Hello,
Try the following.
A <- c(
"10/20/30",
"40/20",
"60/10/10/5",
"80/10")
fun <- function(X){
xname <- deparse(substitute(X))
s <- strsplit(X, "/")
n <- max(sapply(s, length))
tmp <- numeric(n)
f <- function(x){
x <- as.numeric(x)
m <- length(x)
tmp[n] <- x[m]
tmp[seq_len(m - 1)] <- x[seq_len(m - 1)]
tmp
}
res <- do.call(rbind, lapply(s, f))
colnames(res) <- paste0(xname, 1:ncol(res))
data.frame(res)
}
fun(A)
Hope this helps,
Rui Barradas
Em 31-08-2012 17:10, Sapana Lohani escreveu:
Hi,
A column of my df looks like
A
10/20/30
40/20
60/10/10/5
80/10
I want to split it such that the last column has the last composition and if
there are not enough the middle columns get the 0s. That way my df would look
like
A1 A2 A3 A4
10 20 0 30
40 0 0 20
60 10 10 5
80 0 0 10
How can I do that ??
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and provide commented, minimal, self-contained, reproducible code.
