Use 'which(...arr.ind=T)'
> x.1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 6 10 3 4 10 7 9 8 4 10
[2,] 8 7 4 7 4 8 3 NA 3 4
[3,] 7 7 10 10 3 5 3 2 2 2
[4,] 3 4 5 10 10 2 6 9 4 5
[5,] 3 5 9 5 6 NA 3 NA 6 7
[6,] 9 6 10 5 10 4 2 10 NA 5
[7,] 5 2 5 10 3 7 6 4 6 8
[8,] 2 6 1 8 9 2 7 8 3 8
[9,] 9 1 4 9 8 10 2 NA 1 7
[10,] 2 4 8 7 NA 4 3 NA 5 5
> x.4
[1] 5.5 5.5 5.0 7.5 8.0 5.0 3.0 8.0 4.0 6.0
> Med <- apply(x.1, 2, median, na.rm=T) # get median
> Ind <- which(is.na(x.1), arr.ind=T) # determine which are NA
> x.1[Ind] <- Med[Ind[,'col']] # replace with median
> x.1
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 6 10 3 4 10 7 9 8 4 10
[2,] 8 7 4 7 4 8 3 8 3 4
[3,] 7 7 10 10 3 5 3 2 2 2
[4,] 3 4 5 10 10 2 6 9 4 5
[5,] 3 5 9 5 6 5 3 8 6 7
[6,] 9 6 10 5 10 4 2 10 4 5
[7,] 5 2 5 10 3 7 6 4 6 8
[8,] 2 6 1 8 9 2 7 8 3 8
[9,] 9 1 4 9 8 10 2 8 1 7
[10,] 2 4 8 7 8 4 3 8 5 5
>
On 9/27/05, Weiwei Shi <[EMAIL PROTECTED]> wrote:
>
> Hi,
> I have the following codes to replace missing using median, assuming
> missing
> only occurs on continuous variables:
>
> trn1<-read.table('trn1.fv', header=F, na.string='.', sep='|')
>
> # median
> m.trn1<-sapply(1:ncol(trn1), function(i) median(trn1[,i], na.rm=T))
>
> #replace
> trn2<-trn1
> for (each in 1:nrow(trn1)){
> index.missing=which(is.na(trn1[each,]))
> trn2[each,]<-replace(trn1[each,], index.missing, m.trn1[index.missing])
> }
>
>
> Anyone can suggest some ways to improve it since replacing 10 takes 1.5sec:
> > system.time(for (each in 1:10){index.missing=which(is.na(trn1[each,]));
> trn2[each,]<-replace(trn1[each,], index.missing, m.trn1[index.missing]);})
> [1] 1.53 0.00 1.53 0.00 0.00
>
>
> Another general question is
> are there some packages in R doing missing handling?
>
> Thanks,
>
> --
> Weiwei Shi, Ph.D
>
> "Did you always know?"
> "No, I did not. But I believed..."
> ---Matrix III
>
> [[alternative HTML version deleted]]
>
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--
Jim Holtman
Cincinnati, OH
+1 513 247 0281
What the problem you are trying to solve?
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