The problem specification here seems flawed. The normal return values are
unitless, but the abnormal return values have units. I would expect that
returning NA when the standard deviations are zero will give rather less
surprising results.
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"iza.ch1" <iza....@op.pl> wrote:
>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.
>> >
>>
>______________________________________________
>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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