Using the builtin month.abb try this:
regexpr("ov", month.abb) > 0
Although not needed here, if "ov" were a character string that could have
special characters such as . and * that have special meaning in a regular
expression then do this to prevent such interpretation:
regexpr("ov", month.abb, fixed = TRUE) > 0
See ?regexpr
On 1/20/07, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> I'm a relative R novice, and sometimes the simple things trip me up.
>
> Suppose I have
>
> a <- c("apple", "pear")
>
> and I want a logical vector of whether each of these strings contains
> "ear" (in this case, F T). What is the idiom?
>
> Quizzically,
> Mark Lindeman
>
> ______________________________________________
> [email protected] mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
______________________________________________
[email protected] mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
