[R] extractModelParameters HELP!!!
Hello Gurus,
I'm using one of the function, i.e. extractModelParameters, from the
MplusAutomation package to read parameters fromMplus out files into R. I have
360 cells and each cell has several out files (max. 100). I'm extracting the
model parameters from each cell and doing subsequent calculations, primarily,
on
'est.' column. The following code demonstrates how I'm extracting the
parameters
and referencing the est. column:
CurrentDir =
paste("D:/Dissertation/Mplus_Output_2Class_LGMM/",a,"-",b,"-",c,"-",d,"-",e,"-",f,
sep="")
modelResults <- extractModelParameters(CurrentDir, recursive=FALSE,
dropDimensions=TRUE)
numReps = length(list.files(CurrentDir, pattern="*.out"))
after the above code is executed, 'modelResults' has all the data from the out
files as different elements of the returned list. I'm accessing est. column
from
each element in the following way:
for (g in 1:numReps)
MeansTempC1[counterC1] <- modelResults[[g]]$est[31]
I'm able to get the values from the est. column successfully for 359 cells
which
consist of 29532 out files. For just 1 particular cell, when I try to access
the est. column or any other column in that out file, i.e. est_se or
paramHeader, i get the value of NULL and I'm not able to figure out why. The
same code is working for 29532 out files but not for 70 files in that 1 cell. I
would really appreciate if you can assist me troubleshoot this issue. I have
been working on this problem since last Sunday and after spending 6 days I
still
not able to figure out the problem.
Your help is truly appreciated.
Regards
Vaib
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[R] remove quotes from the paste output
Hi,
I'm generating the name of the variable with paste function and then using that
variable name further to get the specific position value from the data.frame,
here is the snippet from my code:
modelResults <- extractModelParameters("C:/PilotStudy/Mplus_Input/Test",
recursive=TRUE)
#extractModelParameters reads all the output files from the Test folder and
create the following variables in R for each file read:
#C..PilotStudy.Mplus_Input.Test.rep1.out.unstandardized.est
#C..PilotStudy.Mplus_Input.Test.rep2.out.unstandardized.est
#C..PilotStudy.Mplus_Input.Test.rep3.out.unstandardized.est
modelResultsTemp <- as.data.frame(modelResults)
MeansTempC1 = rep(NA ,9)
counter = 1
for (f in 1:3)
{
i=31
for (g in 1:3)
{
OutputFileName <-
paste("modelResultsTemp$C..PilotStudy.Mplus_Input.Test.rep",f,".out.unstandardized.est[",i,"]",sep="")
MeansTempC1[counter] = OutputFileName
i=i+1
counter = counter+1
}
}
Its not giving me any error but I'm not getting the desired output in
MeansTempC1 because OutputFileName is a string and thats why its not returning
the OutputFileName[31] etc. Does anybody know how to eliminate the double
quotes
so that I can use the variable name (generated with the paste function) further
in the code
Regards
Bob
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[R] Generate Univariate Normal Mixtures???
Hi, I'm trying to generate a mixture from 2 univariate normal distributions and would like to put a constrained on the means (Equal) of the two distributions. Here is my shot at the code: library(nor1mix) X <- norMix(mu=c(50, 60 ), sig2=c(5,10), w=c(0.5, 0.5)) mixData <- rnorMix(1000, X) plot(mixData, type="l") I did get the 1000 random numbers in equal proportion but how can I confirm if the means was constrained or not in the data generation process? Any help will be highly appreciated VD [[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] Univariate Normal Mixtures - HELP
Hi, I'm generating a mixture of 2 univariate normal distributions using norMix and rnorMix and would like to put a constraint on Mean (Equal). here is my code snippet: library(nor1mix) X <- norMix(mu=c(50, 60 ), sig2=c(5,10), w=c(0.5, 0.5)) mixData <- rnorMix(1000, X) plot(mixData, type="l") I do get the 1000 random numbers in equal proportion but I want to confirm if the Mean is same or not Any help will be highly appreciated Victor [[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 on Non-Linera Mixed-Effect model
Hi,
I'm trying to fit nonlinear mixed effects model using nlme function but getting
an error message. Here is what I have:
fitted_model = nlme(scores~spline(b1,b2,b3,kt,time),
fixed = list(b1~1, b2~1, b3~1, kt~1),
random = b1+b2+b3~1,
groups= ~id,
data = sdat,
start = c(b1=3.5,b2=2,b3=.60,kt=3.5),verbose=T)
Error:
Error in spline(b1, b2, b3, kt, time) : invalid interpolation method
In addition: Warning message:
In if (is.na(method)) stop("invalid interpolation method") :
the condition has length > 1 and only the first element will be used
Any help will be highly appreciated
Victor
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Re: [R] rlaplace using rmutil - HELP
Hi Peter, Thanks for quick reply. Like I said that I'm aware that it needs a vector but I want to use diagonal matrix. As you have noted, It could be possible that rlaplace reading dispDiag as a vector and not matrix like c(3,0,0,0,.20,0,0,0,.1) so I thought to change the code little bit: dispersion = c(3,.20,.10) dispDiag = diag(dispersion) location = c(0,0,0) output = (rlaplace(3, location, as.matrix(dispDiag))) but still get the same output. I want to draw random deviates from multivariate laplace distribution. Do you think rlaplace in VGAM can help? Victor From: Peter Ehlers Cc: r-help@r-project.org Sent: Mon, February 15, 2010 10:52:33 AM Subject: Re: [R] rlaplace using rmutil - HELP On 2010-02-15 8:26, vaibhav dua wrote: > Hi, > > I'm using rlaplace distribution to draw non normal random deviates. I'm using > rmutil package for this. I know rlaplace needs location and dispersion > parameters but I want to pass diag matrix as dispersion parameters and I'm > not getting the correct values. Here is my code snippet: > > dispersion = c(3,.20,.10) > dispDiag = diag(dispersion) > location = c(0,0,0) > output = (rlaplace(3, location, dispDiag)) > > > Here is what I get: > > -0.8475811 0.000 0.000 > > I dont know why it is returning 0's at 2nd and 3rd place. If I change the > values in location vector, it simply picks up the value and replace it at 2nd > and 3rd place. > > You help will be highly appreciated > > Victor If you read the help page carefully, you'll see that rlaplace() wants a *vector* of dispersion parameters. So why not just use rlaplace(3, location, dispersion) With dispDiag, rlaplace is probably taking 's' to be c(3,0,0,0,.20,0,0,0,.1) and is only using the first three elements. (Untested) In case you're not aware: there's also rlaplace() in the VGAM package. -Peter Ehlers > > > > [[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. > > -- Peter Ehlers University of Calgary [[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] rlaplace using rmutil - HELP
Hi, I'm using rlaplace distribution to draw non normal random deviates. I'm using rmutil package for this. I know rlaplace needs location and dispersion parameters but I want to pass diag matrix as dispersion parameters and I'm not getting the correct values. Here is my code snippet: dispersion = c(3,.20,.10) dispDiag = diag(dispersion) location = c(0,0,0) output = (rlaplace(3, location, dispDiag)) Here is what I get: -0.8475811 0.000 0.000 I dont know why it is returning 0's at 2nd and 3rd place. If I change the values in location vector, it simply picks up the value and replace it at 2nd and 3rd place. You help will be highly appreciated Victor [[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.
