Hi Rui
Thanks a lot. It works perfect.
Izzie
> Hello,
>
> Try the following.
>
>
> standardised.abnormal.returns <-
> lapply(seq_len(ncol(abnormal.returns)),function(i) {
> if(standard.deviation[i,1] == 0)
> abnormal.returns[,i]
> else
> abnormal.returns[,i]/standard.deviation[i,1]
> })
>
>
>
> Hope this helps,
>
> Rui Barradas
>
> Em 23-07-2013 22:58, iza.ch1 escreveu:
> > Hi you all
> >
> > I have a question regarding the function. In my function I divide the
> > values by the standard errors and sometimes the standard error is equal to
> > zero and I get the result NA. Can I write the function in the way that if
> > the outcome of the function is zero then the function is not conducted and
> > it stays the value (not divided by standard errors)? the code for my
> > function is the following:
> >
> > standardised.abnormal.returns<-lapply(seq_len(ncol(abnormal.returns)),function(i)
> > {abnormal.returns[,i]/standard.deviation[i,1]})
> >
> > Thank you all for the 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.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
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