Thank you so much, Bill and Arun!
Dimitri
On Wed, Mar 13, 2013 at 9:36 PM, arun wrote:
> HI,
> Try this:
> mydf1<- mydf
> mydf1[]<-lapply(1:3,function(i) {mydf[which(i== myindex),i]<-1; mydf[,i]})
> mydf1
> # c1 c2 c3
> #1 1 NA NA
> #2 NA 1 NA
> #3 NA NA 1
> #4 NA 1 NA
> #5 1 NA NA
>
>
HI,
Try this:
mydf1<- mydf
mydf1[]<-lapply(1:3,function(i) {mydf[which(i== myindex),i]<-1; mydf[,i]})
mydf1
# c1 c2 c3
#1 1 NA NA
#2 NA 1 NA
#3 NA NA 1
#4 NA 1 NA
#5 1 NA NA
identical(mydf1,mygoal)
#[1] TRUE
A.K.
- Original Message -
From: Dimitri Liakhovitski
To: r-help
Cc
Try looping over columns, as in
fDF <- function (x, column)
{
stopifnot(length(dim(x))==2, all(column > 0), all(column <= ncol(x)),
length(column) == nrow(x))
u <- unique(column)
tmp <- split(seq_along(column), factor(column, levels = u))
for (i in seq_along(tmp)) {
x[ tmp
3 matches
Mail list logo
