[R] lmer{lme4}, poisson family and residuals
Hello, Im trying to fit the following model: Dependent variable: MAXDEPTH (the maximum depth reached by a penguin during a given dive) Fixed effects: SUCCESSMN (an index of the individual quality of a bird), STUDYDAY (the day of the study, from -5 to 20, with 0=Dec 20), and the interaction SUCCESSMN*STUDYDAY Random effect: BIRD (the bird id, as each bird is performing several dives) Using lmer with a poisson family, I got a huge amount of warning messages: reml.res - lmer(MAXDEPTH~SUCCESSMN*STUDYDAY+(1|BIRD), family=poisson, method=Laplace) There were 50 or more warnings (use warnings() to see the first 50) warnings() Warning messages: 1: non-integer x = 18.50 2: non-integer x = 20.50 3: non-integer x = 16.50 Whereas it seems to be fine using the quasipoisson family: reml.res - lmer(MAXDEPTH~SUCCESSMN*STUDYDAY+(1|BIRD), family=quasipoisson(link=log), method=Laplace) summary(reml.res) Generalized linear mixed model fit using Laplace Formula: MAXDEPTH ~ SUCCESSMN * STUDYDAY + (1 | BIRD) Family: quasipoisson(log link) AIC BIClogLik deviance 39581.16 39616.86 -19785.58 39571.16 Random effects: Groups NameVariance Std.Dev. BIRD (Intercept) 0.12383 0.3519 Residual 4.05206 2.0130 number of obs: 9324, groups: BIRD, 12 Fixed effects: Estimate Std. Error t value (Intercept)3.6725244 0.1063978 34.517 SUCCESSMN 0.2978806 0.6824748 0.436 STUDYDAY 0.0069852 0.0021121 3.307 SUCCESSMN:STUDYDAY 0.0571982 0.0123831 4.619 Correlation of Fixed Effects: (Intr) SUCCESSMN STUDYD SUCCESSMN -0.276 STUDYDAY-0.099 -0.017 SUCCESSMN:S -0.075 -0.035 0.823 My first question is therefore: can someone tell me why I cant fit the model with family=poisson? Second, I would like to check the residuals but I really dont know how, even after several hours searching the archives and other helps. I tried: residuals(reml.res) Error: 'residuals' is not implemented yet resid(reml.res) Error: 'resid' is not implemented yet names(reml.res) NULL mcmcsamp(reml.res) Error in .local(object, n, verbose, ...) : mcmcsamp for GLMMs not yet implemented in supernodal representation Im supposed to give a talk tomorrow and would be very grateful if someone can help me Im working with the following versions: Package: Matrix Version: 0.995-8 Date: 2006-03-21 Package: lme4 Version: 0.995-2 Date: 2006-01-17 Thank you! Amélie [[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
Re: [R] PCA biplot question
I think you're asking how to plot arrows for variables in one dataset onto the PCA plot for another dataset? You can do this with functions from the vegan package. Particularly, vector and factor fitting can be done with envfit() and surface fitting with ordisurf(). There's a good tutorial at: http://cc.oulu.fi/~jarioksa/softhelp/vegan.html Cheers, Tyler __ 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
[R] the difference between x1 and x1
Hello, I am not sure what to write in the subject line, but I would like to take a character string that is a variable in a data frame and apply a function that takes a numeric argument to this character string. Here is a simplified example that would solve my problem. Imagine I have my data stored in a data frame. x1 - x2 - x3 - x4 - x5 - rnorm(20,0,1); data - as.data.frame(cbind(x1,x2,x3,x4,x5)); I have a vector containing the variables of interest as such. model.list - c(x1,x3,x4); model.list[1] [1] x1 I would like to loop through this vector and apply the floor() function to each variable. In the current form the elements of model.list do not represent the variables in the data frame. floor(model.list[1]) Error in floor(model.list[1]) : Non-numeric argument to mathematical function floor(eval(model.list[1])) Error in floor(eval(model.list[1])) : Non-numeric argument to mathematical function s - expression(paste(floor(,model.list[1],),sep=)) s expression(paste(floor(, model.list[1], ), sep = )) eval(s) [1] floor(x1) I have tried the obvious (to me) without success. Perhaps someone could suggest a solution and some tidbits for me to read up on about the how and why. Thanks, Chad R. Bhatti __ 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
Re: [R] Parallel computing with the snow package: external file I/O possible?
Hello,
After getting help to solve part of my problem and some delay on my
part, I am posting a more refined version to see if someone can help me
further. I am trying to autocalibrate a model in my subject area using
the snow and rgenoud packages. I want to use the key function fn that
is called by genoud() to finalize input, run the model executable, and
compute the objective function on the model output. My problem is that
only in the master node do all of these steps happen. In the slave node
of a cluster of two, the input file gets created but my system() call to
execute the model doesn't happen. I don't understand why the model
doesn't get run in the slave mode, but I noticed something described
below that I thought might be related.
It is hard for me to figure out a short and simple example of my
genoud() problem for posting here, so let me start with some code that
uses clusterCall(). As I understand the snow package, each execution of
the function fun from clusterCall() (or of the function fn() from
genoud()) should be independent. However, in the code below, the
directory created by each node has the same random number in its name.
I was expecting the contents of fun() or fn() to be independent from all
other executions of the same function. What am I missing here?
# Begin code
library(snow)
setDefaultClusterOptions(outfile=/tmp/cluster1)
setDefaultClusterOptions(master=moab)
cl - makeCluster(c(moab, escalante), type=SOCK)
# Define base pathname for output from my.test()
base.dir - ~
# Define a function that is called by clusterCall()
my.test - function(base.dir) {
this.host - as.character(system(hostname, intern=T))
this.rnd - sample(1:1e6, 1)
test.file - paste(sep=, base.dir, this.host, _, this.rnd)
file.create(test.file)
} # end my.test()
clusterCall(cl, my.test, base.dir)
stopCluster(cl)
# End code
For example, the files moab_65835 and escalante_65835 are created.
Regards,
Scott Waichler
Pacific Northwest National Laboratory
[EMAIL PROTECTED]
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Re: [R] the difference between x1 and x1
Try this: # test data set.seed(1) DF - data.frame(x1 = rnorm(5), x2 = rnorm(5), x3 = rnorm(5)) DF model.list - c(x2, x3) # transform for(v in model.list) DF[v] - floor(DF[v]) On 4/20/06, Chad Reyhan Bhatti [EMAIL PROTECTED] wrote: Hello, I am not sure what to write in the subject line, but I would like to take a character string that is a variable in a data frame and apply a function that takes a numeric argument to this character string. Here is a simplified example that would solve my problem. Imagine I have my data stored in a data frame. x1 - x2 - x3 - x4 - x5 - rnorm(20,0,1); data - as.data.frame(cbind(x1,x2,x3,x4,x5)); I have a vector containing the variables of interest as such. model.list - c(x1,x3,x4); model.list[1] [1] x1 I would like to loop through this vector and apply the floor() function to each variable. In the current form the elements of model.list do not represent the variables in the data frame. floor(model.list[1]) Error in floor(model.list[1]) : Non-numeric argument to mathematical function floor(eval(model.list[1])) Error in floor(eval(model.list[1])) : Non-numeric argument to mathematical function s - expression(paste(floor(,model.list[1],),sep=)) s expression(paste(floor(, model.list[1], ), sep = )) eval(s) [1] floor(x1) I have tried the obvious (to me) without success. Perhaps someone could suggest a solution and some tidbits for me to read up on about the how and why. Thanks, Chad R. Bhatti __ 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 __ 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
Re: [R] the difference between x1 and x1
On 4/20/2006 7:02 PM, Chad Reyhan Bhatti wrote: Hello, I am not sure what to write in the subject line, but I would like to take a character string that is a variable in a data frame and apply a function that takes a numeric argument to this character string. Remember that dataframes are lists with named members. So to apply something to the x1 member of a dataframe, you can do the usual data$x1 but you can also do data[[x1]] In addition, dataframes implement the same style of index handling as matrices, so you can refer to the x1 column as data[,x1] You could replace x1 with any expression (e.g. model.list[1]) in either of the latter two examples, e.g. data[[model.list[1]]] or data[,model.list[1]] You say below that you want to loop through the vector; in fact, there's no need to do that. The floor function would work fine on the whole thing at once, e.g. floor(data[,model.list]) Duncan Murdoch Here is a simplified example that would solve my problem. Imagine I have my data stored in a data frame. x1 - x2 - x3 - x4 - x5 - rnorm(20,0,1); data - as.data.frame(cbind(x1,x2,x3,x4,x5)); I have a vector containing the variables of interest as such. model.list - c(x1,x3,x4); model.list[1] [1] x1 I would like to loop through this vector and apply the floor() function to each variable. In the current form the elements of model.list do not represent the variables in the data frame. floor(model.list[1]) Error in floor(model.list[1]) : Non-numeric argument to mathematical function floor(eval(model.list[1])) Error in floor(eval(model.list[1])) : Non-numeric argument to mathematical function s - expression(paste(floor(,model.list[1],),sep=)) s expression(paste(floor(, model.list[1], ), sep = )) eval(s) [1] floor(x1) I have tried the obvious (to me) without success. Perhaps someone could suggest a solution and some tidbits for me to read up on about the how and why. Thanks, Chad R. Bhatti __ 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 __ 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
Re: [R] the difference between x1 and x1
Is this what you are after? floor(data[,model.list]) Or I just didn't understand what you are trying to accomplish? cheers Francisco PS: try to avoid using names that are already reserved to a function like data See ?data From: Chad Reyhan Bhatti [EMAIL PROTECTED] To: R-help@stat.math.ethz.ch Subject: [R] the difference between x1 and x1 Date: Thu, 20 Apr 2006 18:02:02 -0500 (CDT) Hello, I am not sure what to write in the subject line, but I would like to take a character string that is a variable in a data frame and apply a function that takes a numeric argument to this character string. Here is a simplified example that would solve my problem. Imagine I have my data stored in a data frame. x1 - x2 - x3 - x4 - x5 - rnorm(20,0,1); data - as.data.frame(cbind(x1,x2,x3,x4,x5)); I have a vector containing the variables of interest as such. model.list - c(x1,x3,x4); model.list[1] [1] x1 I would like to loop through this vector and apply the floor() function to each variable. In the current form the elements of model.list do not represent the variables in the data frame. floor(model.list[1]) Error in floor(model.list[1]) : Non-numeric argument to mathematical function floor(eval(model.list[1])) Error in floor(eval(model.list[1])) : Non-numeric argument to mathematical function s - expression(paste(floor(,model.list[1],),sep=)) s expression(paste(floor(, model.list[1], ), sep = )) eval(s) [1] floor(x1) I have tried the obvious (to me) without success. Perhaps someone could suggest a solution and some tidbits for me to read up on about the how and why. Thanks, Chad R. Bhatti __ 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 __ 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
[R] R and ViM
To: Martin Maechler [EMAIL PROTECTED] Indeed. Please do check the archives. Yep. Post is there. Now back to the subject: Jose, I think your main contribution is based on autoHotKeys and that only works on Windoze, right? Michael explicitly mentioned he's working in Mac OS X. Martin That's correct, autoHotKeys is a win-only solution. Bill West's solution uses use Win32::OLE, so I guess that means it's win-only solution. I think François Pinard's is the easiest since it uses vim and R only (no 3rd application/language requireed), and that would probably work in all platforms (not sure if the GNU readline inferface is implemented in all builds of R, though). I find autoHotKey very useful (e.g., it fires vim to fill textboxes like this one -gmail editing window-, offering syntax highlighting etc) so I use it all the time and don't mind. I can understand those who don't want to install it just to communicate R and an editor. In that case, the other two solutions are better. I'll refer to them in my Rvim page (archives). However, the new R.vim syntax file has 3-level coloring, and I'm working to get improved indentation, TODO, DEBUG, etc highlighting, and other improvements. The code templates using tSkeleton (not ready yet) may be an added advantage. Those two things you can use even if you don't use autohotkeys of course. -- Cheers, -Jose PS: it seems that google mail (as an email reader) doesn't let you see your own messages when posting to a list you are subscribed to. -- Jose Quesada, PhD. [EMAIL PROTECTED] Dept. of Psychology http://www.andrew.cmu.edu/~jquesada Sussex University Brighton, UK -- Cheers, -Jose -- Jose Quesada, PhD. [EMAIL PROTECTED] Dept. of Psychology http://www.andrew.cmu.edu/~jquesada Sussex University Brighton, UK __ 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
Re: [R] second try; writing user-defined GLM link function
An update for all: Using the combined contributions from Mark and Dr. Ripley, I've been (apparently) successfully formulating both GLM's and GLMM's (using the MASS function glmmPQL) analyzing my nest success data. The beta parameter estimates look reasonable and the top models resemble those from earlier analyses using a different nest survival analysis approach. However, I've now run into problems when trying to predict the daily survival rates from fitted models. For example, for a model considering nest height (NestHtZ) and nest age effects (MeanAge and related terms; there is an overall cubic time trend in this model), I tried to predict the daily survival rate for each day out of a 67 day nest cycle (so MeanAge is a vector of 1 to 67) with mean nest height (also a vector 67 rows in length; both comprise the matrix nestday). Here's what happens: summary(glm.24) Call: glm(formula = Success ~ NestHtZ + MeanAge + I(MeanAge^2) + I(MeanAge^3), family = logexposure(ExposureDays = vc.apfa$days), data = vc.apfa) Deviance Residuals: Min 1Q Median 3Q Max -3.3264 -1.2341 0.6712 0.8905 1.5569 Coefficients: Estimate Std. Error z value Pr(|z|) (Intercept) 6.5742015 1.7767487 3.700 0.000215 *** NestHtZ 0.6205444 0.2484583 2.498 0.012504 * MeanAge -0.6018978 0.2983656 -2.017 0.043662 * I(MeanAge^2) 0.0380521 0.0152053 2.503 0.012330 * I(MeanAge^3) -0.0006349 0.0002358 -2.693 0.007091 ** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 (Dispersion parameter for binomial family taken to be 1) Null deviance: 174.86 on 136 degrees of freedom Residual deviance: 233.82 on 132 degrees of freedom AIC: 243.82 Number of Fisher Scoring iterations: 13 glm.24.pred-predict(glm.24,newdata=nestday, type=response, SE.fit=T) Warning message: longer object length is not a multiple of shorter object length in: plogis(eta)^days Can anyone tell me what I'm doing wrong? cheers, Jessi Brown __ 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
[R] how to control the data type
Hello: I am generating a random number with rnorm(1). The generated number has 8 decimals. I don't want so many decimals. How to control the number of decimals in R? Thanks! Zhongmiao Wang __ 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
Re: [R] how to do Splus compare() function in R
The sign() function requires one argument,
and returns the sign of the value, where I require a number
-1, 0, or 1 depending on if the value is , ==, or the
first argument.
This is the Splus function when listed,
compare
function(e1, e2)
.Internal(compare(e1, e2), do_op, T, 18)
unfortunately I do not understand the .internal function
and cannot find any description of this type of programming
on the CRAN site
This is how I am using it in the source file.
zz - readline()# keyboard input eg: f15
if (charmatch(substring(zz,1,1),f,
nomatch=-1) 0){
ii - as.numeric(substring(zz,2,99))
iii - compare(ii*1e9,f) # f is a vector of length(146) frequencies
# compare() searches thru f and
# creates vector iii with
# -1 if numeric is f[i],
# 0 if numeric is == f[i], and
# 1 if numeric is f[i]
# the end result is a vector(iii) length(146)
# with -1,0,1 if the numeric is in f
i - match(-1,c(iii,-1))-1 # if vector(iii) contains 0 assign the
index
}
# error traps for if 0 is not in vector(iii)
if (i1){cat(Min n = 1\n)
i - 1}
if (ilength(f)){cat(Max n =,length(f),\n)
i - length(f)}
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: Thursday, 20 April 2006 4:02 PM
Subject: Re: [R] how to do Splus compare() function in R
You would need to tell us what it does (and what the inputs are).
I think it is likely that compare(x, y) in S-PLUS is the same as sign(x-y)
in R, at least with numeric vector inputs.
--
Bob Kelly [EMAIL PROTECTED]
Metrologist Microwave Eng.http://www.bellinger.com.au
Bellinger Instruments Pty Ltd Tel: 612 9684 1442
4 Muriel AveFax: 612 9638 4435
Rydalmere NSW
Australia 2116
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Re: [R] how to do Splus compare() function in R
As suggested below,
sign(ii*1e9-f)
will give you the same result as the S-plus
compare(ii*1e-9,f)
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Bellinger Instruments
P/L
Sent: Thursday, April 20, 2006 11:02 PM
To: 'Prof Brian Ripley'
Cc: r-help@stat.math.ethz.ch
Subject: Re: [R] how to do Splus compare() function in R
The sign() function requires one argument, and returns the sign of the
value, where I require a number -1, 0, or 1 depending on if the value is ,
==, or the first argument.
This is the Splus function when listed,
compare
function(e1, e2)
.Internal(compare(e1, e2), do_op, T, 18)
unfortunately I do not understand the .internal function and cannot find any
description of this type of programming on the CRAN site
This is how I am using it in the source file.
zz - readline()# keyboard input eg: f15
if (charmatch(substring(zz,1,1),f,
nomatch=-1) 0){
ii - as.numeric(substring(zz,2,99))
iii - compare(ii*1e9,f) # f is a vector of length(146) frequencies
# compare() searches thru f and
# creates vector iii with
# -1 if numeric is f[i],
# 0 if numeric is == f[i], and
# 1 if numeric is f[i]
# the end result is a vector(iii) length(146)
# with -1,0,1 if the numeric is in f
i - match(-1,c(iii,-1))-1 # if vector(iii) contains 0 assign the
index
}
# error traps for if 0 is not in vector(iii)
if (i1){cat(Min n = 1\n)
i - 1}
if (ilength(f)){cat(Max n =,length(f),\n)
i - length(f)}
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: Thursday, 20 April 2006 4:02 PM
Subject: Re: [R] how to do Splus compare() function in R
You would need to tell us what it does (and what the inputs are).
I think it is likely that compare(x, y) in S-PLUS is the same as sign(x-y)
in R, at least with numeric vector inputs.
--
Bob Kelly [EMAIL PROTECTED]
Metrologist Microwave Eng.http://www.bellinger.com.au
Bellinger Instruments Pty Ltd Tel: 612 9684 1442
4 Muriel AveFax: 612 9638 4435
Rydalmere NSW
Australia 2116
__
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Re: [R] how to do Splus compare() function in R [Broadcast]
You need to tell us what compare() in S-PLUS does; i.e., what does it take
as input and what does it output. Presumably the information would be in
the help page for compare().
Telling us what the function looks like in S-PLUS is of no use, since most
likely the code cannot be used in R without violating S-PLUS copyright
and/or license. Even if that wasn't an obstacle, the fact that it calls
.Internal() says the actual code is written in C, and of course no one
outside of Insightful (and perhaps Bell Labs) has access to code at that
level. And even then, R and S-PLUS are not likely to be compatible at that
level.
Andy
From: Bellinger Instruments P/L
The sign() function requires one argument,
and returns the sign of the value, where I require a number
-1, 0, or 1 depending on if the value is , ==, or the
first argument.
This is the Splus function when listed,
compare
function(e1, e2)
.Internal(compare(e1, e2), do_op, T, 18)
unfortunately I do not understand the .internal function
and cannot find any description of this type of programming
on the CRAN site
This is how I am using it in the source file.
zz - readline()# keyboard input eg: f15
if (charmatch(substring(zz,1,1),f,
nomatch=-1) 0){
ii - as.numeric(substring(zz,2,99))
iii - compare(ii*1e9,f) # f is a vector of
length(146) frequencies
# compare() searches thru f and
# creates vector iii with
# -1 if numeric is f[i],
# 0 if numeric is == f[i], and
# 1 if numeric is f[i]
# the end result is a vector(iii) length(146)
# with -1,0,1 if the numeric is in f
i - match(-1,c(iii,-1))-1 # if vector(iii) contains 0
assign the index
}
# error traps for if 0 is not in vector(iii)
if (i1){cat(Min n = 1\n)
i - 1}
if (ilength(f)){cat(Max n =,length(f),\n)
i - length(f)}
-Original Message-
From: Prof Brian Ripley [mailto:[EMAIL PROTECTED]
Sent: Thursday, 20 April 2006 4:02 PM
Subject: Re: [R] how to do Splus compare() function in R
You would need to tell us what it does (and what the inputs are).
I think it is likely that compare(x, y) in S-PLUS is the same
as sign(x-y) in R, at least with numeric vector inputs.
--
Bob Kelly [EMAIL PROTECTED]
Metrologist Microwave Eng. http://www.bellinger.com.au
Bellinger Instruments Pty Ltd Tel: 612 9684 1442
4 Muriel Ave Fax: 612 9638 4435
Rydalmere NSW
Australia 2116
__
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
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Re: [R] I am surprised (and a little irritated)
Detlef Steuer wrote: Hi again, ... This new rpm R-base-2.3.0-beta should automatically resolve dependencies. At least it did so on my machine. I would be happy to get a report, if you try to install this one and find difficulties or success!. Most important would be to report missing dependencies. As this is my first try to add those, I`m not sure having catched them all. I have not tried it yet, but when I think about it I think that you have resolved the dependencies part already in 2.2.1. When I had set up YaST according to the hacking link and then opened the R-base rpm file everything went smooth with no intervention on my part. Tom ++ | Tom Backer Johnsen, Psychometrics Unit, Faculty of Psychology | | University of Bergen, Christies gt. 12, N-5015 Bergen, NORWAY | | Tel : +47-5558-9185Fax : +47-5558-9879 | | Email : [EMAIL PROTECTED]URL : http://www.galton.uib.no/ | ++ __ 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
[R] Linker incompatability in stalling R on Solaris 2.9
Hi, I am trying to compile R-2.2.1 on Solaris 2.9 with a 64-bit build. Following the instructions in R Installation and Adminstration, I changed the following settings in config.site: [[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
[R] Linker problem in installing 64-bit R
Hi, I am trying to compile R-2.2.1 on Solaris 2.9 with a 64-bit build. Following the instructions in R Installation and Adminstration, I changed the following settings in config.site: CC=gcc -m64 F77=g77 -64 CXX=g++ -m64 LDFLAGS=-L/usr/local/lib/sparcv9 -L/usr/local/lib But I got the following error messages: configure:3987: gcc -m64 -I/usr/local/include -L/usr/local/lib/sparcv9 -L/usr/local/lib conftest.c 5 /usr/ccs/bin/ld: skipping incompatible /work/net-local-b/sparc- sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9/3.4.2/sparcv9/libgcc.a when search ing for -lgcc /usr/ccs/bin/ld: skipping incompatible /work/net-local-b/sparc- sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9/3.4.2/sparcv9/libgcc_eh.a when sea rching for -lgcc_eh /usr/ccs/bin/ld: skipping incompatible /lib/sparcv9/libc.so when searching for -lc /usr/ccs/bin/ld: skipping incompatible /usr/lib/sparcv9/libc.so when searching for -lc /usr/ccs/bin/ld: skipping incompatible /usr/lib/sparcv9/libc.so when searching for -lc /usr/ccs/bin/ld: skipping incompatible /work/net-local-b/sparc- sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9/3.4.2/sparcv9/libgcc.a when search ing for -lgcc /usr/ccs/bin/ld: skipping incompatible /work/net-local-b/sparc- sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9/3.4.2/sparcv9/libgcc_eh.a when sea rching for -lgcc_eh /usr/ccs/bin/ld: skipping incompatible /lib/sparcv9/libc.so when searching for -lc /usr/ccs/bin/ld: skipping incompatible /usr/lib/sparcv9/libc.so when searching for -lc /usr/ccs/bin/ld: skipping incompatible /usr/lib/sparcv9/libc.so when searching for -lc /usr/ccs/bin/ld: warning: sparc:v9 architecture of input file `/work/net-local-b/sparc-sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9 /3.4.2/spar cv9/crt1.o' is incompatible with sparc output /usr/ccs/bin/ld: warning: sparc:v9 architecture of input file `/work/net-local-b/sparc-sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9 /3.4.2/spar cv9/crti.o' is incompatible with sparc output /usr/ccs/bin/ld: warning: sparc:v9 architecture of input file `/usr/ccs/lib/sparcv9/values-Xa.o' is incompatible with sparc output /usr/ccs/bin/ld: warning: sparc:v9 architecture of input file `/work/net-local-b/sparc-sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9 /3.4.2/spar cv9/crtbegin.o' is incompatible with sparc output /usr/ccs/bin/ld: warning: sparc:v9 architecture of input file `/var/tmp//cckWp3Sb.o' is incompatible with sparc output /usr/ccs/bin/ld: warning: sparc:v9 architecture of input file `/work/net-local-b/sparc-sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9 /3.4.2/spar cv9/crtend.o' is incompatible with sparc output /usr/ccs/bin/ld: warning: sparc:v9 architecture of input file `/work/net-local-b/sparc-sun-solaris2.9/bin/../lib/gcc/sparc-sun-solaris2.9 /3.4.2/spar cv9/crtn.o' is incompatible with sparc output configure:3990: $? = 0 configure:4036: result: a.out configure:4041: checking whether the C compiler works configure:4047: ./a.out configure: line 1: 25817 Bus Error (core dumped) ./$ac_file configure:4050: $? = 138 configure:4059: error: cannot run C compiled programs. I would appreciate any help with resolving the problem. Thanks, Min [[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
Re: [R] how to do Splus compare() function in R
You would need to tell us what it does (and what the inputs are). I think it is likely that compare(x, y) in S-PLUS is the same as sign(x-y) in R, at least with numeric vector inputs. On Thu, 20 Apr 2006, Bellinger Instruments P/L wrote: Hi I have a source file in Splus that fails in R as I am using the Splus function compare(). compare( ii * 1e9,f ) where ii is a single variable and f is a vector of length 146 R returns with no function error Can anyone inform me how I can do this in R thanks in advance Bob Kelly Bellinger Instruments Pty Ltd [[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 -- 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
Re: [R] there is no xls reader in R?
Have a look at the read.xls() in gdata package. HTH, Kevin Michael wrote: Currently I have to convert all my xls into csv before I can read it in and process the excel data in R... Is there a way to directly read in xls data? 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 -- Ko-Kang Kevin Wang Homepage: http://wwwmaths.anu.edu.au/~wangk/ Ph (W): +61-2-6125-2431 Ph (H): +61-2-6125-7471 Ph (M): +61-40-451-8301 __ 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
Re: [R] comparing execition time: R vs matlab linear algebra...
You haven't told us your platform. On most platforms, R does not build/ship with an optimized BLAS, and this can make a big difference to execution times of such problems. See the R-admin manual. Beyond that, qr/qr.coef is not the most obvious way to solve linear systems (solve is), and the option LAPACK=TRUE will be faster, not least because it can make better use of an optimzed BLAS. And yes, if `sparse' is a form of sparseness that packages can exploit, it will help to use them, and you must be using sparseness in Matlab to even be able to store the matrices. R is not a numerical linear algbra system and does not intend to compete with say Octave. Rather, it aims to have enough facilities that in almost all uses of R, the slowness of linear algebra is not the limiting factor. I do wonder why you are porting Matlab scripts to R if Matlab can handle them. On Wed, 19 Apr 2006, Rick Reeves wrote: Greetings: We are evaluating the performance of R matrix algebra es as we port a MATLAB R14 script into R. The MATLAB code basically evaluates the AX=B system on sparse matrices that result in output matrices of 100 to 1,000,000 rows/columns. Our R prototype script uses spase base matrices and the methods qr() and qr.coeff(). The following statements are called inside a doubly-nested loop: G is an n x m sparse matrix with most nonzero values near the main diagonal, CURR is an n x 1 vector: the MATALAB script: P = G \ CURR; % matrix left-divide QG = qr(G) P = qr.coef(QG, CURR) The answer we get matches that of the MATLAB code for small (60x60) problems. But, execution times are much longer (40 minutes, compared to 2 minutes for the MATLAB script) Also, the R version cannot accommodate a solution matrix greater than aprox 10,000 x 10,000 elements,while the MATLAB script has generated solutions for 10**6 x 10**6 solution matrices. My questions: 1) Have others noticed this difference in performance between R and MATLAB? 2) Is there R literature (I have searched, not found) that discusses optimizing these solutions in R? such as. 2) Would developing a solution using the Matrix or SparseM classes improve performance? Thanks in advance for any insights! Regards, Rick R -- 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
[R] R-squared for MARS models
Hi, is there an R function which can compute the R-squared for a mars model fitted using the mda package? many thanks in advance best regards Marco Girardello __ 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
Re: [R] R-squared for MARS models
On Wed, 2006-04-19 at 21:18 +0200, Marco Girardello wrote: Hi, is there an R function which can compute the R-squared for a mars model fitted using the mda package? many thanks in advance best regards Marco Girardello Julian Faraway's Extending the Linear Model with R Chapman Hall/CRC (page 248)has a small section on MARS and compares the MARS fit with other methods by using the MARS basis functions in a linear model to get an R^2: # page 248 example library(faraway) library(mda) data(ozone) a - mars(ozone[,-1], ozone[,]) (a.summ - summary(lm(ozone[,1] ~ a$x-1))) a.summ$r.squared HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% *Note new Address and Fax and Telephone numbers from 10th April 2006* %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Gavin Simpson [t] +44 (0)20 7679 0522 ECRC [f] +44 (0)20 7679 0565 UCL Department of Geography Pearson Building [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street London, UK[w] http://www.ucl.ac.uk/~ucfagls/cv/ WC1E 6BT [w] http://www.ucl.ac.uk/~ucfagls/ %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ 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
[R] Need R code
Dear r-users, Suppose I have three datasets: Dataset-1: Date x y Jan-1,2005120 230 Jan-2,2005123 -125 Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-7,200511299 Mar-5,2005200 311 Dataset-2: Date x y Jan-2,2005123 -125 Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-5,200511299 Jan-6,2005-23 12 Mar-5,2005200 311 Dataset-3: Date x y Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-5,200511299 Mar-5,2005200 311 Apl-23,2005 123 200 Now I want to get the common dates along with x and y from this above three datasets keeping the same order in date-variable as it is. For ex. I want to get: Datex y xy x y (from dataset-1) (from dataset-2) (from dataset-3) Jan-3,2005-110 300 -110 300 -110 300 Jan-4,2005 114 -21 114-21 114 -21 Mar-5,2005200 311 200 311 200 311 Can anyone give me any R code to implement this for any number of datasets ? Thanks and regards thanks in advance - [[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
[R] Bootstrap error message: Error in statistic(data, original, ...) : unused argument(s) ( ...)
Dear colleagues,
I've been swamped and fighting with error for a few hours but still
desperately having absolutely no clue:
What's wrong with my bootstraping code?
Thanks a lot!
Error Message:
bootResults=boot(X, myFun, R=1);
Error in statistic(data, original, ...) : unused argument(s) ( ...)
My code is:
X=cbind(column_vector_1, column_vector_2);
myFun=function(X)
{
return(mean(X[, 1])/var(X[, 2]));
}
bootResults=boot(X, myFun, R=1);
-
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Re: [R] A question about nlme
zhongmiao wang [EMAIL PROTECTED] writes: Hello, I have used nlme to fit a model, the R syntax is like fmla0-as.formula(paste(~,paste(colnames(ldata[,9:13]),collapse=+),-1)) fmla1-as.formula(paste(~,paste(colnames(ldata[,14:18]),collapse=+),-1)) fmla2-as.formula(paste(~,paste(colnames(ldata[,19:23]),collapse=+),-1)) Block=pdBlocked(list(pdIdent(fmla0),pdIdent(fmla1),pdIdent(fmla2))) lme(fixed=Score ~ factor(time)-1,data=ldata,random=list(Block), + weights=varIdent(form= ~ 1|time), + correlation=corSymm(form= ~ 1|Block/ID)) and the Error message is Error in parse(file, n, text, prompt) : syntax error in ~ I repeatedly check the ~, however, I cannot find anything wrong with it. Is there anybody can tell what is wrong? Thank you so much! Is the character following ~ _really_ a space in all three cases? Shift-space sometimes catches people off guard. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@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
[R] about vsn
Hi all: As to vsn package,how can I transform my raw data format to the demo data of vsn(named kidney)? My raw data: namesignal dye gene1 815.32 green gene2 671.66 green gene3 713.93 green gene4 703.97 green gene5 493.59 green gene6 477.92 green gene7 346.55 green __ 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
Re: [R] 3D pie
On 19-Apr-06 Peter Ehlers wrote: This discussion of 3-d pie charts comes at an opportune time. I have just formulated a new theory of graphical information transfer which is particularly simple in the case of 3-d pie charts. Let theta denote the angle between the normal to the pie cylinder and the pie-eyed line (connecting eye and centre of pie). Then the information transmitted from pie to viewer is K * (pi/2 - theta)^3 for theta in [0, pi/2]. The normalizing constant may be written in the obvious manner as K = 8 * I_0 / pi^3. I conjecture that I_0 is not large, but I'm still waiting to hear from Microsoft regarding my application for funding to allow me to conduct extensive testing. I think I can confirm your conjecture. With theta = 0, you have in effect a 2-D pie, and then, according to my calculations, if you take I_0 = 3.14159265358979... the information you get is 1 pie. Ted. E-Mail: (Ted Harding) [EMAIL PROTECTED] Fax-to-email: +44 (0)870 094 0861 Date: 20-Apr-06 Time: 09:10:01 -- 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
Re: [R] 3D pie
Hi all, I can understand that it isn't the right way to represent data, i knew that i would hurt some people on this mailling list to discuss about pie. As i've specified on my first message: My only purpose of drawing 3D pie is for customer who don't have to understand what is drawn , but only be impressed by the beauty of the result (like people at commercial division which purchase what you've done but don't use it). Then the expert (who are the ending users) will use the dotchart version (which i've putted too) instead of the pie. Note:it isn't my idea, i wish to follow these advices, but i'm not the buyer... That's all, and for shure i won't forget that pie isn't good, or only at dinner time. Bring me a 3D apple pie and a coffee it's the morning here! (or a sandwich with hamm) Thks for all the answers, Grothendieck, thanks for the R tips with excel,but i'm working under linux, will keep this example in a safe place, could be usefull one day, who knows... Cheers COMTE Guillaume -Message d'origine- De : Frank E Harrell Jr [mailto:[EMAIL PROTECTED] Envoyé : mercredi 19 avril 2006 21:14 À : Rolf Turner Cc : [EMAIL PROTECTED]; COMTE Guillaume; r-help@stat.math.ethz.ch Objet : Re: [R] 3D pie Rolf Turner wrote: Gabor Grothendieck wrote: Since everyone else wimped out with a tedious you-do-not-want-to-do-that, here is a solution that uses R to control Excel and create a 3d chart. . . . People really ***should not*** be encouraged or abetted in wrong-headedness. Excel is terrible. Pie charts are terrible. Don't mess with them. Period. cheers, Rolf Turner [EMAIL PROTECTED] I second that. Helping people do things known to have major problems with the approaches can actually hurt others in the long run. 2-D pie charts are terrible. That makes 3-D pie charts terrible to the 3/2 power. Excel has serious errors and is not a good model for reproducible research. -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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
[R] R-Help
Dear r-users, Suppose I have three datasets: Dataset-1: Date x y Jan-1,2005120 230 Jan-2,2005123 -125 Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-7,200511299 Mar-5,2005200 311 Dataset-2: Date x y Jan-2,2005123 -125 Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-5,200511299 Jan-6,2005-23 12 Mar-5,2005200 311 Dataset-3: Date x y Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-5,200511299 Mar-5,2005200 311 Apl-23,2005 123 200 Now I want to get the common dates along with x and y from this above three datasets keeping the same order in date-variable as it is. For ex. I want to get: Datex y xy x y (from dataset-1) (from dataset-2) (from dataset-3) Jan-3,2005-110 300 -110 300 -110 300 Jan-4,2005 114 -21 114-21 114 -21 Mar-5,2005200 311 200 311 200 311 Can anyone give me any R code to implement this for any number of datasets ? Thanks and regards thanks in advance - [[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
Re: [R] prop.table on three-way table?
Hi marc, I did not manage ctab to do this for me. Again, I am probably using it wrong, but I don't know that the problem is. You asked for a more illustrated example, so here goes: Take this table: ftable(table(sample(paste(dim1_no,1:5,sep=),10,replace=TRUE), sample(paste(dim2_no,1:5,sep=),10,replace=TRUE), sample(paste(dim3_no,1:5,sep=),10,replace=TRUE))) dim3_no1 dim3_no2 dim3_no3 dim3_no5 dim1_no1 dim2_no1 0000 dim2_no2 0020 dim2_no3 0000 dim2_no4 0000 dim2_no5 0000 dim1_no2 dim2_no1 0000 dim2_no2 0000 dim2_no3 0000 dim2_no4 0000 dim2_no5 0101 dim1_no3 dim2_no1 0000 dim2_no2 0000 dim2_no3 0010 dim2_no4 0000 dim2_no5 0000 dim1_no4 dim2_no1 1000 dim2_no2 0000 dim2_no3 0000 dim2_no4 0000 dim2_no5 1000 dim1_no5 dim2_no1 0000 dim2_no2 0100 dim2_no3 1000 dim2_no4 0100 dim2_no5 0000 Now, I would like to get the per cent occurrence of each level of dim3_noX witin the cells by dim1 and dim2. Thus, for this part section of the table above: dim3_no1 dim3_no2 dim3_no3 dim3_no5 dim1_no2 dim2_no1 0000 dim2_no2 0000 dim2_no3 0000 dim2_no4 0000 dim2_no5 0101 I would like to get: dim3_no1 dim3_no2 dim3_no3 dim3_no5 dim1_no2 dim2_no1 0000 dim2_no2 0000 dim2_no3 0000 dim2_no4 0000 dim2_no5 00.500.5 since dim3_no2 represented 50% of the frequency within the cell created by dim1_no2 and dim2_no5. Hope that helped clarify my previous explanation of the problem. /Fredrik 2006/4/19, Marc Schwartz (via MN) [EMAIL PROTECTED]: On Wed, 2006-04-19 at 16:39 +0200, Fredrik Karlsson wrote: Dear list, I am trying to create a three-way table with percent occurrence instead of raw frequencies. However, I cannot get the results I expected: I have the following table: ftable(table( mannerDF$agem, mannerDF$target, mannerDF$manner )) snip 50 bak 0 0 0 0 1 0 pak 0 0 0 0 3 0 sak 0 1 0 0 0 0 spak0 0 0 0 0 0 Now, If I use the prop-table function, I newer get a 1 ratio in any cell: With 'margin=1': 50 bak0. 0. 0. 0. 0.2000 0. pak0. 0. 0. 0. 0.6000 0. sak0. 0.2000 0. 0. 0. 0. spak 0. 0. 0. 0. 0. 0. With 'margin=2': 50 bak 0.0 0.0 0.0 0.0 0.004347826 0.0 pak 0.0 0.0 0.0 0.0 0.010752688 0.0 sak 0.0 0.005747126 0.0 0.0 0.0 0.0 spak 0.0 0.0 0.0 0.0 0.0 0.0 With 'margin=3': 50 bak 0.0 0.0 0.0 0.0 0.001373626 0.0 pak 0.0 0.0 0.0 0.0 0.004120879 0.0 sak 0.0 0.008695652 0.0 0.0 0.0 0.0 spak 0.0 0.0 0.0 0.0 0.0 0.0 What I was looking for is this: 50 bak 0 0 0 0 1 0 pak 0 0 0 0 1 0 sak 0 1 0 0 0 0 spak0 0 0 0 0 0 (With more digits) Am I doing something stupid? I may be missing what you
Re: [R] I am surprised (and a little irritated)
Detlef Steuer wrote: Hi, I`m the one to blame for the readme :-) and for providing the rpms. I am sorry, but the note was not primarily directed at you, perhaps (I am not sure) just as much at Linux in general, or Suse, or OpenSuse, or ... In any case, I am sure you do a great job at providing the rpm's. As to the readme, from my point of view there are some information missing. If you encounter such big problems my readme sucks. But I'm open for critisism and will improve on the current situation for the release of R-2.3.0 next monday. Using the word sucks is too strong. But, so far, and from my point of view, there is room for improvement. It would have helped, if I got the error messages you saw. Now I just have to guess. After installing Linux using the OpenSuse download CD's, when I try to install the rpm for R-base, I am informed that libblas.so.3 is not available and the same with libgfortran.so.0. Neither of the two are in the list in the readme. However, I guess (which should not be necessary) that the first refers to blas-3.0-926 in the list, and the second to gcc-fortran-4.0.2 But I have have been unable to locate these elements, neither on the CD's, nor on the dvd included in the book on SUSE Linux 10 by McAllister which is supposed to have the complete OpenSuse 10. I may be wrong there, there are VERY many rpms's on the dvd, but the task is not very simple. The key is probably (perhaps) in the part where you have written: quoteFrom R-2.2.1 on you can use the CRAN-Mirror near you as YAST installation source. Just add $CRANMIRROR/bin/linux/suse/MAJOR.MINOR as a http source as an installation source for yast. Alternatively you may use the main package repository http://fawn.hsu-hh.de/~steuer/SL-10.0-OSS ./quote But I have been unable to see where to use this information with YAST and how. So, what is needed is information for the uninformed. As far as I can see, a possible solution may be in using the apt approach as suggested in the McAllister book. Therefore I have installed the synaptic program as well, but so far (that was late in the day at the office and I do not have a full connection with the Internet and Linux from home yet, only http), it did not help. I'll try again later today. I think you'll be able to find a step by step instruction by the end of the week on CRAN. I am looking forward to that. I hope that my tale of unsuccessful attempts will be useful. Part of the problem is that R and Suse are moving targets. On the opensuse side there are packages, which are not on the CDs but in the online repositories that have to be downloaded seperately. Ranting alone won't help. I am sure you are right about the problem. As I have said before, I do not intend to rant in any way. But I would very much like to see that R is as available on SUSE Linux (and Linux in general) as it is on Windows for newcomers like me. Look here: http://www.thejemreport.com/mambo/content/view/178/42/ for some introductory material. You can add http://fawn.hsu-hh.de/~steuer/SL-10.0-OSS as installation source for R and ESS. That link looks very useful. Perhaps the readme could be somewhat similar? That the installation procedure is different for any flavour of Linux than the one you know from windows should not surprise you. Whatever distribution you'll end up using: the time invested to learn the respective package management system will pay back. Yes, I am sure you are right, and I am not surprised at the differences between the various flavours of Linux. Still, I think that instructions on installing R on a particular distribution should be oriented towards the completely unintiated user. More experienced users can skip the trivial (in their view) parts. But it should be there by default. I have to add that it is not very constructive to say read the manual as some of the responses (not you) to my mail imply. I do not mind to read the manual, but at least I would like to know where to look in the manuals. In any case, would it be possible to persuade SUSE to include R in the installation, at least as an option? Feel free to ask any question on R on SuSE. I would be happy to send you the next readme for review. Thank you. I look forward to that. Tom ++ | Tom Backer Johnsen, Psychometrics Unit, Faculty of Psychology | | University of Bergen, Christies gt. 12, N-5015 Bergen, NORWAY | | Tel : +47-5558-9185Fax : +47-5558-9879 | | Email : [EMAIL PROTECTED]URL : http://www.galton.uib.no/ | ++ __ 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
[R] Extract AIC, BIC
Hi All, How can extract AIC,BIC from a fitted Garch model? -- SUMANTA BASAK. [[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
Re: [R] I am surprised (and a little irritated)
Gabor Csardi wrote: So you don't want to read that manual after all? :) I did not say so, not at all. The point is that it is not very helpful to say Read the manual when the the problem is very much one of not knowing where to look in the manual. Instructions for installing on a particular distribution should be focused on the uninitiated, not the experiences users. The thing is that for Linux you cannot just develop a program which 'just installs' or 'just works' on any Linux system. (Or even if you could do that, it is not worth the effort.) This is because 'Linux' is just an operating system kernel, shells, compilers, window managers, GUI's are NOT part of it. I know. But it should be possible to do so for a few of the major distros. According to some of the responses, some of them even include R in the installation. This is what Linux disributions are for. They are collections of software containing both the kernel and the user tools (compilers, shells, R, etc.). There are many thousends of linux distribution and they are not at all compatible with each other. It might happen that R is packaged differently in a distribution than in another, etc. This is a tough world, but also gives you the freedom of choice, some people like it, some people don't. For a software package like R, it is absolutely impossible to ensure that it can be installed cleanly without any problems on all the (say) two thousend linux distributions. Usually the people who packaged the piece of software (R in this case) for the specific distribution are the ones who can help you with installation problems. (Not always, but very likely with your problem.) And for most suse packages these people are the suse developers. I am not referring to 2000 distributions. I am referring to one of the most popular ones, SUSE. This is why i suggested to post to the suse list. But you got some useful hints from helpful people here, so after all you are lucky. :) Yes, I am grateful (in the most part) for the responses, where Ditlef's one was in particular very constructive. But (a) I am interested in using R, (b) I want to try R on one of the major distributions of Linux (c) I do not succeed in doing so. What I wonder is then: How many others have made the same attempt, not succeeded, and then given up? Which I do not intend to do. I might have overreacted this, but it is really irritating that you just take the result of the really hard work of many people, the R software, for free, and then complain about something which you could solve easily just by searching on google or in the R mailing list archive, or reading the suse manuals. Yes, I think you are being oversensitive. On the contrary, it is because I DO realize that there is an enormous effort involved in the development of R that I am surprised that the final step, making it available to users is so difficult in this particular case. That is, as far as I can see, something that is (or should be) of concern to the advocates of R (where I regard myself as included). That is the reason for my mail to the list. Tom ++ | Tom Backer Johnsen, Psychometrics Unit, Faculty of Psychology | | University of Bergen, Christies gt. 12, N-5015 Bergen, NORWAY | | Tel : +47-5558-9185Fax : +47-5558-9879 | | Email : [EMAIL PROTECTED]URL : http://www.galton.uib.no/ | ++ __ 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
[R] nls and factor
Is it possible to include a factor in an nls formula? I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ 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
Re: [R] Breakdown a number
On Thu, 20 Apr 2006, Gabor Grothendieck wrote:
On 4/19/06, Paul Roebuck [EMAIL PROTECTED] wrote:
On Wed, 19 Apr 2006, Gabor Grothendieck wrote:
On 4/19/06, Paul Roebuck [EMAIL PROTECTED] wrote:
Isn't there a builtin method for doing this and, if so,
what is it called?
breakdown - function(whole) {
breaks - c(250, 800)
pieces - integer(length(breaks) + 1)
if (whole breaks[2]) {
pieces[3] - whole - breaks[2]
whole - breaks[2]
}
if (whole breaks[1]) {
pieces[2] - whole - breaks[1]
whole - breaks[1]
}
pieces[1] - whole
return(pieces)
}
breakdown(1200) # 250 550 400
Maybe you could discuss in words what you want but perhaps
you are looking for diff:
That was rather my problem searching for my answer as I
was unsure what this was called. I was searching for
'bins' or 'breaks' (like in hist method but not for
plotting) and leading nowhere fast. Alas, I thought
sample code would go further than my description.
bp - c(0, 250, 800, 1200)
diff(bp)
[1] 250 550 400
Don't think diff method is going to work either, at
least in cases where the 'whole' is less than greatest
'break'.
breakdown2 - function(x, breaks = c(250, 800)) {
diff(c(0, breaks, x))
}
breakdown(10) # 10 0 0
breakdown2(10)# 250 550 -790
breakdown(400)# 250 150 0
breakdown2(400) # 250 550 -400
Just sort it:
bd - function(x, breaks = c(250, 800)) diff(sort(c(0, breaks, x)))
Doesn't work either. Tests below didn't fare well.
If you don't want the 0 values in the case that x is one of the
breaks then use diff(sort(unique(c(0, breaks, x
I did want the zeros in the result.
###
check.results - function(input) {
stopifnot(input = 0)
output - bd(input)
expected - breakdown(input)
if (!identical(output, expected)) {
cat(expected:, expected, \n)
cat(output: , output, \n)
return(FALSE)
} else {
return(TRUE)
}
}
check.results(1200) # TRUE
check.results(800) # TRUE
check.results(799) # FALSE
check.results(500) # FALSE
check.results(250) # FALSE
check.results(200) # FALSE
check.results(1)# FALSE
check.results(0)# FALSE
Goal was to split value according to breaks such that each
piece of it could later be multiplied with a corresponding
rate. It's not unlike the type of processing that would go
into figuring out how much a hypothetical out-of-state phone
call would cost (for example, .10 for first five minutes,
0.07 for next ten minutes, and 0.05 for rest of the call)
calc.cost.per.call - function(minutes) {
duration - bd(minutes, breaks = c(5, 10))
rates - c(0.10, 0.07, 0.05)
sum(duration * rates)
}
calc.cost.per.call(24) # would be 1.65
Thanks for your help.
--
SIGSIG -- signature too long (core dumped)
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Re: [R] nls and factor
Manuel, I don't think that it works very easily. Instead, try gnls() in the nlme package. Cheers Andrew On Thu, Apr 20, 2006 at 11:18:02AM +0200, Manuel Gutierrez wrote: Is it possible to include a factor in an nls formula? I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ 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 -- Andrew Robinson Department of Mathematics and StatisticsTel: +61-3-8344-9763 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au __ 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
[R] info : Manova - eigenvector analysis and canonical analysis
Hello everybody ! I try to obtain in R eigenvectors and canonical analysis on MANOVA results, but I don't find how to process? In particular, I would be interesting to obtain standardized canonical coefficients of the canonical variates. There analysis give some information on the correlation between response variates. My data are organised in 2 terms (one is continu, one is a factor) and 3 response variates (continus). summary(manova(cbind(dureeL,dureeI,dureeR)~denst*etatav, data=durt)) Df Pillai approx F num Df den DfPr(F) denst 1 0.3047 10.6629 3 73 6.795e-06 *** etatav 1 0.1070 2.9146 3 73 0.03995 * denst:etatav 1 0.0324 0.8138 3 73 0.49035 Residuals 75 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 I hope you can help me. best regards Céline Hauzy Céline Hauzy Laboratoire Bioemco Ecole Normale Supérieure 46 rue d'Ulm 75230 Paris Cedex 05 Tel : 01 44 32 38 78 Fax : 01 44 32 38 85 info __ 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
Re: [R] nls and factor
Thanks Andrew. I am now trying but without much success. I don't now how to give start values for the factor?. Could you give me an example solution with my toy example? a-as.factor(c(rep(1,50),rep(0,50))) independ-1:100 respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 library(nlme) gnls(respo~independ^b+a,start=list(b=1.8)) Many thanks. Manu --- Andrew Robinson [EMAIL PROTECTED] escribió: Manuel, I don't think that it works very easily. Instead, try gnls() in the nlme package. Cheers Andrew On Thu, Apr 20, 2006 at 11:18:02AM +0200, Manuel Gutierrez wrote: Is it possible to include a factor in an nls formula? I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ 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 -- Andrew Robinson Department of Mathematics and Statistics Tel: +61-3-8344-9763 University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599 Email: [EMAIL PROTECTED] http://www.ms.unimelb.edu.au __ 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
Re: [R] Randomly selecting one row for each factor level [Broadca st]
The following should work: dfr.samp - dfr[tapply(1:nrow(dfr), dfr$x, sample, 1),] dfr.samp x y z 10 a 10 J 2 b 2 B 9 c 9 I Andy From: Kelly Hildner I don't use R much, and I have been unable to figure out how to get the subset of my data frame that I would like. For example, if this were my data frame: dfr - data.frame(x=rep(letters[1:3], 4), y=(1:12), z=(LETTERS[1:12])) dfr x y z 1 a 1 A 2 b 2 B 3 c 3 C 4 a 4 D 5 b 5 E 6 c 6 F 7 a 7 G 8 b 8 H 9 c 9 I 10 a 10 J 11 b 11 K 12 c 12 L I would like to randomly select one row for each level of the factor x and create a new data frame with the results. For example, the result might be: x y z 1 a 1 A 5 b 5 E 6 c 6 F Any help would be greatly appreciated! Thanks, Kelly -- K. Kelly Hildner, Ph.D. NOAA Fisheries Southwest Fisheries Science Center 110 Shaffer Rd. Santa Cruz, CA 95060 __ 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 __ 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
Re: [R] I am surprised (and a little irritated)
Detlef Steuer wrote: On Thu, 20 Apr 2006 10:51:57 +0200 . Sometimes it is very hard to think of the questions a new user has, if you do it yourself on a daily basis for years. I understand that very well. Now I can tell you that I used the info in the Hacking OpenSuse link you provided to the letter, and followed all the instructions in the paragraph with the heading Adding sources to YaST. After that installing R from the R-base rpm was simple. It took a few minutes, but that does not matter. Evidently all the necessary packages were located on from the locations on the net and were installed automatically. So now R is installed and works. The only drawback I can think of is that you have to be online to install. Conclusion: Put something like the info from the link in the readme and other users of R can be recruited from the SUSE world. Thank you! Tom ++ | Tom Backer Johnsen, Psychometrics Unit, Faculty of Psychology | | University of Bergen, Christies gt. 12, N-5015 Bergen, NORWAY | | Tel : +47-5558-9185Fax : +47-5558-9879 | | Email : [EMAIL PROTECTED]URL : http://www.galton.uib.no/ | ++ __ 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
Re: [R] 3D pie
Rolf Turner wrote: People really ***should not*** be encouraged or abetted in wrong-headedness. Excel is terrible. Pie charts are terrible. Don't mess with them. Period. Now I realise the opportunity I missed on April 1st, when I was going to try and (anonymously) post the most flammable R-help posting ever. Something like: I'm trying to make a library with R 1.6.1 to create a 3-d pie chart in excel but seq(0,1,by=0.1)[4]==0.3 is false. Barry __ 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
[R] info : Manova - eigenvector analysis and canonical analysis
Hello everybody ! I try to obtain in R eigenvectors and canonical analysis on MANOVA results, but I don't find how to process? In particular, I would be interesting to obtain standardized canonical coefficients of the canonical variates. There analysis give some information on the correlation between response variates. My data are organised in 2 terms (one is continu, one is a factor) and 3 response variates (continus). summary(manova(cbind(dureeL,dureeI,dureeR)~denst*etatav, data=durt)) Df Pillai approx F num Df den DfPr(F) denst 1 0.3047 10.6629 3 73 6.795e-06 *** etatav 1 0.1070 2.9146 3 73 0.03995 * denst:etatav 1 0.0324 0.8138 3 73 0.49035 Residuals 75 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 I hope you can help me. best regards Céline Hauzy Céline Hauzy Laboratoire Bioemco Ecole Normale Supérieure 46 rue d'Ulm 75230 Paris Cedex 05 Tel : 01 44 32 38 78 Fax : 01 44 32 38 85 info __ 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
Re: [R] 3D pie
On 19-Apr-06 Peter Ehlers wrote: This discussion of 3-d pie charts comes at an opportune time. I have just formulated a new theory of graphical information transfer which is particularly simple in the case of 3-d pie charts. Let theta denote the angle between the normal to the pie cylinder and the pie-eyed line (connecting eye and centre of pie). Then the information transmitted from pie to viewer is K * (pi/2 - theta)^3 for theta in [0, pi/2]. The normalizing constant may be written in the obvious manner as K = 8 * I_0 / pi^3. I conjecture that I_0 is not large, but I'm still waiting to hear from Microsoft regarding my application for funding to allow me to conduct extensive testing. I think I can confirm your conjecture. With theta = 0, you have in effect a 2-D pie, and then, according to my calculations, if you take I_0 = 3.14159265358979... the information you get is 1 pie. Ted. Unless you cut it into quarters, in which case you have that rare situation where pie by two equals pie by four... ... sorry Stuart This message has been checked for viruses but the contents of an attachment may still contain software viruses, which could damage your computer system: you are advised to perform your own checks. Email communications with the University of Nottingham may be monitored as permitted by UK legislation. __ 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
Re: [R] nls and factor
On Thu, 20 Apr 2006, Manuel Gutierrez wrote: Is it possible to include a factor in an nls formula? Yes. What do you intend by it? If you mean what it would mean for a lm formula, you need A[a] and starting values for A. There's an example on p.219 of MASS4. I've searched the help pages without any luck so I guess it is not feasible. I've given it a few attempts without luck getting the message: + not meaningful for factors in: Ops.factor(independ^EE, a) This is a toy example, my realworld case is much more complicated (and can not be solved linearizing an using lm) a-as.factor(c(rep(1,50),rep(0,50))) independ-rnorm(100) respo-rep(NA,100) respo[a==1]-(independ[a==1]^2.3)+2 respo[a==0]-(independ[a==0]^2.1)+3 nls(respo~independ^EE+a,start=list(EE=1.8),trace=TRUE) Any pointers welcomed Many Thanks, Manu __ 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 -- 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
[R] (sem assunto)
Dear R-colleagues, Is it possible to mix TEXT and VALUE of objects in y (or x) label of a plot? For instance: f-2 plot(..., ylab='Axis f', ...) where f means the VALUE of f. Thany you in advance, Eric. -- Barba Departamento de Ciências Exatas Universidade Federal de Lavras Minas Gerais - Brasil [[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
Re: [R] Breakdown a number
Try using breaks[breaks = x] in place of breaks. Your solution
always returns three components if breaks has two whereas this one
does not but you could extend the output it if that is an essential part.
On 4/20/06, Paul Roebuck [EMAIL PROTECTED] wrote:
On Thu, 20 Apr 2006, Gabor Grothendieck wrote:
On 4/19/06, Paul Roebuck [EMAIL PROTECTED] wrote:
On Wed, 19 Apr 2006, Gabor Grothendieck wrote:
On 4/19/06, Paul Roebuck [EMAIL PROTECTED] wrote:
Isn't there a builtin method for doing this and, if so,
what is it called?
breakdown - function(whole) {
breaks - c(250, 800)
pieces - integer(length(breaks) + 1)
if (whole breaks[2]) {
pieces[3] - whole - breaks[2]
whole - breaks[2]
}
if (whole breaks[1]) {
pieces[2] - whole - breaks[1]
whole - breaks[1]
}
pieces[1] - whole
return(pieces)
}
breakdown(1200) # 250 550 400
Maybe you could discuss in words what you want but perhaps
you are looking for diff:
That was rather my problem searching for my answer as I
was unsure what this was called. I was searching for
'bins' or 'breaks' (like in hist method but not for
plotting) and leading nowhere fast. Alas, I thought
sample code would go further than my description.
bp - c(0, 250, 800, 1200)
diff(bp)
[1] 250 550 400
Don't think diff method is going to work either, at
least in cases where the 'whole' is less than greatest
'break'.
breakdown2 - function(x, breaks = c(250, 800)) {
diff(c(0, breaks, x))
}
breakdown(10) # 10 0 0
breakdown2(10)# 250 550 -790
breakdown(400)# 250 150 0
breakdown2(400) # 250 550 -400
Just sort it:
bd - function(x, breaks = c(250, 800)) diff(sort(c(0, breaks, x)))
Doesn't work either. Tests below didn't fare well.
If you don't want the 0 values in the case that x is one of the
breaks then use diff(sort(unique(c(0, breaks, x
I did want the zeros in the result.
###
check.results - function(input) {
stopifnot(input = 0)
output - bd(input)
expected - breakdown(input)
if (!identical(output, expected)) {
cat(expected:, expected, \n)
cat(output: , output, \n)
return(FALSE)
} else {
return(TRUE)
}
}
check.results(1200) # TRUE
check.results(800) # TRUE
check.results(799) # FALSE
check.results(500) # FALSE
check.results(250) # FALSE
check.results(200) # FALSE
check.results(1)# FALSE
check.results(0)# FALSE
Goal was to split value according to breaks such that each
piece of it could later be multiplied with a corresponding
rate. It's not unlike the type of processing that would go
into figuring out how much a hypothetical out-of-state phone
call would cost (for example, .10 for first five minutes,
0.07 for next ten minutes, and 0.05 for rest of the call)
calc.cost.per.call - function(minutes) {
duration - bd(minutes, breaks = c(5, 10))
rates - c(0.10, 0.07, 0.05)
sum(duration * rates)
}
calc.cost.per.call(24) # would be 1.65
Thanks for your help.
--
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Re: [R] R-Help
Read them in as zoo objects (you can replace textConnection(Lines1) with the filename) and then merge them using all = FALSE to retain only common time points. Note that in my English locale I had to modify your Apl to Apr. Lines1 - Date x y Jan-1,2005120 230 Jan-2,2005123 -125 Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-7,200511299 Mar-5,2005200 311 Lines2 - Date x y Jan-2,2005123 -125 Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-5,200511299 Jan-6,2005-23 12 Mar-5,2005200 311 Lines3 - Date x y Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-5,200511299 Mar-5,2005200 311 Apr-23,2005 123 200 library(zoo) DF1 - read.zoo(textConnection(Lines1), header = TRUE, format = %b-%d,%Y) DF2 - read.zoo(textConnection(Lines2), header = TRUE, format = %b-%d,%Y) DF3 - read.zoo(textConnection(Lines3), header = TRUE, format = %b-%d,%Y) merge(DF1, DF2, DF3, all = FALSE) On 4/20/06, stat stat [EMAIL PROTECTED] wrote: Dear r-users, Suppose I have three datasets: Dataset-1: Date x y Jan-1,2005120 230 Jan-2,2005123 -125 Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-7,200511299 Mar-5,2005200 311 Dataset-2: Date x y Jan-2,2005123 -125 Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-5,200511299 Jan-6,2005-23 12 Mar-5,2005200 311 Dataset-3: Date x y Jan-3,2005-110 300 Jan-4,2005114 -21 Jan-5,200511299 Mar-5,2005200 311 Apl-23,2005 123 200 Now I want to get the common dates along with x and y from this above three datasets keeping the same order in date-variable as it is. For ex. I want to get: Datex y xy x y (from dataset-1) (from dataset-2) (from dataset-3) Jan-3,2005-110 300 -110 300 -110 300 Jan-4,2005 114 -21 114-21 114 -21 Mar-5,2005200 311 200 311 200 311 Can anyone give me any R code to implement this for any number of datasets ? Thanks and regards thanks in advance - [[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 __ 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
Re: [R] (sem assunto)
f - 2 plot(runif(10), ylab=paste(Axis, f, collapse= )) ?paste Eric Ferreira wrote: Dear R-colleagues, Is it possible to mix TEXT and VALUE of objects in y (or x) label of a plot? For instance: f-2 plot(..., ylab='Axis f', ...) where f means the VALUE of f. Thany you in advance, Eric. -- Barba Departamento de Ciências Exatas Universidade Federal de Lavras Minas Gerais - Brasil [[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 -- Chuck Cleland, Ph.D. NDRI, Inc. 71 West 23rd Street, 8th floor New York, NY 10010 tel: (212) 845-4495 (Tu, Th) tel: (732) 512-0171 (M, W, F) fax: (917) 438-0894 __ 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
Re: [R] (sem assunto)
Try paste: paste(Axis, f) On 4/20/06, Eric Ferreira [EMAIL PROTECTED] wrote: Dear R-colleagues, Is it possible to mix TEXT and VALUE of objects in y (or x) label of a plot? For instance: f-2 plot(..., ylab='Axis f', ...) where f means the VALUE of f. Thany you in advance, Eric. -- Barba Departamento de Ciências Exatas Universidade Federal de Lavras Minas Gerais - Brasil [[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 __ 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
Re: [R] prop.table on three-way table?
On Thu, 2006-04-20 at 10:28 +0200, Fredrik Karlsson wrote: Hi marc, I did not manage ctab to do this for me. Again, I am probably using it wrong, but I don't know that the problem is. You asked for a more illustrated example, so here goes: Take this table: ftable(table(sample(paste(dim1_no,1:5,sep=),10,replace=TRUE), sample(paste(dim2_no,1:5,sep=),10,replace=TRUE), sample(paste(dim3_no,1:5,sep=),10,replace=TRUE))) dim3_no1 dim3_no2 dim3_no3 dim3_no5 dim1_no1 dim2_no1 0000 dim2_no2 0020 dim2_no3 0000 dim2_no4 0000 dim2_no5 0000 dim1_no2 dim2_no1 0000 dim2_no2 0000 dim2_no3 0000 dim2_no4 0000 dim2_no5 0101 dim1_no3 dim2_no1 0000 dim2_no2 0000 dim2_no3 0010 dim2_no4 0000 dim2_no5 0000 dim1_no4 dim2_no1 1000 dim2_no2 0000 dim2_no3 0000 dim2_no4 0000 dim2_no5 1000 dim1_no5 dim2_no1 0000 dim2_no2 0100 dim2_no3 1000 dim2_no4 0100 dim2_no5 0000 Now, I would like to get the per cent occurrence of each level of dim3_noX witin the cells by dim1 and dim2. Thus, for this part section of the table above: dim3_no1 dim3_no2 dim3_no3 dim3_no5 dim1_no2 dim2_no1 0000 dim2_no2 0000 dim2_no3 0000 dim2_no4 0000 dim2_no5 0101 I would like to get: dim3_no1 dim3_no2 dim3_no3 dim3_no5 dim1_no2 dim2_no1 0000 dim2_no2 0000 dim2_no3 0000 dim2_no4 0000 dim2_no5 00.500.5 since dim3_no2 represented 50% of the frequency within the cell created by dim1_no2 and dim2_no5. Hope that helped clarify my previous explanation of the problem. Fredrik, If I correctly understand what you are doing, which seems to be to calculate row based percentages (actually proportions) within each subgroup, the following should do it. Note that I am using set.seed() so that you can reproduce the data in question here. library(catspec) set.seed(1) x - table(sample(paste(dim1_no, 1:5, sep = ), 10, replace = TRUE), sample(paste(dim2_no, 1:5, sep = ), 10, replace = TRUE), sample(paste(dim3_no, 1:5, sep = ), 10, replace = TRUE)) ctab(x, type = row, percentages = FALSE) Thus, 'x' is: x , , = dim3_no1 dim2_no1 dim2_no3 dim2_no4 dim2_no5 dim1_no10000 dim1_no30000 dim1_no40010 dim1_no50100 , , = dim3_no2 dim2_no1 dim2_no3 dim2_no4 dim2_no5 dim1_no10100 dim1_no30000 dim1_no41000 dim1_no50000 , , = dim3_no3 dim2_no1 dim2_no3 dim2_no4 dim2_no5 dim1_no10100 dim1_no30021 dim1_no40000 dim1_no50000 , , = dim3_no4 dim2_no1 dim2_no3 dim2_no4 dim2_no5 dim1_no10000 dim1_no30000 dim1_no40010 dim1_no50000 , , = dim3_no5 dim2_no1 dim2_no3 dim2_no4 dim2_no5 dim1_no10000 dim1_no30010 dim1_no40000 dim1_no50000 More concisely viewed as: ftable(x) dim3_no1 dim3_no2 dim3_no3 dim3_no4 dim3_no5
Re: [R] there is no xls reader in R?
I like to use the RODBC package for doing this. Here is my code sample: xls - odbcConnectExcel(fname) rawdata.temp - sqlFetch(xls, rawdata, max=2800) close(xls) fname is the full path to the file and rawdata is the name of the excel sheet I want to import. I tried one other approach (I think it was read.xls()) and that approach used perl scripts to read in the xls file and was very slow. RODBC is very fast and has always worked great for me. Haven't tried any of the other ways mentioned. On 4/20/06, Ko-Kang Kevin Wang [EMAIL PROTECTED] wrote: Have a look at the read.xls() in gdata package. HTH, Kevin Michael wrote: Currently I have to convert all my xls into csv before I can read it in and process the excel data in R... Is there a way to directly read in xls data? 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 -- Ko-Kang Kevin Wang Homepage: http://wwwmaths.anu.edu.au/~wangk/ Ph (W): +61-2-6125-2431 Ph (H): +61-2-6125-7471 Ph (M): +61-40-451-8301 __ 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 [[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
Re: [R] R-Help
I think this does what you require.
#Read your data in whatever way you wish:
d1-data.frame(Date=c(2005/1/1,2005/2/1,2005/1/3,2005/1/4,2005/
1/7,2005/3/5),
x=c(119,123,-110,114,11,200),
y=c(230,-125,300,-21,299,311))
d2-data.frame(Date=c(2005/1/3,2005/1/4,2005/1/5,2005/1/6,2005/
3/5),
x=c(-220,116,888,-239,201),
y=c(301,-23,3000,122,312))
d3-data.frame(Date=c(2005/1/4,2005/1/5,2005/3/5,2005/4/23),
x=c(392,511,600,723),
y=c(-81,6699,9311,1200))
#Make a list
listof-list(d1,d2,d3)
#loop over any number of datasets merging as you go
for ( dataset in 1:length(listof)-1) {
if (dataset == 1) {
res-merge(listof[dataset],listof[dataset+1],all=T,by=Date)
} else {
res-merge(res,listof[dataset+1],all=T,by=Date)
}
}
# Hope that helps
JS
---
John Seers
Institute of Food Research
Norwich Research Park
Colney
Norwich
NR4 7UA
tel +44 (0)1603 251490
fax +44 (0)1603 255167
e-mail [EMAIL PROTECTED]
e-disclaimer at http://www.ifr.ac.uk/edisclaimer/
Web sites:
www.ifr.ac.uk
www.foodandhealthnetwork.com
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of stat stat
Sent: 20 April 2006 09:17
To: r-help@stat.math.ethz.ch
Subject: [R] R-Help
Dear r-users,
Suppose I have three datasets:
Dataset-1:
Date x y
Jan-1,2005120 230
Jan-2,2005123 -125
Jan-3,2005-110 300
Jan-4,2005114 -21
Jan-7,200511299
Mar-5,2005200 311
Dataset-2:
Date x y
Jan-2,2005123 -125
Jan-3,2005-110 300
Jan-4,2005114 -21
Jan-5,200511299
Jan-6,2005-23 12
Mar-5,2005200 311
Dataset-3:
Date x y
Jan-3,2005-110 300
Jan-4,2005114 -21
Jan-5,200511299
Mar-5,2005200 311
Apl-23,2005 123 200
Now I want to get the common dates along with x and y from this above
three datasets keeping the same order
in date-variable as it is.
For ex. I want to get:
Datex y xy
x y
(from dataset-1) (from dataset-2) (from
dataset-3)
Jan-3,2005-110 300 -110 300
-110 300
Jan-4,2005 114 -21 114-21 114
-21
Mar-5,2005200 311 200 311 200
311
Can anyone give me any R code to implement this for any number of
datasets ?
Thanks and regards
thanks in advance
-
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Re: [R] I am surprised (and a little irritated)
Detlef: For your information. A few days ago I ordered SuSE 10, the version corresponding to the old Professional. It arrived today, and was installed. When the R-base rpm was downloaded with Mozilla Firebird, it was only to open it when the download was finished, click the install with YaST button, and everything went smooth. I had to insert the DVD, where it looked like all the other components were found. So, most of the necessary modifications in the readme will refer to the OpenSuse version alone. The commercial version is very simple. Tom ++ | Tom Backer Johnsen, Psychometrics Unit, Faculty of Psychology | | University of Bergen, Christies gt. 12, N-5015 Bergen, NORWAY | | Tel : +47-5558-9185Fax : +47-5558-9879 | | Email : [EMAIL PROTECTED]URL : http://www.galton.uib.no/ | ++ __ 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
[R] more on aov contrasts residual error calculation
Hi, I haven't recieved any replies to my last email, so let me be a bit more specific. I have a dataframe and it has the following structure: Condition Mapping SubjectABC 11 510 15 12 13 14 15 16 17 18 29 210 Mapping is a between-subject factor. Condition is a within-subject factor. There are 5 levels of mapping, 8 subjects nested in each level of mapping. For each of the 40 combinations of mapping and subject there are 3 observations, one in each level of the condition factor. I want to estimate the pooled error associated with the following set of 4 orthogonal contrasts: condition.L:mapping.L condition.L:mapping.Q condition.L:mapping.C condition.L:mapping^4 What is the best way to do this? One way is to estimate the linear contrast for condition for each subject, create a 40 row matrix where the measure for each combination of mapping and subject is the linear contrast on condition. If I pass this dataframe to aov, the mse it returns is the value I am looking for. If possible, I would like to obtain the estimate without collapsing the dataframe, but am not sure how to proceed. Suggestions? Steve [[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
[R] Online course - graphics in R
Dr. Paul Murrell will offer his Graphics in R course online at statistics.com May 5 June 2. This course teaches you how to produce publication-quality statistical plots of data using R (a freely available open- source statistical language and environment). It will cover plots such as scatterplots, bar plots, histograms, boxplots and Trellis plots. It will review the underlying model used to produce plots in R so that you can extensively customize these plots. Finally, the course will introduce the grid graphics system and look at producing unique plots from the ground up using basic components. Dr. Murrell, the author of R Graphics, has been a member of the core development team for R since 1999, with a focus on the graphics system in R. He is the Editor-in-Chief of R News, the newsletter of the R project, and an Associate Editor for Computational Statistics. He is also Chair of the American Statistical Association's Statistical Graphics Section, and a Senior Lecturer in the Department of Statistics at the University of Auckland, New Zealand. The course lasts four weeks and consists of a series of four weekly lessons (directed readings and/or notes, plus exercises). Direct interaction (QA) with the instructor throughout the four week period takes place via a private discussion board. Expect to spend about 5-15 hours per week; there are no set hours when you must be online. Details and registration: http://www.statistics.com/content/courses/graphicsR/ Peter Bruce [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
Re: [R] (sem assunto)
Thak you very much indeed! Eric. On 4/20/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: Try paste: paste(Axis, f) On 4/20/06, Eric Ferreira [EMAIL PROTECTED] wrote: Dear R-colleagues, Is it possible to mix TEXT and VALUE of objects in y (or x) label of a plot? For instance: f-2 plot(..., ylab='Axis f', ...) where f means the VALUE of f. Thany you in advance, Eric. -- Barba Departamento de Ciências Exatas Universidade Federal de Lavras Minas Gerais - Brasil [[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 -- Barba Departamento de Ciências Exatas Universidade Federal de Lavras Minas Gerais - Brasil [[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
[R] S4 objects with list of objects as slots: how to subset?
Hello,
I don't manage to see if you have already focussed on this point in some
previous messages so I post my question:
I have a little problem with the S4 style of programming.
I tried to formalize my question: please consider the following example
that you can run I think:
#--
setClass(my.class1,
representation(
my.list= list),
prototype=prototype(
my.list=list()))
#--
setClass(my.class2,
representation(
my.slot= numeric),
prototype=prototype(
my.slot=as.numeric(1:10)))
#--
a.class - new(my.class1)
[EMAIL PROTECTED] - new(my.class2)
[EMAIL PROTECTED] - new(my.class2)
using indexation, using my class definitions, how could I avoid the
following FOR-loop (for performance purpose if it changes something) to
get straightforward for example result = c(1:10,1:10) ?
result -NULL
for (i in 1:length([EMAIL PROTECTED]))
{
result - c(result, [EMAIL PROTECTED]@my.slot)
}
this following code doesn't work:
[EMAIL PROTECTED]:[EMAIL PROTECTED]
I hope I have been clear,
Thank you,
Regards,
--
François Bastardie
French Research Institute for the Development of the Sea (IFREMER)
Ecologie et Modèles pour l'Halieutique
Rue de l'Ile d'Yeu
BP 21105
44311 NANTES Cedex 03 - France
Tél : 02 40 37 41 64
Fax : 02 40 37 40 75
E-mail : [EMAIL PROTECTED]
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Re: [R] S4 objects with list of objects as slots: how to subset?
Try:
c(sapply(1:2, function(i) [EMAIL PROTECTED]@my.slot))
On 4/20/06, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
Hello,
I don't manage to see if you have already focussed on this point in some
previous messages so I post my question:
I have a little problem with the S4 style of programming.
I tried to formalize my question: please consider the following example
that you can run I think:
#--
setClass(my.class1,
representation(
my.list= list),
prototype=prototype(
my.list=list()))
#--
setClass(my.class2,
representation(
my.slot= numeric),
prototype=prototype(
my.slot=as.numeric(1:10)))
#--
a.class - new(my.class1)
[EMAIL PROTECTED] - new(my.class2)
[EMAIL PROTECTED] - new(my.class2)
using indexation, using my class definitions, how could I avoid the
following FOR-loop (for performance purpose if it changes something) to
get straightforward for example result = c(1:10,1:10) ?
result -NULL
for (i in 1:length([EMAIL PROTECTED]))
{
result - c(result, [EMAIL PROTECTED]@my.slot)
}
this following code doesn't work:
[EMAIL PROTECTED]:[EMAIL PROTECTED]
I hope I have been clear,
Thank you,
Regards,
--
François Bastardie
French Research Institute for the Development of the Sea (IFREMER)
Ecologie et Modèles pour l'Halieutique
Rue de l'Ile d'Yeu
BP 21105
44311 NANTES Cedex 03 - France
Tél : 02 40 37 41 64
Fax : 02 40 37 40 75
E-mail : [EMAIL PROTECTED]
__
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[R] parsing arguments of a function
Hi,
I have a simple problem writing a function that is to be called like
Myfunction( Column1 = low, Column2 = high, Column3 = all, Column4 = all,
data = mydata)
{.
contourplot(z ~ mydata$Column3* mydata$Column3)
.
}
Where Column1 and Column1 are the names of the dataframe mydata.
How do I parse the arguments to obtain both, name and value of the
argument.
I tried strsplit(Column1, =) which gives me two strings but then I
fail to access the the corresponding columns of the dataframe with what
I get. I also tried eval(parse(text = .)) but it wont work. What am I
doing wrong. Please, somebody must have done something like that before.
I'm thankful for any hint.
Greetings
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[R] The contrary of command %in%
Dear r-list, I've got a data base: HData[1:10,] NumTree Site Species Date Age DBHH IdentTree 11 Queige Spruce 2002 184 49 33.5 Queige 1 22 Queige Fir 2002 NA 5 4.6 Queige 2 33 Queige Fir 2002 25 8 6.6 Queige 3 44 Queige Spruce 2002 198 47 32.5 Queige 4 55 Queige Fir 2002 200 59 35.3 Queige 5 66 Queige Spruce 2002 80 16 9.4 Queige 6 77 Queige Fir 2002 NA 5 4.2 Queige 7 88 Queige Fir 2002 200 44 32.5 Queige 8 99 Queige Fir 2002 NA 5 3.4 Queige 9 10 10 Queige Spruce 2002 167 48 32.8 Queige 10 ... I want to remove particular points determined by HDataPartP-HData[H1.30,] and HDataPartP2-HData[H8DBH20,] That's why I want to use subset in a close form to: HData2-subset(HData,HData$H1.30HData$IdentTree not%in%HDataPartP2$IdentTree) How should I do that ? Is there any R-syntax saying not element of that object ? Thanks for your help. Ghislain. -- Ghislain Vieilledent 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] The contrary of command %in%
a - 1:10 a [1] 1 2 3 4 5 6 7 8 9 10 c(1,2,3,11,12,13) %in% a [1] TRUE TRUE TRUE FALSE FALSE FALSE ! c(1,2,3,11,12,13) %in% a [1] FALSE FALSE FALSE TRUE TRUE TRUE Is this it? Gabor On Thu, Apr 20, 2006 at 02:54:39PM +0200, Ghislain Vieilledent wrote: Dear r-list, I've got a data base: HData[1:10,] NumTree Site Species Date Age DBHH IdentTree 11 Queige Spruce 2002 184 49 33.5 Queige 1 22 Queige Fir 2002 NA 5 4.6 Queige 2 33 Queige Fir 2002 25 8 6.6 Queige 3 44 Queige Spruce 2002 198 47 32.5 Queige 4 55 Queige Fir 2002 200 59 35.3 Queige 5 66 Queige Spruce 2002 80 16 9.4 Queige 6 77 Queige Fir 2002 NA 5 4.2 Queige 7 88 Queige Fir 2002 200 44 32.5 Queige 8 99 Queige Fir 2002 NA 5 3.4 Queige 9 10 10 Queige Spruce 2002 167 48 32.8 Queige 10 ... I want to remove particular points determined by HDataPartP-HData[H1.30,] and HDataPartP2-HData[H8DBH20,] That's why I want to use subset in a close form to: HData2-subset(HData,HData$H1.30HData$IdentTree not%in%HDataPartP2$IdentTree) How should I do that ? Is there any R-syntax saying not element of that object ? Thanks for your help. Ghislain. -- Ghislain Vieilledent 06 24 62 65 07 [[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 -- Csardi Gabor [EMAIL PROTECTED]MTA RMKI, ELTE TTK __ 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
Re: [R] The contrary of command %in%
Just negate the result of %in%; e.g., ! a %in% b Andy From: Ghislain Vieilledent Dear r-list, I've got a data base: HData[1:10,] NumTree Site Species Date Age DBHH IdentTree 11 Queige Spruce 2002 184 49 33.5 Queige 1 22 Queige Fir 2002 NA 5 4.6 Queige 2 33 Queige Fir 2002 25 8 6.6 Queige 3 44 Queige Spruce 2002 198 47 32.5 Queige 4 55 Queige Fir 2002 200 59 35.3 Queige 5 66 Queige Spruce 2002 80 16 9.4 Queige 6 77 Queige Fir 2002 NA 5 4.2 Queige 7 88 Queige Fir 2002 200 44 32.5 Queige 8 99 Queige Fir 2002 NA 5 3.4 Queige 9 10 10 Queige Spruce 2002 167 48 32.8 Queige 10 ... I want to remove particular points determined by HDataPartP-HData[H1.30,] and HDataPartP2-HData[H8DBH20,] That's why I want to use subset in a close form to: HData2-subset(HData,HData$H1.30HData$IdentTree not%in%HDataPartP2$IdentTree) How should I do that ? Is there any R-syntax saying not element of that object ? Thanks for your help. Ghislain. -- Ghislain Vieilledent 06 24 62 65 07 [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] The contrary of command %in%
A flexible way for doing this is to define logical vectors for the types of samples that you want to include or exclude. You can then use logical negation to select the complementary set: inSet1 - HData$H 1.3 inSet2 - HData$H 8 Hdata$DBH 20 HDataPart1 - Hdata[!inSet1, ] HDataPart2 - Hdata[!inSet2, ] -Christos -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Ghislain Vieilledent Sent: Thursday, April 20, 2006 8:55 AM To: r-help@stat.math.ethz.ch Subject: [R] The contrary of command %in% Dear r-list, I've got a data base: HData[1:10,] NumTree Site Species Date Age DBHH IdentTree 11 Queige Spruce 2002 184 49 33.5 Queige 1 22 Queige Fir 2002 NA 5 4.6 Queige 2 33 Queige Fir 2002 25 8 6.6 Queige 3 44 Queige Spruce 2002 198 47 32.5 Queige 4 55 Queige Fir 2002 200 59 35.3 Queige 5 66 Queige Spruce 2002 80 16 9.4 Queige 6 77 Queige Fir 2002 NA 5 4.2 Queige 7 88 Queige Fir 2002 200 44 32.5 Queige 8 99 Queige Fir 2002 NA 5 3.4 Queige 9 10 10 Queige Spruce 2002 167 48 32.8 Queige 10 ... I want to remove particular points determined by HDataPartP-HData[H1.30,] and HDataPartP2-HData[H8DBH20,] That's why I want to use subset in a close form to: HData2-subset(HData,HData$H1.30HData$IdentTree not%in%HDataPartP2$IdentTree) How should I do that ? Is there any R-syntax saying not element of that object ? Thanks for your help. Ghislain. -- Ghislain Vieilledent 06 24 62 65 07 [[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 __ 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
[R] Count Unique Rows/Values
Hi, I have a dataset which has both numeric and character values with dupllicates. For example: 155 A 138 A 138 B 126 C 126 D 123 A 103 A 103 B 143 D 111 C 111 D 156 C How can I count the number of unqiue entries without counting duplicate entries. Also can I extract the list in a object. What I mean is Col1 unique count = 8 Unique Elements are : 103,111,123,126,138,143,155,156 Col2 unique count = 4 Unique Elements are : A,B,C,D. Any pointers would be of great help. TIA Sachin - [[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
Re: [R] Count Unique Rows/Values
This might help: x - read.table(clipboard, colClasses=c(numeric, character)) (x.unique - lapply(x, unique)) $V1 [1] 155 138 126 123 103 143 111 156 $V2 [1] A B C D sapply(x.unique, length) V1 V2 8 4 Andy From: Sachin J Hi, I have a dataset which has both numeric and character values with dupllicates. For example: 155 A 138 A 138 B 126 C 126 D 123 A 103 A 103 B 143 D 111 C 111 D 156 C How can I count the number of unqiue entries without counting duplicate entries. Also can I extract the list in a object. What I mean is Col1 unique count = 8 Unique Elements are : 103,111,123,126,138,143,155,156 Col2 unique count = 4 Unique Elements are : A,B,C,D. Any pointers would be of great help. TIA Sachin - [[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 __ 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
Re: [R] Bootstrap error message: Error in statistic(data, origina l, ...) : unused argument(s) ( ...)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Michael
Sent: Thursday, April 20, 2006 3:50 AM
To: R-help@stat.math.ethz.ch
Subject: [R] Bootstrap error message: Error in
statistic(data, original, ...) : unused argument(s) ( ...) [Broadcast]
Dear colleagues,
I've been swamped and fighting with error for a few hours but
still desperately having absolutely no clue:
You really don't have to do that, but just RTFM instead. ?boot says:
statistic A function which when applied to data returns a vector
containing the statistic(s) of interest. [...]
In all other cases statistic must take at least two
arguments. The first argument passed will always be
the original data. The second will be a vector of
indices, frequencies or weights which define the
bootstrap sample. [...]
Your myFun clearly does not fit that description. boot() tried to call
myFun with a second argument, but myFun doesn't know what to do with a
second argument, and that's where you get the error.
Andy
What's wrong with my bootstraping code?
Thanks a lot!
Error Message:
bootResults=boot(X, myFun, R=1);
Error in statistic(data, original, ...) : unused argument(s) ( ...)
My code is:
X=cbind(column_vector_1, column_vector_2);
myFun=function(X)
{
return(mean(X[, 1])/var(X[, 2]));
}
bootResults=boot(X, myFun, R=1);
-
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[R] (Fwd) Re: Breakdown a number
--- Forwarded message follows ---
From: Petr Pikal [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Subject:Re: [R] Breakdown a number
Date sent: Thu, 20 Apr 2006 17:22:03 +0200
Hi
try this
fff-function(x,breaks=c(5,10)) {
if (sum(breaksx)==0)
result - c(x,rep(0,length(breaks)))
else
result - c(breaks[breaks=x],x-max(breaks[breaks=x]))
result
}
If you want trailing zeroes you need to add them.
HTH
Petr
Paul Roebuck wrote:
Isn't there a builtin method for doing this and, if so, what is it
called?
breakdown - function(whole) {
breaks - c(250, 800)
pieces - integer(length(breaks) + 1)
if (whole breaks[2]) {
pieces[3] - whole - breaks[2]
whole - breaks[2]
}
if (whole breaks[1]) {
pieces[2] - whole - breaks[1]
whole - breaks[1]
}
pieces[1] - whole
return(pieces)
}
breakdown(1200) # 250 550 400
--
SIGSIG -- signature too long (core dumped)
--- End of forwarded message ---Petr Pikal
[EMAIL PROTECTED]
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[R] Mutually Orthogonal Latin Squares
Hi all, The package crossdes could contruct a complete sets of mutually orthogonal latin squares. The construction works for prime powers only. I hope to know whether there is a way to construct a mutually orthogonal Lation square for 10 or other numbers that could not be prime powers. Thanks for any suggestions. Best wishes, Jinsong Zhao __ 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
Re: [R] Count Unique Rows/Values
Hi, This one is not working for me. It is listing all the rows instead of unique ones. My dataset has 30 odd rows and following is the resulting o/p [[308313]] [1] 126 [[308314]] [1] 126 [[308315]] [1] 126 [[308316]] [1] 126 [[308317]] [1] 126 [[308318]] [1] 126 [[308319]] [1] 126 [[308320]] [1] 126 [[308321]] [1] 126 I used following set of commands. (x.unique - lapply(x$V1, unique)) sapply(x.unique, length) x$V1 is numeric field. where x is my data frame already read (therefore i ignored your first step). Am I missing something. ? Thanks Sachin Liaw, Andy [EMAIL PROTECTED] wrote: This might help: x - read.table(clipboard, colClasses=c(numeric, character)) (x.unique - lapply(x, unique)) $V1 [1] 155 138 126 123 103 143 111 156 $V2 [1] A B C D sapply(x.unique, length) V1 V2 8 4 Andy From: Sachin J Hi, I have a dataset which has both numeric and character values with dupllicates. For example: 155 A 138 A 138 B 126 C 126 D 123 A 103 A 103 B 143 D 111 C 111 D 156 C How can I count the number of unqiue entries without counting duplicate entries. Also can I extract the list in a object. What I mean is Col1 unique count = 8 Unique Elements are : 103,111,123,126,138,143,155,156 Col2 unique count = 4 Unique Elements are : A,B,C,D. Any pointers would be of great help. TIA Sachin - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- -- - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Count Unique Rows/Values
From:Sachin J Hi, This one is not working for me. It is listing all the rows instead of unique ones. My dataset has 30 odd rows and following is the resulting o/p [[308313]] [1] 126 [[308314]] [1] 126 [[308315]] [1] 126 [[308316]] [1] 126 [[308317]] [1] 126 [[308318]] [1] 126 [[308319]] [1] 126 [[308320]] [1] 126 [[308321]] [1] 126 I used following set of commands. (x.unique - lapply(x$V1, unique)) You want x instead of x$V1 as the first argument to lapply(), so that it runs unique() on all columns of x. Andy sapply(x.unique, length) x$V1 is numeric field. where x is my data frame already read (therefore i ignored your first step). Am I missing something. ? Thanks Sachin Liaw, Andy [EMAIL PROTECTED] wrote: This might help: x - read.table(clipboard, colClasses=c(numeric, character)) (x.unique - lapply(x, unique)) $V1 [1] 155 138 126 123 103 143 111 156 $V2 [1] A B C D sapply(x.unique, length) V1 V2 8 4 Andy From: Sachin J Hi, I have a dataset which has both numeric and character values with dupllicates. For example: 155 A 138 A 138 B 126 C 126 D 123 A 103 A 103 B 143 D 111 C 111 D 156 C How can I count the number of unqiue entries without counting duplicate entries. Also can I extract the list in a object. What I mean is Col1 unique count = 8 Unique Elements are : 103,111,123,126,138,143,155,156 Col2 unique count = 4 Unique Elements are : A,B,C,D. Any pointers would be of great help. TIA Sachin - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- -- - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Count Unique Rows/Values
Thanks Andy. That works. Sachin Liaw, Andy [EMAIL PROTECTED] wrote: From:Sachin J Hi, This one is not working for me. It is listing all the rows instead of unique ones. My dataset has 30 odd rows and following is the resulting o/p [[308313]] [1] 126 [[308314]] [1] 126 [[308315]] [1] 126 [[308316]] [1] 126 [[308317]] [1] 126 [[308318]] [1] 126 [[308319]] [1] 126 [[308320]] [1] 126 [[308321]] [1] 126 I used following set of commands. (x.unique - lapply(x$V1, unique)) You want x instead of x$V1 as the first argument to lapply(), so that it runs unique() on all columns of x. Andy sapply(x.unique, length) x$V1 is numeric field. where x is my data frame already read (therefore i ignored your first step). Am I missing something. ? Thanks Sachin Liaw, Andy wrote: This might help: x - read.table(clipboard, colClasses=c(numeric, character)) (x.unique - lapply(x, unique)) $V1 [1] 155 138 126 123 103 143 111 156 $V2 [1] A B C D sapply(x.unique, length) V1 V2 8 4 Andy From: Sachin J Hi, I have a dataset which has both numeric and character values with dupllicates. For example: 155 A 138 A 138 B 126 C 126 D 123 A 103 A 103 B 143 D 111 C 111 D 156 C How can I count the number of unqiue entries without counting duplicate entries. Also can I extract the list in a object. What I mean is Col1 unique count = 8 Unique Elements are : 103,111,123,126,138,143,155,156 Col2 unique count = 4 Unique Elements are : A,B,C,D. Any pointers would be of great help. TIA Sachin - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- -- - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html -- -- - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Count Unique Rows/Values
But it is not giving me the list of unique elements. Count works fine. Sachin Liaw, Andy [EMAIL PROTECTED] wrote: From:Sachin J Hi, This one is not working for me. It is listing all the rows instead of unique ones. My dataset has 30 odd rows and following is the resulting o/p [[308313]] [1] 126 [[308314]] [1] 126 [[308315]] [1] 126 [[308316]] [1] 126 [[308317]] [1] 126 [[308318]] [1] 126 [[308319]] [1] 126 [[308320]] [1] 126 [[308321]] [1] 126 I used following set of commands. (x.unique - lapply(x$V1, unique)) You want x instead of x$V1 as the first argument to lapply(), so that it runs unique() on all columns of x. Andy sapply(x.unique, length) x$V1 is numeric field. where x is my data frame already read (therefore i ignored your first step). Am I missing something. ? Thanks Sachin Liaw, Andy wrote: This might help: x - read.table(clipboard, colClasses=c(numeric, character)) (x.unique - lapply(x, unique)) $V1 [1] 155 138 126 123 103 143 111 156 $V2 [1] A B C D sapply(x.unique, length) V1 V2 8 4 Andy From: Sachin J Hi, I have a dataset which has both numeric and character values with dupllicates. For example: 155 A 138 A 138 B 126 C 126 D 123 A 103 A 103 B 143 D 111 C 111 D 156 C How can I count the number of unqiue entries without counting duplicate entries. Also can I extract the list in a object. What I mean is Col1 unique count = 8 Unique Elements are : 103,111,123,126,138,143,155,156 Col2 unique count = 4 Unique Elements are : A,B,C,D. Any pointers would be of great help. TIA Sachin - [[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 -- -- - [[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 -- -- - [[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
Re: [R] Bootstrap error message: Error in statistic(data, origina l, ...) : unused argument(s) ( ...)
btw, if you change myFun to any R internal function, such as cov, or
corr, it can run successfully...
On 4/20/06, Michael [EMAIL PROTECTED] wrote:
Andy,
I've noticed there should be a weight or frequency somewhere... but my
function does not need it.
I have tried to cheat it by declareing a weight, but not using it:
myFun=function(X, w)
{
return(mean(X[, 1])/var(X[, 2]));
}
bootResults=boot(X, myFun, R=1, stype='w');
The result is zero:
Bootstrap Statistics :
original biasstd. error
t1* 2.305412 0 0
On 4/20/06, Liaw, Andy [EMAIL PROTECTED] wrote:
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] ] On Behalf Of Michael
Sent: Thursday, April 20, 2006 3:50 AM
To: R-help@stat.math.ethz.ch
Subject: [R] Bootstrap error message: Error in
statistic(data, original, ...) : unused argument(s) ( ...) [Broadcast]
Dear colleagues,
I've been swamped and fighting with error for a few hours but
still desperately having absolutely no clue:
You really don't have to do that, but just RTFM instead. ?boot says:
statistic A function which when applied to data returns a vector
containing the statistic(s) of interest. [...]
In all other cases statistic must take at least two
arguments. The first argument passed will always be
the original data. The second will be a vector of
indices, frequencies or weights which define the
bootstrap sample. [...]
Your myFun clearly does not fit that description. boot() tried to
call
myFun with a second argument, but myFun doesn't know what to do with
a
second argument, and that's where you get the error.
Andy
What's wrong with my bootstraping code?
Thanks a lot!
Error Message:
bootResults=boot(X, myFun, R=1);
Error in statistic(data, original, ...) : unused argument(s) ( ...)
My code is:
X=cbind(column_vector_1, column_vector_2);
myFun=function(X)
{
return(mean(X[, 1])/var(X[, 2]));
}
bootResults=boot(X, myFun, R=1);
-
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Re: [R] there is no xls reader in R?
Just one other thought. There is a command line program called xlHtml that you can find via google that will convert xls files to csv and its used like this assuming you have placed it somewhere in your path: read.csv(pipe(xlHtml -te -xc:1-10 -csv myfile.xls)) Its handy since its just a single file and in the situation that there is junk on the first few lines of the spreadsheet you could use the skip= argument to read.csv to skip over that. I don't think the RODBC approach can handle that. On 4/20/06, Gabor Grothendieck [EMAIL PROTECTED] wrote: See: http://finzi.psych.upenn.edu/R/doc/manual/R-data.html#Reading-Excel-spreadsheets Also the RDCOMClient package (find it via google) and rcom (on CRAN) can control Excel from R and can be used. On 4/19/06, Michael [EMAIL PROTECTED] wrote: Currently I have to convert all my xls into csv before I can read it in and process the excel data in R... Is there a way to directly read in xls data? 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 __ 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
[R] nlminb( ) : one compartment open PK model
All,
I have been able to successfully use the optim( ) function with
L-BFGS-B to find reasonable parameters for a one-compartment
open pharmacokinetic model. My loss function in this case was
squared error, and I made no assumptions about the distribution
of the plasma values. The model appeared to fit pretty well.
Out of curiosity, I decided to try to use nlminb( ) applied to
a likelihood function that assumes the plasma values are normally
distributed on the semilog scale (ie, modeling log(conc) over
time). nlminb( ) keeps telling me that it has converged, but
the estimated parameters are always identical to the initial
values I am certain that I have committed ein dummheit
somewhere in the following code, but not sure what... Any help
would be greatly appreciated.
Kind regards,
Greg
model2 - function(parms, dose, time, log.conc)
{
exp1 - exp(-parms[1]*time)
exp2 - exp(-parms[2]*time)
right.hand - log(exp1 - exp2)
numerator - dose*parms[1]*parms[2]
denominator - parms[3]*(parms[2] - parms[1])
left.hand - log(numerator/(denominator))
pred - left.hand + right.hand
# defining the distribution of the values
const - 1/(sqrt(2*pi)*parms[4])
exponent - (-1/(2*(parms[4]^2)))*(log.conc - pred)^2
likelihood - const*exp(exponent)
#defining the merit function
-sum(log(likelihood))
}
deriv2
- deriv( expr = ~ -log(1/(sqrt(2*pi)*S)*exp((-1/(2*(S^2)))*
(log.conc-(log(dose*Ke*Ka/(Cl*(Ka-Ke)))
+log(exp(-Ke*time)-exp(-Ka*time^2)),
namevec = c(Ke,Ka,Cl,S),
function.arg = function(Ke, Ka, Cl, S, dose, time, log.conc) NULL )
gradient2.1compart - function(parms, dose, time, log.conc)
{
Ke - parms[1]; Ka - parms[2]; Cl - parms[3]; S - parms[4]
colSums(attr(deriv2.1compart(Ke, Ka, Cl, S, dose, time, log.conc), gradient))
}
attach(foo.frame)
inits - c(Ke = .5,
Ka = .5,
Cl = 1,
S = 1)
#Trying out the code
nlminb(start = inits,
objective = model2,
gradient = gradient2,
control = list(eval.max = 5000, iter.max = 5000),
lower = rep(0,4),
dose = DOSE,
time = TIME,
log.conc = log(RESPONSE))
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Re: [R] Bootstrap error message: Error in statistic(data, origina l, ...) : unused argument(s) ( ...) [Broadcast]
I quoted the relevant part of the documentation for you. Have you actually
try to read what it says?
Sure, you don't get any error, but have you checked whether any
bootstrapping was actually done? Most of those functions are generics, thus
having the ... argument that can take anything. Doesn't mean they will be
used. See if the following helps:
x - 1:10
myMean - function(x, idx) mean(x[idx])
library(boot)
x.boot - boot(x, mean, R=10)
str(x.boot)
List of 11
$ t0 : num 5.5
$ t: num [1:10, 1] 5.5 5.5 5.5 5.5 5.5 5.5 5.5 5.5 5.5 5.5
$ R: num 10
$ data : int [1:10] 1 2 3 4 5 6 7 8 9 10
$ seed : int [1:626] 403 101 106329237 1120199471 -647917002 -657447328
355480739 291889089 -398656592 -2023350578 ...
$ statistic:function (x, ...)
$ sim : chr ordinary
$ call : language boot(data = x, statistic = mean, R = 10)
$ stype: chr i
$ strata : num [1:10] 1 1 1 1 1 1 1 1 1 1
$ weights : num [1:10] 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
- attr(*, class)= chr boot
x.boot2 - boot(x, myMean, R=10)
str(x.boot2)
List of 11
$ t0 : num 5.5
$ t: num [1:10, 1] 5.7 6.4 5.4 4.9 7.2 4.6 4.4 5.4 5.7 5.9
$ R: num 10
$ data : int [1:10] 1 2 3 4 5 6 7 8 9 10
$ seed : int [1:626] 403 201 106329237 1120199471 -647917002 -657447328
355480739 291889089 -398656592 -2023350578 ...
$ statistic:function (x, idx)
..- attr(*, source)= chr function(x, idx) mean(x[idx])
$ sim : chr ordinary
$ call : language boot(data = x, statistic = myMean, R = 10)
$ stype: chr i
$ strata : num [1:10] 1 1 1 1 1 1 1 1 1 1
$ weights : num [1:10] 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1 0.1
- attr(*, class)= chr boot
Andy
-Original Message-
From: Michael [mailto:[EMAIL PROTECTED]
Sent: Thursday, April 20, 2006 12:19 PM
To: Liaw, Andy
Cc: R-help@stat.math.ethz.ch
Subject: Re: [R] Bootstrap error message: Error in statistic(data, origina
l, ...) : unused argument(s) ( ...) [Broadcast]
btw, if you change myFun to any R internal function, such as cov, or
corr, it can run successfully...
On 4/20/06, Michael [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
wrote:
Andy,
I've noticed there should be a weight or frequency somewhere... but my
function does not need it.
I have tried to cheat it by declareing a weight, but not using it:
myFun=function(X, w)
{
return(mean(X[, 1])/var(X[, 2]));
}
bootResults=boot(X, myFun, R=1, stype='w');
The result is zero:
Bootstrap Statistics :
original biasstd. error
t1* 2.305412 0 0
On 4/20/06, Liaw, Andy [EMAIL PROTECTED] mailto:[EMAIL PROTECTED]
wrote:
-Original Message-
From: [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED]
[mailto: [EMAIL PROTECTED]
mailto:[EMAIL PROTECTED] ] On Behalf Of Michael
Sent: Thursday, April 20, 2006 3:50 AM
To: R-help@stat.math.ethz.ch mailto:R-help@stat.math.ethz.ch
Subject: [R] Bootstrap error message: Error in
statistic(data, original, ...) : unused argument(s) ( ...) [Broadcast]
Dear colleagues,
I've been swamped and fighting with error for a few hours but
still desperately having absolutely no clue:
You really don't have to do that, but just RTFM instead. ?boot says:
statistic A function which when applied to data returns a vector
containing the statistic(s) of interest. [...]
In all other cases statistic must take at least two
arguments. The first argument passed will always be
the original data. The second will be a vector of
indices, frequencies or weights which define the
bootstrap sample. [...]
Your myFun clearly does not fit that description. boot() tried to call
myFun with a second argument, but myFun doesn't know what to do with a
second argument, and that's where you get the error.
Andy
What's wrong with my bootstraping code?
Thanks a lot!
Error Message:
bootResults=boot(X, myFun, R=1);
Error in statistic(data, original, ...) : unused argument(s) ( ...)
My code is:
X=cbind(column_vector_1, column_vector_2);
myFun=function(X)
{
return(mean(X[, 1])/var(X[, 2]));
}
bootResults=boot(X, myFun, R=1);
-
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[R] smooth the ecdf plots
Hi All, I have codes as follows to get the ecdf plots: day.hos2-c(6,4,6,6,4,6,5,4,7,5,6,6,8,6,17,9,8,4,6,3,5,8,7,12,5,10,6,4,6 ,13,7,6,6,25,4,9,96,6,6,6,6,9,4,5,5,4,10,5,7,6) day.hos3-c(5,6,7,6,4,5,6,6,6,6,19,7,5,9,8,8,7,5,6,20,40,5,8,7,7,5,6,13, 11,9,4,6,9,16,6,7,6) f-ecdf(day.hos2) plot(f,col.p='red',col.h='red') g-ecdf(day.hos3) lines(g,lty=2) But in order to compare the two ecdf plots. I want to smooth the ecdf plot, make it like a continuous distribution curve. Could you please help me with it? I try to find some arguments in plot but not successful. Thank you! [[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
Re: [R] smooth the ecdf plots
Better idea: Compare directly. ?qqplot -- Bert Gunter Genentech Non-Clinical Statistics South San Francisco, CA The business of the statistician is to catalyze the scientific learning process. - George E. P. Box -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Linda Lei Sent: Thursday, April 20, 2006 10:08 AM To: r-help@stat.math.ethz.ch Subject: [R] smooth the ecdf plots Hi All, I have codes as follows to get the ecdf plots: day.hos2-c(6,4,6,6,4,6,5,4,7,5,6,6,8,6,17,9,8,4,6,3,5,8,7,12, 5,10,6,4,6 ,13,7,6,6,25,4,9,96,6,6,6,6,9,4,5,5,4,10,5,7,6) day.hos3-c(5,6,7,6,4,5,6,6,6,6,19,7,5,9,8,8,7,5,6,20,40,5,8,7 ,7,5,6,13, 11,9,4,6,9,16,6,7,6) f-ecdf(day.hos2) plot(f,col.p='red',col.h='red') g-ecdf(day.hos3) lines(g,lty=2) But in order to compare the two ecdf plots. I want to smooth the ecdf plot, make it like a continuous distribution curve. Could you please help me with it? I try to find some arguments in plot but not successful. Thank you! [[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 __ 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
Re: [R] there is no xls reader in R?
On Thu, Apr 20, 2006 at 12:29:58PM -0400, Gabor Grothendieck wrote: Just one other thought. There is a command line program called xlHtml that you can find via google that will convert xls files to csv and its used like this assuming you have placed it somewhere in your path: read.csv(pipe(xlHtml -te -xc:1-10 -csv myfile.xls)) Its handy since its just a single file and in the situation that there is junk on the first few lines of the spreadsheet you could use the skip= argument to read.csv to skip over that. I don't think the RODBC approach can handle that. An equivalent approach has long been available in Greg Warnes gregmisc bundle via the gdata package. It uses Perl's SpreadSheet::ParseExcel -- which it even embeds in the R package -- to parse xls, and creates a csv which is then read. Cross-platform, which RODBC's approach isn't, and available directly via install.packages(). 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
[R] Missing p-values using lmer()
Hello, Im trying to perform a REML analysis using the lmer() function (lme4 package). Well, it seems to work well, except that Im not getting any p-value (see example below). Can someone tell me what I did wrong? Thanks for your help, Amélie library(gdata) dive - read.xls(C:/Documents and Settings/Amelie/My Documents/Postdoc/CE 2005-2006/divebydive.xls, perl=C:/perl/bin/perl.exe) library(lme4) Loading required package: Matrix Loading required package: lattice reml.res - lmer(UNDS~SUCCESSMN+(1|BIRD), dive) summary(reml.res) Linear mixed-effects model fit by REML Formula: UNDS ~ SUCCESSMN + (1 | BIRD) Data: dive AIC BIClogLik MLdeviance REMLdeviance 60032.37 60053.8 -30013.1960031.9 60026.37 Random effects: Groups NameVariance Std.Dev. BIRD (Intercept) 4.4504 2.1096 Residual 36.4240 6.0352 number of obs: 9324, groups: BIRD, 12 Fixed effects: Estimate Std. Error t value (Intercept) 13.397640.63887 20.9707 SUCCESSMN4.221974.11527 1.0259 Correlation of Fixed Effects: (Intr) SUCCESSMN -0.276 anova(reml.res) Analysis of Variance Table Df Sum Sq Mean Sq SUCCESSMN 1 38.337 38.337 [[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
Re: [R] [Q] Bayeisan Network with the deal package
Thanks for providing such a simple, complete example. I've never used deal before, but a few minutes with your example led me to something that might help you: First, the documentation for localprior says, x: an object of class 'node' or 'network'. That information led me to checking 'class(ksl)' and 'class(ksl$node[1])': The first is class 'network', while the second is of class 'list'. However, class(ksl$node[[1]]) is of class 'node'. Bingo: localprior(ksl.nw$nodes[[1]]) [[1]] [[1]]$alpha [1] NA [[1]]$nu [1] 64 [[1]]$rho [1] 64 [[1]]$mu [1] 179.289 [[1]]$phi [,1] [1,] 113442.3 [[1]]$tau [1] 64 hope this helps. spencer graves Young-Jin Lee wrote: Dear R-users I am looking for a help in using the deal package. I followed the manual and paper from the author's web site to learn it, as shown below, but I could not figure out how to access the local and posterior probability of the nodes in the constructed network. library(deal) data(ksl) ksl.nw - network(ksl) ksl.prior - jointprior(ksl.nw) banlist - matrix(c(5,5,6,6,7,7,9, 8,9,8,9,8,9,8), ncol = 2) ksl.nw$banlist - banlist ksl.nw - learn(ksl.nw, ksl, ksl.prior)$nw According to the manual, the local prior and posterior can be accessed for each node using localprior() and localposterior(). However, I got null from localprior(ksl.nw$nodes[1]) and localposterior(ksl.nw$nodes[1]). Can anyone tell me what I should do to access the local prior and posterior? Any help would be appreciated. Thanks in advance. Young-Jin [[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 __ 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
Re: [R] Missing p-values using lmer()
You didn't do anything wrong, lmer doesn't give them. And, for good reason. I've been a bit indoctrinated by D. Bates, so let me share what I've learned. With simple analysis of variance models with simple error structures, it is known that the ratio of the variances follow and F distribution. However, with more complex error structures, the null distribution is unknown. Most other multilevel programs accept by analogy that the ratio of the variances do follow an F distribution. That is, it works well for the simple case, therefore it probably is also true for the more complex case. In SAS, one can choose ddf options, such as Kenward-Roger, which hopes that after assuming the ratio of variances follow an F distribution, the only remaining challenge is to properly estimate the denominator degrees of freedom. These kinds of options do not currently exist in lmer and after many discussions on this list Doug Bates decided to remove the p-values for now. This topic has been discussed often on this list and you can see other discussions on the archive which may be more insightful. Harold -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Amelie LESCROEL Sent: Thursday, April 20, 2006 1:41 PM To: r-help@stat.math.ethz.ch Subject: [R] Missing p-values using lmer() Hello, I'm trying to perform a REML analysis using the lmer() function (lme4 package). Well, it seems to work well, except that I'm not getting any p-value (see example below). Can someone tell me what I did wrong? Thanks for your help, Amélie library(gdata) dive - read.xls(C:/Documents and Settings/Amelie/My Documents/Postdoc/CE 2005-2006/divebydive.xls, perl=C:/perl/bin/perl.exe) library(lme4) Loading required package: Matrix Loading required package: lattice reml.res - lmer(UNDS~SUCCESSMN+(1|BIRD), dive) summary(reml.res) Linear mixed-effects model fit by REML Formula: UNDS ~ SUCCESSMN + (1 | BIRD) Data: dive AIC BIClogLik MLdeviance REMLdeviance 60032.37 60053.8 -30013.1960031.9 60026.37 Random effects: Groups NameVariance Std.Dev. BIRD (Intercept) 4.4504 2.1096 Residual 36.4240 6.0352 number of obs: 9324, groups: BIRD, 12 Fixed effects: Estimate Std. Error t value (Intercept) 13.397640.63887 20.9707 SUCCESSMN4.221974.11527 1.0259 Correlation of Fixed Effects: (Intr) SUCCESSMN -0.276 anova(reml.res) Analysis of Variance Table Df Sum Sq Mean Sq SUCCESSMN 1 38.337 38.337 [[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
Re: [R] 3D pie
COMTE == COMTE Guillaume [EMAIL PROTECTED] writes: I know it isn't the best way to represent data, but people are sometimes more interested by the look and feel than by the accuracy of the results... If they aren't intersted in the results, why not just print some random 3D pie chart and use that? Why do you need to produce a new one at all? __ 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
Re: [R] Missing p-values using lmer()
Hi, Harold: Am I correct that the tool currently preferred for estimating p-values for lmer is mcmcsamp? Amélie: My favorite tool for exploring the archives is 'RSiteSearch'. You can also get to it via www.r-project.org, but the last time I tried to copy a web address to paste in an email, the address I got from 'RSiteSearch' worked but the one I got from the web page directly did not. hope this helps, spencer graves Doran, Harold wrote: You didn't do anything wrong, lmer doesn't give them. And, for good reason. I've been a bit indoctrinated by D. Bates, so let me share what I've learned. With simple analysis of variance models with simple error structures, it is known that the ratio of the variances follow and F distribution. However, with more complex error structures, the null distribution is unknown. Most other multilevel programs accept by analogy that the ratio of the variances do follow an F distribution. That is, it works well for the simple case, therefore it probably is also true for the more complex case. In SAS, one can choose ddf options, such as Kenward- Roger, which hopes that after assuming the ratio of variances follow an F distribution, the only remaining challenge is to properly estimate the denominator degrees of freedom. These kinds of options do not currently exist in lmer and after many discussions on this list Doug Bates decided to remove the p-values for now. This topic has been discussed often on this list and you can see other discussions on the archive which may be more insightful. Harold -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Amelie LESCROEL Sent: Thursday, April 20, 2006 1:41 PM To: r-help@stat.math.ethz.ch Subject: [R] Missing p-values using lmer() Hello, I'm trying to perform a REML analysis using the lmer() function (lme4 package). Well, it seems to work well, except that I'm not getting any p-value (see example below). Can someone tell me what I did wrong? Thanks for your help, Amélie library(gdata) dive - read.xls(C:/Documents and Settings/Amelie/My Documents/Postdoc/CE 2005-2006/divebydive.xls, perl=C:/perl/bin/perl.exe) library(lme4) Loading required package: Matrix Loading required package: lattice reml.res - lmer(UNDS~SUCCESSMN+(1|BIRD), dive) summary(reml.res) Linear mixed-effects model fit by REML Formula: UNDS ~ SUCCESSMN + (1 | BIRD) Data: dive AIC BIClogLik MLdeviance REMLdeviance 60032.37 60053.8 -30013.1960031.9 60026.37 Random effects: Groups NameVariance Std.Dev. BIRD (Intercept) 4.4504 2.1096 Residual 36.4240 6.0352 number of obs: 9324, groups: BIRD, 12 Fixed effects: Estimate Std. Error t value (Intercept) 13.397640.63887 20.9707 SUCCESSMN4.221974.11527 1.0259 Correlation of Fixed Effects: (Intr) SUCCESSMN -0.276 anova(reml.res) Analysis of Variance Table Df Sum Sq Mean Sq SUCCESSMN 1 38.337 38.337 [[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 __ 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
Re: [R] smooth the ecdf plots
Linda Lei wrote: Hi All, I have codes as follows to get the ecdf plots: day.hos2-c(6,4,6,6,4,6,5,4,7,5,6,6,8,6,17,9,8,4,6,3,5,8,7,12,5,10,6,4,6 ,13,7,6,6,25,4,9,96,6,6,6,6,9,4,5,5,4,10,5,7,6) day.hos3-c(5,6,7,6,4,5,6,6,6,6,19,7,5,9,8,8,7,5,6,20,40,5,8,7,7,5,6,13, 11,9,4,6,9,16,6,7,6) f-ecdf(day.hos2) plot(f,col.p='red',col.h='red') g-ecdf(day.hos3) lines(g,lty=2) But in order to compare the two ecdf plots. I want to smooth the ecdf plot, make it like a continuous distribution curve. Could you please help me with it? I try to find some arguments in plot but not successful. Thank you! Think about keeping them ECDFs and not distorting them. And you can superimpose two of them using thin blank lines for one and thicker gray scale lines for the other and be able to see them clearly. Use e.g. library(Hmisc) ?ecdf (look at groups argument) Someday I may rename ecdf in Hmisc to avoid confusion with the builtin ecdf function. Frank [[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 -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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
Re: [R] Missing p-values using lmer()
Yes, you're exactly right. You can use the mcmcsamp() function to sample from the posterior of an lmer object. This returns an object of mcmc class and you can do all of your diagnostics using the coda package. -Original Message- From: Spencer Graves [mailto:[EMAIL PROTECTED] Sent: Thursday, April 20, 2006 2:23 PM To: Doran, Harold Cc: Amelie LESCROEL; r-help@stat.math.ethz.ch Subject: Re: [R] Missing p-values using lmer() Hi, Harold: Am I correct that the tool currently preferred for estimating p-values for lmer is mcmcsamp? Amélie: My favorite tool for exploring the archives is 'RSiteSearch'. You can also get to it via www.r-project.org, but the last time I tried to copy a web address to paste in an email, the address I got from 'RSiteSearch' worked but the one I got from the web page directly did not. hope this helps, spencer graves Doran, Harold wrote: You didn't do anything wrong, lmer doesn't give them. And, for good reason. I've been a bit indoctrinated by D. Bates, so let me share what I've learned. With simple analysis of variance models with simple error structures, it is known that the ratio of the variances follow and F distribution. However, with more complex error structures, the null distribution is unknown. Most other multilevel programs accept by analogy that the ratio of the variances do follow an F distribution. That is, it works well for the simple case, therefore it probably is also true for the more complex case. In SAS, one can choose ddf options, such as Kenward- Roger, which hopes that after assuming the ratio of variances follow an F distribution, the only remaining challenge is to properly estimate the denominator degrees of freedom. These kinds of options do not currently exist in lmer and after many discussions on this list Doug Bates decided to remove the p-values for now. This topic has been discussed often on this list and you can see other discussions on the archive which may be more insightful. Harold -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Amelie LESCROEL Sent: Thursday, April 20, 2006 1:41 PM To: r-help@stat.math.ethz.ch Subject: [R] Missing p-values using lmer() Hello, I'm trying to perform a REML analysis using the lmer() function (lme4 package). Well, it seems to work well, except that I'm not getting any p-value (see example below). Can someone tell me what I did wrong? Thanks for your help, Amélie library(gdata) dive - read.xls(C:/Documents and Settings/Amelie/My Documents/Postdoc/CE 2005-2006/divebydive.xls, perl=C:/perl/bin/perl.exe) library(lme4) Loading required package: Matrix Loading required package: lattice reml.res - lmer(UNDS~SUCCESSMN+(1|BIRD), dive) summary(reml.res) Linear mixed-effects model fit by REML Formula: UNDS ~ SUCCESSMN + (1 | BIRD) Data: dive AIC BIClogLik MLdeviance REMLdeviance 60032.37 60053.8 -30013.1960031.9 60026.37 Random effects: Groups NameVariance Std.Dev. BIRD (Intercept) 4.4504 2.1096 Residual 36.4240 6.0352 number of obs: 9324, groups: BIRD, 12 Fixed effects: Estimate Std. Error t value (Intercept) 13.397640.63887 20.9707 SUCCESSMN4.221974.11527 1.0259 Correlation of Fixed Effects: (Intr) SUCCESSMN -0.276 anova(reml.res) Analysis of Variance Table Df Sum Sq Mean Sq SUCCESSMN 1 38.337 38.337 [[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 __ 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
[R] Conditional Row Sum
Hi, How can I accomplish this in R. Example: R1 R2 3 101 4 102 3 102 18102 11101 I want to find Sum(101) = 14 - i.e SUM(R1) where R2 = 101 Sum(102) = 25- SUM(R2) where R2 = 102 TIA Sachin - [[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
Re: [R] Conditional Row Sum
Using the built in data.frame iris sum each of the first 4 columns for each value of the 5th column. rowsum(iris[,-5], iris[,5]) On 4/20/06, Sachin J [EMAIL PROTECTED] wrote: Hi, How can I accomplish this in R. Example: R1 R2 3 101 4 102 3 102 18102 11101 I want to find Sum(101) = 14 - i.e SUM(R1) where R2 = 101 Sum(102) = 25- SUM(R2) where R2 = 102 TIA Sachin - [[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 __ 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
Re: [R] Conditional Row Sum
On Thu, 2006-04-20 at 11:46 -0700, Sachin J wrote: Hi, How can I accomplish this in R. Example: R1 R2 3 101 4 102 3 102 18102 11101 I want to find Sum(101) = 14 - i.e SUM(R1) where R2 = 101 Sum(102) = 25- SUM(R2) where R2 = 102 TIA Sachin Presuming that your data is in a data frame called DF: DF R1 R2 1 3 101 2 4 102 3 3 102 4 18 102 5 11 101 At least three options: with(DF, tapply(R1, R2, sum)) 101 102 14 25 aggregate(DF$R1, list(R2 = DF$R2), sum) R2 x 1 101 14 2 102 25 by(DF$R1, DF$R2, sum) INDICES: 101 [1] 14 -- INDICES: 102 [1] 25 See ?by, ?aggregate and ?tapply and ?with. HTH, Marc Schwartz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Conditional Row Sum
Thanx Marc and Gabor for your help. Sachin Marc Schwartz (via MN) [EMAIL PROTECTED] wrote: On Thu, 2006-04-20 at 11:46 -0700, Sachin J wrote: Hi, How can I accomplish this in R. Example: R1 R2 3 101 4 102 3 102 18 102 11 101 I want to find Sum(101) = 14 - i.e SUM(R1) where R2 = 101 Sum(102) = 25 - SUM(R2) where R2 = 102 TIA Sachin Presuming that your data is in a data frame called DF: DF R1 R2 1 3 101 2 4 102 3 3 102 4 18 102 5 11 101 At least three options: with(DF, tapply(R1, R2, sum)) 101 102 14 25 aggregate(DF$R1, list(R2 = DF$R2), sum) R2 x 1 101 14 2 102 25 by(DF$R1, DF$R2, sum) INDICES: 101 [1] 14 -- INDICES: 102 [1] 25 See ?by, ?aggregate and ?tapply and ?with. HTH, Marc Schwartz - [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Re: [R] Randomly selecting one row for each factor level [Broadca st]
This worked for my example data frame but not when I used it on my
actual data frame because some of the levels of x have only one element.
However, when I replace 'sample' with 'some' (from the car package),
it works like a charm!
Many thanks to Andy and Gabor for their help!
Kelly
Liaw, Andy wrote:
The following should work:
dfr.samp - dfr[tapply(1:nrow(dfr), dfr$x, sample, 1),]
dfr.samp
x y z
10 a 10 J
2 b 2 B
9 c 9 I
Andy
From: Kelly Hildner
I don't use R much, and I have been unable to figure out how
to get the
subset of my data frame that I would like.
For example, if this were my data frame:
dfr - data.frame(x=rep(letters[1:3], 4), y=(1:12),
z=(LETTERS[1:12])) dfr
x y z
1 a 1 A
2 b 2 B
3 c 3 C
4 a 4 D
5 b 5 E
6 c 6 F
7 a 7 G
8 b 8 H
9 c 9 I
10 a 10 J
11 b 11 K
12 c 12 L
I would like to randomly select one row for each level of the
factor x
and create a new data frame with the results. For example,
the result
might be:
x y z
1 a 1 A
5 b 5 E
6 c 6 F
Any help would be greatly appreciated!
Thanks,
Kelly
--
K. Kelly Hildner, Ph.D.
NOAA Fisheries
Southwest Fisheries Science Center
110 Shaffer Rd.
Santa Cruz, CA 95060
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[R] PCA biplot question
Hi everyone, I'd like to project two pcas onto one device window. I plot my first PCA: biplot(prcomp(t(cerebdevmat)), var.axes=FALSE, cex=c(1,.1), pc.biplot=TRUE) Now I'd like to project the features of another PCA onto this graph. Any suggestions? I know this is easily done in MatLab but haven't figured it out in R. Thanks, Tanya [[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
Re: [R] Function to approximate complex integral
On 4/19/06, Doran, Harold [EMAIL PROTECTED] wrote:
I am writing a small function to approximate an integral that cannot be
evaluated in closed form. I am partially successful at this point and am
experiencing one small, albeit important problem. Here is part of my
function below.
This is a psychometric problem for dichotomously scored test items where
x is a vector of 1s or 0s denoting whether the respondent answered the
item correctly (1) or otherwise (0), b is a vector of item difficulties,
and theta is an ability estimate for the individual.
rasch - function(b,theta){
1 / ( 1 + exp(b - theta))
}
The function rasch gives the probability of a correct response to item i
conditional on theta, the individuals ability estimate
myfun - function(x, b, theta){
sum(rasch(b, theta)^x * ( 1 - rasch(b,theta) )^(1-x) * dnorm(theta))
}
This is the likelihood function assuming the data are Bernoulli
distributed multiplied by a population distribution.
Now, when x,b, and theta are of equal length the function works fine as
below
x - c(1,1,0)
b - c(-2,-1,0)
t - c(-2,-1.5,-1)
myfun(x,b,t)
[1] 0.2527884
However, I want theta to be a vector of discrete values that will be
larger than both x and b, something like
t - seq(-5, 0, by = .01)
However, this gives me the error
myfun(x,b,t)
Warning messages:
1: longer object length
is not a multiple of shorter object length in: b - theta
So, for the problem above, I want item 1 to be evaluated at theta 1
through theta q and then item 2 is evaluated at theta 1 and through
theta q and so forth for each item.
Can anyone recommend a way for me to modify my function above to
accomplish this?
I believe (partially based on our off-list discussions of this model)
that you want to use the function outer, as in
rasch - function(b, theta) 1/(1 - exp(outer(theta, b, -)))
If length(theta) == m and length(b) == n then this returns the m by n
matrix of probabilities of subject i correctly answering question j.
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