Re: [R] Integration + Normal Distribution + Directory Browsing Processing Questions
Hi Nils,
I would say, pnorm is faster and has a higher precision.
Best,
Matthias
- original message
Subject: Re: [R] Integration + Normal Distribution + Directory Browsing
Processing Questions
Sent: Mon, 22 Jan 2007
From: Nils Hoeller[EMAIL PROTECTED]
Thank you,
both work fine. Why is pnorm to prefer?
Nils
Matthias Kohl schrieb:
Hi,
why don't you use pnorm?
E.g.,
pnorm(1, mean = 0.1, sd = 1.2) - pnorm(0, mean = 0.1, sd = 1.2)
Matthias
- original message
Subject: Re: [R] Integration + Normal Distribution + Directory Browsing
Processing Questions
Sent: Sun, 21 Jan 2007
From: Dimitris Rizopoulos[EMAIL PROTECTED]
you can use the `...' argument of integrate, e.g.,
integrate(dnorm, 0, 1)
integrate(dnorm, 0, 1, mean = 0.1)
integrate(dnorm, 0, 1, mean = 0.1, sd = 1.2)
look at ?integrate for more info.
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
Quoting Nils Hoeller [EMAIL PROTECTED]:
Hi everyone,
I am new to R, but it's really great and helped me a lot!
But now I have 2 questions. It would be great, if someone can help me:
1. I want to integrate a normal distribution, given a median and sd.
The integrate function works great BUT the first argument has to be a
function
so I do integrate(dnorm,0,1) and it works with standard m. and sd.
But I have the m and sd given.
So for fixed m and sd I work around with a new function mynorm
mynorm - function(n) {
ret - dnorm(n,0.6,0.15)
ret
}
for example.
BUT what can I do for dynamic m and sd?
I want something like integrate(dnorm(,0.6,0.15),0,1), with the first
dnorm parameter open for the
integration but fixed m and sd.
I hope you can help me.
2. I am working with textfiles with rows of measure data.
read.table(file) works fine.
Now I want R to read.table all files within a given directory and
process them one by the other.
for(all files in dir xy) {
x - read.table(nextfile)
process x
}
Is that possible with R? I hope so. Can anyone give me a link to
examples.
Thanks for your help
Nils
__
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--- original message end
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Dr. rer. nat. Matthias Kohl
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PLEASE do read the posting guide
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--- original message end
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Dr. rer. nat. Matthias Kohl
[EMAIL PROTECTED]
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Re: [R] command-line arguments in R
see: ?commandArgs or more detail for R startup mechanisms: ?Startup On 1/22/07, Deepak Chandra [EMAIL PROTECTED] wrote: Hi All, A simple and naive question from a newbie. How can one access command-line arguments in R i.e. equivalent of argv in C? Have spent a lot of time on finding it. Also, it will be useful if someone can point me out to a resource which I could have used to answer this question or similar ones. Thanks, Deepak [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integration + Normal Distribution + Directory Browsing Processing Questions
the reason is that is more natural to use pnorm(), which also should a
more efficient approximation of the Normal integral than intgrate(),
you may even use
diff(pnorm(0:1, mean = 0.5, sd = 1.2))
however, the point I meant to make was to use the '...' argument that
can found in many R functions.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
- Original Message -
From: Nils Hoeller [EMAIL PROTECTED]
To: Matthias Kohl [EMAIL PROTECTED]
Cc: Dimitris Rizopoulos [EMAIL PROTECTED];
r-help@stat.math.ethz.ch
Sent: Monday, January 22, 2007 8:49 AM
Subject: Re: [R] Integration + Normal Distribution + Directory
Browsing Processing Questions
Thank you,
both work fine. Why is pnorm to prefer?
Nils
Matthias Kohl schrieb:
Hi,
why don't you use pnorm?
E.g.,
pnorm(1, mean = 0.1, sd = 1.2) - pnorm(0, mean = 0.1, sd = 1.2)
Matthias
- original message
Subject: Re: [R] Integration + Normal Distribution + Directory
Browsing Processing Questions
Sent: Sun, 21 Jan 2007
From: Dimitris Rizopoulos[EMAIL PROTECTED]
you can use the `...' argument of integrate, e.g.,
integrate(dnorm, 0, 1)
integrate(dnorm, 0, 1, mean = 0.1)
integrate(dnorm, 0, 1, mean = 0.1, sd = 1.2)
look at ?integrate for more info.
I hope it helps.
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.kuleuven.be/~m0390867/dimitris.htm
Quoting Nils Hoeller [EMAIL PROTECTED]:
Hi everyone,
I am new to R, but it's really great and helped me a lot!
But now I have 2 questions. It would be great, if someone can
help me:
1. I want to integrate a normal distribution, given a median and
sd.
The integrate function works great BUT the first argument has to
be a
function
so I do integrate(dnorm,0,1) and it works with standard m. and
sd.
But I have the m and sd given.
So for fixed m and sd I work around with a new function mynorm
mynorm - function(n) {
ret - dnorm(n,0.6,0.15)
ret
}
for example.
BUT what can I do for dynamic m and sd?
I want something like integrate(dnorm(,0.6,0.15),0,1), with the
first
dnorm parameter open for the
integration but fixed m and sd.
I hope you can help me.
2. I am working with textfiles with rows of measure data.
read.table(file) works fine.
Now I want R to read.table all files within a given directory and
process them one by the other.
for(all files in dir xy) {
x - read.table(nextfile)
process x
}
Is that possible with R? I hope so. Can anyone give me a link to
examples.
Thanks for your help
Nils
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible
code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--- original message end
--
Dr. rer. nat. Matthias Kohl
[EMAIL PROTECTED]
www.stamats.de
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Disclaimer: http://www.kuleuven.be/cwis/email_disclaimer.htm
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Combination diffrent variables ,how to calculates
Hi, List , i have 6 predictor variables and i want to make possible combinations of these 6 predictors ,all the data is in matrix form , if i am having 6 predictors than possible combination of sets are 64 2 power 6, or 63 ,whatever it may be i want to store the result in another variable to each combination and that i want to put in some model , i want to put every combination in some model ,please help me how to find suitable combinations of these variables i was not able to do this . Any package is there in R which can help me in this regards any help . thanks in Advance ANIL KUMAR( METEOROLOGIST) LRF SECTION NATIONAL CLIMATE CENTER ADGM(RESEARCH) INDIA METEOROLOGICAL DEPARTMENT SHIVIJI NAGAR PUNE-411005 INDIA MOBILE +919422023277 [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to disable existing menus in tcltk?
Hi! I've constructed a small menu-driven interface to a couple of R functions using the possibilities offered by the tcltk package. When user runs some specific analyses, I would then like to disable some of the menus (or menu choises) that are not applicable after the performed analysis. I tried to modify the state of an existing menu, but it seems that neither tkconfigure nor tkentryconfigure contains the state as one of its options. Here's a snip of the code. How could I disable, for example, the Edit data menu choise after already creating the menu (I want it to be active initially)? gui-tktoplevel() topMenu-tkmenu(gui) tkconfigure(gui,menu=topMenu) editMenu-tkmenu(topMenu, tearoff=FALSE) tkadd(editMenu, command, label=Edit data, command=function() editData()) tkadd(editMenu, command, label=Preferences, command=function() editPref()) tkadd(topMenu, cascade, label=Edit, menu=editMenu) Thanks, Jarno Tuimala __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aov y lme
Dear Tomas You can produce the results in Montgomery Montgomery with lme. Please remark that you should indicate the nesting with the levels in your nested factor. Therefore I recreated your data, but used 1,...,12 for the levels of batch instead of 1,...,4. purity-c(1,-2,-2,1,-1,-3,0,4, 0,-4, 1,0, 1,0,-1,0,-2,4,0,3, -3,2,-2,2,2,-2,1,3,4,0,-1,2,0,2,2,1) suppli-factor(c(rep(1,12),rep(2,12),rep(3,12))) batch-factor(c(rep(1:4,3), rep(5:8,3), rep(9:12,3))) material-data.frame(purity,suppli,batch) As you remarked you can use aov summary(material.aov-aov(purity~suppli+suppli:batch,data=material)) Df Sum Sq Mean Sq F value Pr(F) suppli2 15.056 7.528 2.8526 0.07736 . suppli:batch 9 69.917 7.769 2.9439 0.01667 * Residuals24 63.333 2.639 --- Signif. codes: 0 $,1rx(B***$,1ry(B 0.001 $,1rx(B**$,1ry(B 0.01 $,1rx(B*$,1ry(B 0.05 $,1rx(B.$,1ry(B 0.1 $,1rx(B $,1ry(B 1 Remark that the test of suppli is not the same as in Montgomery. Here it is wrongly tested against the Residuals and you should perform the calculate the test with: (Fsuppi - summary(material.aov)[[1]][1,Mean Sq]/ summary(material.aov)[[1]][2,Mean Sq]) pf(Fsuppi, df1 = 2, df2 = 9) To use lme the correct level numbering is now important to indicate the nesting. You should specify your random component as random=~1|batch If you use random=~1|suppli/batch instead, random components for batch AND suppli are included in the model and you specify a model that incorporates suppli as random and fixed. Therefore the correct syntax is library(nlme) material.lme-lme(purity~suppli,random=~1|batch,data=material) ## You obtain the F-test for suppli using anova anova(material.lme) summary(material.lme) ## Remark that in the summary output, the random effects are ## standard deviations and not variance components and you ## should square them to compare them with Montgomery ## 1.307622^2 = 1.71 1.624466^2 = 2.64 ## Or you can use VarCorr(material.lme) I hope this helps you. Best regards, Christoph Buser -- Credit and Surety PML study: visit our web page www.cs-pml.org -- Christoph Buser [EMAIL PROTECTED] Seminar fuer Statistik, LEO C13 ETH Zurich 8092 Zurich SWITZERLAND phone: x-41-44-632-4673 fax: 632-1228 http://stat.ethz.ch/~buser/ -- Tomas Goicoa writes: Dear R user, I am trying to reproduce the results in Montgomery D.C (2001, chap 13, example 13-1). Briefly, there are three suppliers, four batches nested within suppliers and three determinations of purity (response variable) on each batch. It is a two stage nested design, where suppliers are fixed and batches are random. y_ijk=mu+tau_i+beta_j(nested in tau_i)+epsilon_ijk Here are the data, purity-c(1,-2,-2,1, -1,-3, 0,4, 0,-4, 1, 0, 1,0,-1,0, -2,4,0,3, -3,2,-2,2, 2,-2,1,3, 4,0,-1,2, 0,2,2,1) suppli-factor(c(rep(1,12),rep(2,12),rep(3,12))) batch-factor(rep(c(1,2,3,4),9)) material-data.frame(purity,suppli,batch) If I use the function aov, I get material.aov-aov(purity~suppli+suppli:batch,data=material) summary(material.aov) Df Sum Sq Mean Sq F value Pr(F) suppli2 15.056 7.528 2.8526 0.07736 . suppli:batch 9 69.917 7.769 2.9439 0.01667 * Residuals24 63.333 2.639 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 and I can estimate the variance component for the batches as (7.769- 2.639)/3=1.71 which is the way it is done in Montgomery, D. I want to use the function lme because I would like to make a diagnosis of the model, and I think it is more appropriate. Looking at Pinheiro and Bates, I have tried the following, library(nlme) material.lme-lme(purity~suppli,random=~1|suppli/batch,data=material) VarCorr(material.lme) Variance StdDev suppli =pdLogChol(1) (Intercept) 1.563785 1.250514 batch = pdLogChol(1) (Intercept) 1.709877 1.307622 Residual2.638889 1.624466 material.lme Linear mixed-effects model fit by REML Data: material Log-restricted-likelihood: -71.42198 Fixed: purity ~ suppli (Intercept) suppli2 suppli3 -0.417 0.750 1.583 Random effects: Formula: ~1 | suppli (Intercept) StdDev:1.250514 Formula: ~1 | batch %in% suppli (Intercept) Residual StdDev:1.307622 1.624466 Number of Observations: 36 Number of Groups: suppli batch %in% suppli 312 From VarCorr I obtain the variance component 1.71, but I am not sure if
Re: [R] Scatterplot help
The following three lines will do what you want. You will probably want to change some of the default behaviour; just look at the relevant help pages. plot(x,y) text(x,y,ID) grid(2) On 21/01/07, gnv shqp [EMAIL PROTECTED] wrote: Hi my friends, I'm trying to make a scatterplot like this. 1) I have a 3-variable dataset. They are ID, x, and y. 2) x is for the X-axis, y for the Y-axis, and ID is used to label all the cases in the scatterplot. 3) After creating the scatterplot, I need to add both a X-axis reference line and a Y-axis reference line. The X-axis reference line is a vertical line starting from the center of the X-axis. The Y-axis reference line is a horizontal line starting from the center of the Y-axis. In other words, by creating the two reference lines, the scatterplot is divided into 4 quadrants. Please help me figure out how to do that in R. Many thanks in advance! Feng [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- = David Barron Said Business School University of Oxford Park End Street Oxford OX1 1HP __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to disable existing menus in tcltk?
play below after your code and look at tk window: tkentryconfigure(editMenu,0,state=disable) tkentryconfigure(editMenu,0,state=active) tkentryconfigure(topMenu,1,state=disable) tkentryconfigure(topMenu,1,state=active) HTH On 1/22/07, Jarno Tuimala [EMAIL PROTECTED] wrote: Hi! I've constructed a small menu-driven interface to a couple of R functions using the possibilities offered by the tcltk package. When user runs some specific analyses, I would then like to disable some of the menus (or menu choises) that are not applicable after the performed analysis. I tried to modify the state of an existing menu, but it seems that neither tkconfigure nor tkentryconfigure contains the state as one of its options. Here's a snip of the code. How could I disable, for example, the Edit data menu choise after already creating the menu (I want it to be active initially)? gui-tktoplevel() topMenu-tkmenu(gui) tkconfigure(gui,menu=topMenu) editMenu-tkmenu(topMenu, tearoff=FALSE) tkadd(editMenu, command, label=Edit data, command=function() editData()) tkadd(editMenu, command, label=Preferences, command=function() editPref()) tkadd(topMenu, cascade, label=Edit, menu=editMenu) Thanks, Jarno Tuimala __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Unusual behaviour get.hist.quote
I have been running this script regularly for some time. This morning the following error message appeared. Any suggestions to correct this change would be appreciated. EWL-get.hist.quote(EWL,start=(today - Sys.Date())-350,quote=Cl) trying URL 'http://chart.yahoo.com/table.csv?s=EWLa=1bc 06d=0e!f 07g=dq=qy=0z=EWLx=.csv' Content type 'text/csv' length unknown opened URL downloaded 11Kb Error in if (dat[n] != start) cat(format(dat[n], time series starts %Y-%m-%d\n)) : missing value where TRUE/FALSE needed Regrds Jerry Pressnell [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] ECB/Sidebar/R (Emacs) was: Re: kate editor for R
On 1/22/07, Dirk Eddelbuettel [EMAIL PROTECTED] wrote: On 22 January 2007 at 00:05, Ramon Diaz-Uriarte wrote: | On 1/20/07, Dirk Eddelbuettel [EMAIL PROTECTED] wrote: | Just confirms my suspicion that even after all these years, I barely | scratched the surface of ess. That '2+ years' old feature wouldn't happen to | be documented somewhere, would it? | | Dirk, I must be missing something. All I do is: M-x ecb-activate | Everything works. I do nothing special with ess. For that matter, I do | nothing special when editing LaTeX or Python, and ecb (et al) do work | as intended. I had looked at ECB for C++ programming. It simply hadn't occurred to me that it would plug into ESS. I wasn't aware of it either until I attended Tony Rossini's tutorial at useR! 2006. Score another one for Emacs as an operating system. Oh yes, and almost coffee maker and pizza deliverer :-). R. Dirk -- Hell, there are no rules here - we're trying to accomplish something. -- Thomas A. Edison -- Ramon Diaz-Uriarte Statistical Computing Team Structural Biology and Biocomputing Programme Spanish National Cancer Centre (CNIO) http://ligarto.org/rdiaz __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Unusual behaviour get.hist.quote
Jerry Pressnell wrote: I have been running this script regularly for some time. This morning the following error message appeared. EWL-get.hist.quote(EWL,start=(today - Sys.Date())-350,quote=Cl) Error in if (dat[n] != start) cat(format(dat[n], time series starts %Y-%m-%d\n)) : missing value where TRUE/FALSE needed Looks like this has happened before and spontaneously fixed itself then: http://www.mail-archive.com/r-help@stat.math.ethz.ch/msg77866.html http://www.mail-archive.com/r-help@stat.math.ethz.ch/msg77869.html Now working, must have been a Yahoo! issue. 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 and provide commented, minimal, self-contained, reproducible code.
[R] Combination of variables
Hi, List , i have 6 predictor variables and i want to make possible combinations of these 6 predictors ,all the data is in matrix form , if i am having 6 predictors than possible combination of sets are 64 2 power 6, or 63 ,whatever it may be i want to store the result in another variable to each combination and that i want to put in some model , i want to put every combination in some model ,please help me how to find suitable combinations of these variables i was not able to do this . Any package is there in R which can help me in this regards any help . thanks in Advance ANIL KUMAR( METEOROLOGIST) LRF SECTION NATIONAL CLIMATE CENTER ADGM(RESEARCH) INDIA METEOROLOGICAL DEPARTMENT SHIVIJI NAGAR PUNE-411005 INDIA MOBILE +919422023277 [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] D'Agostino test
Hello,
I'd like to know if the D'Agostino test of normality is reliable,
because somme of our results are not really coherent.
This test seems to be very sensitive. Even compared to a normal
distribution generated by R, the results are not very clear.
thanks for any help
Matthieu.
This e-mail has been scanned for all viruses by Star. The\ s...{{dropped}}
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and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: Re: aov y lme
Dear Prof. Ripley and Christoph, thank you very much for your comments. You have helped me a lot. Thanks, Tomas Goicoa Dear Prof. Ripley Thank you for your email. Yes, this is of course the correct syntax to save us the extra calculation. And I forgot the lower.tail = FALSE for pf() in my example to obtain the p-value. Thank you for the corrections and Best regards, Christoph Buser Prof Brian Ripley writes: On Mon, 22 Jan 2007, Christoph Buser wrote: Dear Tomas You can produce the results in Montgomery Montgomery with lme. Please remark that you should indicate the nesting with the levels in your nested factor. Therefore I recreated your data, but used 1,...,12 for the levels of batch instead of 1,...,4. purity-c(1,-2,-2,1,-1,-3,0,4, 0,-4, 1,0, 1,0,-1,0,-2,4,0,3, -3,2,-2,2,2,-2,1,3,4,0,-1,2,0,2,2,1) suppli-factor(c(rep(1,12),rep(2,12),rep(3,12))) batch-factor(c(rep(1:4,3), rep(5:8,3), rep(9:12,3))) material-data.frame(purity,suppli,batch) As you remarked you can use aov summary(material.aov-aov(purity~suppli+suppli:batch,data=material)) Df Sum Sq Mean Sq F value Pr(F) suppli2 15.056 7.528 2.8526 0.07736 . suppli:batch 9 69.917 7.769 2.9439 0.01667 * Residuals24 63.333 2.639 --- Signif. codes: 0 $,1rx(B***$,1ry(B 0.001 $,1rx(B**$,1ry(B 0.01 $,1rx(B*$,1ry(B 0.05 $,1rx(B.$,1ry(B 0.1 $,1rx(B $,1ry(B 1 Remark that the test of suppli is not the same as in Montgomery. Here it is wrongly tested against the Residuals and I don't think so: aov() is doing the correct thing for the model specified. The aov() model you want is probably aov(purity ~ suppli + Error(suppli:batch), data=material) and this gives summary(.Last.value) Error: suppli:batch Df Sum Sq Mean Sq F value Pr(F) suppli 2 15.056 7.528 0.969 0.4158 Residuals 9 69.917 7.769 Error: Within Df Sum Sq Mean Sq F value Pr(F) Residuals 24 63.333 2.639 you should perform the calculate the test with: (Fsuppi - summary(material.aov)[[1]][1,Mean Sq]/ summary(material.aov)[[1]][2,Mean Sq]) pf(Fsuppi, df1 = 2, df2 = 9) You want the other tail. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Combination of variables
You don't say what model you want to do. It isn't necessary to store each combination of predictors in a separate matrix unless you really need to do this for some other reason, in which case I imagine you could adopt this idea. I dare say there are better ways, but this should work (assuming you want a linear model): x-matrix(runif(30*6),ncol=6) y - rnorm(30) cmbs - list() for (i in 1:6) cmbs - c(cmbs,combn(6,i,simplify=FALSE)) for (i in 1:length(cmbs)) print(summary(lm(y ~ x[,cmbs[[i]]]))) On 22 Jan 2007 08:51:22 -, anil kumar rohilla [EMAIL PROTECTED] wrote: Hi, List , i have 6 predictor variables and i want to make possible combinations of these 6 predictors ,all the data is in matrix form , if i am having 6 predictors than possible combination of sets are 64 2 power 6, or 63 ,whatever it may be i want to store the result in another variable to each combination and that i want to put in some model , i want to put every combination in some model ,please help me how to find suitable combinations of these variables i was not able to do this . Any package is there in R which can help me in this regards any help . thanks in Advance ANIL KUMAR( METEOROLOGIST) LRF SECTION NATIONAL CLIMATE CENTER ADGM(RESEARCH) INDIA METEOROLOGICAL DEPARTMENT SHIVIJI NAGAR PUNE-411005 INDIA MOBILE +919422023277 [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- = David Barron Said Business School University of Oxford Park End Street Oxford OX1 1HP __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Combination of variables
You might find the following code useful. It's part of a package I'm
developing for interactive model exploration.
Hadley
# Generate all models
# Fit all combinations of x variables ($2^p$)
#
# This technique generalises \code{\link{fitbest}}. While it is much
# slower it will work for any type of model.
#
# @arguments vector y values
# @arguments matrix of x values
# @arguments method used to fit the model, eg
\code{\link{lm}},\code{\link[MASS]{rlm}}
# @keyword regression
fitall - function(y, x, method=lm, ...) {
data - cbind(y=y, x)
combs - do.call(expand.grid, rep(list(c(FALSE, TRUE)), ncol(x)))[-1, ]
vars - apply(combs, 1, function(i) names(x)[i])
form - paste(y ~ , lapply(vars, paste, collapse= + ), sep = )
form - lapply(form, as.formula)
models - lapply(form, function(f) eval(substitute(method(f,
data=data, ...), list(f=f, data=data, method=method
names(models) - 1:length(models)
class(models) - c(ensemble, class(models))
models
}
# Generate best linear models
# Use the leaps package to generate the best subsets.
#
# @arguments model formula
# @arguments data frame
# @arguments number of subsets of each size to record
# @arguments other arguments passed to \code{\link[leaps]{regsubsets}}
# @keyword regression
fitbest - function(formula, data, nbest=10, ...) {
b - regsubsets(formula, data=data, nbest=nbest, ...)
mat - summary(b, matrix.logical = TRUE)$which
intercept - c(, -1)[as.numeric(mat[,1])]
vars - apply(mat[,-1], 1, function(x) colnames(mat[, -1])[x])
form - paste(formula[[2]], ~ , lapply(vars, paste, collapse= +
), sep = )
form - lapply(form, as.formula)
models - lapply(form, function(f) eval(substitute(lm(f, data=data),
list(f=f, data=data
names(models) - 1:length(models)
class(models) - c(ensemble, class(models))
models
}
On 22 Jan 2007 08:51:22 -, anil kumar rohilla
[EMAIL PROTECTED] wrote:
Hi,
List , i have 6 predictor variables and i want to make possible combinations
of these 6 predictors ,all the data is in matrix form ,
if i am having 6 predictors than possible combination of sets are 64 2 power
6, or 63 ,whatever it may be i want to store the result in another variable
to each combination and that i want to put in some model ,
i want to put every combination in some model ,please help me how to find
suitable combinations of these variables
i was not able to do this .
Any package is there in R which can help me in this regards
any help .
thanks in Advance
ANIL KUMAR( METEOROLOGIST)
LRF SECTION
NATIONAL CLIMATE CENTER
ADGM(RESEARCH)
INDIA METEOROLOGICAL DEPARTMENT
SHIVIJI NAGAR
PUNE-411005 INDIA
MOBILE +919422023277
[EMAIL PROTECTED]
[[alternative HTML version deleted]]
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[R] Time-varying correlation calculation
Dear R useres, I'm interested in getting a series of time-varying correlation, simply between two random variables. Could you please introduce a package to do this task? Thank you so much for any help. Amir - Don't pick lemons. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient code. how to reduce running time?
On Jan 21, 2007, at 8:11 PM, John Fox wrote: Dear Haris, Using lapply() et al. may produce cleaner code, but it won't necessarily speed up a computation. For example: X - data.frame(matrix(rnorm(1000*1000), 1000, 1000)) y - rnorm(1000) mods - as.list(1:1000) system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) [1] 40.53 0.05 40.61NANA system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) [1] 53.29 0.37 53.94NANA Interesting, in my system the results are quite different: system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) [1] 192.035 12.601 797.094 0.000 0.000 system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) [1] 59.913 9.918 289.030 0.000 0.000 Regular MacOSX install with ~760MB memory. In cases such as this, I don't even find the code using *apply() easier to read. Regards, John Haris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Compare effects between lm-models
Dear helpeRs, I'm estimating a series of linear models (using lm) in which in every new model variables are added. I want to test to what degree the new variables can explain the effects of the variables already present in the models. In order to do that, I simply observe wether these effects decrease in strength and / or lose their significance. My question is: does any of you know a package / function in R that can test whether these changes in effects between models are significant? I figure these effects follow a T-distribution and I know the std. devs., so it must be easy to do manually. But I would like not to invent the wheel, when the function is already present. Below is an example of what I mean. In model2, the variable z is added, which is hypothesized to partly explain the effect of x. Indeed, the effect of x decreases in model2, compared to model1. What I want to find out, is if this decrease is statistically significant. Many thanks, Rense x - c(1,1,1,1,1,2,2,2,2,2,3,4,4,4,5) z - c(2,2,2,2,2,2,2,2,3,3,3,3,4,4,5) y - c(1,2,2,2,3,3,3,3,4,4,4,5,5,5,5) model1 - lm(y~x) model2 - lm(y~x+z) [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] naiveBayes question
Aimin: The problem is that the columns you choose for training (only 4 variables) do not match the ones used for prediction (all except y). David I try to use naiveBayes p.nb.90-naiveBayes(y~aa_three+bas+bcu+aa_ss,data=training) pr.nb.90-table(predict(p.nb.90,training[,-13],type=class),training[,13]) bur I get this error Error in object$tables[[v]] : subscript out of bounds head is data set head(training) pr aa_three aa_one aa_ss aa_posaas bas ams bmsacu bcu omega y index 1 1acx ALA A C 1 127.71 0 69.99 0 -0.2498560 0 79.91470 outward TRUE 2 1acx PRO P C 2 68.55 0 55.44 0 -0.0949008 0 76.60380 outward TRUE 3 1acx ALA A E 3 52.72 0 47.82 0 -0.0396550 0 52.19970 outward TRUE 4 1acx PHE F E 4 22.62 0 31.21 0 0.1270330 0 169.52500 inward TRUE 5 1acx SER S E 5 71.32 0 52.84 0 -0.1312380 0 7.47528 outward TRUE 6 1acx VAL V E 6 12.92 0 22.40 0 0.1728390 0 149.09400 inward TRUE anyone know why? Aimin __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compare effects between lm-models
You can use the anova function a la:
anova(model1, model2)
Analysis of Variance Table
Model 1: y ~ x
Model 2: y ~ x + z
Res.DfRSS Df Sum of Sq F Pr(F)
1 13 4.4947
2 12 4.4228 10.0720 0.1952 0.6665
I would suggest getting a copy of MASS and/or reading
http://www.stats.ox.ac.uk/pub/MASS3/Exegeses.pdf
Max
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Re: [R] How to get correct integration in C for step function?
On Sun, 21 Jan 2007, Lynette wrote: Dear all, I am using Rdqags in C to realize the integration. It seems for the continous C function I can get correct results. However, for step functions, the results are not correct. For example, the following one, when integrated from 0 to 1 gives 1 instead of the correct 1.5 Using integrate() in R for an R-defined step function gives the right answer (eg on the example in ?ecdf). This suggests a problem in your C code, since integrate() just calls dqags. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] predict.survreg() with frailty term and newdata
It can't be done with the current code. In a nutshell, you are trying to use a feature that I never got around to coding. It's been on my to do list, but may never make it to the top. Terry __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Finding the effect of Box-Cox transformation using vis.boxcoxu
?box.cox ?boxcox On Jan 22, 2007, at 2:25 AM, Arun Kumar Saha wrote: I have a dataset 'data' and I want to see the effect of Box-Cox transformation on it Interactively for different lambda values. I already got a look on function vis.boxcoxu in package TeachingDemos. But I didn't find any option to put user's own dataset. Can anyone tell me how to put my own dataset here i.e. data? _ Professor Michael Kubovy University of Virginia Department of Psychology USPS: P.O.Box 400400Charlottesville, VA 22904-4400 Parcels:Room 102Gilmer Hall McCormick RoadCharlottesville, VA 22903 Office:B011+1-434-982-4729 Lab:B019+1-434-982-4751 Fax:+1-434-982-4766 WWW:http://www.people.virginia.edu/~mk9y/ __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient code. how to reduce running time?
On Mon, 22 Jan 2007, Charilaos Skiadas wrote: On Jan 21, 2007, at 8:11 PM, John Fox wrote: Dear Haris, Using lapply() et al. may produce cleaner code, but it won't necessarily speed up a computation. For example: X - data.frame(matrix(rnorm(1000*1000), 1000, 1000)) y - rnorm(1000) mods - as.list(1:1000) system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) [1] 40.53 0.05 40.61NANA system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) [1] 53.29 0.37 53.94NANA Interesting, in my system the results are quite different: system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) [1] 192.035 12.601 797.094 0.000 0.000 system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) [1] 59.913 9.918 289.030 0.000 0.000 Regular MacOSX install with ~760MB memory. But MacOS X is infamous for having rather specific speed problems with its malloc, and so gives different timing results from all other platforms. We are promised a solution in MacOS 10.5. Both of your machines seem very slow compared to mine: system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) user system elapsed 11.011 0.250 11.311 system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) user system elapsed 13.463 0.260 13.812 and that on a 64-bit platform (AMD64 Linux FC5). -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient code. how to reduce running time?
Dear Haris, My timings were on a 3 GHz Pentium 4 system with 1 GB of memory running Win XP SP2 and R 2.4.1. I'm no expert on these matters, and I wouldn't have been surprised by qualitatively different results on different systems, but this difference is larger than I would have expected. One thing that seems particularly striking in your results is the large difference between elapsed time and user CPU time, making me wonder what else was going on when you ran these examples. Regards, John John Fox Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604 http://socserv.mcmaster.ca/jfox -Original Message- From: Charilaos Skiadas [mailto:[EMAIL PROTECTED] Sent: Monday, January 22, 2007 10:00 AM To: John Fox Cc: r-help@stat.math.ethz.ch; 'miraceti' Subject: Re: [R] efficient code. how to reduce running time? On Jan 21, 2007, at 8:11 PM, John Fox wrote: Dear Haris, Using lapply() et al. may produce cleaner code, but it won't necessarily speed up a computation. For example: X - data.frame(matrix(rnorm(1000*1000), 1000, 1000)) y - rnorm(1000) mods - as.list(1:1000) system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) [1] 40.53 0.05 40.61NANA system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) [1] 53.29 0.37 53.94NANA Interesting, in my system the results are quite different: system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) [1] 192.035 12.601 797.094 0.000 0.000 system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) [1] 59.913 9.918 289.030 0.000 0.000 Regular MacOSX install with ~760MB memory. In cases such as this, I don't even find the code using *apply() easier to read. Regards, John Haris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Latin hyper cube sampling from expand.grid()
Dear R experts I am looking for a package which gives me latin hyper cube samples from the grid of values produced from the command expand.grid. Any pointers to this issue might be very useful. Basically, I am doing the following: a-(1:10) b-(20:30) dataGrid-expand.grid(a,b) Now, is there a way to use this dataGrid in the package lhs to get latin hyper cube samples. Thanking you Prasanna __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Compare effects between lm-models
Thank you Alain and Max for your swift responses. It might be that I'm misunderstanding your responses, but aren't you testing if there is a difference between the two full models? What I want to know, os whether the effect of a specific predictor (x) differs between model1 and model2. I'm not interested (presently) if the fit of model 2 is better than that of model 1 (for instance). thanks again, Rense On Jan 22, 2007, at 16:26 , Kuhn, Max wrote: You can use the anova function a la: anova(model1, model2) Analysis of Variance Table Model 1: y ~ x Model 2: y ~ x + z Res.DfRSS Df Sum of Sq F Pr(F) 1 13 4.4947 2 12 4.4228 10.0720 0.1952 0.6665 I would suggest getting a copy of MASS and/or reading http://www.stats.ox.ac.uk/pub/MASS3/Exegeses.pdf Max -- LEGAL NOTICE Unless expressly stated otherwise, this message is confidential and may be privileged. It is intended for the addressee(s) only. Access to this E-mail by anyone else is unauthorized. If you are not an addressee, any disclosure or copying of the contents of this E-mail or any action taken (or not taken) in reliance on it is unauthorized and may be unlawful. If you are not an addressee, please inform the sender immediately. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient code. how to reduce running time?
Dear Brian, -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Prof Brian Ripley Sent: Monday, January 22, 2007 11:06 AM To: Charilaos Skiadas Cc: John Fox; r-help@stat.math.ethz.ch Subject: Re: [R] efficient code. how to reduce running time? On Mon, 22 Jan 2007, Charilaos Skiadas wrote: On Jan 21, 2007, at 8:11 PM, John Fox wrote: Dear Haris, Using lapply() et al. may produce cleaner code, but it won't necessarily speed up a computation. For example: X - data.frame(matrix(rnorm(1000*1000), 1000, 1000)) y - rnorm(1000) mods - as.list(1:1000) system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) [1] 40.53 0.05 40.61NANA system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) [1] 53.29 0.37 53.94NANA Interesting, in my system the results are quite different: system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) [1] 192.035 12.601 797.094 0.000 0.000 system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) [1] 59.913 9.918 289.030 0.000 0.000 Regular MacOSX install with ~760MB memory. But MacOS X is infamous for having rather specific speed problems with its malloc, and so gives different timing results from all other platforms. We are promised a solution in MacOS 10.5. Thanks for the clarification. Both of your machines seem very slow compared to mine: system.time(for (i in 1:1000) mods[[i]] - lm(y ~ X[,i])) user system elapsed 11.011 0.250 11.311 system.time(mods - lapply(as.list(X), function(x) lm(y ~ x))) user system elapsed 13.463 0.260 13.812 and that on a 64-bit platform (AMD64 Linux FC5). As you can see from the specs (in a previous message), my system is quite old, which probably accounts for at least part of the difference. The ratios of the user times for my and your system aren't too different though: 53.29/40.53 # mine [1] 1.314829 13.463/11.011 # yours [1] 1.222686 Regards, John -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with C extension
Hello,
thanks for help and code.
We did a lot of work to speedup our function in R. We have a nested
loop, vectorizing is the fastest way. But there we got a very big matrix
and problems with memory. So we want to stay by loops an speedup with C.
My code is similar to this. (my_c is code from Brian D. Ripley)
SEXP test(SEXP a, SEXP b, SEXP in)
{
SEXP ans, new;
int n=INTEGER(in)[0],i,j;
PROTECT(ans = allocVector(REALSXP, 1));
REAL(ans)[0]=REAL(a)[0];
/*for(j = 0; i m; j++)*/
for(i = 0; i n; i++)
{
/* b= ... ^i *j*/
PROTECT(new = allocVector(REALSXP, i+2));
new = my_c(ans,b);
PROTECT(ans = allocVector(REALSXP, i+2));
ans = new;
UNPROTECT(2);
}
UNPROTECT(1);
return ans;
}
We get an error by in=1300
.Call(test,1,3,as.integer(1300));
Fehler: type mismatch
.Call(test,1,3,as.integer(1300));
Speicherzugriffsfehler
Is there a possibility to free allocated memory? free(...) does not work.
Is there a possibility to reallocate a vector?
Thanks a lot
Markus
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get correct integration in C for step function?
Well, I have no idea either. I can get correct answers for continous functions but incorrect for step functions. Sign, I have been trying to realize the integration in C for long time. Thank you for your answering. Best, Lynette - Original Message - From: Thomas Lumley [EMAIL PROTECTED] To: Lynette [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch; AJ Rossini [EMAIL PROTECTED]; [EMAIL PROTECTED] Sent: Monday, January 22, 2007 10:48 AM Subject: Re: [R] How to get correct integration in C for step function? On Sun, 21 Jan 2007, Lynette wrote: Dear all, I am using Rdqags in C to realize the integration. It seems for the continous C function I can get correct results. However, for step functions, the results are not correct. For example, the following one, when integrated from 0 to 1 gives 1 instead of the correct 1.5 Using integrate() in R for an R-defined step function gives the right answer (eg on the example in ?ecdf). This suggests a problem in your C code, since integrate() just calls dqags. -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lnme convergence
Dear R-user, I am trying to use the R nlme function to fit a non linear mixed effects model. The method has some problem to reach the convergence. I am trying to understand causes of the problem by following step by step evolution of the iterative algorithm (verbose=TRUE command). However, I don't understand the meaning of some information given while the algorithm is running, especialy the last part referring to convergence : Convergence: fixed reStruct 0.01168598 0.82634050 Futhermore, is the tolerance criteria (nlmeControl) related to any of the these two values? Thank you for any help Thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] efficient code. how to reduce running time?
On Jan 22, 2007, at 10:39 AM, John Fox wrote: One thing that seems particularly striking in your results is the large difference between elapsed time and user CPU time, making me wonder what else was going on when you ran these examples. Yes, indeed there were a lot of other things going on, this is the only machine I have and I use it continuously. I'll try to run another test tonight when the machine is not in use. It did seem a very striking difference though. But am I wrong in thinking that these measurements should be independent of what other applications are running at the same time, and should measure exactly the time in terms of CPU cycles needed to finish this task, regardless of how often the process got to use the CPU? I guess I was working under that assumption, which indeed makes the above comparison a very unfair one, because there was a lot more going on during the first system.time call. Still, the difference is quite large, which of course could simply have to do with the internals of the two commands, coupled with Prof. Ripley's comments about malloc in Mac OS X. Regards, John Haris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with C extension
On Mon, 22 Jan 2007, Markus Schmidberger wrote:
Hello,
thanks for help and code.
We did a lot of work to speedup our function in R. We have a nested loop,
vectorizing is the fastest way. But there we got a very big matrix and
problems with memory. So we want to stay by loops an speedup with C.
My code is similar to this. (my_c is code from Brian D. Ripley)
And not in many ways similar to mine, hence your error message.
You really do have to handle all the types, as I did most.
SEXP test(SEXP a, SEXP b, SEXP in)
{
SEXP ans, new;
int n=INTEGER(in)[0],i,j;
PROTECT(ans = allocVector(REALSXP, 1));
REAL(ans)[0]=REAL(a)[0];
/*for(j = 0; i m; j++)*/
for(i = 0; i n; i++)
{
/* b= ... ^i *j*/
PROTECT(new = allocVector(REALSXP, i+2));
new = my_c(ans,b);
PROTECT(ans = allocVector(REALSXP, i+2));
ans = new;
UNPROTECT(2);
}
UNPROTECT(1);
return ans;
}
We get an error by in=1300
.Call(test,1,3,as.integer(1300));
Fehler: type mismatch
.Call(test,1,3,as.integer(1300));
Speicherzugriffsfehler
Is there a possibility to free allocated memory? free(...) does not work.
No, but garbage collection does.
Is there a possibility to reallocate a vector?
Yes, sort of. See lengthgets().
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] [ESS] How to get correct integration in C for step function?
On Mon, 22 Jan 2007, Lynette wrote:
Well,
I have no idea either. I can get correct answers for continous functions but
incorrect for step functions.
I have just tried using Rdqags from C for the function x0 and it worked
fine (once I had declared all the arguments correctly). The code is below.
-thomas
#include Rinternals.h
#include R_ext/Applic.h
double stepfn(double x){
return (x0);
}
SEXP call_stepfn(SEXP x){
SEXP answer;
int i,n;
n=length(x);
PROTECT(answer=allocVector(REALSXP,n));
for(i=0;in;i++){
REAL(answer)[i]=stepfn(REAL(x)[i]);
}
UNPROTECT(1);
return answer;
}
void vec_stepfn(double *x, int n, void *ex){
int i;
for (i=0;in;i++) x[i]=stepfn(x[i]);
return;
}
void Cvec_stepfn(double *x, double *n){
vec_stepfn(x, *n, (void *) NULL);
return;
}
SEXP int_stepfn(SEXP lower, SEXP upper){
SEXP answer;
double result, abserr;
int last, neval, ier;
int lenw;
int *iwork;
double *work;
int limit=100;
double reltol=0.1;
double abstol=0.1;
lenw = 4 * limit;
iwork = (int *) R_alloc(limit, sizeof(int));
work = (double *) R_alloc(lenw, sizeof(double));
Rdqags(vec_stepfn, (void *) NULL, REAL(lower), REAL(upper),
abstol, reltol,
result, abserr, neval, ier,
limit, lenw, last,
iwork, work);
printf(%f %f %d\n, result,abserr, ier);
PROTECT(answer=allocVector(REALSXP,1));
REAL(answer)[0]=result;
UNPROTECT(1);
return answer;
}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Example function for bigglm (biglm) data input from file
This is to submit a commented example function for use in the data
argument to the bigglm(biglm) function, when you want to read the data
from a file (instead of a URL), or rescale or modify the data before
fitting the model. In the hope that this may be of help to someone out
there.
make.data - function (filename, chunksize, ...) {
conn-NULL;
function (reset=FALSE) {
if (reset) {
if (!is.null(conn)) {
close(conn);
};
# This is for a file.
# For other methods, see: help(connections)
# and replace the following definition of conn
# (and possibly the read.table call).
conn - file (description=filename, open=r);
} else {
# It's best that the file you use has no header
# line, because when you use the connection to
# read each excerpt, any header won't get re-read.
# If you choose to skip the first line, then the
# first line of each excerpt will be skipped.
rval - read.table (conn, nrows=chunksize,
skip=0, header=FALSE,...);
if (nrow(rval)==0) {
# Then we have reached the end of the input.
# Clean up:
close(conn);
conn-NULL;
rval-NULL;
} else {
# We did not reach the end of the input,
# so this function will return data.
# Here, you can define any derived fields
# or put instructions to rescale input data
# that you want done after the data are read
# but before they are used for fitting.
# For example:
rval$rescaled_column - rval$original_column / 100.0;
# If you don't want to do anything like this,
# then delete this else clause, and make
# the end of the function resemble the URL
# example in bigglm.
};
return(rval);
}
}
};
a - make.data ( filename = myfile, chunksize = 100,
# In our definition of make.data, any remaining
# arguments get passed to the read.table function by
# the ... argument.
# Define column types:
colClasses = list (character, character,
integer, numeric, numeric),
# Define the column names in the call:
# (recall that we cannot rely on the file header)
col.names = c(fromState, toState,
first, original_column, second)
);
library(biglm);
bigglm (formula = toState ~ 1 + first + rescaled_column,
data = a, family = binomial(link='logit'),
weights = ~second);
summary(.Last.value)
NOTICE TO RECIPIENTS: Any information contained in or attach...{{dropped}}
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get correct integration in C for step function?
Dear all, especially to Thomas,
I have figured out the problem. For the step function, something wrong with
my C codes. I should use the expression ((x=0.25)(x=0.75)) ? 2:1 instead
of ((x=1/4)(x=3/4)) ? 2:1 ). Have no idea why 0.25 makes difference from
1/4 in C. But now I can go ahead with the correct integration in C. Thank
you all. And hope this helps to others.
Best wishes,
Lynette
- Original Message -
From: Lynette [EMAIL PROTECTED]
To: r-help@stat.math.ethz.ch
Cc: AJ Rossini [EMAIL PROTECTED]; [EMAIL PROTECTED]
Sent: Sunday, January 21, 2007 6:24 PM
Subject: [R] How to get correct integration in C for step function?
Dear all,
I am using Rdqags in C to realize the integration. It seems for the
continous C function I can get correct results. However, for step
functions,
the results are not correct. For example, the following one, when
integrated
from 0 to 1 gives 1 instead of the correct 1.5
void func( double *x, int n, void *ex )
{
int i;
for(i=0;in;i++) { x[i]=( ((x=1/4)(x=3/4)) ? 2:1 ) ; }
return;
}
while the following one when integrated from 0 to 1 gives the correct
0.7853983
void func( double *x, int n, void *ex )
{
int i;
for(i=0;in;i++) { x[i]= pow(1-x[i]*x[i],.5); }
return;
}
Please advise the problems. Thanks a lot.
Best,
Lynette
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http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to get correct integration in C for step function?
On Mon, 22 Jan 2007, Lynette wrote: Dear all, especially to Thomas, I have figured out the problem. For the step function, something wrong with my C codes. I should use the expression ((x=0.25)(x=0.75)) ? 2:1 instead of ((x=1/4)(x=3/4)) ? 2:1 ). Have no idea why 0.25 makes difference from 1/4 in C. But now I can go ahead with the correct integration in C. Thank you all. And hope this helps to others. 1 and 4 are ints in C, so 1/4 is 0 (integer division). -thomas __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Integration + Normal Distribution + Directory Browsing Processing Questions
Nils Hoeller wrote: for example. BUT what can I do for dynamic m and sd? I want something like integrate(dnorm(,0.6,0.15),0,1), with the first dnorm parameter open for the integration but fixed m and sd. integrate(function(x)dnorm(x,0.1,1.2), 0, 1) is a way of fixing additional parameters. -- Erich Neuwirth, University of Vienna Faculty of Computer Science Computer Supported Didactics Working Group Visit our SunSITE at http://sunsite.univie.ac.at Phone: +43-1-4277-39464 Fax: +43-1-4277-39459 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R interpretation
Hi, I am new to R (and not really a stats expert) and am having trouble interpreting its output. I am running a human learning experiment, with 6 scores per subject in both the pretest and the posttest. I believe I have fitted the correct model for my data- a mixed-effects design, with subject as a random factor and session (pre vs post) nested within group (trained vs control). I am confused about the output. The summary command gives me this table: D.lme- lme(score~GROUP/session, random=~1|subject, data=ILD4L ) summary(D.lme) Linear mixed-effects model fit by REML Data: ILD4L Subset: EXP == F AIC BIC logLik -63.69801 -45.09881 37.84900 Random effects: Formula: ~1 | subject (Intercept) Residual StdDev: 0.1032511 0.1727145 Fixed effects: score ~ GROUP/session Value Std.Error DF t-value p-value (Intercept) 0.10252778 0.05104328 152 2.008644 0.0463 GROUPT 0.09545347 0.06752391 12 1.413625 0.1829 GROUPC:sessionpost -0.00441389 0.04070919 152 -0.108425 0.9138 GROUPT:sessionpost -0.23586042 0.03525520 152 -6.690090 0. Correlation: (Intr) GROUPT GROUPC GROUPT -0.756 GROUPC:sessionpost -0.399 0.301 GROUPT:sessionpost 0.000 -0.261 0.000 Standardized Within-Group Residuals: Min Q1 Med Q3 Max -2.66977386 -0.52935645 -0.08616759 0.57215015 3.26532101 Number of Observations: 168 Number of Groups: 14 I believe the fixed-effects section of this output to be telling me that my model intercept (which I assume to be the control group pretest?) is significantly different from 0, and that GROUPT (i.e. the trained group) does not differ significantly from the intercept- therefore no pretest difference between groups? The next line is, I believe showing that the GROUPC x sessionpost interaction (i.e., control posttest scores?) is not significantly different from the intercept (i.e. control pretest scores). Finally, I am interpreting the final line as indicating that the GROUPT x sessionpost interaction (ie, trained posttest scores?) is significantly different from the trained pretest scores (GROUPT). A treatment contrast that I would like to apply would be for Control-post vs Trained-post, to see if the groups differ after training, but I'm not sure how to do this- and I feel I am probably overcomplicating the matter. also, I am confused about how to report this output in my publication. For instance, what should I be reporting for df? Those found on the output of the anova table? Would it be possible to look through this for me and indicate how to interpret the R output, and also how I should be reporting this? Apologies for asking such basic questions, but I would like to start using R more regularly and to make sure I am doing so correctly. Many thanks, Dan __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Repeated measures
In the two solutions for the repeated measures problem given in the original reply below, the F and p values given by aov() with the error strata defined by Error() are different from those given by lme(). However, when one does the problem by hand using the standard split plot model, the results agree with those of nlme(). The difference between the two aov() solutions is in the partitioning of sums of squares. Is there a ready explanation for this discrepancy? Thanks, Richard Plant tolerance - tolerance - + read.table(http://www.ats.ucla.edu/stat/Splus/examples/alda/tolerance1. txt, + sep=,, header=TRUE) tolerance.long - reshape(tolerance, + varying = list(c(tol11,tol12,tol13, +tol14, tol15)), + v.names = c(tol), timevar = time, + times = 11:15, direction = long) tolerance.aov2 - aov(tol ~ factor(male) + factor(male):factor(id) + factor(time) + factor(time):male, data = tolerance.long) tolerance.sum - summary(tolerance.aov2) tolerance.sum Df Sum Sq Mean Sq F valuePr(F) factor(male) 1 0.3599 0.3599 2.6077 0.111967 factor(time) 4 2.8326 0.7081 5.1309 0.001358 ** factor(male):factor(id) 14 8.2990 0.5928 4.2951 4.295e-05 *** factor(time):male4 0.1869 0.0467 0.3386 0.850786 Residuals 56 7.7289 0.1380 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 tolerance.list - tolerance.sum[[1]] tolerance.mat - as.matrix(tolerance.list[3]) tolerance.F.male - tolerance.mat[1,1]/tolerance.mat[3,1] tolerance.F.male [1] 0.607137 tolerance.df - as.matrix(tolerance.list[1]) tolerance.p.male - 1 - pf(tolerance.F.male,tolerance.df[1,1],tolerance.df[3,1]) tolerance.p.male [1] 0.4488394 Message: 68 Date: Wed, 17 Jan 2007 05:45:01 -0500 From: Chuck Cleland [EMAIL PROTECTED] Subject: Re: [R] Repeated measures To: Tom Backer Johnsen [EMAIL PROTECTED] Cc: r-help@stat.math.ethz.ch Message-ID: [EMAIL PROTECTED] Content-Type: text/plain; charset=ISO-8859-1 Tom Backer Johnsen wrote: I am having a hard time understanding how to perform a repeated measures type of ANOVA with R. When reading the document found here: http://cran.r-project.org/doc/contrib/Lemon-kickstart/kr_repms.html I find that there is a reference to a function make.rm () that is supposed to rearrange a one row per person type of frame to a one row per observation type of frame. But that function does not seem to be there. Nor does the help.search suggest anything. Is that function buried in some package? I'm not able to find that function. Perhaps that document is out of date. Is there some simple documentation that might be useful somewhere? Starting with a really simple problem (one group, two observations)? Here is an example showing the use of reshape() and analysis via aov() and lme() in the nlme package. tolerance - read.table(http://www.ats.ucla.edu/stat/Splus/examples/alda/tolerance1. tx t, sep=,, header=TRUE) tolerance.long - reshape(tolerance, varying = list(c(tol11,tol12,tol13, tol14, tol15)), v.names = c(tol), timevar = time, times = 11:15, direction = long) tolerance.aov - aov(tol ~ as.factor(time) * male + Error(id), data = tolerance.long) summary(tolerance.aov) Error: id Df Sum Sq Mean Sq male 1 0.085168 0.085168 Error: Within Df Sum Sq Mean Sq F value Pr(F) as.factor(time) 4 2.8326 0.7081 3.0538 0.02236 * male 1 0.3024 0.3024 1.3039 0.25745 as.factor(time):male 4 0.1869 0.0467 0.2015 0.93670 Residuals69 16.0002 0.2319 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 library(nlme) tolerance.lme - lme(tol ~ as.factor(time) * male, random = ~ 1 | id, data = tolerance.long) anova(tolerance.lme) numDF denDF F-value p-value (Intercept) 156 353.9049 .0001 as.factor(time) 456 5.1309 0.0014 male 114 0.6071 0.4488 as.factor(time):male 456 0.3386 0.8508 RSiteSearch(repeated measures) points to other examples, functions, and documentation. Tom __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. -- 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] Query about using optimizers in R without causing program to crash
Hi
I am a newbie to R and am using the lm function to
fit my data.
This optimization is to be performed for around 45000
files not all of which lend themselves to
optimization. Some of these will and do crash.
However, How do I ensure that the program simply goes
to the next file in line without exiting the code with
the error
Error in lm.fit(x, y, offset = offset, singular.ok =
singular.ok, ...) :
NA/NaN/Inf in foreign function call (arg 4)
everytime it encounters troublesome data?
I would greatly appreciate your input as it would
avoid me having to manually type
for fileId in (c(4351:46000)) { ... }
for fileId in (c(5761:46000)) { ... }, etc...
Thanks
Lalitha
Now that's room service! Choose from over 150,000 hotels
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Recursive-SVM (R-SVM)
I am trying to implement a simple r-svm example using the iris data (only two
of the classes are taken and data is within the code). I am running into some
errors. I am not an expert on svm's. If any one has used it, I would appreciate
their help. I am appending the code below.
Thanks../Murli
###
### R-code for R-SVM
### use leave-one-out / Nfold or bootstrape to permute data for external CV
### build SVM model and use mean-balanced weight to sort genes on training set
### and recursive elimination of least important genes
### author: Dr. Xin Lu, Research Scientist
### Biostatistics Department, Harvard School of Public Health
library(e1071)
## read in SVM formated data in filename
## the format following the defination of SVMTorch
## the first line contains 2 integer: nSample nFeature+1
## followed by a matrix, each row for one sample, with the last column being
+/1 1 for class label
ReadSVMdata - function(filename)
{
dd - read.table( filename, header=F, skip=1)
x - as.matrix( dd[, 1:(ncol(dd)-1)] )
y - factor( dd[, ncol(dd)] )
ret - list(x=x, y=y)
}
## create a decreasing ladder for recursive feature elimination
CreatLadder - function( Ntotal, pRatio=0.75, Nmin=5 )
{
x - vector()
x[1] - Ntotal
for( i in 1:100 )
{
pp - round(x[i] * pRatio)
if( pp == x[i] )
{
pp - pp-1
}
if( pp = Nmin )
{
x[i+1] - pp
} else
{
break
}
}
x
}
## R-SVM core code
## input:
## x: row matrix of data
## y: class label: 1 / -1 for 2 classes
## CVtype:
## integer: N fold CV
## LOO: leave-one-out CV
## bootstrape: bootstrape CV
## CVnum: number of CVs
## LOO: defined as sample size
## Nfold and bootstrape: user defined, default as sample size
## output: a named list
## Error: a vector of CV error on each level
## SelFreq: a matrix for the frequency of each gene being selected in each level
## with each column corresponds to a level of selection
## and each row for a gene
## The top important gene in each level are those high-freqent ones
RSVM - function(x, y, ladder, CVtype, CVnum=0 )
{
## check if y is binary response
Ytype - names(table(y))
if( length(Ytype) != 2)
{
print(ERROR!! RSVM can only deal with 2-class problem)
return(0)
}
## class mean
m1 - apply(x[ which(y==Ytype[1]), ], 2, mean)
m2 - apply(x[ which(y==Ytype[2]), ], 2, mean)
md - m1-m2
yy - vector( length=length(y))
yy[which(y==Ytype[1])] - 1
yy[which(y==Ytype[2])] - -1
y - yy
## check ladder
if( min(diff(ladder)) = 0 )
{
print(ERROR!! ladder must be monotonously decreasing)
return(0);
}
if( ladder[1] != ncol(x) )
{
ladder - c(ncol(x), ladder)
}
nSample - nrow(x)
nGene - ncol(x)
SampInd - seq(1, nSample)
if( CVtype == LOO )
{
CVnum - nSample
} else
{
if( CVnum == 0 )
{
CVnum - nSample
}
}
## vector for test error and number of tests
ErrVec - vector( length=length(ladder))
names(ErrVec) - paste(Lev_, ladder, sep=)
nTests - 0
SelFreq - matrix( 0, nrow=nGene, ncol=length(ladder))
colnames(SelFreq) - paste(Lev_, ladder, sep=)
## for each CV
for( i in 1:CVnum )
{
## split data
if( CVtype == LOO )
{
TestInd - i
TrainInd - SampInd[ -TestInd]
} else
{
if( CVtype == bootstrape )
{
TrainInd - sample(SampInd, nSample, replace=T )
TestInd - SampInd[ which(!(SampInd %in% TrainInd ))]
} else
{
## Nfold
TrainInd - sample(SampInd, nSample*(CVtype-1)/CVtype )
TestInd - SampInd[ which(!(SampInd %in% TrainInd ))]
}
}
nTests - nTests + length(TestInd)
## in each level, train a SVM model and record test error
xTrain - x[TrainInd, ]
yTrain - y[TrainInd]
xTest - x[TestInd,]
yTest - y[TestInd]
## index of the genes used in the
SelInd - seq(1, nGene)
for( gLevel in 1:length(ladder) )
{
## record the genes selected in this ladder
SelFreq[SelInd, gLevel] - SelFreq[SelInd, gLevel] +1
## train SVM model and test error
svmres - svm(xTrain[, SelInd], yTrain, scale=F, type=C-classification,
kernel=linear )
if( CVtype == LOO )
{
svmpred - predict(svmres, matrix(xTest[SelInd], nrow=1) )
} else
{
svmpred - predict(svmres, xTest[, SelInd] )
}
ErrVec[gLevel] - ErrVec[gLevel] + sum(svmpred != yTest )
## weight vector
W - t(svmres$coefs*yTrain[svmres$index]) %*% svmres$SV * md[SelInd]
rkW - rank(W)
if( gLevel length(ladder) )
{
SelInd - SelInd[which(rkW (ladder[gLevel] - ladder[gLevel+1]))]
}
}
}
ret - list(ladder=ladder, Error=ErrVec/nTests, SelFreq=SelFreq)
}
SummaryRSVM - function( RSVMres )
{
ERInd - max( which(RSVMres$Error == min(RSVMres$Error)) )
MinLevel - RSVMres$ladder[ERInd]
FreqVec - RSVMres$SelFreq[, ERInd]
SelInd - which( rank(FreqVec) = (ladder[1]-MinLevel) )
# print(MinCV error of, min(RSVMres$Error), at, MinLevel, genes )
ret - list( MinER=min(RSVMres$Error), MinLevel=MinLevel, SelInd=SelInd)
}
###
#my code starts below
#data-ReadSVMdata(iris_r-svm.txt)
#The data read from the file is given below.
Re: [R] Query about using try block
Hi
Thanks for your response.
However I seem to be doing something wrong regarding
the try block resulting in yet another error described
below.
I have a function that takes in a file name and
does the fit for the data in that file.
Hence based on your input, I tried
try ( (fit = lm(y~x, data = data_fitting)), silent =
T);
I left the subsequent lines of my code unchanged.
coeffs = as.list(coef(fit);
lambda = exp(coeffs$x)
After the change using try, when I tried to resume
processing under R as follows
source(fitting.R)
for filename in list { process(filename); }
It says Cannot find object fit ...(in the line
trying to get the coefficients...)
Am I closing the try block in the wrong place?
This function does some post processing on the
coefficients returned by coef(fit), puts them in a
list and sends it to another function.
(i.e. around 6 lines of code after the call to fit).
Thanks
Lalitha
--- Andreas Hary [EMAIL PROTECTED] wrote:
Look at ?try
Your code will probably need to be something like
the following:
fit - list()
for(fileId in 1:n){
try(fit[i] - lm(formula,data=???,...), silent=F)
#or silent=T if you would like to be made aware
of problems
}
Best wishes,
Andreas
lalitha viswanath wrote:
Hi
I am a newbie to R and am using the lm function
to
fit my data.
This optimization is to be performed for around
45000
files not all of which lend themselves to
optimization. Some of these will and do crash.
However, How do I ensure that the program simply
goes
to the next file in line without exiting the code
with
the error
Error in lm.fit(x, y, offset = offset,
singular.ok =
singular.ok, ...) :
NA/NaN/Inf in foreign function call (arg
4)
everytime it encounters troublesome data?
I would greatly appreciate your input as it would
avoid me having to manually type
for fileId in (c(4351:46000)) { ... }
for fileId in (c(5761:46000)) { ... }, etc...
Thanks
Lalitha
Now that's room service! Choose from over 150,000
hotels
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained,
reproducible code.
--
=
Andreas Hary
Flat 5, 70 Finsbury Park Road
Lond, N4 2JX, UK
Email:[EMAIL PROTECTED]
Mobile: 07906 860 987
Want to start your own business?
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query about using optimizers in R without causing program to crash
Hi Lalitha,
Use
try()
or
tryCatch()
Cheers
Andrew
On Mon, Jan 22, 2007 at 12:43:28PM -0800, lalitha viswanath wrote:
Hi
I am a newbie to R and am using the lm function to
fit my data.
This optimization is to be performed for around 45000
files not all of which lend themselves to
optimization. Some of these will and do crash.
However, How do I ensure that the program simply goes
to the next file in line without exiting the code with
the error
Error in lm.fit(x, y, offset = offset, singular.ok =
singular.ok, ...) :
NA/NaN/Inf in foreign function call (arg 4)
everytime it encounters troublesome data?
I would greatly appreciate your input as it would
avoid me having to manually type
for fileId in (c(4351:46000)) { ... }
for fileId in (c(5761:46000)) { ... }, etc...
Thanks
Lalitha
Now that's room service! Choose from over 150,000 hotels
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Andrew Robinson
Department of Mathematics and StatisticsTel: +61-3-8344-9763
University of Melbourne, VIC 3010 Australia Fax: +61-3-8344-4599
http://www.ms.unimelb.edu.au/~andrewpr
http://blogs.mbs.edu/fishing-in-the-bay/
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] sequential processing
One option for processing very large files with R is split:
## split a large file into pieces
#--parameters: the folder, file and number of parts
FLD=/home/user/data
F=very_large_file.dat
parts=50
#---split
cd $FLD
fn=`echo $F | awk -F\. '{print $1}'` #file name without extension
ln=`wc -l $F | awk '{print $1}'` #number of lines in the file
forsplit=`expr $ln / $parts + 1` #number of lines in each part
echo == $F will be split in $parts parts of $forsplit lines each.
split -l $forsplit $F $fn
You could also load the entire file into a DBMS then pull parts of it
into R, or read specific lines through a pipe e.g.
readLines(pipe(sed, grep, python... command)).
Don't try to replicate the SAS processing into R. The exact
translations of the SAS DATA STEP usage of _N_, first., last., retain
etc into R would be: inefficient, ugly, retrogressive, wrong, rigid,
complicated, silly and so on. For a start, read up on indexing - this
seemingly simple and innocuous R feature is in fact far more powerful
than the entire DATA STEP with its whole bag of tricks. Then search
the list for similar questions, for example
http://thread.gmane.org/gmane.comp.lang.r.general/44332/focus=44343
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Gerard Smits
Sent: Sunday, January 21, 2007 2:22 PM
To: r-help@stat.math.ethz.ch
Subject: [R] sequential processing
Like many others, I am new to R but old to SAS.
Am I correct in understanding that R processes a data frame in a
sequential ly? This would imply that large input files could be
read, without the need to load the entire file into memory.
Related to the manner of reading a frame, I have been looking for the
equivalent of SAS _n_ (I realize that I can use a variant of which to
identify an index value) as well as useful SAS features such as
first., last., retain, etc. Any help with this conversion
appreciated.
Thanks,
Gerard Smits
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PLEASE do read the posting guide
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[R] lattice: put key where unused panel would have been
Hi, Say I have z-data.frame(y=runif(190), x=runif(190), f=gl(5,38), g=gl(19,10)) plot-xyplot(y~x|g, data=z, layout=c(5,4), groups=f, auto.key=TRUE) How might one place the key in the empty space where the twentieth panel would have been? Thanks, Ben __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Question about rpart and regression trees
Dear All I would like to use rpart to obtain a regression tree for a dataset like the following: Y X1 X2 X3 X4 5.500033B A 3 2 0.35625148 D B 6 5 0.8062546 E C 4 3 5.100014C A 3 2 5.7000422 A A 3 2 0.76875436 C A 6 5 1.0312537 D A 4 1 Y is the objective variable. X1, X2, X3 and X4 can take, respectively, the following values: X1: A,B,C,D,E X2: A,B,C,D,E X3: 3,4,5,6 X4. 1,2,3,4,5 Should I convert X3 and X4 to factor before running rpart? Thanks in advance, Paul __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] sequential processing
So, I take it, given that the use of a pipe is suggested for sequential reading, that the standard approach to processing a data frame is to load the entire file? Please correct if wrong. BTW, I am not interested in finding direct translations of SAS data step statements to R, but instead in finding an approach by which I can address the type of problems I consistent have to deal with (grouped processing with retention of baseline records, etc.). I'll read more on the indexing as a means of dealing with relative position issues Thanks, Gerard You could also load the entire file into a DBMS then pull parts of it into R, or read specific lines through a pipe e.g. readLines(pipe(sed, grep, python... command)). Don't try to replicate the SAS processing into R. The exact translations of the SAS DATA STEP usage of _N_, first., last., retain etc into R would be: inefficient, ugly, retrogressive, wrong, rigid, complicated, silly and so on. For a start, read up on indexing - this seemingly simple and innocuous R feature is in fact far more powerful than the entire DATA STEP with its whole bag of tricks. Then search the list for similar questions, for example http://thread.gmane.org/gmane.comp.lang.r.general/44332/focus=44343 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Gerard Smits Sent: Sunday, January 21, 2007 2:22 PM To: r-help@stat.math.ethz.ch Subject: [R] sequential processing Like many others, I am new to R but old to SAS. Am I correct in understanding that R processes a data frame in a sequential ly? This would imply that large input files could be read, without the need to load the entire file into memory. Related to the manner of reading a frame, I have been looking for the equivalent of SAS _n_ (I realize that I can use a variant of which to identify an index value) as well as useful SAS features such as first., last., retain, etc. Any help with this conversion appreciated. Thanks, Gerard Smits __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice: put key where unused panel would have been
Benjamin Tyner said the following on 1/22/2007 3:18 PM: Hi, Say I have z-data.frame(y=runif(190), x=runif(190), f=gl(5,38), g=gl(19,10)) plot-xyplot(y~x|g, data=z, layout=c(5,4), groups=f, auto.key=TRUE) How might one place the key in the empty space where the twentieth panel would have been? Thanks, Ben __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. You can try supplying x and y coordinates to auto.key xyplot(y ~ x | g, data = z, layout = c(5, 4), groups = f, auto.key = list(x = .85, y = .9)) HTH, --sundar __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] logistic regression model + Cross-Validation
why not use lda{MASS} and it has cv=T option; it does loo, though.
Or use randomForest.
if you have to use lrm, then the following code might help:
n.fold - 5 # 5-fold cv
n.sample - 50 # assumed 50 samples
s - sample(1:n.fold, size=n.sample, replace=T)
for (i in 1:n.fold){
# create your training data and validation data for each fold
trn - YOURWHOLEDATAFRAME[s!=i,]
val - YOURWHOLEDATAFRAME[s==i,]
# now do your own modeling using lrm
# todo
}
HTH,
weiwei
On 1/21/07, nitin jindal [EMAIL PROTECTED] wrote:
If validate.lrm does not has this option, do any other function has it.
I will certainly look into your advice on cross validation. Thnx.
nitin
On 1/21/07, Frank E Harrell Jr [EMAIL PROTECTED] wrote:
nitin jindal wrote:
Hi,
I am trying to cross-validate a logistic regression model.
I am using logistic regression model (lrm) of package Design.
f - lrm( cy ~ x1 + x2, x=TRUE, y=TRUE)
val - validate.lrm(f, method=cross, B=5)
val - validate(f, ...)# .lrm not needed
My class cy has values 0 and 1.
val variable will give me indicators like slope and AUC. But, I also
need
the vector of predicted values of class variable cy for each record
while
cross-validation, so that I can manually look at the results. So, is
there
any way to get those probabilities assigned to each class.
regards,
Nitin
No, validate.lrm does not have that option. Manually looking at the
results will not be easy when you do enough cross-validations. A single
5-fold cross-validation does not provide accurate estimates. Either use
the bootstrap or repeat k-fold cross-validation between 20 and 50 times.
k is often 10 but the optimum value may not be 10. Code for averaging
repeated cross-validations is in
http://biostat.mc.vanderbilt.edu/twiki/pub/Main/RmS/logistic.val.pdf
along with simulations of bootstrap vs. a few cross-validation methods
for binary logistic models.
Frank
--
Frank E Harrell Jr Professor and Chair School of Medicine
Department of Biostatistics Vanderbilt University
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Weiwei Shi, Ph.D
Research Scientist
GeneGO, Inc.
Did you always know?
No, I did not. But I believed...
---Matrix III
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[R] Loess with more than 4 predictors / offsets
Hello, Does anyone know of an R version of loess that allows more than 4 predictors and/or allows the specification of offsets? For that matter, does anyone know of _any_ version of loess that does either of the things I mention? Thanks, Paul Louisell 650-833-6254 [EMAIL PROTECTED] Research Associate (Statistician) Modeling Data Analytics ARPC [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Questions about xtable and print.xtable
I have been using the wonderful xtable package lately, in combination with Sweave, and I have a couple of general questions along with a more particular one. I'll start with the particular question. I basically have a 1x3 array with column names but no row names. I want to create a latex table with column setting set to |rrr|. I want the column names to appear, but the row names not to appear. The code I am trying is this: library(xtable) x - matrix(c(1:3), c(1,3), dimnames=list(NULL,c(1:3))) tab - xtable(x, align=||) print.xtable(tab, include.rownames=FALSE) print.xtable(tab) The problem here is that the xtable call requires an align value that has one extra row setting, I suppose to account for a possible row name. However, the first print.xtable call seems to ignore the align argument set in the xtable call, when include.rownames is included. Any workarounds will be most welcomed. More generally, I have the following questions: 1) Why are the include.rownames and include.colnames parameters not appearing in the xtable call, but only in the print.xtable call instead? Why do I need to specify n+1 arguments for things like align and digits, when I don't want the row names to be printed? In general, why are the align and digits calls not setable in print.xtable, but only in xtable? 2) I like to enclose my tabular environments in a center environment, instead of a table environment. Unless I've missed it, I don't see how I can do that from within the xtable package. Is this really not possible, and if so why not? The latex.environments setting seems to only be allowed when floating=TRUE, which is exactly what I want to avoid. Any particular reason it is not allowed when floating=FALSE as well? That's it really, thanks in advance for any responses. Haris __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Query about using try block
Usually (that is, not limited in R language), when error occurs in
try, stacks are rollbacked, so the variables defined in try no longer
exists after calling try.
One non-elegant solution is:
fit-NULL
try ( (fit = lm(y~x, data = data_fitting)), silent =T)
if(!is.null(fit)){
coeffs = as.list(coef(fit))
## other subsequent processes
}
On 1/23/07, lalitha viswanath [EMAIL PROTECTED] wrote:
Hi
Thanks for your response.
However I seem to be doing something wrong regarding
the try block resulting in yet another error described
below.
I have a function that takes in a file name and
does the fit for the data in that file.
Hence based on your input, I tried
try ( (fit = lm(y~x, data = data_fitting)), silent =
T);
I left the subsequent lines of my code unchanged.
coeffs = as.list(coef(fit);
lambda = exp(coeffs$x)
After the change using try, when I tried to resume
processing under R as follows
source(fitting.R)
for filename in list { process(filename); }
It says Cannot find object fit ...(in the line
trying to get the coefficients...)
Am I closing the try block in the wrong place?
This function does some post processing on the
coefficients returned by coef(fit), puts them in a
list and sends it to another function.
(i.e. around 6 lines of code after the call to fit).
Thanks
Lalitha
--- Andreas Hary [EMAIL PROTECTED] wrote:
Look at ?try
Your code will probably need to be something like
the following:
fit - list()
for(fileId in 1:n){
try(fit[i] - lm(formula,data=???,...), silent=F)
#or silent=T if you would like to be made aware
of problems
}
Best wishes,
Andreas
lalitha viswanath wrote:
Hi
I am a newbie to R and am using the lm function
to
fit my data.
This optimization is to be performed for around
45000
files not all of which lend themselves to
optimization. Some of these will and do crash.
However, How do I ensure that the program simply
goes
to the next file in line without exiting the code
with
the error
Error in lm.fit(x, y, offset = offset,
singular.ok =
singular.ok, ...) :
NA/NaN/Inf in foreign function call (arg
4)
everytime it encounters troublesome data?
I would greatly appreciate your input as it would
avoid me having to manually type
for fileId in (c(4351:46000)) { ... }
for fileId in (c(5761:46000)) { ... }, etc...
Thanks
Lalitha
Now that's room service! Choose from over 150,000
hotels
__
R-help@stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained,
reproducible code.
--
=
Andreas Hary
Flat 5, 70 Finsbury Park Road
Lond, N4 2JX, UK
Email: [EMAIL PROTECTED]
Mobile: 07906 860 987
Want to start your own business?
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Re: [R] factor()
TeamInfo TEAMNAME LEVEL WORKTIME BONUS 1 batch sunan B 135 9,818 2 batch Chenqi E 121 6,050 3 batch jiangxu F 97 4,189 4 online zhouxi F 63 2,720 5 online chenhe H 36 1,064 ## try: factor(TeamInfo$TEAM) [1] batch batch batch online online Levels: batch online ## or attach(TeamInfo) factor(TEAM) [1] batch batch batch online online Levels: batch online On 1/23/07, qing [EMAIL PROTECTED] wrote: Dear All, I am running Windows XP, R 2.4.1, and doing an example about factor(). read.csv(c:\\TeamInfo.csv)-TeamInfo; TeamInfo; TEAMNAME LEVEL WORKTIME BONUS 1 batch sunan B 135 9,818 2 batch Chenqi E 121 6,050 3 batch jiangxu F 97 4,189 4 online zhouxi F 63 2,720 5 online chenhe H 36 1,064 6 online NA 7 online NA 8 online NA 9 online NA 10 client NA 11 client NA 12 client NA 13 client NA 14 client NA factor(TEAM)-Teamactor; Error in typeof(x) : object TEAM not found Any help or suggestions that you can provide will be greatly appreciated. Qing __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Question about rpart and regression trees
On Mon, 22 Jan 2007, Paul Smith wrote: Dear All I would like to use rpart to obtain a regression tree for a dataset like the following: Y X1 X2 X3 X4 5.500033 B A 3 2 0.35625148D B 6 5 0.8062546 E C 4 3 5.100014 C A 3 2 5.7000422 A A 3 2 0.76875436C A 6 5 1.0312537 D A 4 1 Y is the objective variable. X1, X2, X3 and X4 can take, respectively, the following values: X1: A,B,C,D,E X2: A,B,C,D,E X3: 3,4,5,6 X4. 1,2,3,4,5 Should I convert X3 and X4 to factor before running rpart? If they really are factors, yes. If they are ordered factors, no. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loess with more than 4 predictors / offsets
On Mon, 22 Jan 2007, Louisell, Paul wrote: Hello, Does anyone know of an R version of loess that allows more than 4 predictors and/or allows the specification of offsets? For that matter, does anyone know of _any_ version of loess that does either of the things I mention? Why would you want offsets in a regression?: just subtract them from the lhs. (R's lm has gained offsets by analogy with glm, but the S original did not have them). If you would be more comfortable working with them, it would be very easy to create a modified version that supports them. Also, have you heard of the 'curse of dimensionality'? Localization even to 4 dimensions is no longer really an appropriate term, and Euclidean distance will be the main determinant of 'local' and is quite arbitrary. Thanks, Paul Louisell 650-833-6254 [EMAIL PROTECTED] Research Associate (Statistician) Modeling Data Analytics ARPC [[alternative HTML version deleted]] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] D'Agostino test
Am Montag, 22. Januar 2007 12:33 schrieb Matthieu Mourroux: Hello, I'd like to know if the D'Agostino test of normality is reliable, The test is not consistent. The test statistic can be used for testing the hypothesis of uniformity. See the paper Baringhaus, L.; Henze, N. A test for uniformity with unknown limits based on D'Agostino's $D$. Statist. Probab. Lett. 9 (1990), no. 4, 299--304. for details. L. Baringhaus Leibniz Universitaet Hannover Institut fuer Mathematische Stochastik __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] error in arules package
Hi, we noticed there was a error in the arules package. After reading the source code, we saw that the Dice similarity index was miscalculated in dissimilarity function : an copy-paste from Jaccard Index was not corrected (2* a_b_c, ie 2*(a+b+c) in the code instead of 2*a +b + c !!!). After our mail to R-help (21/11/2006), we thought the authors could do something but I just try the function and the error is still there. I hope the authors will read my mail. Sincerely yours, Fred. -- Dr. Frédéric Chiroleu Biométricien UMR 53 PVBMT (Peuplements Végétaux et Bio-agresseurs en Milieu Tropical) CIRAD-AMIS Pôle de Protection des Plantes (3P) Laboratoire d'Écologie Terrestre et de Lutte Intégrée (LETLI) 7, chemin de l'IRAT Ligne Paradis 97410 Saint-Pierre Île de la Réunion - France Tél. : 02 62 49 92 30 Standard : 02 62 49 92 00 Fax : 02 62 49 92 93 courriel : [EMAIL PROTECTED] __ R-help@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
