[R] closed testing procedure
Hi, are the methods for multiple testing p value adjustment (Shaffer, Westfall, free) implemented in the function adjusted() in multcomp package so called closed testing procedure? what about those methods (holm, hochberg, hommel, BH, BY) implemented in the p.adjust() in the stats package? Thanks John __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Multi-output regression
Hi,
I have a question regarding the modeling methodology of the following problem:
* I have two data sets {X_i,y_i} {X_i,z_i}, i=1..N,
where y_i = f(X_i) + i.i.d. Gaussian noise
and z_i = g(X_i) + i.i.d. Gaussian noise
* I apply bayesian linear regression to each of them and obtain
p(y|X) and p(z|X)
I would like to improve the prediction of the two models using the knowledge
that
f and g are related (for example f(X_i) = g(X_i) - 1). I can obtain a model
p(y,z).
I know that there are methods for multi-output regression, but I hope both can
be modeled independently and then combined.
Thank you for your help!
Steffan
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Re: [R] Multi-output regression
I am not a Bayesian. In the non-Bayesian case you would use SUR to model both
equations simultaneously. If both use the exact same matrix of data, X
(i.e., the value are numerically absolutely identical), then SUR will
collapse to OLS. In that sense you get a combined estimate using SUR that
respects the correlation of the error terms between equations. However, it
will be efficient to estimate each equation individually if the Xs are
exactly the same.
Daniel
zeec wrote:
Hi,
I have a question regarding the modeling methodology of the following
problem:
* I have two data sets {X_i,y_i} {X_i,z_i}, i=1..N,
where y_i = f(X_i) + i.i.d. Gaussian noise
and z_i = g(X_i) + i.i.d. Gaussian noise
* I apply bayesian linear regression to each of them and obtain
p(y|X) and p(z|X)
I would like to improve the prediction of the two models using the
knowledge that
f and g are related (for example f(X_i) = g(X_i) - 1). I can obtain a
model p(y,z).
I know that there are methods for multi-output regression, but I hope both
can be modeled independently and then combined.
Thank you for your help!
Steffan
--
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[R] how to remove NA/NaN/Inf in a matrix??
Hi all.. I'm very new R, and i'm analyzing microarray data using Bioconductor.. Recently i was given microarray data to analyze. The problem is whenever i run MAS5 presence calls algorithm, it throws an error saying NA/NaN/Inf in foreign function. How do i remove such NA/NaN/Inf's ?? I tried na.omit(dataframe) but stil problem exists. dimension of matrix (data) is 35556 7. data.mas5calls=mas5calls(data) Getting probe level data... Computing p-values Error in FUN(1:6[[1L]], ...) : NA/NaN/Inf in foreign function call (arg 2) Thank you. -- ** Anand M.T School of Biotechnology (Bio-Informatics), International Instituteof Information Technology (I2IT), P-14, Rajiv Gandhi Infotech park, Hinjewadi, Pune-411 057. INDIA. The secret of success comprised in three words.. Work, Finish Publish - Michael Faraday [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] function censReg in panel data setting
Dear Igors On 9 September 2011 22:00, Igors igors.lahanc...@gmail.com wrote: Thank you for your answer. However it doesn't solve my problem fully. The problem is that I have much bigger data set than I sent to you (it was only a small part : 874 obs.). My full data set is 546718 obs. If I try to use censReg on full data set, then it still gives me the same already mentioned error about Na's in the initial gradient. I have sent you an e-mail with full dataset and the code. Thanks! I would really appreciate if you could check how it works and suggest me how to solve this problem. Now, I get the same error message. The problem is that the log-likelihood function becomes minus infinity. I guess that this problem occurs, because a likelihood contribution of (at least) one individual observation is so small that it is rounded zero and hence, its logarithm is minus infinity. However, I have to take a deeper look into this issue before I can give a definite answer and hopefully find a solution. I have noticed that you use iterlim argumet to specify maximum number of iterations. How this argument could affect possibility of obtaining estimates? Using your small data set, the maximization of the (log) likelihood function in censReg() did not converge within 100 iterations. Therefore, I increased the maximum number of iterations from 100 to 200 -- and then the maximization converged :-( /Arne -- Arne Henningsen http://www.arne-henningsen.name __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to remove NA/NaN/Inf in a matrix??
On Sep 11, 2011, at 5:38 AM, anand m t wrote: Hi all.. I'm very new R, and i'm analyzing microarray data using Bioconductor.. Recently i was given microarray data to analyze. The problem is whenever i run MAS5 presence calls algorithm, it throws an error saying NA/NaN/Inf in foreign function. How do i remove such NA/NaN/Inf's ?? I tried na.omit(dataframe) but stil problem exists. dimension of matrix (data) is 35556 7. data.mas5calls=mas5calls(data) Getting probe level data... Computing p-values Error in FUN(1:6[[1L]], ...) : NA/NaN/Inf in foreign function call (arg 2) Besides the possibility that your data has NAs there is the possibilty that the authors of whatever package has that function did not anticipate some negative values or some other condition that your data led to. Maybe the foreign function, i.e. not 'mas5calls' tried to take the log of 0 or some such. It's not possible to saymuch more given the vagueness of your question. This is also not the Bioconductor mailing list. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] On-line machine learning packages?
What R packages are available for performing classification tasks? That is, when the predictor has done its job on the dataset (based on the training set and a range of variables), feedback about the true label will be available and this information should be integrated for the next classification round. //Jay __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How download to spreadsheet?
Hello all: As everyone knows the stock quotes in Yahoo can be downloaded directly on EXCEL spreadsheet. we can also use the simple sentence of package tseries of R to download Yahoo quote: x - get.hist.quote(instrument = ibm, start = 2010-01-01, quote = close) x The result will display on the R console. BUT how can we also download them onto spreadsheet? Thank! -- View this message in context: http://r.789695.n4.nabble.com/How-download-to-spreadsheet-tp3805182p3805182.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] On-line machine learning packages?
On Sep 11, 2011, at 11:42 AM, Jay wrote: What R packages are available for performing classification tasks? That is, when the predictor has done its job on the dataset (based on the training set and a range of variables), feedback about the true label will be available and this information should be integrated for the next classification round. You should look at CRAN Task Views. Extremely easy to find from the main R-project page. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How download to spreadsheet?
require(quantmod)
getSymbols('IBM', from='2010-01-01')
write.csv(Cl(IBM), file='ibm-2010-present.csv')
I can open the resulting file in Excel without problems. Depending on
your preference, you may want to use write.csv2 instead.
On 11 September 2011 06:59, Yumin zpx...@gmail.com wrote:
The result will display on the R console. BUT how can we also download
them onto
spreadsheet?
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Envoyait de mon telephone mobil
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Re: [R] On-line machine learning packages?
Hi, I used the rseek search engine to look for suitable solutions, however as I was unable to find anything useful, I'm asking for help. Anybody have experience with these kinds of problems? I looked into dynaTree, but as information is a bit scares and as I understand it, it might not be what I'm looking for..(?) BR, Jay On Sep 11, 7:15 pm, David Winsemius dwinsem...@comcast.net wrote: On Sep 11, 2011, at 11:42 AM, Jay wrote: What R packages are available for performing classification tasks? That is, when the predictor has done its job on the dataset (based on the training set and a range of variables), feedback about the true label will be available and this information should be integrated for the next classification round. You should look at CRAN Task Views. Extremely easy to find from the main R-project page. -- David Winsemius, MD West Hartford, CT __ r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] On-line machine learning packages?
Try this: http://cran.r-project.org/web/views/MachineLearning.html On 09/11/2011 12:43 PM, Jay wrote: Hi, I used the rseek search engine to look for suitable solutions, however as I was unable to find anything useful, I'm asking for help. Anybody have experience with these kinds of problems? I looked into dynaTree, but as information is a bit scares and as I understand it, it might not be what I'm looking for..(?) BR, Jay On Sep 11, 7:15 pm, David Winsemiusdwinsem...@comcast.net wrote: On Sep 11, 2011, at 11:42 AM, Jay wrote: What R packages are available for performing classification tasks? That is, when the predictor has done its job on the dataset (based on the training set and a range of variables), feedback about the true label will be available and this information should be integrated for the next classification round. You should look at CRAN Task Views. Extremely easy to find from the main R-project page. -- David Winsemius, MD West Hartford, CT __ r-h...@r-project.org mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to remove NA/NaN/Inf in a matrix??
Try: library(Hmisc) ?na.delete Ken Hutchison On Sep 11, 2554 BE, at 5:38 AM, anand m t anandro...@gmail.com wrote: Hi all.. I'm very new R, and i'm analyzing microarray data using Bioconductor.. Recently i was given microarray data to analyze. The problem is whenever i run MAS5 presence calls algorithm, it throws an error saying NA/NaN/Inf in foreign function. How do i remove such NA/NaN/Inf's ?? I tried na.omit(dataframe) but stil problem exists. dimension of matrix (data) is 35556 7. data.mas5calls=mas5calls(data) Getting probe level data... Computing p-values Error in FUN(1:6[[1L]], ...) : NA/NaN/Inf in foreign function call (arg 2) Thank you. -- ** Anand M.T School of Biotechnology (Bio-Informatics), International Instituteof Information Technology (I2IT), P-14, Rajiv Gandhi Infotech park, Hinjewadi, Pune-411 057. INDIA. The secret of success comprised in three words.. Work, Finish Publish - Michael Faraday [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Emacs ESS finding all versions of R
Dear R-help, Apologies if this isn't exactly the right place for this question. I am trying to run R in emacs using ESS. I have done this successfully many times - it works right now on my home computer in windows 7 and in ubuntu. The problem is that when I start emacs the regular splash screen doesn't show. In the mini-buffer I get the message Finding all versions of R on your system. And then emacs just freezes up. I am running windows xp. emacs is located at C:\Program Files\emacs-23.3\bin\emacs I have an ess-5.14 folder at C:\Program Files\emacs-23.3\site-lisp\ess-5.14 R is located at C:\Program Files\R\R-2.13.1\bin\R or C:\Program Files\R\R-2.13.1\bin\i386\Rterm ? In my ~/.emacs file I have tried included only the following lines (require 'ess-site) and (require 'ess-site) (setq inferior-R-program-name C:\\Program Files\\R\\R-2.13\\bin\\R) as well as many variations on the R path, including \, /, and going to Rterm instead. Any help would be greatly appreciated. Thanks, Alex Olssen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] (no subject)
Dear all,
Can anyone take a look at my program below?
There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu).
I fixed p1=0.15 for both functions. For any fixed value of lambda (between
0.01 and 0.99),
I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu
values.
Then I plug the calculated cl and cu back into the function f2.
Eventually, I want to find the lambda value and the corresponding cl and cu
values that would
make f2=0.1.
The result of this program does not seem to match the answer I have. Can
some one give me
some hint? Thank you very much.
Hannah
u1 - -3
u2 - 4
f1 - function(lambda,z,p1){
lambda*(p1*exp(u1*z-u1^2/2)+(0.2-p1)*exp(u2*z-u2^2/2))-(1-lambda)*0.8}
f2 - function(p1,cl, cu){
0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl-
u1)+(1-pnorm(cu-u1)))+(0.2-p1)*(pnorm(cl-u2)+(1-pnorm(cu-u2}
p1 - 0.15
lam - seq(0.01,0.99, by=0.001)
x1 - numeric(length(lam))
for (i in 1:length(lam)){
cl - uniroot(f1, lower =-10, upper = 0,
tol = 1e-10,p1=p1,lambda=lam[i])$root
cu - uniroot(f1, lower =0, upper = 10,
tol = 1e-10,p1=p1,lambda=lam[i])$root
x1[i]- f2(p1=p1, cl=cl, cu=cu) }
k - 1
while(klength(lam) x1[k]=0.1){
k=k+1
}
k-k-1;k
lower - uniroot(f1, lower =-10, upper = 0,
tol = 1e-10,p1=p1,lambda=lam[k])$root
upper - uniroot(f1, lower =0, upper = 10,
tol = 1e-10,p1=p1,lambda=lam[k])$root
res - c(lower, upper)
[[alternative HTML version deleted]]
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Re: [R] Emacs ESS finding all versions of R
Alex, This query belongs on the ESS list ess-h...@r-project.org Please place any followup questions there. You seem to have accepted the default location of R, so ESS should just work. The only line you need in ~/.emacs is (require 'ess-site) The other lines you mention cause interference and should not be there. I assume you have 32-bit windows since you mentioned i386/ but did not mention x64/ The i386/Rterm.exe and x64/Rterm.exe are the relevant executable files. ESS will find them. It is possible that you have other lines in your .emacs or in your site-start.el files and they are causing trouble. Start emacs without any initialization files emacs -Q And then manually load the ess-site.el file. Please report back to the ESS mailing list Rich On Sun, Sep 11, 2011 at 4:23 PM, Alex Olssen alex.ols...@gmail.com wrote: Dear R-help, Apologies if this isn't exactly the right place for this question. I am trying to run R in emacs using ESS. I have done this successfully many times - it works right now on my home computer in windows 7 and in ubuntu. The problem is that when I start emacs the regular splash screen doesn't show. In the mini-buffer I get the message Finding all versions of R on your system. And then emacs just freezes up. I am running windows xp. emacs is located at C:\Program Files\emacs-23.3\bin\emacs I have an ess-5.14 folder at C:\Program Files\emacs-23.3\site-lisp\ess-5.14 R is located at C:\Program Files\R\R-2.13.1\bin\R or C:\Program Files\R\R-2.13.1\bin\i386\Rterm ? In my ~/.emacs file I have tried included only the following lines (require 'ess-site) and (require 'ess-site) (setq inferior-R-program-name C:\\Program Files\\R\\R-2.13\\bin\\R) as well as many variations on the R path, including \, /, and going to Rterm instead. Any help would be greatly appreciated. Thanks, Alex Olssen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variable scoping question
Bill,
Thanks for the improvment. I had not been concerned in looking at the
type; was primarily interested in the sizes. Will go back and update
the function.
On Thu, Sep 8, 2011 at 11:53 AM, William Dunlap wdun...@tibco.com wrote:
In my.ls() you ought to convert the pos argument
to an environment and consistently use that environment
in the calls to eval, get, and ls in the function.
E.g., with the following modification
my.ls1 - function (pos = 1, sorted = FALSE, envir = as.environment(pos))
{
.result - sapply(ls(envir = envir, all.names = TRUE), function(..x)
object.size(eval(as.symbol(..x),
envir = envir)))
if (sorted) {
.result - rev(sort(.result))
}
.ls - as.data.frame(rbind(as.matrix(.result), `**Total` = sum(.result)))
names(.ls) - Size
.ls$Size - formatC(.ls$Size, big.mark = ,, digits = 0,
format = f)
.ls$Mode - c(unlist(lapply(rownames(.ls)[-nrow(.ls)], function(x)
mode(eval(as.symbol(x),
envir = envir, ---)
.ls
}
we get:
sorted - 10:1 # put a variable in .GlobalEnv that is also in my.ls*
my.ls1()
Size Mode
my.ls 12,576 function
my.ls1 23,424 function
sorted 88 numeric
**Total 36,088 ---
my.ls()
Size Mode
my.ls 12,576 function
my.ls1 23,424 function
sorted 48 logical
**Total 36,048 ---
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Nordlund, Dan
(DSHS/RDA)
Sent: Thursday, September 08, 2011 8:33 AM
To: r-help@r-project.org
Subject: Re: [R] Variable scoping question
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of jim holtman
Sent: Thursday, September 08, 2011 6:52 AM
To: Bos, Roger
Cc: r-help@r-project.org
Subject: Re: [R] Variable scoping question
Here is a function I use to look at the sizes of objects:
my.ls -
function (pos = 1, sorted = FALSE)
{
.result - sapply(ls(pos = pos, all.names = TRUE), function(..x)
object.size(eval(as.symbol(..x
if (sorted) {
.result - rev(sort(.result))
}
.ls - as.data.frame(rbind(as.matrix(.result), `**Total` =
sum(.result)))
names(.ls) - Size
.ls$Size - formatC(.ls$Size, big.mark = ,, digits = 0,
format = f)
.ls$Mode - c(unlist(lapply(rownames(.ls)[-nrow(.ls)], function(x)
mode(eval(as.symbol(x),
---)
.ls
}
Jim,
I thought I would try out your function above. I copy-and-pasted the
function into a newly started,
vanilla R session, then ran the following code.
x - 1:1000
my.ls()
Size Mode
my.ls 12,576 function
x 4,040 character
**Total 16,616 ---
I don't understand the character mode for the vector x. Any thoughts? Here
is my sessionInfo().
sessionInfo()
R version 2.13.1 (2011-07-08)
Platform: x86_64-pc-mingw32/x64 (64-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2] LC_CTYPE=English_United States.1252
[3] LC_MONETARY=English_United States.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United States.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
loaded via a namespace (and not attached):
[1] tools_2.13.1
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
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--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
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[R] how to quote factors in a function?
### code ###
x=sample(LETTERS[1:26],100,T)
prob=function(y){
count=sum(x==y)
total=length(x)
count/total
}
### end ###
How do I quote the letters A,B,C,D ect, in order to make the prob function
return the weights of the desired Letter?
Thanks!
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Re: [R] (no subject)
Please include a meaningful 'Subject' because these conversations are
archived, and serve as a valuable help resource for the community. I
don't believe I understand what you want.
Stephen
On 09/11/2011 03:32 PM, li li wrote:
Dear all,
Can anyone take a look at my program below?
There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu).
I fixed p1=0.15 for both functions. For any fixed value of lambda (between
0.01 and 0.99),
I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu
values.
Then I plug the calculated cl and cu back into the function f2.
Eventually, I want to find the lambda value and the corresponding cl and cu
values that would
make f2=0.1.
The result of this program does not seem to match the answer I have. Can
some one give me
some hint? Thank you very much.
Hannah
u1- -3
u2- 4
f1- function(lambda,z,p1){
lambda*(p1*exp(u1*z-u1^2/2)+(0.2-p1)*exp(u2*z-u2^2/2))-(1-lambda)*0.8}
f2- function(p1,cl, cu){
0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl-
u1)+(1-pnorm(cu-u1)))+(0.2-p1)*(pnorm(cl-u2)+(1-pnorm(cu-u2}
p1- 0.15
lam- seq(0.01,0.99, by=0.001)
x1- numeric(length(lam))
for (i in 1:length(lam)){
cl- uniroot(f1, lower =-10, upper = 0,
tol = 1e-10,p1=p1,lambda=lam[i])$root
cu- uniroot(f1, lower =0, upper = 10,
tol = 1e-10,p1=p1,lambda=lam[i])$root
x1[i]- f2(p1=p1, cl=cl, cu=cu) }
k- 1
while(klength(lam) x1[k]=0.1){
k=k+1
}
k-k-1;k
lower- uniroot(f1, lower =-10, upper = 0,
tol = 1e-10,p1=p1,lambda=lam[k])$root
upper- uniroot(f1, lower =0, upper = 10,
tol = 1e-10,p1=p1,lambda=lam[k])$root
res- c(lower, upper)
[[alternative HTML version deleted]]
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Re: [R] how to quote factors in a function?
You quote them in quotes. A
Also: bad scoping.
Michael Weylandt
On Sep 11, 2011, at 6:11 PM, casperyc caspe...@hotmail.co.uk wrote:
### code ###
x=sample(LETTERS[1:26],100,T)
prob=function(y){
count=sum(x==y)
total=length(x)
count/total
}
### end ###
How do I quote the letters A,B,C,D ect, in order to make the prob function
return the weights of the desired Letter?
Thanks!
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-quote-factors-in-a-function-tp3805913p3805913.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] how to quote factors in a function?
Hi casperyc,
Try
# option 1
x - sample(LETTERS[1:26], 100, TRUE)
prob - function(y){
count=sum(x==y)
total=length(x)
count/total
}
prob('A')
prob('B')
prob('Z')
# option 2
tx - prop.table(table(x))
tx
tx['A']
# option 2.1
foo - function(x, l = NULL){
tx - prop.table(table(x))
if(is.null(l)){
r - tx
}
else r - tx[l]
r
}
foo(x)
foo(x, 'A')
HTH,
Jorge
On Sun, Sep 11, 2011 at 6:11 PM, casperyc wrote:
### code ###
x=sample(LETTERS[1:26],100,T)
prob=function(y){
count=sum(x==y)
total=length(x)
count/total
}
### end ###
How do I quote the letters A,B,C,D ect, in order to make the prob
function
return the weights of the desired Letter?
Thanks!
--
View this message in context:
http://r.789695.n4.nabble.com/how-to-quote-factors-in-a-function-tp3805913p3805913.html
Sent from the R help mailing list archive at Nabble.com.
__
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
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Re: [R] (no subject)
Hello,
you would get more answers if you code had proper indentation and comments.
Also, please provide a meaningful topic. You should also explain how
this is an R question and not just a debug my code request. What are
are you trying to achieve? Which of the numerous variables you
declared should we look at? What was the result you expected and what
did you get?
JC
2011/9/11 li li hannah@gmail.com:
Dear all,
Can anyone take a look at my program below?
There are two functions: f1 (lambda,z,p1) and f2(p1,cl, cu).
I fixed p1=0.15 for both functions. For any fixed value of lambda (between
0.01 and 0.99),
I solve f1(p1=0.15, lambda=lambda, z)=0 for the corresponding cl and cu
values.
Then I plug the calculated cl and cu back into the function f2.
Eventually, I want to find the lambda value and the corresponding cl and cu
values that would
make f2=0.1.
The result of this program does not seem to match the answer I have. Can
some one give me
some hint? Thank you very much.
Hannah
u1 - -3
u2 - 4
f1 - function(lambda,z,p1){
lambda*(p1*exp(u1*z-u1^2/2)+(0.2-p1)*exp(u2*z-u2^2/2))-(1-lambda)*0.8}
f2 - function(p1,cl, cu){
0.8*(pnorm(cl)+(1-pnorm(cu)))/(0.8*(pnorm(cl)+(1-pnorm(cu)))+p1*(pnorm(cl-
u1)+(1-pnorm(cu-u1)))+(0.2-p1)*(pnorm(cl-u2)+(1-pnorm(cu-u2}
p1 - 0.15
lam - seq(0.01,0.99, by=0.001)
x1 - numeric(length(lam))
for (i in 1:length(lam)){
cl - uniroot(f1, lower =-10, upper = 0,
tol = 1e-10,p1=p1,lambda=lam[i])$root
cu - uniroot(f1, lower =0, upper = 10,
tol = 1e-10,p1=p1,lambda=lam[i])$root
x1[i]- f2(p1=p1, cl=cl, cu=cu) }
k - 1
while(klength(lam) x1[k]=0.1){
k=k+1
}
k-k-1;k
lower - uniroot(f1, lower =-10, upper = 0,
tol = 1e-10,p1=p1,lambda=lam[k])$root
upper - uniroot(f1, lower =0, upper = 10,
tol = 1e-10,p1=p1,lambda=lam[k])$root
res - c(lower, upper)
[[alternative HTML version deleted]]
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[R] coxreg vs coxph: time-dependent treatment
Dear List, After including cluster() option the coxreg (from eha package) produces results slightly different than that of coxph (from survival) in the following time-dependent treatment effect calculation (example is used just to make the point). Will appreciate any explaination / comment. cheers, Ehsan require(survival) require(eha) data(heart) # create weights follow - heart$stop - heart$start fit - glm(transplant ~ age + surgery + year + follow, data=heart, family = binomial) heart$iptw - ifelse(heart$transplant == 0, 1 - predict(fit, type = response), predict(fit, type = response)) summary(heart$iptw) # no weights (basic calculation) fit0 - coxph(Surv(start,stop,event)~transplant, data=heart) fit0 # fit with coxph without case-weights fit1 - coxreg(Surv(start,stop,event)~transplant, data=heart) fit1 # fit with coxreg from eha without case-weights # coxph fit2 - coxph(Surv(start,stop,event)~transplant + cluster(id), data=heart, weights = iptw, robust = T) fit2 # fit with coxph having robust and cluster option fit3 - coxph(Surv(start,stop,event)~transplant + cluster(id), data=heart, weights = iptw) fit3 # fit with coxph having cluster option fit4 - coxph(Surv(start,stop,event)~transplant, data=heart, weights = iptw) fit4 # fit with coxph # coxreg fit5 - coxreg(Surv(start,stop,event)~transplant + cluster(id), data=heart, weights = iptw) fit5 # fit with coxreg from eha having cluster option fit6 - coxreg(Surv(start,stop,event)~transplant, data=heart, weights = iptw) fit6 # fit with coxreg from eha exp(coef(fit3))# HR from coxph having cluster option transplant1 0.3782417 exp(coef(fit4))# HR from coxph transplant1 0.3782417 exp(coef(fit5))[1] # HR from coxreg having cluster option transplant1 0.3645507 exp(coef(fit6))[1] # HR from coxreg transplant1 0.3782417 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] coxreg vs coxph: time-dependent treatment
Sorry: there was an error in the weight calculation, fixed version is the following, but still the final estimates differ as explained in the original email: # require(survival) require(eha) data(heart) head(heart) follow - heart$stop - heart$start fit - glm(transplant ~ age + surgery + year + follow, data=heart, family = binomial) heart$wt - ifelse(heart$transplant == 0, (1 - predict(fit, type = response)), (predict(fit, type = response))) heart$iptw - unlist(tapply(1/heart$wt, heart$id, cumprod)) summary(heart$iptw) # no weights fit0 - coxph(Surv(start,stop,event)~transplant, data=heart) fit0 # fit with coxph without case-weights fit1 - coxreg(Surv(start,stop,event)~transplant, data=heart) fit1 # fit with coxreg from eha without case-weights # coxph fit2 - coxph(Surv(start,stop,event)~transplant + cluster(id), data=heart, weights = iptw, robust = T) fit2 # fit with coxph having robust and cluster option fit3 - coxph(Surv(start,stop,event)~transplant + cluster(id), data=heart, weights = iptw) fit3 # fit with coxph having cluster option fit4 - coxph(Surv(start,stop,event)~transplant, data=heart, weights = iptw) fit4 # fit with coxph # coxreg fit5 - coxreg(Surv(start,stop,event)~transplant + cluster(id), data=heart, weights = iptw) fit5 # fit with coxreg from eha having cluster option fit6 - coxreg(Surv(start,stop,event)~transplant, data=heart, weights = iptw) fit6 # fit with coxreg from eha exp(coef(fit3))# HR from coxph having cluster option exp(coef(fit4))# HR from coxph exp(coef(fit5))[1] # HR from coxreg having cluster option exp(coef(fit6))[1] # HR from coxreg # exp(coef(fit3))# HR from coxph having cluster option transplant1 17.94681 exp(coef(fit4))# HR from coxph transplant1 17.94681 exp(coef(fit5))[1] # HR from coxreg having cluster option transplant1 20.06519 exp(coef(fit6))[1] # HR from coxreg transplant1 17.94681 # __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding groups to regression line panel function in Lattice
On Fri, Sep 9, 2011 at 9:38 PM, Bigelow, Seth sbige...@fs.fed.us wrote:
I wish to display a single-panel Lattice figure with grouped data and fitted
regression lines. I don't seem to be able to get the
individual regression lines to display, e.g.;
d - data.frame(q = rep(1:6, each=10), cc = rep(seq(10,100, 10),6))
# create data frame with group identifier 'q', independent
variable cc
d$ba = d$q*0.1*d$cc + 5*rnorm(nrow(d))
# create dependent variable 'ba'
mypanel - function(...){
# panel function
panel.lmline(d$cc, d$ba, groups = d$q)
#
panel.xyplot(...)
}
But panel.lmline() does not honour a 'groups' argument, so there is no
reason for this to work.
xyplot(ba~cc,d,
groups=q,
panel = panel.superpose,
panel.groups=mypanel
)
Can anyone suggest how to get lines to display by group?
How about
xyplot(ba~cc,d, groups=q, type=c(p, r))
(and how to get lmline panel to fit line without estimating an intercept?)
xyplot(ba~cc,d, groups=q, panel = panel.superpose,
panel.groups = function(x, y, ...) {
panel.xyplot(x, y, ...)
panel.abline(0, coef(lm(y ~ 0 + x)), ...)
})
-Deepayan
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[R] Open a file to APPEND
Hi, I want to store the output of my program to a file. However, With subsequent runs of my code, I'd like to append to the same log file. Currently, I'm using: outfile - file(log.txt, open=w) cat(results, file=outfile) This works, but will overwrite the log file each time. Is there a way to open a file and have R append to the end? Thanks! -- Noah Silverman UCLA Department of Statistics 8117 Math Sciences Building #8208 Los Angeles, CA 90095 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Hourly data with zoo
I have date data as a numeric and hourly data in 0 to 2300 hours in a dataframe. d - rep(20110101,24) h - seq(from = 0, to = 2300, by = 100) df - data.frame(LST_DATE = d, LST_TIME = h, data = rnorm(24, 0, 1)) S - chron(dates. = as.character(df$LST_DATE), times. = paste(as.character(df$LST_TIME/100), :0:0, sep = ), format = c(dates = Ymd, times = h:m:s)) X - zoo(df$data, order.by = S) And I want to create a regular zoo series, The above works but its pretty ugly. Is there a more elegant way to do this. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
