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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