Hello,

Try the following.


x <- c("slope (60/25/15)",
"slope (90/10)",
"slope (40/35/15/10)",
"slope (40/25/25/10)")
dat <- data.frame(X = x)

lst <- unlist(lapply(dat, function(.x) gsub("slope \\((.*)\\)$", "\\1", .x)))
lst <- strsplit(lst, "/")
keep <- seq_len(max(unlist(lapply(lst, length))))
vec <- rep("0", length(keep))
result <- do.call(rbind, lapply(lst, function(.x) as.integer(c(.x, vec)[keep])))


Now, in your case, just substitute 'dat' by the name of your vector.
(I think this solution is too complicated but it predicts for the case where the input vector is in fact a list.)

Hope this helps,

Rui Barradas

Em 17-08-2012 06:05, Sapana Lohani escreveu:
Hi I am new to R so am struggling with the commands here

I have one column in a table that looks like


slope (60/25/15)
slope (90/10)
slope (40/35/15/10)
slope (40/25/25/10)
I want to have 4 columns with just the number inside the parenthesis. when 
there is no number that cell can have 0. I want the output like this

60 25 15 0
90 10 0 0
40 35 15 10
40 25 25 10

Can somebody help me??
Thanks
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