.comm Yr 1 15-Jan-86 673.25175645 65910 2842537485 86 2 31-Jan-86 677.00167350 54060 2712026940 86 3 14-Feb-86 680.25157985 37955 2542512530 86 4 28-Feb-86 691.75162775 49760 1603033730 86 5 14-Mar-86 706.50163495 54120 2799526125 86 6 31-Mar-86 709.75164120 54715 3039024325 86 I can subset for a single year e.g: head(subset(df, Yr ==00) But how can I subset for multiple periods e.g 00- 05? The following won't work: head(subset(df, Yr ==00 Yr==01) or head(subset(df, Yr = c(00,01,02,03) I can't help but feeling that I am missing something and there is a simpler route. I leafed through R newletter 4.1 which deals with dates and times but it seemed that strptime and POSIXct / POSIXlt are not what I need either. Can anybody help me? Regards Alex __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem you are trying to solve? __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help with heatmap - how to remove annoying X before numeric values?
Hello All,
I have a simple question based on how things are labeled on my heat map;
particularly, there is this annoying X that appears before the numeric
value of all the labels of my columns.
Let's say I have the following silly data, stored in temp.txt
190519101950199220112020
Gnat0.080.290.290.370.390.43
Snake 0.160.340.320.400.410.53
Bat 0.400.540.520.600.600.63
Cat 0.160.270.290.390.370.41
Dog 0.430.540.520.610.600.62
Lynx0.500.570.540.590.5 0.59
I use the following commands to generate my heatmap:
heat - read.table('temp.txt')
x - as.matrix(heat)
heatmap.2(x, keysize=1.2, dendrogram=none, trace=none, Colv = FALSE,
main = Silly Data, labCol=
NULL, margin=c(7,8))
This generates a very nice heatmap, but there is one thing I have an issue
with: How do I get rid of the 'X' that seems to come automatically before my
numeric column values? I just want those columns to be labeled 1905, 1910,
1950, and so on. I cannot find anything in the heatmap.2 documentation that
suggests how I should do this.
Thank you very much for your time, and patience in reading this!
Sincerely,
Suzanne
[[alternative HTML version deleted]]
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with heatmap - how to remove annoying X before numeric values?
read.table is doing that, not heatmap.2. Use
read.table(temp.txt, header = TRUE, check.names = FALSE)
On 7/18/07, Suzanne Matthews [EMAIL PROTECTED] wrote:
Hello All,
I have a simple question based on how things are labeled on my heat map;
particularly, there is this annoying X that appears before the numeric
value of all the labels of my columns.
Let's say I have the following silly data, stored in temp.txt
190519101950199220112020
Gnat0.080.290.290.370.390.43
Snake 0.160.340.320.400.410.53
Bat 0.400.540.520.600.600.63
Cat 0.160.270.290.390.370.41
Dog 0.430.540.520.610.600.62
Lynx0.500.570.540.590.5 0.59
I use the following commands to generate my heatmap:
heat - read.table('temp.txt')
x - as.matrix(heat)
heatmap.2(x, keysize=1.2, dendrogram=none, trace=none, Colv = FALSE,
main = Silly Data, labCol=
NULL, margin=c(7,8))
This generates a very nice heatmap, but there is one thing I have an issue
with: How do I get rid of the 'X' that seems to come automatically before my
numeric column values? I just want those columns to be labeled 1905, 1910,
1950, and so on. I cannot find anything in the heatmap.2 documentation that
suggests how I should do this.
Thank you very much for your time, and patience in reading this!
Sincerely,
Suzanne
[[alternative HTML version deleted]]
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with heatmap - how to remove annoying X before numeric values?
Hi Suzanne,
My solution (which I am sure is not the best) would
be:
heat - read.table('temp.txt')
heat
X1905 X1910 X1950 X1992 X2011 X2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat0.40 0.54 0.52 0.60 0.60 0.63
Cat0.16 0.27 0.29 0.39 0.37 0.41
Dog0.43 0.54 0.52 0.61 0.60 0.62
Lynx 0.50 0.57 0.54 0.59 0.50 0.59
a-names(heat)
b-strsplit(a,split=X)
w-unlist(b)
w
[1] 1905 1910 1950
1992 2011 2020
z - w[seq(2,length(w),by=2)]
z
[1] 1905 1910 1950 1992 2011 2020
names(heat) - z
heat
1905 1910 1950 1992 2011 2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat 0.40 0.54 0.52 0.60 0.60 0.63
Cat 0.16 0.27 0.29 0.39 0.37 0.41
Dog 0.43 0.54 0.52 0.61 0.60 0.62
Lynx 0.50 0.57 0.54 0.59 0.50 0.59
Regards,
Moshe.
--- Suzanne Matthews [EMAIL PROTECTED]
wrote:
Hello All,
I have a simple question based on how things are
labeled on my heat map;
particularly, there is this annoying X that
appears before the numeric
value of all the labels of my columns.
Let's say I have the following silly data, stored in
temp.txt
190519101950199220112020
Gnat0.080.290.290.370.390.43
Snake 0.160.340.320.400.410.53
Bat 0.400.540.520.600.600.63
Cat 0.160.270.290.390.370.41
Dog 0.430.540.520.610.600.62
Lynx0.500.570.540.590.5 0.59
I use the following commands to generate my heatmap:
heat - read.table('temp.txt')
x - as.matrix(heat)
heatmap.2(x, keysize=1.2, dendrogram=none,
trace=none, Colv = FALSE,
main = Silly Data, labCol=
NULL, margin=c(7,8))
This generates a very nice heatmap, but there is one
thing I have an issue
with: How do I get rid of the 'X' that seems to come
automatically before my
numeric column values? I just want those columns to
be labeled 1905, 1910,
1950, and so on. I cannot find anything in the
heatmap.2 documentation that
suggests how I should do this.
Thank you very much for your time, and patience in
reading this!
Sincerely,
Suzanne
[[alternative HTML version deleted]]
__
R-help@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
and provide commented, minimal, self-contained,
reproducible code.
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with heatmap - how to remove annoying X before numeric values?
read.table('temp.txt', check.names = FALSE)
would be easier (and more general, since make.names can do more than
prepend an 'X').
On Wed, 18 Jul 2007, Moshe Olshansky wrote:
Hi Suzanne,
My solution (which I am sure is not the best) would
be:
heat - read.table('temp.txt')
heat
X1905 X1910 X1950 X1992 X2011 X2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat0.40 0.54 0.52 0.60 0.60 0.63
Cat0.16 0.27 0.29 0.39 0.37 0.41
Dog0.43 0.54 0.52 0.61 0.60 0.62
Lynx 0.50 0.57 0.54 0.59 0.50 0.59
a-names(heat)
b-strsplit(a,split=X)
w-unlist(b)
w
[1] 1905 1910 1950
1992 2011 2020
z - w[seq(2,length(w),by=2)]
z
[1] 1905 1910 1950 1992 2011 2020
names(heat) - z
heat
1905 1910 1950 1992 2011 2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat 0.40 0.54 0.52 0.60 0.60 0.63
Cat 0.16 0.27 0.29 0.39 0.37 0.41
Dog 0.43 0.54 0.52 0.61 0.60 0.62
Lynx 0.50 0.57 0.54 0.59 0.50 0.59
Regards,
Moshe.
--- Suzanne Matthews [EMAIL PROTECTED]
wrote:
Hello All,
I have a simple question based on how things are
labeled on my heat map;
particularly, there is this annoying X that
appears before the numeric
value of all the labels of my columns.
Let's say I have the following silly data, stored in
temp.txt
190519101950199220112020
Gnat0.080.290.290.370.390.43
Snake 0.160.340.320.400.410.53
Bat 0.400.540.520.600.600.63
Cat 0.160.270.290.390.370.41
Dog 0.430.540.520.610.600.62
Lynx0.500.570.540.590.5 0.59
I use the following commands to generate my heatmap:
heat - read.table('temp.txt')
x - as.matrix(heat)
heatmap.2(x, keysize=1.2, dendrogram=none,
trace=none, Colv = FALSE,
main = Silly Data, labCol=
NULL, margin=c(7,8))
This generates a very nice heatmap, but there is one
thing I have an issue
with: How do I get rid of the 'X' that seems to come
automatically before my
numeric column values? I just want those columns to
be labeled 1905, 1910,
1950, and so on. I cannot find anything in the
heatmap.2 documentation that
suggests how I should do this.
Thank you very much for your time, and patience in
reading this!
Sincerely,
Suzanne
[[alternative HTML version deleted]]
__
R-help@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
and provide commented, minimal, self-contained,
reproducible code.
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
R-help@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
and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with heatmap - how to remove annoying X before numeric values?
I was right saying that my solution was not the best
possible!
--- Prof Brian Ripley [EMAIL PROTECTED] wrote:
read.table('temp.txt', check.names = FALSE)
would be easier (and more general, since make.names
can do more than
prepend an 'X').
On Wed, 18 Jul 2007, Moshe Olshansky wrote:
Hi Suzanne,
My solution (which I am sure is not the best)
would
be:
heat - read.table('temp.txt')
heat
X1905 X1910 X1950 X1992 X2011 X2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat0.40 0.54 0.52 0.60 0.60 0.63
Cat0.16 0.27 0.29 0.39 0.37 0.41
Dog0.43 0.54 0.52 0.61 0.60 0.62
Lynx 0.50 0.57 0.54 0.59 0.50 0.59
a-names(heat)
b-strsplit(a,split=X)
w-unlist(b)
w
[1] 1905 1910 1950
1992 2011 2020
z - w[seq(2,length(w),by=2)]
z
[1] 1905 1910 1950 1992 2011 2020
names(heat) - z
heat
1905 1910 1950 1992 2011 2020
Gnat 0.08 0.29 0.29 0.37 0.39 0.43
Snake 0.16 0.34 0.32 0.40 0.41 0.53
Bat 0.40 0.54 0.52 0.60 0.60 0.63
Cat 0.16 0.27 0.29 0.39 0.37 0.41
Dog 0.43 0.54 0.52 0.61 0.60 0.62
Lynx 0.50 0.57 0.54 0.59 0.50 0.59
Regards,
Moshe.
--- Suzanne Matthews
[EMAIL PROTECTED]
wrote:
Hello All,
I have a simple question based on how things are
labeled on my heat map;
particularly, there is this annoying X that
appears before the numeric
value of all the labels of my columns.
Let's say I have the following silly data, stored
in
temp.txt
19051910195019922011
2020
Gnat0.080.290.290.370.39
0.43
Snake 0.160.340.320.400.41
0.53
Bat 0.400.540.520.600.60
0.63
Cat 0.160.270.290.390.37
0.41
Dog 0.430.540.520.610.60
0.62
Lynx0.500.570.540.590.5
0.59
I use the following commands to generate my
heatmap:
heat - read.table('temp.txt')
x - as.matrix(heat)
heatmap.2(x, keysize=1.2, dendrogram=none,
trace=none, Colv = FALSE,
main = Silly Data, labCol=
NULL, margin=c(7,8))
This generates a very nice heatmap, but there is
one
thing I have an issue
with: How do I get rid of the 'X' that seems to
come
automatically before my
numeric column values? I just want those columns
to
be labeled 1905, 1910,
1950, and so on. I cannot find anything in the
heatmap.2 documentation that
suggests how I should do this.
Thank you very much for your time, and patience
in
reading this!
Sincerely,
Suzanne
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http://www.stats.ox.ac.uk/~ripley/
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Re: [R] HELP FOR BUGS
You might find the 'arm' package useful. For a good introduction to heirarchical modeling, using 'arm' and also WinBUGS and R2WinBUGS, read Gelman, A; J Hill 2007. Data analysis using regression and multilevel/hierarchical models. Cambridge University Press. Cheers, Mike. Ali raza-4 wrote: Hi Sir I am very new user of R for the research project on multilevel logistic regression. There is confusion about bugs() function in R and BUGS software. Is there any relation between these two? Is there any comprehensive package for Multilevel Logistic modelling in R? Please guide in this regard. Thank You RAZA - Boardwalk for $500? In 2007? Ha! [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. -- View this message in context: http://www.nabble.com/HELP-FOR-BUGS-tf4078749.html#a11605645 Sent from the R help mailing list archive at Nabble.com. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] HELP FOR BUGS
Hi Sir I am very new user of R for the research project on multilevel logistic regression. There is confusion about bugs() function in R and BUGS software. Is there any relation between these two? Is there any comprehensive package for Multilevel Logistic modelling in R? Please guide in this regard. Thank You RAZA - Boardwalk for $500? In 2007? Ha! [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] HELP FOR BUGS
Ali raza wrote: Hi Sir I am very new user of R for the research project on multilevel logistic regression. There is confusion about bugs() function in R Do you mean bugs() from package R2WinBUGS? Yes, it is related to the software WinBUGS 1.4.x (and OpenBUGS 2.x with package BRugs). Uwe Ligges and BUGS software. Is there any relation between these two? Is there any comprehensive package for Multilevel Logistic modelling in R? Please guide in this regard. Thank You RAZA - Boardwalk for $500? In 2007? Ha! [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help with handling replicates before reshaping data
Dear list,
I have a dataset consists of duplicated sequences within day for each patient
(see below data) and I want to reshape the data with patient as time variable.
However the reshape function only takes the first sequence of the replicates
and ignores the second. How can I 1) average the duplicates and 2) give the
duplicated sequences unique names before reshaping the data ?
data
patient day seq y
1 10 1 acdf -0.52416066
2 10 1 cdsv 0.62551539
3 10 1 dlfg -1.54668047
4 10 1 acdf 0.82404978
5 10 1 cdsv -1.17459914
6 10 2 acdf 0.47238216
7 10 2 cdsv -0.92364896
8 10 2 dlfg 1.19273992
9 10 2 acdf 0.03759663
10 10 2 cdsv 1.05106783
11 12 1 acdf 0.43575105
12 12 1 cdsv 1.01675547
13 12 1 dlfg -1.54601413
14 12 1 acdf 1.03384654
15 12 1 cdsv 0.32197671
16 12 2 acdf 0.37355285
17 12 2 cdsv -0.39780850
18 12 2 dlfg -0.37693499
19 12 2 acdf -1.28989165
20 12 2 cdsv -0.06938098
21 23 1 acdf -0.68486972
22 23 1 cdsv -1.08035660
23 23 1 dlfg 0.93124685
24 23 1 acdf -0.78737514
25 23 1 cdsv -1.56315904
26 23 2 acdf -2.30913270
27 23 2 cdsv -1.64583577
28 23 2 dlfg 1.87435485
29 23 2 acdf -1.99671825
30 23 2 cdsv 0.62995993
redata-reshape(data,idvar=c(day,seq),timevar=patient,direction=wide)
The reshaped data has only three sequences for each day and didn't take into
account the value of the second replicate.
redata
day seq y.10 y.12 y.23
1 1 acdf -0.5241607 0.4357510 -0.6848697
2 1 cdsv 0.6255154 1.0167555 -1.0803566
3 1 dlfg -1.5466805 -1.5460141 0.9312469
6 2 acdf 0.4723822 0.3735529 -2.3091327
7 2 cdsv -0.9236490 -0.3978085 -1.6458358
8 2 dlfg 1.1927399 -0.3769350 1.8743548
Another problem I have is that I want to check for duplicates in the dataset.
If there are duplicates then print out the sequences. I tried with the code
below but got not so nice output. How can I make the output look nicer or is
there better way to do this?
pat-subset(data, data$patien==10 data$day==1)
if(any(duplicated(pat$seq,MARGIN=1)) ==FALSE)
cat(No duplicates,\n, sep=) else {cat (duplicates ,\n,sep=)
print(pat$seq[duplicated(pat$seq)]) }
I got this output:
duplicates
[1] acdf cdsv
Levels: acdf cdsv dlfg
[1] NA NA
Warning message:
not meaningful for factors in: Ops.factor(cat(duplicates, \n, sep =
), print(pat$seq[duplicated(pat$seq)]))
But would like the output to be something like:
duplicates
[1] acdf cdsv
Thanks alot for any help,
Have a nice weekend !
Tom
-
Jämför pris på flygbiljetter och hotellrum:
http://shopping.yahoo.se/b/a/c_169901_resor_biljetter.html
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] help with handling replicates before reshaping data
Hi Tom, I have a dataset consists of duplicated sequences within day for each patient (see below data) and I want to reshape the data with patient as time variable. However the reshape function only takes the first sequence of the replicates and ignores the second. How can I 1) average the duplicates and 2) give the duplicated sequences unique names before reshaping the data ? data patient day seq y 1 10 1 acdf -0.52416066 2 10 1 cdsv 0.62551539 3 10 1 dlfg -1.54668047 4 10 1 acdf 0.82404978 5 10 1 cdsv -1.17459914 6 10 2 acdf 0.47238216 You mind find that the functions in the reshape package give you a bit more flexibility. # The reshape package expects data like to have # the value variable named value d2 - rename(data, c(y = value)) # I think this is the format you want, which will average over the reps cast(d2, day + seq ~ patient, mean) Hadley __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help Needed!!
Hi, Can anyone help me with repeated meausres MANOVA in R ? For repeated measures ANOVA I used function aov. Is there something like this exists for MANOVA? Thanks, Deepa - [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help Needed!!
See ?summary.manova -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O On 10/07/07, deepa gupta [EMAIL PROTECTED] wrote: Hi, Can anyone help me with repeated meausres MANOVA in R ? For repeated measures ANOVA I used function aov. Is there something like this exists for MANOVA? Thanks, Deepa - [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help with write.foreign (exporting data to Stata)
Hi. I'm trying to export a dataframe from R into Stata to use a statistical function I have there. I attached library write.foreign and renamed my variables to get them to match Stata's required format, and now have the following error: file /tmp/Rtmps7rmrM/file1c06dac8.raw not found Other than typing write.foreign, do I need to do something in R to get it to save the file on my hard drive? When I search for the file name on my computer nothing comes up. I'm using a Mac in case that makes a difference. Thanks, Kate -- View this message in context: http://www.nabble.com/Help-with-write.foreign-%28exporting-data-to-Stata%29-tf4057346.html#a11525796 Sent from the R help mailing list archive at Nabble.com. __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with write.foreign (exporting data to Stata)
I am not sure what you are doing there but what you need is library(foreign) and write.dta() see ?write.dta once you have loaded the foreign package Stefan Original Message Subject: [R] Help with write.foreign (exporting data to Stata) From: kdestler [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Date: Tue Jul 10 2007 19:37:54 GMT+0200 Hi. I'm trying to export a dataframe from R into Stata to use a statistical function I have there. I attached library write.foreign and renamed my variables to get them to match Stata's required format, and now have the following error: file /tmp/Rtmps7rmrM/file1c06dac8.raw not found Other than typing write.foreign, do I need to do something in R to get it to save the file on my hard drive? When I search for the file name on my computer nothing comes up. I'm using a Mac in case that makes a difference. Thanks, Kate __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with write.foreign (exporting data to Stata)
On Tue, 10 Jul 2007, Stefan Grosse wrote: I am not sure what you are doing there but what you need is library(foreign) and write.dta() write.foreign should also work, though. My guess is that Kate used tempfile() to specify the filenames, and that the data file would then have been deleted on leaving R. This is only a guess, of course. The syntax for write.dta is write.dta(the.data.set, file=dataset.dta) and for write.foreign is write.foreign(the.data.set,codefile=dataset.do, datafile=dataset.raw, package=Stata) -thomas see ?write.dta once you have loaded the foreign package Stefan Original Message Subject: [R] Help with write.foreign (exporting data to Stata) From: kdestler [EMAIL PROTECTED] To: r-help@stat.math.ethz.ch Date: Tue Jul 10 2007 19:37:54 GMT+0200 Hi. I'm trying to export a dataframe from R into Stata to use a statistical function I have there. I attached library write.foreign and renamed my variables to get them to match Stata's required format, and now have the following error: file /tmp/Rtmps7rmrM/file1c06dac8.raw not found Other than typing write.foreign, do I need to do something in R to get it to save the file on my hard drive? When I search for the file name on my computer nothing comes up. I'm using a Mac in case that makes a difference. Thanks, Kate __ R-help@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 and provide commented, minimal, self-contained, reproducible code. Thomas Lumley Assoc. Professor, Biostatistics [EMAIL PROTECTED] University of Washington, Seattle __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help on fisher.test(stats)?
Dear friends, My dataset have many zeros, so i must use fisher exact test . Unfortunately, the fisher.test(stats) function fail to do it. Anybody knows how to do the fisher exact test with many zeros in the dataset? My dataset is: a-matrix(c(0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,1,1,0,2,1,5,1,1,6,4,4,1,17,2,8,5,7,1,1,24,3,6,1,1,3,2,16,7,4,0,2,4,0,17,0,1,0,0,0,1,2),nrow=8,byrow=TRUE) data.frame(a) b-a[,-7] as.matrix(b) c-as.matrix(b) c [,1] [,2] [,3] [,4] [,5] [,6] [1,]010000 [2,]010000 [3,]011021 [4,]116441 [5,]285711 [6,]361132 [7,]740240 [8,]010001 fisher.test(c,workspace=20) ´íÎóÓÚfisher.test(c, workspace = 2e+17) : Íâ½Óº¯Êýµ÷ÓÃʱ²»ÄÜÓÐNA(arg10) ´ËÍâ: Warning message: Ç¿ÖƸıä¹ý³ÌÖвúÉúÁËNA Any suggestion or help are greatly appreciated. -- With Kind Regards, oooO: (..): :\.(:::Oooo:: ::\_)::(..):: :::)./::: ::(_/ : [***] Zhi Jie,Zhang ,PHD Tel:86-21-54237149 Dept. of Epidemiology,School of Public Health,Fudan University Address:No. 138 Yi Xue Yuan Road,Shanghai,China Postcode:200032 Email:[EMAIL PROTECTED] Website: www.statABC.com [***] oooO: (..): :\.(:::Oooo:: ::\_)::(..):: :::)./::: ::(_/ : [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] help on fisher.test(stats)?
Original Message Subject: [R] help on fisher.test(stats)? From: zhijie zhang [EMAIL PROTECTED] To: R-help@stat.math.ethz.ch Date: 09.07.2007 09:03 Dear friends, My dataset have many zeros, so i must use fisher exact test . Unfortunately, the fisher.test(stats) function fail to do it. Anybody knows how to do the fisher exact test with many zeros in the dataset? My dataset is: a-matrix(c(0,1,0,0,0,0,1,0,1,0,0,0,0,1,0,1,1,0,2,1,5,1,1,6,4,4,1,17,2,8,5,7,1,1,24,3,6,1,1,3,2,16,7,4,0,2,4,0,17,0,1,0,0,0,1,2),nrow=8,byrow=TRUE) data.frame(a) b-a[,-7] as.matrix(b) c-as.matrix(b) c [,1] [,2] [,3] [,4] [,5] [,6] [1,]010000 [2,]010000 [3,]011021 [4,]116441 [5,]285711 [6,]361132 [7,]740240 [8,]010001 fisher.test(c,workspace=20) ŽíÎóÓÚfisher.test(c, workspace = 2e+17) : ÍâœÓº¯Êýµ÷ÓÃʱ²»ÄÜÓÐNA(arg10) ŽËÍâ: Warning message: Ç¿Öƞıä¹ý³ÌÖвúÉúÁËNA Any suggestion or help are greatly appreciated. Your workspace is by far to large. I have done it with fisher.test(c,workspace=4000) Fisher's Exact Test for Count Data data: c p-value = 0.01548 alternative hypothesis: two.sided (btw. it took half an hour...) Simulation would also be an alternative approach: fisher.test(c,simulate=T) Fisher's Exact Test for Count Data with simulated p-value (based on 2000 replicates) data: c p-value = 0.01349 alternative hypothesis: two.sided As you see the p-value is not that different, you could use more replications: fisher.test(c,simulate=T,B=100) Fisher's Exact Test for Count Data with simulated p-value (based on 1e+06 replicates) data: c p-value = 0.01514 alternative hypothesis: two.sided and it is still much faster... Stefan -=-=- ... The most incomprehensible thing about the world is that it is comprehensible. (A. Einstein) __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in installing rggobi in ubuntu linux
Looks like rggobi can't find GGobi. Make sure that PKG_CONFIG_PATH contains the path to your ggobi.pc file. For example: export PKG_CONFIG_PATH=/usr/local/lib/pkgconfig I would have assumed, however, that the ggobi package would have installed to the /usr prefix, in which case pkg-config should have no problem finding GGobi. On 7/8/07, Kenneth Cabrera [EMAIL PROTECTED] wrote: Hi R users. I am experimenting with ubuntu 7.04 Feisty. I install the ggobi package with apt-get. I got almost all the packages, but when I try to obtain rggobi, I got this message: - install.packages(rggobi) Aviso en install.packages(rggobi) : argument 'lib' is missing: using '/usr/local/lib/R/site-library' --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done probando la URL 'http://cran.at.r-project.org/src/contrib/rggobi_2.1.4-4.tar.gz' Content type 'application/x-gzip' length 401451 bytes URL abierta == downloaded 392Kb * Installing *source* package 'rggobi' ... checking for pkg-config... /usr/bin/pkg-config checking pkg-config is at least version 0.9.0... yes checking for GGOBI... configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/usr/share/R/include -I/usr/share/R/include -g -DUSE_EXT_PTR=1 -D_R_=1 -fpic -g -O2 -c brush.c -o brush.o En el fichero incluÃdo de brush.c:1: RSGGobi.h:5:22: error: GGobiAPI.h: No existe el fichero ó directorio In file included from RSGGobi.h:6, from brush.c:1: conversion.h:174: error: expected â=â, â,â, â;â, âasmâ or â__attribute__â before âasCLogicalâ conversion.h:176: error: expected â=â, â,â, â;â, âasmâ or â__attribute__â before âasCRawâ --- snip --- brush.c:124: error: âtâ no se declaró aquà (primer uso en esta función) brush.c:124: error: âsâ no se declaró aquà (primer uso en esta función) brush.c:124: error: el objeto âGGOBI(erroneous-expression)â llamado no es una función brush.c: En el nivel principal: brush.c:135: error: expected â)â before âcidâ make: *** [brush.o] Error 1 chmod: no se puede acceder a `/usr/local/lib/R/site-library/rggobi/libs/*': No existe el fichero ó directorio ERROR: compilation failed for package 'rggobi' ** Removing '/usr/local/lib/R/site-library/rggobi' The downloaded packages are in /tmp/RtmpVCacJd/downloaded_packages Warning message: installation of package 'rggobi' had non-zero exit status in: install.packages(rggobi) --- What am I doing wrong? Thank you for your help. -- Kenneth Roy Cabrera Torres Cel 315 504 9339 __ R-help@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 and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help in installing rggobi in ubuntu linux
On 9 July 2007 at 22:38, Michael Lawrence wrote: | Looks like rggobi can't find GGobi. Make sure that PKG_CONFIG_PATH contains | the path to your ggobi.pc file. For example: | | export PKG_CONFIG_PATH=/usr/local/lib/pkgconfig | | I would have assumed, however, that the ggobi package would have installed | to the /usr prefix, in which case pkg-config should have no problem finding | GGobi. | | On 7/8/07, Kenneth Cabrera [EMAIL PROTECTED] wrote: | | Hi R users. | | I am experimenting with ubuntu 7.04 Feisty. | | I install the ggobi package with apt-get. | | I got almost all the packages, but | when I try to obtain rggobi, I got | this message: Why don;t you install the Rggobi that is provided via Ubuntu? It is version 2.1.4-4-1 and it corresponds to the 2.1.4-2 version of Ggobi you just installed. Just do 'sudo apt-get install r-omegahat-ggobi' On Debian, we are now at 2.1.5-* for both (and we renamed it r-cran-rggobi as it now resides on CRAN). Hth, Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] Help in installing rggobi in ubuntu linux
Hi R users. I am experimenting with ubuntu 7.04 Feisty. I install the ggobi package with apt-get. I got almost all the packages, but when I try to obtain rggobi, I got this message: - install.packages(rggobi) Aviso en install.packages(rggobi) : argument 'lib' is missing: using '/usr/local/lib/R/site-library' --- Please select a CRAN mirror for use in this session --- Loading Tcl/Tk interface ... done probando la URL 'http://cran.at.r-project.org/src/contrib/rggobi_2.1.4-4.tar.gz' Content type 'application/x-gzip' length 401451 bytes URL abierta == downloaded 392Kb * Installing *source* package 'rggobi' ... checking for pkg-config... /usr/bin/pkg-config checking pkg-config is at least version 0.9.0... yes checking for GGOBI... configure: creating ./config.status config.status: creating src/Makevars ** libs gcc -std=gnu99 -I/usr/share/R/include -I/usr/share/R/include -g -DUSE_EXT_PTR=1 -D_R_=1 -fpic -g -O2 -c brush.c -o brush.o En el fichero incluÃdo de brush.c:1: RSGGobi.h:5:22: error: GGobiAPI.h: No existe el fichero ó directorio In file included from RSGGobi.h:6, from brush.c:1: conversion.h:174: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘asCLogical’ conversion.h:176: error: expected ‘=’, ‘,’, ‘;’, ‘asm’ or ‘__attribute__’ before ‘asCRaw’ --- snip --- brush.c:124: error: ‘t’ no se declaró aquà (primer uso en esta función) brush.c:124: error: ‘s’ no se declaró aquà (primer uso en esta función) brush.c:124: error: el objeto ‘GGOBI(erroneous-expression)’ llamado no es una función brush.c: En el nivel principal: brush.c:135: error: expected ‘)’ before ‘cid’ make: *** [brush.o] Error 1 chmod: no se puede acceder a `/usr/local/lib/R/site-library/rggobi/libs/*': No existe el fichero ó directorio ERROR: compilation failed for package 'rggobi' ** Removing '/usr/local/lib/R/site-library/rggobi' The downloaded packages are in /tmp/RtmpVCacJd/downloaded_packages Warning message: installation of package 'rggobi' had non-zero exit status in: install.packages(rggobi) --- What am I doing wrong? Thank you for your help. -- Kenneth Roy Cabrera Torres Cel 315 504 9339 __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
[R] help with vector construction
Hi all, I want to make a vector with the third column of a matrix, but only for the 2+3n rows of the matrix, with n being an entire number from 0 to a million. How can I do that in an easy way? Thanks in advance, Juan Pablo [[alternative HTML version deleted]] __ R-help@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 and provide commented, minimal, self-contained, reproducible code.
