Gabor Grothendieck escribió:
> Try this where m is the matrix:
>
> 100 * colMeans(m > 5 & m < 9, na.rm = TRUE)
Dear Gabor,
Just perfect!
Thanks a lot,
Antonio
>
>
> On 11/1/06, antonio rodriguez <[EMAIL PROTECTED]> wrote:
>> Hi,
>>
>> Having a matrix F.zoo (6575,189) with NA's in some columns I'm trying to
>> extract from each column the percent of days within an specific range,
>> so I've wrote this procedure:
>>
>> length(subset(F.zoo[,86],(F.zoo[,86]>=5) & (F.zoo[,86]<=
>> 9)))/(length(F.zoo[,86])-length(subset(F.zoo[,86],is.na(F.zoo[,86]))))*100
>>
>>
>> But to do this for each column (189) is pretty hard, so I want to write
>> a function in order to perform this automatically, such I have the
>> percent value corresponding to a specific column. I' tried these two
>> formulas but I can't get it. I think the problem is how to set the
>> initial values for the loop:
>>
>> Formula1:
>>
>> nnn<-function(x){for (i in F.zoo[,i]){
>> print(length(subset(F.zoo[,i],(F.zoo[,i]>=5) & (F.zoo[,i]<=
>> 9)))/(length(F.zoo[,i])-length(subset(F.zoo[,i],is.na(F.zoo[,i]))))*100)
>> }
>> }
>>
>> Formula 2:
>>
>> H<-t(matrix(1,189))
>>
>> nnn<-function(x){for (i in col(H){
>> print(length(subset(F.zoo[,i],(F.zoo[,i]>=5) & (F.zoo[,i]<=
>> 9)))/(length(F.zoo[,i])-length(subset(F.zoo[,i],is.na(F.zoo[,i]))))*100)
>> }
>> }
>>
>> Thanks,
>>
>> Antonio
>>
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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.
>>
>
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