gsub('.+; (.+);.+','\\1',x)
or if you just want the value out:
gsub('.+; Surv\\(months\\): ([0-9]+);.+','\\1',x)
You can also look at strsplit:
> strsplit(x,';')
[[1]]
[1] "99-625: Cell type: S" " Surv(months): 21" "
STATUS(0=alive, 1=dead): 1"
> lapply(strsplit(x,';'),'[',2)
[[1]]
[1] " Surv(months): 21"
But i would follow David's second suggestion and just read them in with
sep=';' instead.
Justin
On Wed, Feb 29, 2012 at 11:24 AM, Fred G <bayespoker...@gmail.com> wrote:
> Computer Friends,
>
> with the following example lines:
>
> [107] "98-610: Cell type: S; Surv(months): 6; STATUS(0=alive, 1=dead): 1"
>
> [108] "99-625: Cell type: S; Surv(months): 21; STATUS(0=alive, 1=dead): 1"
>
> i want to be able to isolate the number of months of survival for each row.
>
> is there a regular expression that can find the first instance of a ";",
> delete everything in front of it-- and find the second instance of an ";"
> and delete everything behind it? in python there is a function line.find(),
> would be grateful to hear the R equiv; or, any other better alternatives to
> get the number of months of survival stored as a variable.
>
> Much Thank You!
>
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>
> ______________________________________________
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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