This should do the same thing
random.del <- function (x, n.keeprows, del.percent){
del<-function(col){
col[sample.int(length(col),length(col)*del.percent/100)]<-NA
col
}
change<-n.keeprows:nrow(x)
x[change,]<-lapply(x[change,],del)
x
}
This is faster because it's vectorized.
[1] "Mine"
user system elapsed
0.004 0.000 0.002
[1] "Yours"
user system elapsed
1.172 0.020 1.193
Tom
On Sat, Apr 23, 2011 at 8:37 PM, sneaffer <sneaf...@mail.ru> wrote:
>
> Hello R-world,
> Please, help me to get round my little mess
> I have a data.frame in which I'd rather like some values to be NA for the
> future imputation process.
>
> I've come up with the following piece of code:
>
> random.del <- function (x, n.keeprows, del.percent){
> n.items <- ncol(x)
> k <- n.items*(del.percent/100)
> x.del <- x
> for (i in (n.keeprows+1):nrow(x)){
> j <- sample(1:n.items, k)
> x.del[i,j] <- NA
> }
> return (x.del)
> }
>
> The problems is that random.del turns out to be slow on huge samples.
> Is there any other more effective/charming way to do the same?
>
> Thanks,
> Sergey
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/How-to-erase-replace-certain-elements-in-the-data-frame-tp3470883p3470883.html
> Sent from the R help mailing list archive at Nabble.com.
>
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