Dear list, I try to update the prabclus package. R CMD check works nicely, no warnings, good results in all tests. However, building the package fails: ginkgo:/disk5/home/chrish/RAusw/libsrc R CMD build prabclus * checking for file 'prabclus/DESCRIPTION' ... OK * preparing 'prabclus': * checking whether 'INDEX' is up-to-date ... OK * removing junk files * building 'prabclus_2.0-1.tar.gz' tar: prabclus/man/.#prabinit.Rd: Cannot stat: No such file or directory tar: Error exit delayed from previous errors Does anybody have an idea why tar is looking for prabclus/man/.#prabinit.Rd (which indeed doesn't exist, but why should it?) Thank you, Christian *** NEW ADDRESS! *** Christian Hennig University College London, Department of Statistical Science Gower St., London WC1E 6BT, phone +44 207 679 1698 chrish at stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche
I have a large vector of around 12597 elements and I wish to calculate
p-value for each element using a formula of something like:
p-value= 1- exp^(kexp^(-labda))
I was wondering someone could give some ideas how to implement for each element.
thankyou very much
Rangesh.K
if labda is the elements of the vector and you know what kexp is , you can use apply apply(your_vector, 1, function(x) 1- exp^(kexp^(-x))) HTH Jean On Wed, 3 Aug 2005, Rangesh Kunnavakkam wrote:> I have a large vector of around 12597 elements and I wish to calculate > p-value for each element using a formula of something like: > p-value= 1- exp^(kexp^(-labda)) > I was wondering someone could give some ideas how to implement for each element. > thankyou very much > Rangesh.K > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html >
Hi, I tried to do a logistic regression with polr(MASS). I thought I already converted the response to factor, but obvious I was wrong. Could anyone tell me what I did wrong and how to correct it? Thank you very much!> Lease=read.csv("LeaseDummy.csv", header=TRUE) > Lease$ID <-as.integer(factor(Lease$EarlyTermination))> >RegA=polr(ID~1+MSA+SIC.Code+TenantOption+LLOption+TOExercised, + data=Lease, method=c("logistic")) Error in polr(ID ~ 1 + MSA + SIC.Code + TenantOption + LLOption + TOExercised, : response must be a factor> summary(RegA)Best, Ed. --- Jean Eid <jeaneid at chass.utoronto.ca> wrote:> > if labda is the elements of the vector and you know > what kexp is , you can > use apply > > apply(your_vector, 1, function(x) 1- > exp^(kexp^(-x))) > > > HTH > > Jean > On Wed, 3 Aug 2005, Rangesh Kunnavakkam wrote: > > > I have a large vector of around 12597 elements > and I wish to calculate > > p-value for each element using a formula of > something like: > > p-value= 1- > exp^(kexp^(-labda)) > > I was wondering someone could give some ideas how > to implement for each element. > > thankyou very much > > Rangesh.K > > > > ______________________________________________ > > R-help at stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html > > > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html >
Why are you adding as.integer before the factor statement? You are forcing the variable to be an integer even tough you are passing factor within the statement. Try Lease$ID <-factor(Lease$EarlyTermination) Cheers Francisco>From: Haibo Huang <edhuang00 at yahoo.com> >To: r-help at r-project.org >Subject: [R] Convert numeric to factor >Date: Wed, 3 Aug 2005 13:52:13 -0700 (PDT) > >Hi, > >I tried to do a logistic regression with polr(MASS). I >thought I already converted the response to factor, >but obvious I was wrong. Could anyone tell me what I >did wrong and how to correct it? Thank you very much! > > > Lease=read.csv("LeaseDummy.csv", header=TRUE) > > Lease$ID <- >as.integer(factor(Lease$EarlyTermination)) > > > > >RegA=polr(ID~1+MSA+SIC.Code+TenantOption+LLOption+TOExercised, > >+ data=Lease, method=c("logistic")) >Error in polr(ID ~ 1 + MSA + SIC.Code + TenantOption + >LLOption + TOExercised, : > response must be a factor > > summary(RegA) > > >Best, >Ed. > > > > >--- Jean Eid <jeaneid at chass.utoronto.ca> wrote: > > > > > if labda is the elements of the vector and you know > > what kexp is , you can > > use apply > > > > apply(your_vector, 1, function(x) 1- > > exp^(kexp^(-x))) > > > > > > HTH > > > > Jean > > On Wed, 3 Aug 2005, Rangesh Kunnavakkam wrote: > > > > > I have a large vector of around 12597 elements > > and I wish to calculate > > > p-value for each element using a formula of > > something like: > > > p-value= 1- > > exp^(kexp^(-labda)) > > > I was wondering someone could give some ideas how > > to implement for each element. > > > thankyou very much > > > Rangesh.K > > > > > > ______________________________________________ > > > R-help at stat.math.ethz.ch mailing list > > > https://stat.ethz.ch/mailman/listinfo/r-help > > > PLEASE do read the posting guide! > > http://www.R-project.org/posting-guide.html > > > > > > > ______________________________________________ > > R-help at stat.math.ethz.ch mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide! > > http://www.R-project.org/posting-guide.html > > > >______________________________________________ >R-help at stat.math.ethz.ch mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide! >http://www.R-project.org/posting-guide.html