In that case just add fixed = TRUE
On Sun, Jan 18, 2009 at 2:58 PM, Wacek Kusnierczyk
<waclaw.marcin.kusnierc...@idi.ntnu.no> wrote:
> Gabor Grothendieck wrote:
>> Try this:
>>
>>
>> # values
>> setdiff(x, grep("abc", x, value = TRUE))
>>
>> Another possibility is:
>>
>> z <- "abc"
>> x0 <- c(x, z) # to handle no match case
>> x0[- grep(z, x0)] # values
>>
>
> on quick testing, these two and the if-based version have comparable
> runtime, with a minor win for the last one, and if the input is moderate
> this makes no real difference.
>
> however, the second solution above is likely to fail if the pattern is
> more complex, e.g., contains a character class or a wildcard:
>
> strings = c("xyz")
> pattern = "a[a-z]"
> strings[-grep(pattern, c(strings, pattern))]
> # character(0)
>
>
> vQ
>
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