[R] Reg GARCH+ARIMA
Hi, Although my doubt is pretty,as i m not from stats background i am not sure how to proceed on this. Currently i am doing a forecasting.I used ARIMA to forecast and time series was volatile i used garchFit for residuals. How to use the output of Garch to correct the forecasted values from ARIMA. Here is my code: ###delta is the data fit-arima(delta,order=c(2,,0,1)) fit.res - resid(fit) ##Check for Residuals acf((fit.res-mean(fit.res))/sd(fit.res)) acf(((fit.res-mean(fit.res))/sd(fit.res))^2) fit.fore = predict(fit, n.ahead=test) ##Use ARIMA GARCH To fit residuals from ARIMA Model 1. fitted.gar-garchFit(formula =~arma(2,1)+garch(1,1),data=*fit.res*,cond.dist = std,trace=FALSE) sresi=fitted@residuals/fitted@sigma.t ###Standardised Residuals acf(sresi) acf(sresi^2) fore.res-predict(fitted.ga, n.ahead=test) OR 2. fitted.gar-garchFit(formula =~arma(2,1)+garch(1,1),data=*delta*,cond.dist = std,trace=FALSE) sresi=fitted@residuals/fitted@sigma.t ###Standardised Residuals acf(sresi) acf(sresi^2) fore.res-predict(fitted.ga, n.ahead=test) My Question is 1. How to use fore.res(Result from Garch Model) to change fit.fore (Forecasted values from ARIMA) 2.Out of 1 and 2 for GARCH which one is correct.Pretty confused.Shud we need to use the residuals got from ARIMA Model or the series directly ? Regards, Raghav [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Effect Size Formaula and Calulations in R
pwr package: ES.h and ES.w1, ES.w2, cohen.ES On Wed, Mar 17, 2010 at 1:28 PM, Jim Silverton jim.silver...@gmail.com wrote: Hello, I would like to find out how to use R to compute the effect size of two samples for a two sample t test. Is there a formula for the fisher's exact test? Any R code and/or formula would be greatly appreciated. Thanks, Jim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- CH Chan Research Assistant - KWH http://www.macgrass.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Summing daily data for each year
Hello there I have a DAILY data set (as shown below) with Date and Streamflow from April 1995 to Aug 2006. I want a yearly sum the streamflow value (for each year - first and last incomplete year). Does anyone know how to do this in R? I have tried everything I could think of, e.g. imported the streamflow data as time series using: flowdata - ts(ZVo[,2],start = c(1995,4,1), frequency = 7) # perhaps frequency = 365 is the better one and tried to aggregate using aggregate() without success ZVo V1 V2 1 1995-04-01 0.002766309 2 1995-04-02 0.002402973 3 1995-04-03 0.002254335 4 1995-04-04 0.002221305 5 1995-04-05 0.002180017 6 1995-04-06 0.002031379 7 1995-04-07 0.001957060 8 1995-04-08 0.001940545 9 1995-04-09 0.001924030 10 1995-04-10 0.001783650 . . . 4156 2006-08-16 0.02861272 4157 2006-08-17 0.03652353 4158 2006-08-18 0.05372419 4159 2006-08-19 0.05630058 4160 2006-08-20 0.06274154 4161 2006-08-21 0.06981833 Many thanks Santosh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] C# DLL Library
The link works fine for me on firefox, chrome and safari. Other ways to get to it are: - search rcpp mailing list on google. - go to r-forge (again google knows where), click on the Rcpp - R/C++ interface, click on the Lists tab, click on subscribe, etc ... Le 17/03/10 04:42, Jeremie Smaga a écrit : I found the problem for the package that wasn't found... My R version was 2.9. Sorry about that. However, I would really appreciate it if you could let me know where I could find the mailing list... Thanks, Jeremie On Tue, Mar 16, 2010 at 6:56 PM, Romain Francois rom...@r-enthusiasts.comwrote: Hello, disclaimer: I don't know C# at all and how it might connect to c++, etc ... For an introductory ride about Rcpp, you can consult the Rcpp-introduction vignette which you can download from the cran page of Rcpp or if you have it installed, you can just do: vignette( Rcpp-introduction, package = Rcpp ) For semi-self-explanatory code examples, you can consult our unit tests: system.file( unitTests, package = Rcpp ) For more questions, we have a dedicated mailing list: https://lists.r-forge.r-project.org/cgi-bin/mailman/listinfo/rcpp-devel Romain Le 16/03/10 11:43, Jeremie Smaga a écrit : Hello Richard, Thanks for the tips. I was aware of the R(D)COM. In the last public version, there is now splash screen appearing which is kind of boring so I think I need to buy it or something. Anyway, I think the idea was to create a common CORE library that can be used from C# and from R. So I think I'll have a look at RCPP which looks to be the best solution. Have you tried it already? Do you know any good tutorial for this package? Thanks, Jeremie On Tue, Mar 16, 2010 at 5:23 PM,richard.cot...@hsl.gov.uk wrote: I would like to develop a core library which I will be using both from R and from C#. - Writing the DLL freely in C# and then create a wrapper? - Writing it in C++ - Writing it in C, the other options are really not good ideas. As far as I know, there currently is no way to call .NET code from R. If you want a library that can be called from both a .NET environment and from R, then writing it in C or C++ is likely your best bet. Alternatively, you can run R code from within .NET using the rcom package. There's an example in F# here (what's true for F# is true for C#). http://cs.hubfs.net/blogs/thepopeofthehub/archive/2007/11/06/FSharpWithR.aspx See ?.C for calling C code, and the rcpp package for an interface to C++ code. Choosing between those two languages mostly depends on whether on not your library is especially suited to object oriented programming or not. Regards, Richie. Mathematical Sciences Unit* **HSL*http://www.hsl.gov.uk/contact-us.htm r-help-boun...@r-project.org wrote on 16/03/2010 09:04:54: Good afternoon everybody, I am sorry, this question might look trivial to some of you, but I read quite a lot of stuff about package creation and I would like a bit of you advices. I read that it was possible to import DLL to R. The thing is, I am not sure that the DLL created with C# will be compatible. I still have not implemented anything, so if it is only a matter of method signatures, I can make sure everything fits. Otherwise, I could use a C++ DLL, but I don't know if it is really recommended. (In fact, I would love to be able to develop it in Visual Studio because it is where I developed the rest of my platform). So, what would you advise? Thanks, -- Jeremie Smaga -- Romain Francois Professional R Enthusiast +33(0) 6 28 91 30 30 http://romainfrancois.blog.free.fr |- http://tr.im/OIXN : raster images and RImageJ |- http://tr.im/OcQe : Rcpp 0.7.7 `- http://tr.im/O1wO : highlight 0.1-5 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Summing daily data for each year
HI Santosh, The simplest way I can think of is (after you import the data) to do: tapply(Zvo$V2, YEAR, sum) The question is how to get YEAR. You could do that if you know how to play with date objects. Another option is to do something like: YEAR - as.data.frame(strsplit(Zvo$V2, -))[1,] For example: as.data.frame(strsplit(c( 1995-04-01 ,1995-04-02), -))[1,] I am sure others can think of better ways. Tal Contact Details:--- Contact me: tal.gal...@gmail.com | 972-52-7275845 Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) | www.r-statistics.com (English) -- On Wed, Mar 17, 2010 at 8:20 AM, santosh.ar...@csiro.au wrote: Hello there I have a DAILY data set (as shown below) with Date and Streamflow from April 1995 to Aug 2006. I want a yearly sum the streamflow value (for each year - first and last incomplete year). Does anyone know how to do this in R? I have tried everything I could think of, e.g. imported the streamflow data as time series using: flowdata - ts(ZVo[,2],start = c(1995,4,1), frequency = 7) # perhaps frequency = 365 is the better one and tried to aggregate using aggregate() without success ZVo V1 V2 1 1995-04-01 0.002766309 2 1995-04-02 0.002402973 3 1995-04-03 0.002254335 4 1995-04-04 0.002221305 5 1995-04-05 0.002180017 6 1995-04-06 0.002031379 7 1995-04-07 0.001957060 8 1995-04-08 0.001940545 9 1995-04-09 0.001924030 10 1995-04-10 0.001783650 . . . 4156 2006-08-16 0.02861272 4157 2006-08-17 0.03652353 4158 2006-08-18 0.05372419 4159 2006-08-19 0.05630058 4160 2006-08-20 0.06274154 4161 2006-08-21 0.06981833 Many thanks Santosh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Summing daily data for each year
Hello there I have a DAILY data set (as shown below) with Date and Streamflow from April 1995 to Aug 2006. I want a yearly sum the streamflow value (for each year - first and last incomplete year). Does anyone know how to do this in R? I have tried everything I could think of, e.g. imported the streamflow data as time series using: flowdata - ts(ZVo[,2],start = c(1995,4,1), frequency = 7) # perhaps frequency = 365 is the better one and tried to aggregate using aggregate() without success ZVo V1 V2 1 1995-04-01 0.002766309 2 1995-04-02 0.002402973 3 1995-04-03 0.002254335 4 1995-04-04 0.002221305 5 1995-04-05 0.002180017 6 1995-04-06 0.002031379 7 1995-04-07 0.001957060 8 1995-04-08 0.001940545 9 1995-04-09 0.001924030 10 1995-04-10 0.001783650 . . . 4156 2006-08-16 0.02861272 4157 2006-08-17 0.03652353 4158 2006-08-18 0.05372419 4159 2006-08-19 0.05630058 4160 2006-08-20 0.06274154 4161 2006-08-21 0.06981833 Many thanks Santosh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Summing daily data for each year
You might try something like this:
ZVo1 - with(ZVo, {
Year - substring(as.character(V1), 1, 4)
Count - table(Year) ## for a count of days
TFlow - tapply(V2, Year, sum) ## total flow
data.frame(Year=Year, No=Count, TotalFlow=TFlow)
})
Bill Venables
CSIRO/CMIS Cleveland Laboratories
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of santosh.ar...@csiro.au
Sent: Wednesday, 17 March 2010 3:22 PM
To: r-help@r-project.org
Subject: [ExternalEmail] [R] Summing daily data for each year
Hello there
I have a DAILY data set (as shown below) with Date and Streamflow from April
1995
to Aug 2006. I want a yearly sum the streamflow value (for each year - first
and last incomplete year).
Does anyone know how to do this in R?
I have tried everything I could think of, e.g. imported the streamflow
data as time series using:
flowdata - ts(ZVo[,2],start = c(1995,4,1), frequency = 7) # perhaps frequency
= 365 is the better one
and tried to aggregate using aggregate() without success
ZVo
V1 V2
1 1995-04-01 0.002766309
2 1995-04-02 0.002402973
3 1995-04-03 0.002254335
4 1995-04-04 0.002221305
5 1995-04-05 0.002180017
6 1995-04-06 0.002031379
7 1995-04-07 0.001957060
8 1995-04-08 0.001940545
9 1995-04-09 0.001924030
10 1995-04-10 0.001783650
.
.
.
4156 2006-08-16 0.02861272
4157 2006-08-17 0.03652353
4158 2006-08-18 0.05372419
4159 2006-08-19 0.05630058
4160 2006-08-20 0.06274154
4161 2006-08-21 0.06981833
Many thanks
Santosh
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Re: [R] odfWeave Error
I am following an example from online and am getting the following error: .Error unizipping fileunzip not found. Are zip and unzip in your path? If not you could use odfWeaveControl. Example: odfctrl - odfWeaveControl( zipCmd = c(h:/bin/zip.exe -r $$file$$ ., h:/bin/unzip.exe -o $ $file$$), ) odfWeave(template, outfile, control = odfctrl) __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Why eval(parse(text=var(vec))) return a matrix but NOT a number?
Dear List
I am getting a problem when using eval(parse).
Code below sketchs what I am trying to do:
For each row of a N*K dataframe (I use a 2*2 dataframe in the example below),
applying a number of functions and get the outputs (two functions,
sum and var are used in the example below).
The problem is eval(parse(text=sum(para))) works fine but not when
sum is replaced by var.
in the later case, a matrix instead of a number is returned.
Any suggestion highly appreciated.
Thank you
#===The function
myloop -function(datfra,funs) {
rows-dim(datfra)[1];
totfunnum-length(funs);
for (i in 1:rows) {
vec-datfra[i,];
for(k in 1:totfunnum) {
print(funs[k]);
x-eval(parse(text=funs[k]));
print(x);
}
}
}
#Experiemental run
workport-data.frame(matrix(1:4,2,2))
funs-c(sum(vec,na.rm=T),var(vec,na.rm=T))
myloop(workport,funs)
# Outputs of the
Experimental run
[1] sum(vec,na.rm=T)
[1] 4
[1] var(vec,na.rm=T)
X1 X2
X1 NA NA
X2 NA NA
[1] sum(vec,na.rm=T)
[1] 6
[1] var(vec,na.rm=T)
X1 X2
X1 NA NA
X2 NA NA
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Re: [R] C# DLL Library
JSmaga wrote: I was aware of the R(D)COM. In the last public version, there is now splash screen appearing which is kind of boring so I think I need to buy it or something. I would pay for RDCOM, but it requires that you explain the details of you application spread (which is mostly 1 in my case, since I only work on request), and that's unacceptable to me. I think there is a way to get around that using the Rcommander (?) version, but I found a better solution: I have moved to using RServe with c# recently; it has nice support for all types of structures, and works under Linux (at server side) or Windows. While RServe is Java, you can use IKVM to make it accessible to c#, and it works out of the box. Getting this translation to work within an hour was one of the biggest miracles in my 40-year programmer's life; kudos to the IKVM (and to rserve, clearly). You can download my simple test application (no support, and it may not even work) for Visual Studio from http://www.menne-biomed.de/uni/rserve.zip. You can use it with a local, Windows rserve, or, more reliably, with a virtual linux box running under VMWare. The latter works great as a development environment, and I can create complex installations, for example with NONMEM, and provide my colleagues the virtual machine as a no-brainer. Dieter -- View this message in context: http://n4.nabble.com/C-DLL-Library-tp1594583p1596029.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] C# DLL Library
Hello, Also, if you go for socket connection, you could give a try to svSocket. Best, Philippe ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons University, Belgium ( ( ( ( ( .. On 17/03/10 09:16, Dieter Menne wrote: JSmaga wrote: I was aware of the R(D)COM. In the last public version, there is now splash screen appearing which is kind of boring so I think I need to buy it or something. I would pay for RDCOM, but it requires that you explain the details of you application spread (which is mostly 1 in my case, since I only work on request), and that's unacceptable to me. I think there is a way to get around that using the Rcommander (?) version, but I found a better solution: I have moved to using RServe with c# recently; it has nice support for all types of structures, and works under Linux (at server side) or Windows. While RServe is Java, you can use IKVM to make it accessible to c#, and it works out of the box. Getting this translation to work within an hour was one of the biggest miracles in my 40-year programmer's life; kudos to the IKVM (and to rserve, clearly). You can download my simple test application (no support, and it may not even work) for Visual Studio from http://www.menne-biomed.de/uni/rserve.zip. You can use it with a local, Windows rserve, or, more reliably, with a virtual linux box running under VMWare. The latter works great as a development environment, and I can create complex installations, for example with NONMEM, and provide my colleagues the virtual machine as a no-brainer. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] define F-ratio computations with aov
Greetings to all, This is my model: aov.fit-aov(Y~A+B+C+D+E+A:C+A:E) In summary(aov.fit) all F values are comptuted by eg MS(A)/MS(Residuals). This is not correct (or what I want), except for F(B) and F(A:E). I suppose P values are not correct either. Is it possible with aov to define the way F computations will be done? I 'd like them to be like this: F(A)=MS(A)/MS(E), F(C)=MS(C)/MS(E), F(D)=MS(D)/MS(E), F(E)=MS(E)/MS(A:E), F(A:C)=MS(A:C)/MS(A:E) thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] C# DLL Library
Hi guys, Correct me if I am wrong, but R(D)COM (and hence RServe and so on) allow you to run R code from C# but is useless to use C(# or ++) code in R right? Because it is in fact what I am trying to find out in this post. Thanks for the tips though, I was also looking for some information about alternative to R(D)COM. Best, Jeremie On Wed, Mar 17, 2010 at 4:26 PM, Philippe Grosjean phgrosj...@sciviews.orgwrote: Hello, Also, if you go for socket connection, you could give a try to svSocket. Best, Philippe ..°})) ) ) ) ) ) ( ( ( ( (Prof. Philippe Grosjean ) ) ) ) ) ( ( ( ( (Numerical Ecology of Aquatic Systems ) ) ) ) ) Mons University, Belgium ( ( ( ( ( .. On 17/03/10 09:16, Dieter Menne wrote: JSmaga wrote: I was aware of the R(D)COM. In the last public version, there is now splash screen appearing which is kind of boring so I think I need to buy it or something. I would pay for RDCOM, but it requires that you explain the details of you application spread (which is mostly 1 in my case, since I only work on request), and that's unacceptable to me. I think there is a way to get around that using the Rcommander (?) version, but I found a better solution: I have moved to using RServe with c# recently; it has nice support for all types of structures, and works under Linux (at server side) or Windows. While RServe is Java, you can use IKVM to make it accessible to c#, and it works out of the box. Getting this translation to work within an hour was one of the biggest miracles in my 40-year programmer's life; kudos to the IKVM (and to rserve, clearly). You can download my simple test application (no support, and it may not even work) for Visual Studio from http://www.menne-biomed.de/uni/rserve.zip. You can use it with a local, Windows rserve, or, more reliably, with a virtual linux box running under VMWare. The latter works great as a development environment, and I can create complex installations, for example with NONMEM, and provide my colleagues the virtual machine as a no-brainer. Dieter __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jeremie Smaga [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] PCA - blank loadings
Which principal component function are you using?
Check the documentation for that and look for the part of the object
that provides the PC's. Those are your loadings.
Xanthe Walker xanthe.wal...@gmail.com 16/03/2010 23:16:47
Hi,
I have successfully completed a PCA and printed the loadings, however,
numerous values are blank. I know that this means the values are just
very
small but not equal to zero.
Is there a way to print out the loadings, including the very small
values, I
need them for graphing purposes.
Thanks,
Xan
[[alternative HTML version deleted]]
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Re: [R] C# DLL Library
JSmaga wrote: Correct me if I am wrong, but R(D)COM (and hence RServe and so on) allow you to run R code from C# but is useless to use C(# or ++) code in R right? Correct. I use a c wrapper if I want to run c# code from R similar, but it can get nasty to cross the marshalling barrier. Dieter -- View this message in context: http://n4.nabble.com/C-DLL-Library-tp1594583p1596092.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to parse a string (by a new markup) with R ?
Wow, Thank you very much Andrej!
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (English)
--
2010/3/17 Andrej Blejec andrej.ble...@nib.si
A version using regular expressions, lot of regexpr() and substr()
functions is attached.
Finally everything is packed into splitSeq() function
Andrej
--
Andrej Blejec
National Institute of Biology
Vecna pot 111 POB 141
SI-1000 Ljubljana
SLOVENIA
e-mail: andrej.ble...@nib.si
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tel: + 386 (0)59 232 789
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International Conference on Teaching Statistics
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-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Gabor Grothendieck
Sent: Tuesday, March 16, 2010 3:24 PM
To: Tal Galili
Cc: r-help@r-project.org; seqinr-fo...@r-forge.wu-wien.ac.at
Subject: Re: [R] How to parse a string (by a new markup) with R ?
We show how to use the gsubfn package to parse this.
The rules are not entirely clear so we will assume the following:
- there is a fixed template for the output which is the same as your
output but possibly with different character strings filled in. This
implies, for example, that there are exactly Stem0, Stem1, Stem2 and
Stem3 and no fewer or more stems.
- the sequence always starts with the open of Stem0, at least one dot
and the open of Stem1. There are no dots prior to the open of Stem0.
This seems to be implicit in your sample output since there is no zero
length string in your sample output corresponding to dots prior to
Stem0.
- Stem0 closes with the same number of as there are to open it
You can modify this yourself to take into account the actual rules
whatever they are.
We first calculate, k, the number of leading 's using strapply.
Then we replace the leading k 's with }'s and the trailing k 's with
{'s giving us Str3:
}}}..{
{{.
We again use strapply, this time to get the lengths of the runs. Note
that
zero length runs are possible so we cannot, for example, use rle for
this. For
example there is a zero length run of dots between the last and the
first {.
read.fwf is used to actually parse out the strings using the lengths we
just
calculated.
Finally we fill in the template using relist.
# inputs
Seq -
GCCTCGATAGCTCAGTTGGGAGAGCGTACGACTGAAGATCGTAAGGtCACCAGTTCGATCCTGGTTCGGG
GCA
Str -
..
.
template -
list(
Stem 0 opening = ,
before Stem 1 = ,
Stem 1 = list(opening = ,
inside = ,
closing =
),
between Stem 1 and 2 = ,
Stem 2 = list(opening = ,
inside = ,
closing =
),
between Stem 2 and 3 = ,
Stem 3 = list(opening = ,
inside = ,
closing =
),
After Stem 3 = ,
Stem 0 closing =
)
# processing
# create string made by repeating string s k times followed by more
reps - function(s, k, more = ) {
paste(paste(rep(s, k), collapse = ), more, sep = )
}
library(gsubfn)
k - nchar(strapply(Str, ^+, c)[[1]])
Str2 - sub(^+, reps(}, k), Str)
Str3 - sub(reps(, k, ([^]*)$), reps({, k, \\1), Str2)
pat -
^(}*)([.]*)(*)([.]*)(*)([.]*)(*)([.]*)(*)([.]*)(*)([.]*)(*)([.]*
)({*)([.]*)$
lens - sapply(strapply(Str3, pat, c)[[1]], nchar)
tokens - unlist(read.fwf(textConnection(Seq), lens, as.is = TRUE))
closeAllConnections()
tokens[is.na(tokens)] -
out - relist(tokens, template)
out
Here is the str of the output for your sample input:
str(out)
List of 9
$ Stem 0 opening : chr GCCTCGA
$ before Stem 1 : chr TA
$ Stem 1 :List of 3
..$ opening: chr GCTC
..$ inside : chr AGTTGGGA
..$ closing: chr GAGC
$ between Stem 1 and 2: chr G
$ Stem 2 :List of 3
..$ opening: chr TACGA
..$ inside : chr CTGAAGA
..$ closing: chr TCGTA
$ between Stem 2 and 3: chr AGGtC
$ Stem 3 :List of 3
..$ opening: chr ACCAG
..$ inside : chr TTCGATC
..$ closing: chr CTGGT
$ After Stem 3: chr
$ Stem 0 closing : chr TCC
On Tue, Mar 16, 2010 at 6:10 AM, Tal Galili tal.gal...@gmail.com
wrote:
Hello all,
For some work I am doing on RNA, I want to use R to do string parsing
that
(I think) is like a simplistic HTML parsing.
For example, let's say we have the following two variables:
Seq -
GCCTCGATAGCTCAGTTGGGAGAGCGTACGACTGAAGATCGTAAGGtCACCAGTTCGATCCTGGTTCGGG
GCA
Str -
..
.
[R] Transform contingency table into data.frame ?
Dear list, I have a contingency table : a - letters[1:3] t - table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df - as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cut out single lines out of matrix
Hi I am a bit curious why one shall do such a twisted construction. Accessing list is basically the same as accessing corresponding matrix row, you only need to remember drop=FALSE option. Regards Petr r-help-boun...@r-project.org napsal dne 17.03.2010 00:58:54: Here is a way of creating a list of the matrices: x - matrix(1:(12*30), nrow=30) # create a list of single row matrices x.l - lapply(seq(nrow(x)), function(a) x[a,, drop=FALSE]) x.l [[1]] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,]1 31 61 91 121 151 181 211 241 271 301 331 [[2]] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,]2 32 62 92 122 152 182 212 242 272 302 332 [[3]] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,]3 33 63 93 123 153 183 213 243 273 303 333 # access the 5th element x.l[[5]] [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [1,]5 35 65 95 125 155 185 215 245 275 305 335 On Tue, Mar 16, 2010 at 6:20 PM, Schmidt Martin m.schm...@students.unibe.ch wrote: Hey dear users I'm trying to kind of split my matrix which looks as follows: dim(out) [1] 30 12 What I finally want is each line as it's own matrix which I can handle then separately. Like, say: out1- [1,] out2-[2,] .. Would you do that with a for() loop or does exist an other appropriate solution? I unfortunately couldn't find any solution! Thanks for help __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/ posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Transform contingency table into data.frame ?
Hi, I don't really understand what the problem is... There is no duplicated column... Maybe you mean the row names? If it is so, then just don't use row names. Ivan Le 3/17/2010 10:45, Carlos Petti a écrit : Dear list, I have a contingency table : a- letters[1:3] t- table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df- as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Transform contingency table into data.frame ?
Hi r-help-boun...@r-project.org napsal dne 17.03.2010 10:45:48: Dear list, I have a contingency table : a - letters[1:3] t - table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df - as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? You probably do not distinct between columns of data frame and row names of data frame. From what you told us it is not clear which one you want to keep if row names use as.data.frame(as.matrix(t, row.names = names(t))) if you want to keep column as.data.frame(t) shall suffice. Row names is not a data frame column. Regards Petr Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Constrained non linear regression using ML
Dear Gabor, dear R users, I had already read the betareg documentation. As far as I can understand from the help, it does not allow for constrained regression. Regards Gabor Grothendieck wrote: Check out the betareg package. On Tue, Mar 16, 2010 at 2:58 PM, Corrado ct...@york.ac.uk wrote: Dear R users, I have to fit the non linear regression: y~1-exp(-(k0+k1*p1+k2*p2+ +kn*pn)) where ki=0 for each i in [1 n] and pi are on R+. I am using, at the moment, nls, but I would rather use a Maximum Likelhood based algorithm. The error is not necessarily normally distributed. y is approximately beta distributed, and the volume of data is medium to large (the y,pi may have ~ 40,000 elements). I have studied the packages in the task views Optimisation and Robust Statistical Methods, but I did look like what I was looking for was there. Maybe I am wrong. The nearest thing was nlrob, but even that does not allow for constraints, as far as I can understand. Any suggestion? Regards -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Transform contingency table into data.frame ?
Dear list, Sorry, I did not explain myself very well. I want to obtain a data.frame like this : Freq a1 b1 c1 This data.frame contains just one column (Freq) and each row is named. But when I use this code : df - as.data.frame(t) or this code : df - as.data.frame(t, row.names(t)) The a column remains. Thanks, Carlos 2010/3/17 Petr PIKAL petr.pi...@precheza.cz Hi r-help-boun...@r-project.org napsal dne 17.03.2010 10:45:48: Dear list, I have a contingency table : a - letters[1:3] t - table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df - as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? You probably do not distinct between columns of data frame and row names of data frame. From what you told us it is not clear which one you want to keep if row names use as.data.frame(as.matrix(t, row.names = names(t))) if you want to keep column as.data.frame(t) shall suffice. Row names is not a data frame column. Regards Petr Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Missing index in vector assignment
Hi r-help-boun...@r-project.org napsal dne 17.03.2010 00:03:21: Dear r-helpers, I am getting a mismatch error between two variables: svp - ksvm(x, y, type=nu-svc) Error in .local(x, ...) : x and y don't match. and I suspect that it might be due to missing index in the y variable which Why suspect? It is probably better to read the docs. Try ?ksvm and see how x or y shall be defined. Besides why do you expect everybody knows what ksvm is I get ??ksvm No help files found with alias or concept or title matching ‘ksvm’ using regular expression matching. therefore I the function is not a part of my installed packages. I defined as: y - (LVvar[,1]) I tried various methods to make the y assignment in the same format as x, which is a dataframe x - (LVvar[,-1]) and looks like x rCoordCap rKnowGrow rGoalcom rSupport rOpcomm rT2Cadap 1 4.979167 4.50 5.812500 6.145833 5.979167 5.031250 ... but I still get y without the indexes as a vector: y [1] -1. -6.91193182 -1. 0.74431818 -6.91193182 Why are the results different for x and y, even though the assignment is the same except I exclude the columns for y? LVvar is probably data frame. So x is data frame too but y lost dimension so it is a vector. Help page for subsetting is a bit complicated as it has to cover various object types. ?[ Basically if you want to keep dimensions = you has to use drop=FALSE option in subsetting. Default is FALSE so whenever result is only one column of data frame it is coerced to vector. drop: logical. If ‘TRUE’ the result is coerced to the lowest possible dimension. The default is to drop if only one column is left, but *not* to drop if only one row is left. Regards Petr Cheers, Chaehan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Transform contingency table into data.frame ?
Hi and did you try my suggestion? as.data.frame(as.matrix(t, row.names = names(t))) Regards Petr r-help-boun...@r-project.org napsal dne 17.03.2010 11:16:43: Dear list, Sorry, I did not explain myself very well. I want to obtain a data.frame like this : Freq a1 b1 c1 This data.frame contains just one column (Freq) and each row is named. But when I use this code : df - as.data.frame(t) or this code : df - as.data.frame(t, row.names(t)) The a column remains. Thanks, Carlos 2010/3/17 Petr PIKAL petr.pi...@precheza.cz Hi r-help-boun...@r-project.org napsal dne 17.03.2010 10:45:48: Dear list, I have a contingency table : a - letters[1:3] t - table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df - as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? You probably do not distinct between columns of data frame and row names of data frame. From what you told us it is not clear which one you want to keep if row names use as.data.frame(as.matrix(t, row.names = names(t))) if you want to keep column as.data.frame(t) shall suffice. Row names is not a data frame column. Regards Petr Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Transform contingency table into data.frame ?
Hi Carlos, try as.data.frame.table(t) hth. Carlos Petti schrieb: Dear list, I have a contingency table : a - letters[1:3] t - table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df - as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Transform contingency table into data.frame ?
I would then just change it using names() For example: names(df) - c(letter, Freq) df letter Freq a a1 b b1 c c1 If you prefer having nothing, I don't really know how to do it because I couldn't use my dataframes without column names! HTH Ivan Le 3/17/2010 11:16, Carlos Petti a écrit : Dear list, Sorry, I did not explain myself very well. I want to obtain a data.frame like this : Freq a1 b1 c1 This data.frame contains just one column (Freq) and each row is named. But when I use this code : df- as.data.frame(t) or this code : df- as.data.frame(t, row.names(t)) The a column remains. Thanks, Carlos 2010/3/17 Petr PIKALpetr.pi...@precheza.cz Hi r-help-boun...@r-project.org napsal dne 17.03.2010 10:45:48: Dear list, I have a contingency table : a- letters[1:3] t- table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df- as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? You probably do not distinct between columns of data frame and row names of data frame. From what you told us it is not clear which one you want to keep if row names use as.data.frame(as.matrix(t, row.names = names(t))) if you want to keep column as.data.frame(t) shall suffice. Row names is not a data frame column. Regards Petr Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Transform contingency table into data.frame ?
Now I get it, I was still thinking you wanted two columns. I was confused by the print example. Petr's suggestion works well then Ivan Le 3/17/2010 11:27, Petr PIKAL a écrit : Hi and did you try my suggestion? as.data.frame(as.matrix(t, row.names = names(t))) Regards Petr r-help-boun...@r-project.org napsal dne 17.03.2010 11:16:43: Dear list, Sorry, I did not explain myself very well. I want to obtain a data.frame like this : Freq a1 b1 c1 This data.frame contains just one column (Freq) and each row is named. But when I use this code : df- as.data.frame(t) or this code : df- as.data.frame(t, row.names(t)) The a column remains. Thanks, Carlos 2010/3/17 Petr PIKALpetr.pi...@precheza.cz Hi r-help-boun...@r-project.org napsal dne 17.03.2010 10:45:48: Dear list, I have a contingency table : a- letters[1:3] t- table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df- as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? You probably do not distinct between columns of data frame and row names of data frame. From what you told us it is not clear which one you want to keep if row names use as.data.frame(as.matrix(t, row.names = names(t))) if you want to keep column as.data.frame(t) shall suffice. Row names is not a data frame column. Regards Petr Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] hexadecimal colors
Hi I would like to produce a red shading I figured the easiest way to do that would be to use rgb in the following way: a- seq(0,0.9,by=0.1) redshade - rgb(red=1,green=a, blue=a) However, I don't really know how to plot things using hexadecimal colors. I used a function which tries to find the closest color to the rgb shades but it didn't work very well. Any comments would be appreciated. Thanks, Hadassa [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] save data to an R object
Dear R people, Is it possible to save three data sets in an R object and to call each data from this object independently! Regards, Cheba [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] hexadecimal colors
To plot, just use 'redshade' plot(1:10, col=redshade, cex=4,pch=16) On Wed, Mar 17, 2010 at 6:55 AM, Hadassa Brunschwig hadassa.brunsch...@mail.huji.ac.il wrote: Hi I would like to produce a red shading I figured the easiest way to do that would be to use rgb in the following way: a- seq(0,0.9,by=0.1) redshade - rgb(red=1,green=a, blue=a) However, I don't really know how to plot things using hexadecimal colors. I used a function which tries to find the closest color to the rgb shades but it didn't work very well. Any comments would be appreciated. Thanks, Hadassa [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] save data to an R object
Hi, Here is one option: #some sample dataframes df1 - data.frame(letter=letters[1:3], number=1:3) df1 - data.frame(letter=letters[4:6], number=4:6) #save the dataframes save(df1, df2, file=test.Rda) #load them in another session load(test.Rda) df1 letter number 1 a 1 2 b 2 3 c 3 df2 letter number 1 d 4 2 e 5 3 f 6 Is that what you're looking for? Ivan Le 3/17/2010 11:56, cheba meier a écrit : Dear R people, Is it possible to save three data sets in an R object and to call each data from this object independently! Regards, Cheba [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] save data to an R object
?list Intro to R That is what a list is for. On Wed, Mar 17, 2010 at 6:56 AM, cheba meier cheba.me...@googlemail.comwrote: Dear R people, Is it possible to save three data sets in an R object and to call each data from this object independently! Regards, Cheba [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help running a Fortran subroutine from R
dc896148 wrote: Then I pass the parameters according to the order they are specified in the filter: array - data.matrix(read.table(time702.txt,header=F)) array1 - array nx - 60 ny - 120 halfintx - 3 halfinty - 3 mask - matrix(array(rep(1.0,25)),5,5) subarray - matrix(0,5,5) subarray1 - matrix(0,5,5) Then I run the Fortran subroutine... out - .Fortran(filter2d, as.single(array), as.single(array), as.integer(nx), as.integer(ny), as.integer(halfintx), as.integer(halfinty), as.single(mask), as.single(subarray), as.single(subarray1)) The smoothed output is 'array1', which I just passed as 'array' in the specification. It can be any matrix, but must be the same dimension as 'array', which is given as nx by ny. Missing values in time702.txt are denoted by 999.00, and are defined that way in the subroutine. Can anyone see where I may be doing something wrong? I am not good with fortran, as this code was previously written and now I am trying to adapt it to R on the fly. First of all, if array1 is the output array of the fortran subroutine, then you should do as.single(array1) instead of as.single(array) as the second argument in the call of filter2d. Furthermore you have set both halfintx and halfinty to 3. Looking in the fortran code this means that subarray and subarray1 are fortran arrays with dimension (7,7). You should parametrize the definition of these arrays in the R code: subarray - matrix(0, 2*halfintx+1,2*halfinty+1) subarray1 - matrix(0, 2*halfintx+1,2*halfinty+1) Similar for mask since it is indexed in fortran with indices -halfintx : halfintx which means 2*halfintx+1 elements. mask - matrix(array(rep(1.0,(2*halfintx+1)*(2*halfinty+1))),2*halfintx+1,2*halfinty+1) (Hopefully no typos here; I haven't tested this). Finally, I am not sure whether you can get array1 returned by the fortran as a single precision real array. I would consider using R numeric doubles. You would only have to change the real in the Fortran with double precision and the 999 with 999D0. And make a small example with required output to make testing easier. Good luck Berend -- View this message in context: http://n4.nabble.com/Help-running-a-Fortran-subroutine-from-R-tp1595641p1596231.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Transform contingency table into data.frame ?
Dear list, Thank you for your answers. Petr, your solution works great :-) Peter's solution too. Carlos 2010/3/17 Ivan Calandra ivan.calan...@uni-hamburg.de Now I get it, I was still thinking you wanted two columns. I was confused by the print example. Petr's suggestion works well then Ivan Le 3/17/2010 11:27, Petr PIKAL a écrit : Hi and did you try my suggestion? as.data.frame(as.matrix(t, row.names = names(t))) Regards Petr r-help-boun...@r-project.org napsal dne 17.03.2010 11:16:43: Dear list, Sorry, I did not explain myself very well. I want to obtain a data.frame like this : Freq a1 b1 c1 This data.frame contains just one column (Freq) and each row is named. But when I use this code : df- as.data.frame(t) or this code : df- as.data.frame(t, row.names(t)) The a column remains. Thanks, Carlos 2010/3/17 Petr PIKALpetr.pi...@precheza.cz Hi r-help-boun...@r-project.org napsal dne 17.03.2010 10:45:48: Dear list, I have a contingency table : a- letters[1:3] t- table(a) I'm looking for a way to transform this table into data frame, as follows : Freq a1 b1 c1 I used : df- as.data.frame(t, row.names = names(t)) But, this function do not remove the duplicated column. Do you know the solution ? You probably do not distinct between columns of data frame and row names of data frame. From what you told us it is not clear which one you want to keep if row names use as.data.frame(as.matrix(t, row.names = names(t))) if you want to keep column as.data.frame(t) shall suffice. Row names is not a data frame column. Regards Petr Thanks in advance, Carlos [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] testing parallelism of does-response curves using package drc
Using just nls() you can set the complete model and the alternative model in this way da - expand.grid(x=20:50, trat=c(0.9,1)) da$y - 10*da$trat*da$x/(12+da$x)+rnorm(da$x,0,0.1) plot(y~x, da) da$trat - as.factor(da$trat) n0 - nls(y~A[trat]*x/(B[trat]+x), data=da, start=list(A=c(9,10), B=c(12,12))) n1 - nls(y~A*x/(B[trat]+x), data=da, start=list(A=c(9.5), B=c(12,12))) anova(n1,n0) Look at gnls() function in the nlme package for a easier way to specify the model. Walmes Zeviani. - ..ooo0 ... ..()... 0ooo... Walmes Zeviani ...\..(.(.)... Master in Statistics and Agricultural Experimentation \_). )../ walmeszevi...@hotmail.com, Lavras - MG, Brasil (_/ -- View this message in context: http://n4.nabble.com/testing-parallelism-of-does-response-curves-using-nls-tp1591356p1596256.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: save data to an R object
Hi r-help-boun...@r-project.org napsal dne 17.03.2010 11:56:45: Dear R people, Is it possible to save three data sets in an R object and to call each data from this object independently! ?list Regards Petr Regards, Cheba [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mmap error-12, macbook pro
Dear List, I am trying to read some files using read.csv and total size of those files are 3.99 GB. I am using MacBook Pro with 4GB RAM(snow leopard). I also tried to run a chunk from those files and altogether the size was 1.33 GB. But every time I was getting the following error R(1200) malloc: *** mmap(size=16777216) failed (error code=12) *** error: can't allocate region *** set a breakpoint in malloc_error_break to debug I managed to run smaller chunks (400MB) and I also saved the object. But problem is when I try to load some of those objects(which is not more than 1.5 GB altogether), I get same error again. Can someone please help? why am I getting this error? does R need more space than the actual file size? I may buy new machine if its something related to RAM size. but if it is some R problem then it will be very useless to buy new machine. So, I really need to know why this is happening. Any help is appreciated. Thanks in advance. regards, Mamun -- View this message in context: http://n4.nabble.com/mmap-error-12-macbook-pro-tp1596234p1596234.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mmap error-12, macbook pro
Hi, I am trying to read some files using read.csv and total size of those files are 3.99 GB. I am using MacBook Pro with 4GB RAM(snow leopard). I also tried to run a chunk from those files and altogether the size was 1.33 GB. But every time I was getting the following error R(1200) malloc: *** mmap(size=16777216) failed (error code=12) *** error: can't allocate region *** set a breakpoint in malloc_error_break to debug I managed to run smaller chunks (400MB) and I also saved the object. But problem is when I try to load some of those objects(which is not more than 1.5 GB altogether), I get same error again. Can someone please help? why am I getting this error? does R need more space than the actual file size? I may buy new machine if its something related to RAM size. but if it is some R problem then it will be very useless to buy new machine. So, I really need to know why this is happening. Thanks. -- View this message in context: http://n4.nabble.com/mmap-error-12-macbook-pro-tp1596208p1596208.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mmap error, macbook pro
Hi, I am trying to read some files using read.csv and total size of those files are 3.99 GB. I am using MacBook Pro with 4GB RAM(snow leopard). I also tried to run a chunk from those files and altogether the size was 1.33 GB. But every time I was getting the following error R(1200) malloc: *** mmap(size=16777216) failed (error code=12) *** error: can't allocate region *** set a breakpoint in malloc_error_break to debug I managed to run smaller chunks (400MB) and I also saved the object. But problem is when I try to load some of those objects(which is not more than 1.5 GB altogether), I get same error again. Can someone please help? why am I getting this error? does R need more space than the actual file size? I may buy new machine if its something related to RAM size. but if it is some R problem then it will be very useless to buy new machine. So, I really need to know why this is happening. Thanks. Shyama -- View this message in context: http://n4.nabble.com/mmap-error-macbook-pro-tp1596091p1596091.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] accessing info in object slots from listed objects using loops
Hey,
I have stacked a couple of garchFit objects in a list with names $fit1,
$fit2, ..., $fiti assigning objects names using a loop, i.e. after running
the loop modelStack = list($fit1, $fit2,...,$fiti).
Thus the following apply;
a = modelStack$fit2, then a is the second garchFit object of formal class
'fGarch' with 11 slots, @call, @formula... etc.
I then want to extract information in the 11 slots using another for loop,
say i want to access the slot a...@fit$coef. In order for this to happen I need
to combine modelStack$fit2, where the $fit2 object name must be constructed
in the loop, with the slot name @fit$coef.
In order to construct the proper list names one could use the following in
the for loop;
name = paste(modelStack$fit,i,@fit$coef,sep=) #
modelstack$f...@fit$coef for i = 2,
and then use something like
get(name)
But this returns; Error in get(name) : object 'modelstack$f...@fit$coef' not
found
If I just type modelstack$f...@fit$coef in the command window it returns the
proper info from the object list. I figure this is because R somehow
interpret $ and @ in get() differently than $ and @ as list
separators.
Does anyone know how to extract information in slots of listed objects using
a loop and on the run generated variable/object names?
JT
sample code
nAssets = length(modelStack)
for(i in 1:nAssets){
name = paste(modelStack[,i,]...@name$series$h,sep=)
a = get(name)
t = length(a)
} # end for loop
--
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http://n4.nabble.com/accessing-info-in-object-slots-from-listed-objects-using-loops-tp1596135p1596135.html
Sent from the R help mailing list archive at Nabble.com.
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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] Missing index in vector assignment
Jim Petr, Thank you for your hint - I am really grateful, because they helped me to get one step further, and although now the problem lies somewhere else, you encouraged that we can find the solution soon! 1. To Petr's comments Petr, your hint to define y: y - LVvar[,1, drop=FALSE] did solve the problem, so I got a data.frame with the indexes. Yet, then I turned to the call svp - ksvm(x, y, type=nu-svc) Error in .local(x, ...) : y must be a vector or a factor. So then I followed your second advice, looking up the additional information from help file: == x is defined as: a symbolic description of the model to be fit. When not using a formula x can be a matrix or vector containing the training data or a kernel matrix of class kernelMatrix of the training data or a list of character vectors (for use with the string kernel). Note, that the intercept is always excluded, whether given in the formula or not. y is defined as a response vector with one label for each row/component of x. Can be either a factor (for classification tasks) or a numeric vector (for regression). == So I tried to convert LVvar into a matrix via as.matrix() but didn't make a difference. 2. To Jim's comments On Wed, Mar 17, 2010 at 1:10 AM, jim holtman jholt...@gmail.com wrote: Please provide what LVvar is. LVvar is a dataframe At least provide str(LVvar), or preferably a 'dput' of the object. str(LVvar) returns: 'data.frame': 55 obs. of 7 variables: $ rPerform : num 0.0682 -0.0682 -0.7443 0.7443 0.2619 ... $ rCoordCap: num 4.98 6.08 5.73 5.92 4.96 ... $ rKnowGrow: num 4.5 5.92 5.23 6.08 4.38 ... $ rGoalcom : num 5.81 6.58 6 5.75 5.29 ... $ rSupport : num 6.15 6.92 6.6 4.92 6 ... $ rOpcomm : num 5.98 6.25 6.33 6.5 5.29 ... $ rT2Cadap : num 5.03 6.12 4.9 6.25 5.12 ... - attr(*, na.action)=Class 'omit' Named int 40 .. ..- attr(*, names)= chr 40 == dput(LVvar) returns (abbreviated with ...): structure(list(rPerform = c(0.0681818181818183, -0.0681818181818183, -0.744318181818182, 0.744318181818182, 0.261931818181818, -0.900568181818182, ... rCoordCap = c(4.979167, 6.08, 5.73, 5.916667, 4.958333, ... .Names = c(rPerform, rCoordCap, rKnowGrow, rGoalcom, rSupport, rOpcomm, rT2Cadap), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, 36L, 37L, 38L, 39L, 41L, 42L, 43L, 44L, 45L, 46L, 47L, 48L, 49L, 50L, 51L, 52L, 53L, 54L, 55L, 56L), na.action = structure(40L, .Names = 40, class = omit), class = data.frame) PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Here's my try (please have mercy for a complete R beginner): library(kernlab) library(methods) # Data Definitions LV - c(rPerform,rCoordCap, rKnowGrow, rGoalcom, rSupport, rOpcomm, rT2Cadap ) # creates a dataframe LVvar - na.omit(loopLV_IndexScores(LV, u_proj)) x - (LVvar[,-1]) y - (LVvar[,1]) svp - ksvm(x, y, type=nu-svc) svp === Thanks so much for not giving up. Cheers, Chaehan On Tue, Mar 16, 2010 at 6:03 PM, Chaehan So chaehan...@gmail.com wrote: Dear r-helpers, I am getting a mismatch error between two variables: svp - ksvm(x, y, type=nu-svc) Error in .local(x, ...) : x and y don't match. and I suspect that it might be due to missing index in the y variable which I defined as: y - (LVvar[,1]) I tried various methods to make the y assignment in the same format as x, which is a dataframe x - (LVvar[,-1]) and looks like x rCoordCap rKnowGrow rGoalcom rSupport rOpcomm rT2Cadap 1 4.979167 4.50 5.812500 6.145833 5.979167 5.031250 ... but I still get y without the indexes as a vector: y [1] -1. -6.91193182 -1. 0.74431818 -6.91193182 Why are the results different for x and y, even though the assignment is the same except I exclude the columns for y? Cheers, Chaehan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve? -- Humboldt University Berlin, Germany Institute of Psychology Rudower Chaussee 18, Room 1221 12489 Berlin Germany Office: +49 30 2093 - 9337 Mobile: +49 171- 626 9373 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the
Re: [R] Constrained non linear regression using ML
Try it anyways -- maybe none of your constraints are active. On Wed, Mar 17, 2010 at 6:01 AM, Corrado ct...@york.ac.uk wrote: Dear Gabor, dear R users, I had already read the betareg documentation. As far as I can understand from the help, it does not allow for constrained regression. Regards Gabor Grothendieck wrote: Check out the betareg package. On Tue, Mar 16, 2010 at 2:58 PM, Corrado ct...@york.ac.uk wrote: Dear R users, I have to fit the non linear regression: y~1-exp(-(k0+k1*p1+k2*p2+ +kn*pn)) where ki=0 for each i in [1 n] and pi are on R+. I am using, at the moment, nls, but I would rather use a Maximum Likelhood based algorithm. The error is not necessarily normally distributed. y is approximately beta distributed, and the volume of data is medium to large (the y,pi may have ~ 40,000 elements). I have studied the packages in the task views Optimisation and Robust Statistical Methods, but I did look like what I was looking for was there. Maybe I am wrong. The nearest thing was nlrob, but even that does not allow for constraints, as far as I can understand. Any suggestion? Regards -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Antw: odfWeave Error
Hi, I am using 7-zip for unzipping (http://www.7-zip.org/). Then odfWeaveControl needs to be set to odfWeaveControl(zipCmd=c(c:/programme/7-zip/7z a -tzip $$file$$,c:/programme/7-zip/7z x $$file$$ -aoa)) - control where 'c:/programme/7-zip/' is the path to my 7-zip installation. You can then use control to set the control-parameter in odfWeave. Frank Btibert3 btibe...@gmail.com 17.03.2010 03:02 Hi All, This is probably basic, but I am just starting with R and very interested in learning how to automate my reports. I am just learning about Sweave and LaTex. I am following an example from online and am getting the following error: .Error unizipping fileunzip not found. I am not a programmer by any stretch, but I realize that the package is either not recognizing my native unzip utility, have to configure one in the package, or both. Additionally, I saw in the tutorial that they recommend a zip program. I tried to install it, but I couldn't figure out what to do. Any help you can provide will be greatly appreciated. Brock -- View this message in context: http://n4.nabble.com/odfWeave-Error-tp1595848p1595848.html Sent from the R help mailing list archive at Nabble.com. Universitätsklinikum Jena Körperschaft des öffentlichen Rechts und Teilkörperschaft der Friedrich-Schiller-Universität Jena BachstraÃe 18, 07743 Jena Verwaltungsratsvorsitzender: Prof. Dr. Thomas Deufel; Medizinischer Vorstand: Prof. Dr. Klaus Höffken; Wissenschaftlicher Vorstand: Prof. Dr. Klaus Benndorf; Kaufmännischer Vorstand und Sprecher des Klinikumsvorstandes Rudolf Kruse Bankverbindung: Sparkasse Jena; BLZ: 830 530 30; Kto.: 221; Gerichtsstand Jena Steuernummer: 161/144/02978; USt.-IdNr. : DE 150545777 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Are loops handled differently in newer versions of R?
Thanks everyone for your assistance with this! Very much appreciated,
and the students of my statistics course will also be pleased to have a
fix ;)
Mike
Don MacQueen wrote:
Joshua's explanation of rownames makes a lot more sense than my
speculation about conversion to numeric. Rownames of dataframes is an
area in which there have definitely been changes in R in the last year
or two, give or take. (I don't recall details or timing)
Therefore, I find it very plausible that in R 2.8.x the rownames of
your dataframe were different than they are now in R 2.10.x, given how
you constructed the dataframe. This then would be the explanation of
why the script worked in 2.8.x and not in 2.10.x.
-Don
At 8:57 PM -0700 3/16/10, Joshua Wiley wrote:
Michael,
I have to agree with Don that using a factor as a loop index seems
like a risky choice. At any rate, part of the problem is that you are
referencing a nonexistant part of your dataframe. i is an index of
characters, but your rownames are 1:5, not LETTERS[1:5]. If you give
your dataframe rownames, you can then use your loop, see below.
locn-c(A, B, C, D, E)
n-c(28, 14, 21, 52, 35)
corr.r-c(0.40, 0.63, 0.38, 0.44, 0.35)
lab8.dat-data.frame(locn, n, corr.r)
lab8.dat
calc.prob.t-function(n, r)
{
df-n-2
t-(r-0)/(sqrt((1-r^2)/df)) # I'm assuming you mean r^2 here not r2
probt-2*(pt(t, df, lower.tail=FALSE))
probt
}
p_unadj-NULL # since you assign it to null anyways, there's not real
point in the other assignment
p_unadj
for ( i in lab8.dat[,1] )
p_unadj[i] - calc.prob.t(lab8.dat[i,2], lab8.dat[i,3])
p_unadj # all NAs as you noticed
rownames(lab8.dat) - lab8.dat$locn
for ( i in lab8.dat[,1] )
p_unadj[i] - calc.prob.t(lab8.dat[i,2], lab8.dat[i,3])
p_unadj # now lab8.dat[A,2] etc. means something, and it works
##
On Tue, Mar 16, 2010 at 7:53 PM, Michael Rennie mdren...@gmail.com
wrote:
Hi gang,
I'm perplexed- I have some code that uses for() loops that works
fine in R
version 2.8 on my mac, worked fine in version 2.8 on my old windows
machine,
but doesn't work in version 2.10 on windows.
The loop implements a function over a data frame (code is included
below).
In Mac (running version 2.8), the results of the loop are what I
expect:
p_unadj
[1] 0.034939481 0.015743706 0.089287030 0.001098538 0.039290594
But in Windows (running version 2.10.1), I get a bunch of NA's...
p_unadj
A B CDE
NA NA NA NA NA
If I had to guess, I'd say that R v. 2.10 is handling the i in
lab8.dat[,1]
differently, given that it's keeping the row names in the output for
p_unadj... but why would that stop it from applying the function?
Any thoughts or suggestions are welcome.
Cheers,
Mike
Here's the code...
#build the dataset
locn-c(A, B, C, D, E)
n-c(28, 14, 21, 52, 35)
corr.r-c(0.40, 0.63, 0.38, 0.44, 0.35)
lab8.dat-data.frame(locn, n, corr.r)
lab8.dat
attach(lab8.dat)
#write the function
calc.prob.t-function(n, r)
#given a sample size (n) and correlation coefficient (r), returns the
probability for that test
{
df-n-2
t-(r-0)/(sqrt((1-r2)/df))
probt-2*(pt(t, df, lower.tail=FALSE))
probt
}
#try out the function...
calc.prob.t(lab8.dat$n[1], lab8.dat$corr.r[1])
#it works.
#write a loop to implement that function for every correlation in your
dataset...
p_unadj-numeric(length(lab8.dat[,1]))
p_unadj-NULL
p_unadj
#all this just built an empty vector to store the results of our
loop...
for ( i in lab8.dat[,1] )
p_unadj[i]-calc.prob.t(lab8.dat[i,2], lab8.dat[i,3])
p_unadj
#if executed on my Mac, running R v.2.8, this works (and did using
2.8 on my
old windows machine). Running v. 2.10 in Windows, I get NAs.
--
Michael D. Rennie, Ph.D.
Postdoctoral Fellow, Environmental and Life Sciences Program
Trent University
2140 East Bank Drive, DNA Building (2nd Floor)
Peterborough, Ontario K9J 7B8
Vox:705.755.2287 Fax:705.755.1559
www.*people.trentu.ca/michaelrennie
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University of California, Riverside
http://*www.*joshuawiley.com/
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[R] Is there any R package that can find the maxima of a 1-D time series
Is there any R package that can help me with digging out the maxima of a 1-D trajectory ? I have 975 1-D curves. They are only known as time series. That is a set of points ordered with respect to time. Some curves exhibit one only peak. Others have two peaks of different height. We wish to find the number of peaks and their position along the time axis. Apparently it's a trivial problem solved looking at the zeros and the change of sign of the 1st derivative. In practice it is necessary to apply some criteria (which ones ?) to discriminate between real peaks and noise oscillations. Presumably I ought to define the noise level with respect to peaks height in this application ... Maybe wavelets can help ? Thank you in advance, Maura tutti i telefonini TIM! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] define F-ratio computations with aov
Howdy, In the past, I've just run the ANOVA as normal, and then just grabbed the appropriate MS for the estimation of F ratios. Eg, this will get you the MS in your anova object: summary(obj.aov)[[1]][3] or summary(obj.aov)$Mean And if you want a specific MS, summary(obj.aov)[[1]][[1,3]] or summary(obj.aov)[[1]]$Mean[1] Then you can just put whichever MS over whichever other MS, estimate your F-ratios, with something like: Ffact- summary(obj.aov)[[1]]$Mean[1]/summary(obj.aov)[[1]]$Mean[3] estimate the p-values with: pFfact-1-pf(Ffact, summary(obj.aov)[[1]]$Df[1], summary(obj.aov)[[1]]$Df[3]) and you're off to the races. You can also specify error strata in the aov() model, but then all you get is the MS and you have to estimate your F-ratios anyway (though the indexing is a little different). E.g., if you had a nested anova, you could specify it as: ex.aov-aov(Fixed ~ Nested + Error(Nested/Fixed)) At least this way, the summary() doesn't give you the wrong F-ratios, so you aren't temped to interpret them incorrectly (as you would in the previous example). HTH, Mike Galanidis Alexandros wrote: Greetings to all, This is my model: aov.fit-aov(Y~A+B+C+D+E+A:C+A:E) In summary(aov.fit) all F values are comptuted by eg MS(A)/MS(Residuals). This is not correct (or what I want), except for F(B) and F(A:E). I suppose P values are not correct either. Is it possible with aov to define the way F computations will be done? I 'd like them to be like this: F(A)=MS(A)/MS(E), F(C)=MS(C)/MS(E), F(D)=MS(D)/MS(E), F(E)=MS(E)/MS(A:E), F(A:C)=MS(A:C)/MS(A:E) thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] accessing info in object slots from listed objects using loops
Try this:
sapply(modelStack, slot, 'fit')['coef',]
On Wed, Mar 17, 2010 at 7:11 AM, torgrims torgr...@stud.ntnu.no wrote:
Hey,
I have stacked a couple of garchFit objects in a list with names $fit1,
$fit2, ..., $fiti assigning objects names using a loop, i.e. after running
the loop modelStack = list($fit1, $fit2,...,$fiti).
Thus the following apply;
a = modelStack$fit2, then a is the second garchFit object of formal class
'fGarch' with 11 slots, @call, @formula... etc.
I then want to extract information in the 11 slots using another for loop,
say i want to access the slot a...@fit$coef. In order for this to happen I
need
to combine modelStack$fit2, where the $fit2 object name must be constructed
in the loop, with the slot name @fit$coef.
In order to construct the proper list names one could use the following in
the for loop;
name = paste(modelStack$fit,i,@fit$coef,sep=) #
modelstack$f...@fit$coef for i = 2,
and then use something like
get(name)
But this returns; Error in get(name) : object 'modelstack$f...@fit$coef' not
found
If I just type modelstack$f...@fit$coef in the command window it returns the
proper info from the object list. I figure this is because R somehow
interpret $ and @ in get() differently than $ and @ as list
separators.
Does anyone know how to extract information in slots of listed objects using
a loop and on the run generated variable/object names?
JT
sample code
nAssets = length(modelStack)
for(i in 1:nAssets){
name = paste(modelStack[,i,]...@name$series$h,sep=)
a = get(name)
t = length(a)
} # end for loop
--
View this message in context:
http://n4.nabble.com/accessing-info-in-object-slots-from-listed-objects-using-loops-tp1596135p1596135.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Constrained non linear regression using ML
Dear Gabor, 1) The constraints are active, at least from a formal point view. 3) I have tried several times to run betareg.fit on the data, and the only thing I can obtain is the very strange error: Error in dimnames(x) - dn : length of 'dimnames' [2] not equal to array extent The error is strange because, because the function dimnames is not called anywhere. Regards Gabor Grothendieck wrote: Try it anyways -- maybe none of your constraints are active. On Wed, Mar 17, 2010 at 6:01 AM, Corrado ct...@york.ac.uk wrote: Dear Gabor, dear R users, I had already read the betareg documentation. As far as I can understand from the help, it does not allow for constrained regression. Regards Gabor Grothendieck wrote: Check out the betareg package. On Tue, Mar 16, 2010 at 2:58 PM, Corrado ct...@york.ac.uk wrote: Dear R users, I have to fit the non linear regression: y~1-exp(-(k0+k1*p1+k2*p2+ +kn*pn)) where ki=0 for each i in [1 n] and pi are on R+. I am using, at the moment, nls, but I would rather use a Maximum Likelhood based algorithm. The error is not necessarily normally distributed. y is approximately beta distributed, and the volume of data is medium to large (the y,pi may have ~ 40,000 elements). I have studied the packages in the task views Optimisation and Robust Statistical Methods, but I did look like what I was looking for was there. Maybe I am wrong. The nearest thing was nlrob, but even that does not allow for constraints, as far as I can understand. Any suggestion? Regards -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Summing daily data for each year
Try this. We read in the data using read.zoo and then aggregate it. read.zoo assumes Date class unless you indicate otherwise. Then we compute the year and aggregate over that. as.numeric(format(time(z), %Y)) would be an alternate way to compute the year. Lines - 1995-04-01 0.002766309 1995-04-02 0.002402973 1995-04-03 0.002254335 1995-04-04 0.002221305 1995-04-05 0.002180017 1995-04-06 0.002031379 1995-04-07 0.001957060 1995-04-08 0.001940545 1995-04-09 0.001924030 1995-04-10 0.001783650 library(zoo) # z - read.zoo(myfile.dat) z - read.zoo(textConnection(Lines)) aggregate(z, as.numeric(floor(as.yearmon(time(z, sum) The result of the last statement is (since our test data above only has one year): 1995 0.02146160 See ?read.zoo and ?aggregate.zoo in the zoo package and also the three vignettes (pdf documents) that come with the package. On Wed, Mar 17, 2010 at 2:20 AM, santosh.ar...@csiro.au wrote: Hello there I have a DAILY data set (as shown below) with Date and Streamflow from April 1995 to Aug 2006. I want a yearly sum the streamflow value (for each year - first and last incomplete year). Does anyone know how to do this in R? I have tried everything I could think of, e.g. imported the streamflow data as time series using: flowdata - ts(ZVo[,2],start = c(1995,4,1), frequency = 7) # perhaps frequency = 365 is the better one and tried to aggregate using aggregate() without success ZVo V1 V2 1 1995-04-01 0.002766309 2 1995-04-02 0.002402973 3 1995-04-03 0.002254335 4 1995-04-04 0.002221305 5 1995-04-05 0.002180017 6 1995-04-06 0.002031379 7 1995-04-07 0.001957060 8 1995-04-08 0.001940545 9 1995-04-09 0.001924030 10 1995-04-10 0.001783650 . . . 4156 2006-08-16 0.02861272 4157 2006-08-17 0.03652353 4158 2006-08-18 0.05372419 4159 2006-08-19 0.05630058 4160 2006-08-20 0.06274154 4161 2006-08-21 0.06981833 Many thanks Santosh __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Constrained non linear regression using ML
Contact the maintainer regarding problems with the package. Not sure if this is acceptable but if you get it to run you could consider just dropping the variables from your model that correspond to active constraints. Also try the maxLik package. You will have to define the likelihood yourself but it does support constraints. On Wed, Mar 17, 2010 at 9:07 AM, Corrado ct...@york.ac.uk wrote: Dear Gabor, 1) The constraints are active, at least from a formal point view. 3) I have tried several times to run betareg.fit on the data, and the only thing I can obtain is the very strange error: Error in dimnames(x) - dn : length of 'dimnames' [2] not equal to array extent The error is strange because, because the function dimnames is not called anywhere. Regards Gabor Grothendieck wrote: Try it anyways -- maybe none of your constraints are active. On Wed, Mar 17, 2010 at 6:01 AM, Corrado ct...@york.ac.uk wrote: Dear Gabor, dear R users, I had already read the betareg documentation. As far as I can understand from the help, it does not allow for constrained regression. Regards Gabor Grothendieck wrote: Check out the betareg package. On Tue, Mar 16, 2010 at 2:58 PM, Corrado ct...@york.ac.uk wrote: Dear R users, I have to fit the non linear regression: y~1-exp(-(k0+k1*p1+k2*p2+ +kn*pn)) where ki=0 for each i in [1 n] and pi are on R+. I am using, at the moment, nls, but I would rather use a Maximum Likelhood based algorithm. The error is not necessarily normally distributed. y is approximately beta distributed, and the volume of data is medium to large (the y,pi may have ~ 40,000 elements). I have studied the packages in the task views Optimisation and Robust Statistical Methods, but I did look like what I was looking for was there. Maybe I am wrong. The nearest thing was nlrob, but even that does not allow for constraints, as far as I can understand. Any suggestion? Regards -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How can I save the result for goodness of fit test
pinusan wrote: Dear All, I run the goodness of fit test using goodfit() in vcd package. The result is as follow: Goodness-of-fit test for poisson distribution X^2 df P( X^2) Pearson 1.053348 2 0.5905661 Warning message: In summary.goodfit(gf) : Chi-squared approximation may be incorrect I want to save the the test statistics(X^2), df, and p-value. How can I save the result. Actually, I want to make a table. The usual general answer to this question is to read the documentation, and look for the Value section. ?goodfit gives Value A list of class goodfit with elements: observed observed frequencies. countcorresponding counts. fitted expected frequencies (fitted by ML). type a character string indicating the distribution fitted. method a character string indicating the fitting method (can be either ML, MinChisq or fixed if the parameters were specified). df degrees of freedom. par a named list of the (estimated) distribution parameters. So, the quantities you want are not returned in the resulting object. But, they *are* returned from the summary method that you used to print the result. [That isn't documented, so perhaps that could be added.] So, the 2nd general answer to your question is to use str() on the result of the summary() function. ## Simulated data examples: dummy - rnbinom(200, size = 1.5, prob = 0.8) gf - goodfit(dummy, type = nbinomial, method = MinChisq) summary(gf) Goodness-of-fit test for nbinomial distribution X^2 df P( X^2) Pearson 0.5865029 1 0.4437746 Warning message: In summary.goodfit(gf) : Chi-squared approximation may be incorrect result - summary(gf) Goodness-of-fit test for nbinomial distribution X^2 df P( X^2) Pearson 0.5865029 1 0.4437746 Warning message: In summary.goodfit(gf) : Chi-squared approximation may be incorrect str(result) num [1, 1:3] 0.587 1 0.444 - attr(*, dimnames)=List of 2 ..$ : chr Pearson ..$ : chr [1:3] X^2 df P( X^2) result X^2 df P( X^2) Pearson 0.5865029 1 0.4437746 So, there you go! In addition, there is warning message In summary.goodfit(gf) : Chi-squared approximation may be incorrect. How can I interpret this result. Read the Details section Have a nice day. -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Streethttp://www.math.yorku.ca/SCS/friendly.html Toronto, ONT M3J 1P3 CANADA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Missing index in vector assignment
Hi r-help-boun...@r-project.org napsal dne 17.03.2010 13:04:05: Jim Petr, Thank you for your hint - I am really grateful, because they helped me to get one step further, and although now the problem lies somewhere else, you encouraged that we can find the solution soon! 1. To Petr's comments Petr, your hint to define y: y - LVvar[,1, drop=FALSE] did solve the problem, so I got a data.frame with the indexes. Going through help page y - as.matrix(LVvar[,1, drop=FALSE]) x - as.matrix(LVvar[,-1] svp - ksvm(x, y, type=nu-svc) shall work. However as I never used this package and function I am not sure if it is used correctly. Regards Petr Yet, then I turned to the call svp - ksvm(x, y, type=nu-svc) Error in .local(x, ...) : y must be a vector or a factor. So then I followed your second advice, looking up the additional information from help file: == x is defined as: a symbolic description of the model to be fit. When not using a formula x can be a matrix or vector containing the training data or a kernel matrix of class kernelMatrix of the training data or a list of character vectors (for use with the string kernel). Note, that the intercept is always excluded, whether given in the formula or not. y is defined as a response vector with one label for each row/component of x. Can be either a factor (for classification tasks) or a numeric vector (for regression). == So I tried to convert LVvar into a matrix via as.matrix() but didn't make a difference. 2. To Jim's comments On Wed, Mar 17, 2010 at 1:10 AM, jim holtman jholt...@gmail.com wrote: Please provide what LVvar is. LVvar is a dataframe At least provide str(LVvar), or preferably a 'dput' of the object. str(LVvar) returns: 'data.frame': 55 obs. of 7 variables: $ rPerform : num 0.0682 -0.0682 -0.7443 0.7443 0.2619 ... $ rCoordCap: num 4.98 6.08 5.73 5.92 4.96 ... $ rKnowGrow: num 4.5 5.92 5.23 6.08 4.38 ... $ rGoalcom : num 5.81 6.58 6 5.75 5.29 ... $ rSupport : num 6.15 6.92 6.6 4.92 6 ... $ rOpcomm : num 5.98 6.25 6.33 6.5 5.29 ... $ rT2Cadap : num 5.03 6.12 4.9 6.25 5.12 ... - attr(*, na.action)=Class 'omit' Named int 40 .. ..- attr(*, names)= chr 40 == dput(LVvar) returns (abbreviated with ...): structure(list(rPerform = c(0.0681818181818183, -0.0681818181818183, -0.744318181818182, 0.744318181818182, 0.261931818181818, -0.900568181818182, ... rCoordCap = c(4.979167, 6.08, 5.73, 5.916667, 4.958333, ... .Names = c(rPerform, rCoordCap, rKnowGrow, rGoalcom, rSupport, rOpcomm, rT2Cadap), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, 36L, 37L, 38L, 39L, 41L, 42L, 43L, 44L, 45L, 46L, 47L, 48L, 49L, 50L, 51L, 52L, 53L, 54L, 55L, 56L), na.action = structure(40L, .Names = 40, class = omit), class = data.frame) PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/ posting-guide.html and provide commented, minimal, self-contained, reproducible code. Here's my try (please have mercy for a complete R beginner): library(kernlab) library(methods) # Data Definitions LV - c(rPerform,rCoordCap, rKnowGrow, rGoalcom, rSupport, rOpcomm, rT2Cadap ) # creates a dataframe LVvar - na.omit(loopLV_IndexScores(LV, u_proj)) x - (LVvar[,-1]) y - (LVvar[,1]) svp - ksvm(x, y, type=nu-svc) svp === Thanks so much for not giving up. Cheers, Chaehan On Tue, Mar 16, 2010 at 6:03 PM, Chaehan So chaehan...@gmail.com wrote: Dear r-helpers, I am getting a mismatch error between two variables: svp - ksvm(x, y, type=nu-svc) Error in .local(x, ...) : x and y don't match. and I suspect that it might be due to missing index in the y variable which I defined as: y - (LVvar[,1]) I tried various methods to make the y assignment in the same format as x, which is a dataframe x - (LVvar[,-1]) and looks like x rCoordCap rKnowGrow rGoalcom rSupport rOpcomm rT2Cadap 1 4.979167 4.50 5.812500 6.145833 5.979167 5.031250 ... but I still get y without the indexes as a vector: y [1] -1. -6.91193182 -1. 0.74431818 -6.91193182 Why are the results different for x and y, even though the assignment is the same except I exclude the columns for y? Cheers, Chaehan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/ posting-guide.html and provide commented, minimal, self-contained,
Re: [R] Constrained non linear regression using ML
On 17 March 2010 14:22, Gabor Grothendieck ggrothendi...@gmail.com wrote: Contact the maintainer regarding problems with the package. Not sure if this is acceptable but if you get it to run you could consider just dropping the variables from your model that correspond to active constraints. Also try the maxLik package. You will have to define the likelihood yourself but it does support constraints. Yes. And specifying the likelihood function is probably (depending on your distributional assumptions) not too complicated. BTW: Even if your y follows a beta distribution, it does not mean that your error term also follows a beta distribution. And it the distribution of the error term which is crucial for specifying the likelihood function. /Arne On Wed, Mar 17, 2010 at 9:07 AM, Corrado ct...@york.ac.uk wrote: Dear Gabor, 1) The constraints are active, at least from a formal point view. 3) I have tried several times to run betareg.fit on the data, and the only thing I can obtain is the very strange error: Error in dimnames(x) - dn : length of 'dimnames' [2] not equal to array extent The error is strange because, because the function dimnames is not called anywhere. Regards Gabor Grothendieck wrote: Try it anyways -- maybe none of your constraints are active. On Wed, Mar 17, 2010 at 6:01 AM, Corrado ct...@york.ac.uk wrote: Dear Gabor, dear R users, I had already read the betareg documentation. As far as I can understand from the help, it does not allow for constrained regression. Regards Gabor Grothendieck wrote: Check out the betareg package. On Tue, Mar 16, 2010 at 2:58 PM, Corrado ct...@york.ac.uk wrote: Dear R users, I have to fit the non linear regression: y~1-exp(-(k0+k1*p1+k2*p2+ +kn*pn)) where ki=0 for each i in [1 n] and pi are on R+. I am using, at the moment, nls, but I would rather use a Maximum Likelhood based algorithm. The error is not necessarily normally distributed. y is approximately beta distributed, and the volume of data is medium to large (the y,pi may have ~ 40,000 elements). I have studied the packages in the task views Optimisation and Robust Statistical Methods, but I did look like what I was looking for was there. Maybe I am wrong. The nearest thing was nlrob, but even that does not allow for constraints, as far as I can understand. Any suggestion? Regards -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Arne Henningsen http://www.arne-henningsen.name __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plm within models: is the correct F-statistic reported?
Dear Achim On 3/16/10, Achim Zeileis achim.zeil...@uibk.ac.at wrote: Hence, when saying summary() different models with no effects are assumed. For gr_fe the model without effects just omits value/capital but keeps the firm-specific interecepts. For gr_lm not even the intercept is kept in the model. Thus: gr_fe_null - lm(invest ~ 0 + firm, data = pgr) gr_lm_null - lm(invest ~ 0, data = pgr) What would be the more useful no effects model in the plm(..., effect=twoways) case? Considering the same setting, library(AER) data(Grunfeld, package = AER) library(plm) gr - subset(Grunfeld, firm %in% c(General Electric, General Motors, IBM)) pgr - plm.data(gr, index = c(firm, year)) I am fitting a twoways model and an individual with manually specified time effects. gr_fe1 - plm(invest ~ value + capital, data = pgr, +model = within, effect=twoways) summary(gr_fe1) Twoways effects Within Model Call: plm(formula = invest ~ value + capital, data = pgr, effect = twoways, model = within) Balanced Panel: n=3, T=20, N=60 Residuals : Min. 1st Qu. Median 3rd Qu.Max. -153.00 -29.102.23 34.80 125.00 Coefficients : Estimate Std. Error t-value Pr(|t|) value 0.1295 0.02245.77 1.4e-06 *** capital 0.4184 0.0353 11.85 5.5e-14 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Total Sum of Squares:957000 Residual Sum of Squares: 138000 F-statistic: 107.246 on 2 and 36 DF, p-value: 6.84e-16 gr_fe2 - plm(invest ~ value + capital + year, data = pgr, +model = within, effect=individual) summary(gr_fe2) Oneway (individual) effect Within Model Call: plm(formula = invest ~ value + capital + year, data = pgr, effect = individual, model = within) Balanced Panel: n=3, T=20, N=60 Residuals : Min. 1st Qu. Median 3rd Qu.Max. -153.00 -29.102.23 34.80 125.00 Coefficients : Estimate Std. Error t-value Pr(|t|) value 0.1295 0.02245.77 1.4e-06 *** capital 0.4184 0.0353 11.85 5.5e-14 *** year1936 -83.962553.6143 -1.57 0.1261 year1937 -150.920658.3282 -2.59 0.0139 * year1938 -81.234350.7175 -1.60 0.1180 year1939 -137.457953.4385 -2.57 0.0144 * year1940 -96.358453.9837 -1.78 0.0827 . year1941 -56.558753.0089 -1.07 0.2931 year1942 -36.653950.9966 -0.72 0.4769 year1943 -78.079452.0249 -1.50 0.1421 year1944 -66.472552.5047 -1.27 0.2136 year1945 -89.556254.2876 -1.65 0.1077 year1946 -59.114755.3115 -1.07 0.2923 year1947 -87.544452.6530 -1.66 0.1051 year1948 -119.912553.3167 -2.25 0.0307 * year1949 -167.955254.1999 -3.10 0.0038 ** year1950 -172.767655.0212 -3.14 0.0034 ** year1951 -191.136957.6114 -3.32 0.0021 ** year1952 -195.450359.6377 -3.28 0.0023 ** year1953 -174.663966.3451 -2.63 0.0124 * year1954 -181.127368.5794 -2.64 0.0121 * --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Total Sum of Squares:189 Residual Sum of Squares: 138000 F-statistic: 21.8327 on 21 and 36 DF, p-value: 1.32e-14 Following the reasoning in your previous e-mail, I assume that the (more useful) no effects model used in the twoways case is gr_fe1_null - lm(invest ~ 0 + firm + year, data = pgr) However I cannot replicate the F-statistic: 107.246. anova(gr_fe1_null, gr_fe1) Analysis of Variance Table Response: invest Df Sum Sq Mean Sq F value Pr(F) firm 3 7664439 2554813 101.46 2e-16 *** year 19 932060 490561.95 0.040 * Residuals 38 956886 25181 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Warning message: In anova.lmlist(object, ...) : models with response NULL removed because response differs from model 1 anova(gr_fe1_null, gr_fe2) Analysis of Variance Table Response: invest Df Sum Sq Mean Sq F value Pr(F) firm 3 7664439 2554813 101.46 2e-16 *** year 19 932060 490561.95 0.040 * Residuals 38 956886 25181 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Warning message: In anova.lmlist(object, ...) : models with response NULL removed because response differs from model 1 In the case of individual with manually introduced time effects, I assume the following null is used: gr_fe2_null - lm(invest ~ 0 + firm, data = pgr) But even here I cannot replicate the F-statistic: 21.8327. anova(gr_fe2_null, gr_fe2) Analysis of Variance Table Response: invest Df Sum Sq Mean Sq F value Pr(F) firm 3 7664439 255481377.1 2e-16 *** Residuals 57 1888946 33139 --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Warning message: In anova.lmlist(object, ...) : models with response NULL removed because response differs from model 1 Am I doing something wrong? Liviu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help
Re: [R] Is there any R package that can find the maxima of a 1-D
On 17-Mar-10 12:31:43, mau...@alice.it wrote: Is there any R package that can help me with digging out the maxima of a 1-D trajectory ? I have 975 1-D curves. They are only known as time series. That is a set of points ordered with respect to time. Some curves exhibit one only peak. We wish to find the number of peaks and their position along the time axis. Apparently it's a trivial problem solved looking at the zeros and the change of sign of the 1st derivative. In practice it is necessary to apply some criteria (which ones?) to discriminate between real peaks and noise oscillations. Presumably I ought to define the noise level with respect to peaks height in this application ... Maybe wavelets can help ? Thank you in advance, Maura Precisely. You need to define what you wish peak to mean. Then you can implement your wish in code. The most inclusive definition: (X[n] = X[n-1])(X[n] = X[n+1]) will of course catch everything, including noisy fluctuations (and as a result may hide real (underlying) peaks hidden by the noise). You might extend the above: (X[n]=X[n-1])(X[n-1]=X[n-2])(X[n]=X[n+1])(X[n+1]=X[n+2]) or you might apply a smoother (possibly wavelets) to reduce the noise and then find the peaks of that. And so on ... Apparently you already have some notion of what you want peak to mean, since you say Others have two peaks of different height, and you also recognise an effect of noise. But the possibilities are endless! Sir Hector Munro's classic Tables of the 3000-feet Mountains of Scotland (first published 1891) did not give a formal definition owing to the impossibility of deciding what should be considered separate mountains. On the other hand, J. Rooke Corbett's later Scottish Mountains 2500 Feet And Under 3000 Feet In Height With Re-Ascent Of 500 Feet On All Sides did use the re-ascent definition given in the title: it is a separate mountain if you have to climb at least 500 feet from any other peak to reach its summit. However, a single mountain may have more than one peak. For example, the mountain of Lochnagar (overlooking the Balmoral Estate and the theme of a novel by Prince Charles) is held to have two separate peaks, marked on the Ordnance Survey Map as Cac Carn Beag and Cac Carn Mor (don't ask ... ), at 3789 feet and 3768 feet respectively, separated by a ridge of about 1/4 mile which dips by about 100 feet. They can be seen at the right-hand end of the photo of Lochnagar shown in http://en.wikipedia.org/wiki/Lochnagar (to the right of the notch just right of centre) On the other hand, look at the photo of part of Cairngorm mountain at http://www.scotclimb.org.uk/gallery.php?id=200 and ask: Is there a peak here, or is it all noise? However, here: http://www.scotclimb.org.uk/gallery.php?id=126 you can rather clearly distinguish between peak and noise! So we can (more or less) make the distinction when we look. But how to define this sort of thing so that R can understand? Up to you! Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 17-Mar-10 Time: 14:02:13 -- XFMail -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Constrained non linear regression using ML
Dear Arne, Gabor, I solved the problem with betareg (downloaded the package). I run it on my data, and unfortunately the constraint is definitively active, if I remove the active variables, I then remove the most significant variables! Of course the error is important, not the distribution of the variable. In this case, one of the assumptions is that the error may be distributed ~ beta. I think that betareg makes this assumption, am I right? I am finding it difficult to solve two problems: 1) write the maximum likelihood function (what do you suggest?) 2) deal with the fact that a few factors actually have values of y (the response) at the extremes: that is 0 and 1. But that mean that the link function returns Infinite values in that case 3) the error is dependent on E(y). PS: Additional silly question: what is the discrete equivalent of beta? binomial? Arne Henningsen wrote: On 17 March 2010 14:22, Gabor Grothendieck ggrothendi...@gmail.com wrote: Contact the maintainer regarding problems with the package. Not sure if this is acceptable but if you get it to run you could consider just dropping the variables from your model that correspond to active constraints. Also try the maxLik package. You will have to define the likelihood yourself but it does support constraints. Yes. And specifying the likelihood function is probably (depending on your distributional assumptions) not too complicated. BTW: Even if your y follows a beta distribution, it does not mean that your error term also follows a beta distribution. And it the distribution of the error term which is crucial for specifying the likelihood function. /Arne -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] define F-ratio computations with aov
Please look at the Error term in ?aov On Wed, Mar 17, 2010 at 4:25 AM, Galanidis Alexandros a...@env.aegean.grwrote: Greetings to all, This is my model: aov.fit-aov(Y~A+B+C+D+E+A:C+A:E) In summary(aov.fit) all F values are comptuted by eg MS(A)/MS(Residuals). This is not correct (or what I want), except for F(B) and F(A:E). I suppose P values are not correct either. Is it possible with aov to define the way F computations will be done? I 'd like them to be like this: F(A)=MS(A)/MS(E), F(C)=MS(C)/MS(E), F(D)=MS(D)/MS(E), F(E)=MS(E)/MS(A:E), F(A:C)=MS(A:C)/MS(A:E) For your example, it looks like mydata.aov - aov(Y ~ A+B+C+D + Error(E/A), data=mydata) might be what you need. It isn't possible to be sure of the correct statement for your example without seeing the actual treatment plan. The standard split-plot design is specified with splplot - data.frame(y=rnorm(72), blocks=factor(rep(1:6, each=12)), plots=factor(rep(rep(1:3, each=4), 6)), subplots=factor(rep(1:4, 18)), A=factor(rep(c(3,1,2, 3,1,2, 2,3,1, 3,2,1, 2,1,3, 1,2,3), each=4)), B=factor( c(4, 3, 2, 1, 1, 2, 4, 3, 1, 2, 3, 4, 3, 1, 2, 4, 4, 1, 2, 3, 2, 1, 3, 4, 2, 3, 4, 1, 4, 2, 3, 1, 1, 4, 2, 3, 3, 4, 1, 2, 1, 3, 4, 2, 2, 3, 4, 1, 4, 1, 3, 2, 3, 4, 1, 2, 3, 4, 2, 1, 3, 1, 4, 2, 4, 3, 1, 2, 1, 2, 3, 4))) splplot.aov - aov(y ~ A*B + Error(blocks/plots/subplots), data=splplot) summary(splplot.aov) summary(splplot.aov) Error: blocks Df Sum Sq Mean Sq F value Pr(F) Residuals 5 2.0231 0.40463 Error: blocks:plots Df Sum Sq Mean Sq F value Pr(F) A 2 0.2876 0.1438 0.1135 0.8938 Residuals 10 12.6646 1.2665 Error: blocks:plots:subplots Df Sum Sq Mean Sq F value Pr(F) B 3 2.814 0.93797 0.8987 0.4493 A:B6 3.358 0.55964 0.5362 0.7778 Residuals 45 46.965 1.04367 As you see, the correct F tests are automatically determined and displayed. Rich [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Adding minutes to 24 hour time
Hi r-help-boun...@r-project.org napsal dne 17.03.2010 14:57:32: Hi, Does anyone know how to add minutes (up to 100 min) to a 24 hour time, to create a new 24 hour time? I can't seem to find any documentation or examples explaining how to do this. The variables of interest are 'ARRIVE','WAIT', and 'DEPART' in the attached partial dataframe. I want 'DEPART' to be the sum of 'ARRIVE' and 'WAIT' in 24 hour format. Also, can anyone direct me to some relevant documentation? You can make a date/time object from YEAR, WEEK, ARRIVE by mydate - strptime(paste()) see ?strptime then you can add ARRIVE just by mydate+ARRIVE*60 (You did not say explicitly what values are ARRIVE so assume minutes) You shall probably also consult zoo library, especially na.locf function. Regards Petr Thank you, Mike [příloha Dataframe.pdf odstraněna uživatelem Petr PIKAL/CTCAP] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Adding minutes to 24 hour time
On Wed, Mar 17, 2010 at 2:57 PM, Hosack, Michael mhos...@state.pa.us wrote:
Hi,
Does anyone know how to add minutes (up to 100 min) to a 24 hour time, to
create a new 24 hour time? I can't seem to find any documentation or examples
explaining how to do this. The variables of interest are 'ARRIVE','WAIT', and
'DEPART' in the attached partial dataframe. I want 'DEPART' to be the sum
of 'ARRIVE' and 'WAIT' in 24 hour format. Also, can anyone direct me to some
relevant documentation?
Thank you,
Mike
If you convert all data to a date-and-time ?POSIXlt object, you can
just convert the minutes to seconds and add together with +.
Another way would be the something like this:
addTime-function(timeTxt,mins){
start.time-strsplit(timeTxt,:)
start.time-do.call(rbind,start.time)
storage.mode(start.time)-numeric
hours-mins%/%60
mins.left-mins%%60
end.mins-(start.time[,2]+mins.left)%%60
end.hours-(start.time[,1]+hours+(start.time[,2]+mins.left)%/%60)%%24
end.time-paste(end.hours,end.mins,sep=:)
return(end.time)
}
addTime(c(15:23,7:00),c(70,100))
or this:
addTime2-function(timeTxt,mins){
orig.date-as.POSIXct(paste(2001-01-01,timeTxt))
new.Date-orig.date+mins*60
new.Date-strsplit(as.character(new.Date), )
new.Time-(sapply(new.Date,[,2))
return(new.Time)
}
addTime2(c(15:23,7:00),c(70,100))
Regards,
Gustaf
--
Gustaf Rydevik, M.Sci.
tel: +46(0)703 051 451
address:Essingetorget 40,112 66 Stockholm, SE
skype:gustaf_rydevik
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Missing index in vector assignment
Petr - your suggestion WORKS! Thank you so much, really! happy-Chaehan On Wed, Mar 17, 2010 at 2:22 PM, Petr PIKAL petr.pi...@precheza.cz wrote: Hi r-help-boun...@r-project.org napsal dne 17.03.2010 13:04:05: Jim Petr, Thank you for your hint - I am really grateful, because they helped me to get one step further, and although now the problem lies somewhere else, you encouraged that we can find the solution soon! 1. To Petr's comments Petr, your hint to define y: y - LVvar[,1, drop=FALSE] did solve the problem, so I got a data.frame with the indexes. Going through help page y - as.matrix(LVvar[,1, drop=FALSE]) x - as.matrix(LVvar[,-1] svp - ksvm(x, y, type=nu-svc) shall work. However as I never used this package and function I am not sure if it is used correctly. Regards Petr Yet, then I turned to the call svp - ksvm(x, y, type=nu-svc) Error in .local(x, ...) : y must be a vector or a factor. So then I followed your second advice, looking up the additional information from help file: == x is defined as: a symbolic description of the model to be fit. When not using a formula x can be a matrix or vector containing the training data or a kernel matrix of class kernelMatrix of the training data or a list of character vectors (for use with the string kernel). Note, that the intercept is always excluded, whether given in the formula or not. y is defined as a response vector with one label for each row/component of x. Can be either a factor (for classification tasks) or a numeric vector (for regression). == So I tried to convert LVvar into a matrix via as.matrix() but didn't make a difference. 2. To Jim's comments On Wed, Mar 17, 2010 at 1:10 AM, jim holtman jholt...@gmail.com wrote: Please provide what LVvar is. LVvar is a dataframe At least provide str(LVvar), or preferably a 'dput' of the object. str(LVvar) returns: 'data.frame': 55 obs. of 7 variables: $ rPerform : num 0.0682 -0.0682 -0.7443 0.7443 0.2619 ... $ rCoordCap: num 4.98 6.08 5.73 5.92 4.96 ... $ rKnowGrow: num 4.5 5.92 5.23 6.08 4.38 ... $ rGoalcom : num 5.81 6.58 6 5.75 5.29 ... $ rSupport : num 6.15 6.92 6.6 4.92 6 ... $ rOpcomm : num 5.98 6.25 6.33 6.5 5.29 ... $ rT2Cadap : num 5.03 6.12 4.9 6.25 5.12 ... - attr(*, na.action)=Class 'omit' Named int 40 .. ..- attr(*, names)= chr 40 == dput(LVvar) returns (abbreviated with ...): structure(list(rPerform = c(0.0681818181818183, -0.0681818181818183, -0.744318181818182, 0.744318181818182, 0.261931818181818, -0.900568181818182, ... rCoordCap = c(4.979167, 6.08, 5.73, 5.916667, 4.958333, ... .Names = c(rPerform, rCoordCap, rKnowGrow, rGoalcom, rSupport, rOpcomm, rT2Cadap), row.names = c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 20L, 21L, 22L, 23L, 24L, 25L, 26L, 27L, 28L, 29L, 30L, 31L, 32L, 33L, 34L, 35L, 36L, 37L, 38L, 39L, 41L, 42L, 43L, 44L, 45L, 46L, 47L, 48L, 49L, 50L, 51L, 52L, 53L, 54L, 55L, 56L), na.action = structure(40L, .Names = 40, class = omit), class = data.frame) PLEASE do read the posting guide http://www.R-project.org/posting-guide.htmlhttp://www.r-project.org/ posting-guide.html and provide commented, minimal, self-contained, reproducible code. Here's my try (please have mercy for a complete R beginner): library(kernlab) library(methods) # Data Definitions LV - c(rPerform,rCoordCap, rKnowGrow, rGoalcom, rSupport, rOpcomm, rT2Cadap ) # creates a dataframe LVvar - na.omit(loopLV_IndexScores(LV, u_proj)) x - (LVvar[,-1]) y - (LVvar[,1]) svp - ksvm(x, y, type=nu-svc) svp === Thanks so much for not giving up. Cheers, Chaehan On Tue, Mar 16, 2010 at 6:03 PM, Chaehan So chaehan...@gmail.com wrote: Dear r-helpers, I am getting a mismatch error between two variables: svp - ksvm(x, y, type=nu-svc) Error in .local(x, ...) : x and y don't match. and I suspect that it might be due to missing index in the y variable which I defined as: y - (LVvar[,1]) I tried various methods to make the y assignment in the same format as x, which is a dataframe x - (LVvar[,-1]) and looks like x rCoordCap rKnowGrow rGoalcom rSupport rOpcomm rT2Cadap 1 4.979167 4.50 5.812500 6.145833 5.979167 5.031250 ... but I still get y without the indexes as a vector: y [1] -1. -6.91193182 -1. 0.74431818 -6.91193182 Why are the results different for x and y, even though the assignment is the same except I exclude the columns for y? Cheers, Chaehan [[alternative HTML version deleted]]
Re: [R] Constrained non linear regression using ML
For specific questions on the betareg package contact the maintainer. If the likelihood based approaches are giving too much difficulty try moving to a Bayesian framework (WinBUGS/R2WinBUGS, JAGS/r2jags, etc.) On Wed, Mar 17, 2010 at 10:03 AM, Corrado ct...@york.ac.uk wrote: Dear Arne, Gabor, I solved the problem with betareg (downloaded the package). I run it on my data, and unfortunately the constraint is definitively active, if I remove the active variables, I then remove the most significant variables! Of course the error is important, not the distribution of the variable. In this case, one of the assumptions is that the error may be distributed ~ beta. I think that betareg makes this assumption, am I right? I am finding it difficult to solve two problems: 1) write the maximum likelihood function (what do you suggest?) 2) deal with the fact that a few factors actually have values of y (the response) at the extremes: that is 0 and 1. But that mean that the link function returns Infinite values in that case 3) the error is dependent on E(y). PS: Additional silly question: what is the discrete equivalent of beta? binomial? Arne Henningsen wrote: On 17 March 2010 14:22, Gabor Grothendieck ggrothendi...@gmail.com wrote: Contact the maintainer regarding problems with the package. Not sure if this is acceptable but if you get it to run you could consider just dropping the variables from your model that correspond to active constraints. Also try the maxLik package. You will have to define the likelihood yourself but it does support constraints. Yes. And specifying the likelihood function is probably (depending on your distributional assumptions) not too complicated. BTW: Even if your y follows a beta distribution, it does not mean that your error term also follows a beta distribution. And it the distribution of the error term which is crucial for specifying the likelihood function. /Arne -- Corrado Topi PhD Researcher Global Climate Change and Biodiversity Area 18,Department of Biology University of York, York, YO10 5YW, UK Phone: + 44 (0) 1904 328645, E-mail: ct...@york.ac.uk __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] haarfisz
Dear all, The package haarfisz, for computing Haar-Fisz variance stabilization transforms for Poisson-like data has just been submitted to CRAN. This was previously bundled with wavethresh, but has now been unbundled. Best wishes, Guy Nason ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] odfWeave Error
Thanks for your help! Here is what I tried with no luck. I use 7-zip on my machine. Just in case it helps, I am running XP Pro 32 at the office. odfctrl - odfWeaveControl(zipCmd = c(C:/Program Files/7-Zip/7zG.exe -r $$file$$ ., C:/Program Files/7-Zip/7zG.exe -o $$file$$)) odfWeave(example01_in.odt, example01.odt, control = odfctrl) I even tried to use 7z.exe, as well as 7zFM.exe. Essentially, I attempted to use all 3 exectuable files that are located in the install directory. Any ideas? Thanks again! -- View this message in context: http://n4.nabble.com/odfWeave-Error-tp1595848p1596418.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Converting factors to numeric in a dataframe
I am currently trying to write a program that minimises the amount of work required for auditable qPCR data. At the moment I am using an Excel (.csv) spreadsheet as source data that has been transposed to the column format required for R to read. Unfortunately, this means I have* *to manually confirm the whole data set prior to doing any analysis, which is taking a considerable amount of time! My idea now is to read the raw data in directly and get R to do the transformation prior to analysis. The problem I now have is that, upon transposition, the data are converted to character in a matrix, rather than factor and numeric in a dataframe. I have succeeded in changing the matrix to a dataframe (via as.data.frame(object)), but this then converts all the data to factor which I cant use for my analysis since, other than the column headings, I need the data to be numeric. I have tried coercing the data to numeric using the as() and as.numeric() commands, but this has no effect on the data format. I have no experience in programming and so am at a loss as to what to do: am I making a basic error in my programming or missing something essential (or both!)? I am using R version 2.9.0 at the moment, but this will change as soon as I have sorted this issue out. Below is the code I have put together, as you can see it is VERY brief but essential to allow my analysis to proceed: pcrdata-read.csv(File_path,header=FALSE) pcrdata-as.data.frame(t(pcrdata)) pcrdata[2:51]-as.numeric(as.character(pcrdata)) Any help would be gratefully appreciated, Mike Glanville [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] waveband
Dear all, The package waveband, for computing wavelet shrinkage credible intervals, has been uploaded to CRAN. This was previously bundled with wavethresh, but has now been unbundled. Best wishes, Guy Nason -- Professor Guy Nason Head of Department Department of Mathematics University of Bristol http://www.stats.bris.ac.uk/~magpn PA: Helen Craven helen.cra...@bristol.ac.uk, +44 (0) 117 928 7978 [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mmap error on macbook pro
Dear List, I am trying to read some files using read.csv and total size of those files are 3.99 GB. I am using MacBook Pro with 4GB RAM(snow leopard). I also tried to run a chunk from those files and altogether the size was 1.33 GB. But every time I was getting the following error R(1200) malloc: *** mmap(size=16777216) failed (error code=12) *** error: can't allocate region *** set a breakpoint in malloc_error_break to debug I managed to run smaller chunks (400MB) and I also saved the object. But problem is when I try to load some of those objects(which is not more than 1.5 GB altogether), I get same error again. Can someone please help? why am I getting this error? does R need more space than the actual file size? I may buy new machine if its something related to RAM size. but if it is some R problem then it will be very useless to buy new machine. So, I really need to know why this is happening. Any help is appreciated. Thanks in advance. regards, Shyamasree __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] wavethresh
Dear all, A new version of wavethresh has just been submitted to CRAN. This is version 4.5 All best wishes, Guy Nason ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How good is R at making publication quality tables?
Hello Everyone, I have just started learning R and am in the process of figuring out what it can and can't do. I must say I am very impressed with R so far and am amazed that something this good can actually be free. Recently, I finished reading R for SAS and SPSS Users and have begun reading SAS and R and Data Manipulation with R. Based on what I've read in these books and elsewhere, I get the impression that R is very good at drawing high quality graphs but maybe not so good at creating nice looking tables of the sort I'm used to getting through SAS ODS. Am I right or wrong about this? If I am wrong, can anyone show me some examples of how R can be used to create really nice looking tables? I often make tables of adverse events in clinical trials that have n(%) values in the cells. I'd love to see an example that does a nice job of making that sort of table but would be happy to see any examples that someone might be willing to send to me. Thanks, Paul __ Looking for the perfect gift? Give the gift of Flickr! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] odfWeave Error
Can you use 7zG to unzip an odt file from a dos prompt? Max On Wed, Mar 17, 2010 at 9:46 AM, Btibert3 btibe...@gmail.com wrote: Thanks for your help! Here is what I tried with no luck. I use 7-zip on my machine. Just in case it helps, I am running XP Pro 32 at the office. odfctrl - odfWeaveControl(zipCmd = c(C:/Program Files/7-Zip/7zG.exe -r $$file$$ ., C:/Program Files/7-Zip/7zG.exe -o $$file$$)) odfWeave(example01_in.odt, example01.odt, control = odfctrl) I even tried to use 7z.exe, as well as 7zFM.exe. Essentially, I attempted to use all 3 exectuable files that are located in the install directory. Any ideas? Thanks again! -- View this message in context: http://n4.nabble.com/odfWeave-Error-tp1595848p1596418.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Max __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting factors to numeric in a dataframe
Hi! I don't really understand why you do pcrdata-as.data.frame(t(pcrdata)) Do you need to transpose the dataset? Because read.csv() creates a dataframe already. Something I found really useful recently is the package xlsReadWrite where the function read.xls() has an argument colClasses (read.table() and read.csv() have it too, but it never worked fine for me) which would allow you to specify the class of each column at reading. Is this what you were looking for or am I completely wrong? By the way, you could send us the output from str(pcrdata), I mean just after reading in the data Ivan Le 3/17/2010 15:23, Michael Glanville a écrit : I am currently trying to write a program that minimises the amount of work required for auditable qPCR data. At the moment I am using an Excel (.csv) spreadsheet as source data that has been transposed to the column format required for R to read. Unfortunately, this means I have* *to manually confirm the whole data set prior to doing any analysis, which is taking a considerable amount of time! My idea now is to read the raw data in directly and get R to do the transformation prior to analysis. The problem I now have is that, upon transposition, the data are converted to character in a matrix, rather than factor and numeric in a dataframe. I have succeeded in changing the matrix to a dataframe (via as.data.frame(object)), but this then converts all the data to factor which I can't use for my analysis since, other than the column headings, I need the data to be numeric. I have tried coercing the data to numeric using the as() and as.numeric() commands, but this has no effect on the data format. I have no experience in programming and so am at a loss as to what to do: am I making a basic error in my programming or missing something essential (or both!)? I am using R version 2.9.0 at the moment, but this will change as soon as I have sorted this issue out. Below is the code I have put together, as you can see it is VERY brief but essential to allow my analysis to proceed: pcrdata-read.csv(File_path,header=FALSE) pcrdata-as.data.frame(t(pcrdata)) pcrdata[2:51]-as.numeric(as.character(pcrdata)) Any help would be gratefully appreciated, Mike Glanville [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] email id
this is my mail id abraufs...@gmail.com [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] question about multinom function (nnet)
Dear All. I have the following table that I want to analyze using multinom function freq segments sample 4271 Seg1 tumour 4311 Seg2 tumour 3515 Seg1 normal 3561 Seg2 normal I want to compare model with both factors to the one where only sample is present. model1=multinom(freq~segments+sample,data=table) model2=multinom(freq~ sample,data=table) anova(model2,model1) Likelihood ratio tests of Multinomial Models Response: freq Model Resid. df Resid. Dev TestDf LR stat. Pr(Chi) 1sample 6 5.5452609828 2 segments + sample 3 0.0001600513 1 vs 2 3 5.545101 0.1359658 What I do not understand is where df of 6 and 3 come from? Shouldn't they be 2 and 1? Many thanks for your help in advance Sincerely Sergii -- Sergii Ivakhno PhD student Computational Biology Group Cancer Research UK Cambridge Research Institute Li Ka Shing Centre Robinson Way Cambridge CB2 0RE England +44 (0)1223 404293 (O) +44 (0)1223 404128 (F) http://www.compbio.group.cam.ac.uk http://www.compbio.group.cam.ac.uk/ / This communication is from Cancer Research UK. Our website is at www.cancerresearchuk.org. We are a charity registered under number 1089464 and a company limited by guarantee registered in England Wales under number 4325234. Our registered address is 61 Lincoln's Inn Fields, London WC2A 3PX. Our central telephone number is 020 7242 0200. This communication and any attachments contain information which is confidential and may also be privileged. It is for the exclusive use of the intended recipient(s). If you are not the intended recipient(s) please note that any form of disclosure, distribution, copying or use of this communication or the information in it or in any attachments is strictly prohibited and may be unlawful. If you have received this communication in error, please notify the sender and delete the email and destroy any copies of it. E-mail communications cannot be guaranteed to be secure or error free, as information could be intercepted, corrupted, amended, lost, destroyed, arrive late or incomplete, or contain viruses. We do not accept liability for any such matters or their consequences. Anyone who communicates with us by e-mail is taken to accept the risks in doing so. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: boxplot, vertical position of x-axis labels
thanks a lot! both ways work. greetings - kay -- View this message in context: http://n4.nabble.com/boxplot-vertical-position-of-x-axis-labels-tp1594857p1596456.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How good is R at making publication quality tables?
Please look at the latex function in the Hmisc package. The default display is very good. And there are many optional arguments that give you very fine control over the appearance of the table. Rich [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Converting factors to numeric in a dataframe
Why use a csv dataset as an intermediary? Use RExcel and get the dataset directly from your Excel source. See http://rcom.univie.ac.at for full details. You can download the RExcelInstaller package from CRAN with install.packages(RExcelInstaller) library(RExcelInstaller) installRExcel() Rich [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Converting factors to numeric in a dataframe
Hi r-help-boun...@r-project.org napsal dne 17.03.2010 15:23:34: I am currently trying to write a program that minimises the amount of work required for “auditable” qPCR data. At the moment I am using an Excel (.csv) spreadsheet as source data that has been transposed to the column format required for R to read. Unfortunately, this means I have* *to manually confirm the whole data set prior to doing any analysis, which is taking a considerable amount of time! My idea now is to read the raw data in directly and get R to do the transformation prior to analysis. The problem I now have is that, upon transposition, the data are converted to “character” in a matrix, rather than “factor” and “numeric” in a dataframe. I have succeeded in changing the matrix to a dataframe (via as.data.frame(object)), but this then converts all the data to “factor” which I can’t use for my analysis since, other than the column headings, I need the data to be numeric. I have tried coercing the data to numeric using the as() and as.numeric() commands, but this has no effect on the data format. I have no experience in programming and so am at a loss as to what to do: am I making a basic error in my programming or missing something essential (or both!)? I am using R version 2.9.0 at the moment, but this will change as soon as I have sorted this issue out. Below is the code I have put together, as you can see it is VERY brief but essential to allow my analysis to proceed: pcrdata-read.csv(File_path,header=FALSE) This is supposed to be data frame already. As you did not show us any of possible clues of data type like str(pcrdata) it is difficult to say. However from your description your original data are in columns which have numeric and character data together which is not possible. I believe that there are options for reading such data. pcrdata-as.data.frame(t(pcrdata)) OK. Here you say you get data in columns but they are all character. pcrdata[2:51]-as.numeric(as.character(pcrdata)) Here it depends whether they are all numeric or if some of them shall be character (factor). Functions like those above can not be used directly on data frames. You need to use apply. apply(pcrdata, 1, as.character) Exact sequence of required functions is impossible to guess without knowing structure of your objects. You shall also consult R intro and R data manuals. Regards Petr Any help would be gratefully appreciated, Mike Glanville [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Using nrow with summaryBy
Hello Everyone- I'm calculating summary statistics on a dataset (~4000 records, observations are not uniformly distributed) using summaryBy and trying to add a column with the number of observations to the output as well. What occurs to me is to use nrow(), but this doesn't appear to be working I'm able to replicate the same results with an example from the summaryBy docs: data(dietox) dietox12- subset(dietox,Time==12) library(doBy) #this one works summaryBy(Weight+Feed~Evit+Cu,data=dietox12,FUN=c(mean,var,length)) #adding nrow doesn't give the number of rows summaryBy(Weight+Feed~Evit+Cu,data=dietox12,FUN=c(mean,var,length,nrow)) There must be a way to do this, but I can't figure it out. I suspect there is another function that would be compatible with summaryBy. Thanks in advance. -Tony [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How good is R at making publication quality tables?
Paul Miller wrote: Hello Everyone, I have just started learning R and am in the process of figuring out what it can and can't do. I must say I am very impressed with R so far and am amazed that something this good can actually be free. Recently, I finished reading R for SAS and SPSS Users and have begun reading SAS and R and Data Manipulation with R. Based on what I've read in these books and elsewhere, I get the impression that R is very good at drawing high quality graphs but maybe not so good at creating nice looking tables of the sort I'm used to getting through SAS ODS. You're really only limited by your imagination here. I have written several custom table functions to output LaTeX, but you can output whatever you like (HTML, plain-text, org-mode files...), you're in complete control with R. I can second the Hmisc package though. I often use a combination of summary.formula and the latex function to output really nice looking tables that get put into a long PDF report for a study. I can say that both of these functions, summary.formula and latex, in Hmisc have a LOT of arguments, and almost every time I said I wish it looked a little different, there was an option to control it. Specifically, I found the options: exclude1, long, longtable, combine, test, do be very useful. I often make tables by some treatment group, so all these are using method = reverse to accomplish that. And if you don't like the output, latex.summary.formula.reverse is a good function to make your own version of, to output exactly what you want. I have a local copy that augments the tables that contain unadjusted p-values with adjusted p-values from a model. But apart from Hmisc, just realize that with R you have a nice programming language to produce any type of output you want, you're not limited to what someone else gave you. --Erik __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Adding minutes to 24 hour time
Thank you Gustaf for your help! It worked perfectly. Now I just need to figure out how you did it. Petr, thank you for assisting me, but I could not get your approach to work. Mike __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] save data to an R object
Yes, that was very useful, thank you, Cheba 2010/3/17 Ivan Calandra ivan.calan...@uni-hamburg.de Hi, Here is one option: #some sample dataframes df1 - data.frame(letter=letters[1:3], number=1:3) df1 - data.frame(letter=letters[4:6], number=4:6) #save the dataframes save(df1, df2, file=test.Rda) #load them in another session load(test.Rda) df1 letter number 1 a 1 2 b 2 3 c 3 df2 letter number 1 d 4 2 e 5 3 f 6 Is that what you're looking for? Ivan Le 3/17/2010 11:56, cheba meier a écrit : Dear R people, Is it possible to save three data sets in an R object and to call each data from this object independently! Regards, Cheba [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How good is R at making publication quality tables?
Hi Paul, For instructions and examples using the Hmisc latex() function you might want to take a look at http://lib.stat.cmu.edu/S/Harrell/doc/summary.pdf. -Best, Ista On Wed, Mar 17, 2010 at 10:51 AM, Paul Miller pjmiller...@yahoo.com wrote: Hello Everyone, I have just started learning R and am in the process of figuring out what it can and can't do. I must say I am very impressed with R so far and am amazed that something this good can actually be free. Recently, I finished reading R for SAS and SPSS Users and have begun reading SAS and R and Data Manipulation with R. Based on what I've read in these books and elsewhere, I get the impression that R is very good at drawing high quality graphs but maybe not so good at creating nice looking tables of the sort I'm used to getting through SAS ODS. Am I right or wrong about this? If I am wrong, can anyone show me some examples of how R can be used to create really nice looking tables? I often make tables of adverse events in clinical trials that have n(%) values in the cells. I'd love to see an example that does a nice job of making that sort of table but would be happy to see any examples that someone might be willing to send to me. Thanks, Paul __ Looking for the perfect gift? Give the gift of Flickr! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How good is R at making publication quality tables?
On Wed, Mar 17, 2010 at 10:25 AM, Erik Iverson er...@ccbr.umn.edu wrote: Paul Miller wrote: Hello Everyone, I have just started learning R and am in the process of figuring out what it can and can't do. I must say I am very impressed with R so far and am amazed that something this good can actually be free. Recently, I finished reading R for SAS and SPSS Users and have begun reading SAS and R and Data Manipulation with R. Based on what I've read in these books and elsewhere, I get the impression that R is very good at drawing high quality graphs but maybe not so good at creating nice looking tables of the sort I'm used to getting through SAS ODS. You're really only limited by your imagination here. I have written several custom table functions to output LaTeX, but you can output whatever you like (HTML, plain-text, org-mode files...), you're in complete control with R. I can second the Hmisc package though. I often use a combination of summary.formula and the latex function to output really nice looking tables that get put into a long PDF report for a study. I can say that both of these functions, summary.formula and latex, in Hmisc have a LOT of arguments, and almost every time I said I wish it looked a little different, there was an option to control it. Specifically, I found the options: exclude1, long, longtable, combine, test, do be very useful. I often make tables by some treatment group, so all these are using method = reverse to accomplish that. And if you don't like the output, latex.summary.formula.reverse is a good function to make your own version of, to output exactly what you want. I have a local copy that augments the tables that contain unadjusted p-values with adjusted p-values from a model. But apart from Hmisc, just realize that with R you have a nice programming language to produce any type of output you want, you're not limited to what someone else gave you. Another option to consider is the xtable package. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using nrow with summaryBy
On Mar 17, 2010, at 11:23 AM, Tony Laidig wrote: Hello Everyone- I'm calculating summary statistics on a dataset (~4000 records, observations are not uniformly distributed) using summaryBy and trying to add a column with the number of observations to the output as well. What occurs to me is to use nrow(), but this doesn't appear to be working I'm able to replicate the same results with an example from the summaryBy docs: data(dietox) dietox12- subset(dietox,Time==12) library(doBy) #this one works summaryBy(Weight+Feed~Evit+Cu,data=dietox12,FUN=c(mean,var,length)) #adding nrow doesn't give the number of rows summaryBy(Weight+Feed~Evit +Cu,data=dietox12,FUN=c(mean,var,length,nrow)) I'm a bit puzzled. One of my many newbie mistakes was to assume that length() applied to dataframes would tell me how many rows it had. It appears that the authors of summaryBy have figured out how to get length() to tell you the number of observations, presumably on a subsetted vector where length would make sense. So ... it's not clear why you also want nrow (which would not make sense for a subsetted vector). There must be a way to do this, but I can't figure it out. I suspect there is another function that would be compatible with summaryBy. Thanks in advance. -Tony David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How good is R at making publication quality tables?
On Wed, Mar 17, 2010 at 2:51 PM, Paul Miller pjmiller...@yahoo.com wrote: Am I right or wrong about this? If I am wrong, can anyone show me some examples of how R can be used to create really nice looking tables? I often make tables of adverse events in clinical trials that have n(%) values in the cells. I'd love to see an example that does a nice job of making that sort of table but would be happy to see any examples that someone might be willing to send to me. To complement all said up to now, see this documentation [1]. For clinical trials you might also want to check rreport [2]. There was also a recent thread [3] on the topic. Liviu [1] http://cran.r-project.org/web/packages/RcmdrPlugin.Export/RcmdrPlugin.Export.pdf [2] http://biostat.mc.vanderbilt.edu/wiki/Main/Rreport [3] [R] Use of R in clinical trials __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Why eval(parse(text=var(vec))) return a matrix but NOT a number?
On Wed, 17 Mar 2010, Yong Wang wrote:
Dear List
I am getting a problem when using eval(parse).
Code below sketchs what I am trying to do:
For each row of a N*K dataframe (I use a 2*2 dataframe in the example below),
applying a number of functions and get the outputs (two functions,
sum and var are used in the example below).
The problem is eval(parse(text=sum(para))) works fine but not when
sum is replaced by var.
in the later case, a matrix instead of a number is returned.
Any suggestion highly appreciated.
Thank you
#===The function
myloop -function(datfra,funs) {
rows-dim(datfra)[1];
totfunnum-length(funs);
for (i in 1:rows) {
vec-datfra[i,];
# I suggest:
browser()
# Now try print(vec), str(vec), etc till you understand
#why the results do not agree with your expectation
# HTH,
# Chuck
for(k in 1:totfunnum) {
print(funs[k]);
x-eval(parse(text=funs[k]));
print(x);
}
}
}
#Experiemental run
workport-data.frame(matrix(1:4,2,2))
funs-c(sum(vec,na.rm=T),var(vec,na.rm=T))
myloop(workport,funs)
# Outputs of the
Experimental run
[1] sum(vec,na.rm=T)
[1] 4
[1] var(vec,na.rm=T)
X1 X2
X1 NA NA
X2 NA NA
[1] sum(vec,na.rm=T)
[1] 6
[1] var(vec,na.rm=T)
X1 X2
X1 NA NA
X2 NA NA
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Charles C. Berry(858) 534-2098
Dept of Family/Preventive Medicine
E mailto:cbe...@tajo.ucsd.edu UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/ La Jolla, San Diego 92093-0901
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Re: [R] Using nrow with summaryBy
Hi David, I have probably 2 stupid questions regarding what you said but it might be important to understand: - why nrow() would not make sens for a subsetted vector? On the help page of nrow(), it's written that we can apply it on a vector, array or dataframe (basically everything...?). So what's the difference between a normal vector (for which it would make sense and work) and a subsetted vector? - why assuming that length() applied to dataframes would tell me how many rows it had would be a mistake? I mean in this case, length() is calculated for each numerical variable (which are vectors, aren't they?). I think these questions concern the way R handle the data and that's why I think it might be important for me to understand these issues. Thanks for your input. Regards, Ivan Le 3/17/2010 16:39, David Winsemius a écrit : On Mar 17, 2010, at 11:23 AM, Tony Laidig wrote: Hello Everyone- I'm calculating summary statistics on a dataset (~4000 records, observations are not uniformly distributed) using summaryBy and trying to add a column with the number of observations to the output as well. What occurs to me is to use nrow(), but this doesn't appear to be working I'm able to replicate the same results with an example from the summaryBy docs: data(dietox) dietox12- subset(dietox,Time==12) library(doBy) #this one works summaryBy(Weight+Feed~Evit+Cu,data=dietox12,FUN=c(mean,var,length)) #adding nrow doesn't give the number of rows summaryBy(Weight+Feed~Evit+Cu,data=dietox12,FUN=c(mean,var,length,nrow)) I'm a bit puzzled. One of my many newbie mistakes was to assume that length() applied to dataframes would tell me how many rows it had. It appears that the authors of summaryBy have figured out how to get length() to tell you the number of observations, presumably on a subsetted vector where length would make sense. So ... it's not clear why you also want nrow (which would not make sense for a subsetted vector). There must be a way to do this, but I can't figure it out. I suspect there is another function that would be compatible with summaryBy. Thanks in advance. -Tony David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using nrow with summaryBy
On Mar 17, 2010, at 12:10 PM, Ivan Calandra wrote: Hi David, I have probably 2 stupid questions regarding what you said but it might be important to understand: - why nrow() would not make sens for a subsetted vector? On the help page of nrow(), it's written that we can apply it on a vector, array or dataframe (basically everything...?). So what's the difference between a normal vector (for which it would make sense and work) and a subsetted vector? nrow(c(0,1,3,4)) NULL nrow(1:12) NULL (It did not throw an error but the help page does say the value could be NULL.) length(1:12) [1] 12 - why assuming that length() applied to dataframes would tell me how many rows it had would be a mistake? I mean in this case, length() is calculated for each numerical variable (which are vectors, aren't they?). length applied to any list is the number of elements at the first level. Dataframes are lists of vectors so length applied to data.frames gives you the number of columns, not the length of an individual vector in the dataframe. I think these questions concern the way R handle the data and that's why I think it might be important for me to understand these issues. It's important, fur sure. Thanks for your input. Regards, Ivan Le 3/17/2010 16:39, David Winsemius a écrit : On Mar 17, 2010, at 11:23 AM, Tony Laidig wrote: Hello Everyone- I'm calculating summary statistics on a dataset (~4000 records, observations are not uniformly distributed) using summaryBy and trying to add a column with the number of observations to the output as well. What occurs to me is to use nrow(), but this doesn't appear to be working I'm able to replicate the same results with an example from the summaryBy docs: data(dietox) dietox12- subset(dietox,Time==12) library(doBy) #this one works summaryBy(Weight+Feed~Evit+Cu,data=dietox12,FUN=c(mean,var,length)) #adding nrow doesn't give the number of rows summaryBy(Weight+Feed~Evit +Cu,data=dietox12,FUN=c(mean,var,length,nrow)) I'm a bit puzzled. One of my many newbie mistakes was to assume that length() applied to dataframes would tell me how many rows it had. It appears that the authors of summaryBy have figured out how to get length() to tell you the number of observations, presumably on a subsetted vector where length would make sense. So ... it's not clear why you also want nrow (which would not make sense for a subsetted vector). There must be a way to do this, but I can't figure it out. I suspect there is another function that would be compatible with summaryBy. Thanks in advance. -Tony David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 ivan.calan...@uni-hamburg.de ** http://www.for771.uni-bonn.de http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Writing dataframes to SAS XPT format
Colleagues, On occasion, I need to output an R dataframe to a file in SAS XPT format. Although the foreign package supports reading of XPT files and writing to a format that SAS can read, it does not support writing to XPT format (confirmed with Thomas Lumley, the author of write.foreign). Has anyone developed such a function? If so, would you be willing to share it? Better yet, post it to CRAN so that it is available to all users. Dennis Dennis Fisher MD P (The P Less Than Company) Phone: 1-866-PLessThan (1-866-753-7784) Fax: 1-866-PLessThan (1-866-753-7784) www.PLessThan.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mmap error-12, macbook pro
There seem to have been three almost identical reports of this from three different email addresses, including two widely-separate UK academic institutions, and two user names, which is a bit weird. However: R will need more than 4GB memory to handle a 3.99GB file, and probably more than 4GB memory to handle a 1.33GB file. On 32-bit R I think the recommendation is that the data should occupy no more than 10% of the address space, ie, about 400MB. For 64-bit R, you are likely to still find R very slow if the data set is more than about 1/3 of physical memory. For a 4GB data file I would recommend a computer with at least 16GB memory. Now, it is often possible to load only a small fraction of the data at one time and so to analyse large data sets on smaller computers. Three examples that I have worked on - analysing data from a survey data set about the same size as your data, by keeping most of the data in a SQLite database and just loading a few variables at a time. - fitting linear regression models to large data sets with the biglm package, by keeping the data in a SQLite database and just loading a few rows at a time - analysing whole-genome genetic data (40GB or so) by storing the data in a netCDF file and reading appropriate chunks with the ncdf package. -thomas On Wed, 17 Mar 2010, Mamun wrote: Dear List, I am trying to read some files using read.csv and total size of those files are 3.99 GB. I am using MacBook Pro with 4GB RAM(snow leopard). I also tried to run a chunk from those files and altogether the size was 1.33 GB. But every time I was getting the following error R(1200) malloc: *** mmap(size=16777216) failed (error code=12) *** error: can't allocate region *** set a breakpoint in malloc_error_break to debug I managed to run smaller chunks (400MB) and I also saved the object. But problem is when I try to load some of those objects(which is not more than 1.5 GB altogether), I get same error again. Can someone please help? why am I getting this error? does R need more space than the actual file size? I may buy new machine if its something related to RAM size. but if it is some R problem then it will be very useless to buy new machine. So, I really need to know why this is happening. Any help is appreciated. Thanks in advance. regards, Mamun -- View this message in context: http://n4.nabble.com/mmap-error-12-macbook-pro-tp1596234p1596234.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.eduUniversity of Washington, Seattle __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Retrieving latitude and longitude via Google Maps API
You may want to look at http://www.openstreetmap.org/ or
http://maps.cloudmade.com/ instead of google maps. I don't know if they give
the same detail/searches, but open source sites tend to be easier to work with
and have fewer restrictions on use. While I like the google maps, I tend to
not use them for anything other than personal browsing since I am not sure
about the legal/copyright issues.
Hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
project.org] On Behalf Of Nutter, Benjamin
Sent: Tuesday, March 16, 2010 1:42 PM
To: r-help@r-project.org
Subject: [R] Retrieving latitude and longitude via Google Maps API
Does anyone have any experience retrieving latitutde and longitude for
an address from the Google Maps API?
I'd like to have an R script that submits a street address, city,
state,
and zip code and returns the coordinates. So far, I've been submitting
the coordinates from another program, then loading the coordinates in R
and merging them back into the data frame I want to use. It'd be nice
to be able to do it all in one script, but I'm not comprehending the
API
thing very well.
I'm using R 2.9.1 on Windows XP. Any suggestions or pointers?
Benjamin
Benjamin Nutter | Biostatistician | Quantitative Health Sciences
Cleveland Clinic | 9500 Euclid Ave. | Cleveland, OH 44195 |
(216) 445-1365
===
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Cleveland Clinic is ranked one of the top hospitals
in America by U.S.News World Report (2009).
Visit us online at http://www.clevelandclinic.org for
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Re: [R] plm within models: is the correct F-statistic reported?
On Wed, 17 Mar 2010, Liviu Andronic wrote: Dear Achim On 3/16/10, Achim Zeileis achim.zeil...@uibk.ac.at wrote: Hence, when saying summary() different models with no effects are assumed. For gr_fe the model without effects just omits value/capital but keeps the firm-specific interecepts. For gr_lm not even the intercept is kept in the model. Thus: gr_fe_null - lm(invest ~ 0 + firm, data = pgr) gr_lm_null - lm(invest ~ 0, data = pgr) What would be the more useful no effects model in the plm(..., effect=twoways) case? Considering the same setting, library(AER) data(Grunfeld, package = AER) library(plm) gr - subset(Grunfeld, firm %in% c(General Electric, General Motors, IBM)) pgr - plm.data(gr, index = c(firm, year)) I am fitting a twoways model and an individual with manually specified time effects. gr_fe1 - plm(invest ~ value + capital, data = pgr, +model = within, effect=twoways) [snip] Following the reasoning in your previous e-mail, I assume that the (more useful) no effects model used in the twoways case is gr_fe1_null - lm(invest ~ 0 + firm + year, data = pgr) Yes. However I cannot replicate the F-statistic: 107.246. anova(gr_fe1_null, gr_fe1) Here, you compare apples and oranges. gr_fe1 is of class plm and gr_fe1_null is of class lm. This does not fly. gr_fe1_lm - lm(invest ~ 0 + value + capital + firm + year, data = pgr) anova(gr_fe1_lm, gr_fe1_null) which does the job. In short: - plm(..., model = within) offers a convenient approach of what is usually done in this kind of analysis. - You can replicate everything by hand using lm() but have to take care of everything yourself. But you do get the same results. - Don't mix the two approaches. Best, Z __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Writing dataframes to SAS XPT format
Dennis Fisher wrote: Colleagues, On occasion, I need to output an R dataframe to a file in SAS XPT format. Although the foreign package supports reading of XPT files and writing to a format that SAS can read, it does not support writing to XPT format (confirmed with Thomas Lumley, the author of write.foreign). Has anyone developed such a function? If so, would you be willing to share it? Better yet, post it to CRAN so that it is available to all users. http://cran.r-project.org/web/packages/SASxport/index.html HTH, Tobias __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] replace NA value with 0
Building on the question how to replace NA with 0. My data set below has date, station 1, flags for station 1, station 2, flags for station 2, etc... I would like to make the values in the station columns equal to 1 and the NA in the station columns equal to 0 and then sum each row for the number of 1 present in the row. head(data.matrix, n=10) date 05AE005 flg_05AE005 05AF010 flg_05AF010 05BM014 flg_05BM014 1 1900-01-01 NANA NANA NANA 2 1900-01-02 NANA NANA .23NA 3 1900-01-03 NANA NANA .45NA 4 1900-01-04 NANA NANA NANA 5 1900-01-05 NANA NANA NANA 6 1900-01-06 NANA NANA NANA 7 1900-01-07 0.75 NA .09NA NANA 8 1900-01-08 0.87 NA .23NA NANA 9 1900-01-09 0.26 NA .78NA NANA 10 1900-01-10 0.23 NA NANA NANA # figure out which columns the data are in colpos - seq(2, by = 2, length.out = n) # make value 1 and NA 0 colpos[!is.na(colpos)] - 1.0 colpos[is.na(colpos)] - 0.0 # sum number of values on a given day rowsum(colpos, ) The above script is what i have tried - and does not work. The values are not being replaced with 1s and the NAs with 0s and the rows are not being summed. any help would be greatly appreciated. E -- View this message in context: http://n4.nabble.com/replace-NA-value-with-0-tp834446p1596779.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How good is R at making publication quality tables?
If you are working in Windows and want Word output, it might be worthwhile to have a look at SWord available from the download section on rcom.univie.ac.at. Warning: the program is free for noncommercial use, but NOT licensed under GPL or LGPL. On 3/17/2010 3:51 PM, Paul Miller wrote: Hello Everyone, I have just started learning R and am in the process of figuring out what it can and can't do. I must say I am very impressed with R so far and am amazed that something this good can actually be free. Recently, I finished reading R for SAS and SPSS Users and have begun reading SAS and R and Data Manipulation with R. Based on what I've read in these books and elsewhere, I get the impression that R is very good at drawing high quality graphs but maybe not so good at creating nice looking tables of the sort I'm used to getting through SAS ODS. Am I right or wrong about this? If I am wrong, can anyone show me some examples of how R can be used to create really nice looking tables? I often make tables of adverse events in clinical trials that have n(%) values in the cells. I'd love to see an example that does a nice job of making that sort of table but would be happy to see any examples that someone might be willing to send to me. Thanks, Paul __ Looking for the perfect gift? Give the gift of Flickr! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: Loop overwrite and data output problems
Hi Petr, Thanks again!!! model is a list. So your suggestion: mod - matrix(NA, 1000, ncols) doesn't work. I thought that do.call and rbind would be the best for these data? Cheers, Ross -- View this message in context: http://n4.nabble.com/Loop-overwrite-and-data-output-problems-tp1570593p1596889.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using nrow with summaryBy
Use NROW rather than nrow. On Wed, Mar 17, 2010 at 11:23 AM, Tony Laidig c...@mit.edu wrote: Hello Everyone- I'm calculating summary statistics on a dataset (~4000 records, observations are not uniformly distributed) using summaryBy and trying to add a column with the number of observations to the output as well. What occurs to me is to use nrow(), but this doesn't appear to be working I'm able to replicate the same results with an example from the summaryBy docs: data(dietox) dietox12- subset(dietox,Time==12) library(doBy) #this one works summaryBy(Weight+Feed~Evit+Cu,data=dietox12,FUN=c(mean,var,length)) #adding nrow doesn't give the number of rows summaryBy(Weight+Feed~Evit+Cu,data=dietox12,FUN=c(mean,var,length,nrow)) There must be a way to do this, but I can't figure it out. I suspect there is another function that would be compatible with summaryBy. Thanks in advance. -Tony [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How good is R at making publication quality tables?
Hi Paul, Sorry I didn't get to that subject in the first edition of R for SAS and SPSS Users. Several of the options people have mentioned will be in the second edition, although that's about a year off. I did get them added to R for Stata Users, due out in early April. Cheers, Bob -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Paul Miller Sent: Wednesday, March 17, 2010 10:51 AM To: r-help@r-project.org Subject: [R] How good is R at making publication quality tables? Hello Everyone, I have just started learning R and am in the process of figuring out what it can and can't do. I must say I am very impressed with R so far and am amazed that something this good can actually be free. Recently, I finished reading R for SAS and SPSS Users and have begun reading SAS and R and Data Manipulation with R. Based on what I've read in these books and elsewhere, I get the impression that R is very good at drawing high quality graphs but maybe not so good at creating nice looking tables of the sort I'm used to getting through SAS ODS. Am I right or wrong about this? If I am wrong, can anyone show me some examples of how R can be used to create really nice looking tables? I often make tables of adverse events in clinical trials that have n(%) values in the cells. I'd love to see an example that does a nice job of making that sort of table but would be happy to see any examples that someone might be willing to send to me. Thanks, Paul __ Looking for the perfect gift? Give the gift of Flickr! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plm within models: is the correct F-statistic reported?
On 3/17/10, Achim Zeileis achim.zeil...@uibk.ac.at wrote:
Here, you compare apples and oranges. gr_fe1 is of class plm and
gr_fe1_null is of class lm. This does not fly.
gr_fe1_lm - lm(invest ~ 0 + value + capital + firm + year, data = pgr)
anova(gr_fe1_lm, gr_fe1_null)
which does the job.
Indeed, I see.
In short:
- plm(..., model = within) offers a convenient approach of what
is usually done in this kind of analysis.
Unfortunately plm(..., effect=twoways, model = within) fails on my
particular unbalanced panel data (100% CPU and the task never
finishes); there are no such issues with individual or time. Worse
is that I cannot replicate the issue on dummy data.
- You can replicate everything by hand using lm() but have to take
care of everything yourself. But you do get the same results.
- Don't mix the two approaches.
I wanted to avoid displaying 2000 individuals in the regression
coefficients, the reason for trying the mix-up. I also tried to do the
trick using a plm() model for the null, too, but there is no anova
method for these.
gr_fe1 - plm(invest ~ value + capital, data = pgr,
+model = within, effect=twoways)
gr_fe1_null - plm(invest ~ 0 + firm + year, data = pgr, model = pooling)
anova(gr_fe1_null, gr_fe1)
Error in UseMethod(anova) :
no applicable method for 'anova' applied to an object of class
c('plm', 'panelmodel')
And having checked the source, I wouldn't venture to implement one.
I'm still a bit stuck on how to proceed. Thank you
Liviu
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[R] Retaining variable name in a function
Hi All,
Im interested in creating a function that will convert a variable within a
data.frame to a factor while retaining the original name (yes, I know that I
can just: var -factor(var) but I need it as a function for other
purposes). e.g.:
# this was an attempt but fails.
facts - function(meta, mod, modname = spec) {
meta$mod - factor(meta$mod)
colnames(meta)['mod'] - modname
return(meta)
}
# ideally, would like to just specify the data.frame (=meta) and
# variable to convert to factor (=mod) (similar to function input below).
But am
# also interested in having the option to create a new variable and
# name (similar to the function input above).
facts - function(meta, mod) {
meta$mod - factor(meta$mod)
return(meta)
}
Thanks for any help!
AC
[[alternative HTML version deleted]]
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Re: [R] Retaining variable name in a function
How about this? It changes a column to a factor, and
optionally renames it.
facts - function(meta, mod, newmodname) {
meta[,mod] - factor(meta[,mod])
if(!missing(newmodname)) {
colnames(meta)[colnames(meta) == mod] - newmodname
}
meta
}
testdata - data.frame(a=c(1,2,3), b=c(1,1,2))
str(testdata)
'data.frame': 3 obs. of 2 variables:
$ a: num 1 2 3
$ b: num 1 1 2
test1 - facts(testdata, a)
str(test1)
'data.frame': 3 obs. of 2 variables:
$ a: Factor w/ 3 levels 1,2,3: 1 2 3
$ b: num 1 1 2
test2 - facts(testdata, b, newb)
str(test2)
'data.frame': 3 obs. of 2 variables:
$ a : num 1 2 3
$ newb: Factor w/ 2 levels 1,2: 1 1 2
Sarah
On Wed, Mar 17, 2010 at 3:06 PM, AC Del Re de...@wisc.edu wrote:
Hi All,
Im interested in creating a function that will convert a variable within a
data.frame to a factor while retaining the original name (yes, I know that I
can just: var -factor(var) but I need it as a function for other
purposes). e.g.:
# this was an attempt but fails.
facts - function(meta, mod, modname = spec) {
meta$mod - factor(meta$mod)
colnames(meta)['mod'] - modname
return(meta)
}
# ideally, would like to just specify the data.frame (=meta) and
# variable to convert to factor (=mod) (similar to function input below).
But am
# also interested in having the option to create a new variable and
# name (similar to the function input above).
facts - function(meta, mod) {
meta$mod - factor(meta$mod)
return(meta)
}
Thanks for any help!
AC
--
Sarah Goslee
http://www.functionaldiversity.org
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