Try this:
grep("^([^a]|a[^b]|ab[^c])*.{0,2}$", x, perl = TRUE)
On Sun, Jan 18, 2009 at 2:37 PM, Rau, Roland <r...@demogr.mpg.de> wrote:
> Thank you very much to all of you for your fast and excellent help.
> Since the "-grep(...)" solution seems to be favored by most of the answers,
> I just wonder if there is really no regular expression which does the job?!?
>
> Thanks again,
> Roland
>
>
>
> -----Original Message-----
> From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
> Sent: Sun 1/18/2009 8:28 PM
> To: Rau, Roland
> Cc: r-help@r-project.org
> Subject: Re: [R] regex -> negate a word
>
> Try this:
>
> # indexes
> setdiff(seq_along(x), grep("abc", x))
>
> # 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 Sun, Jan 18, 2009 at 1:35 PM, Rau, Roland <r...@demogr.mpg.de> wrote:
>> Dear all,
>>
>> let's assume I have a vector of character strings:
>>
>> x <- c("abcdef", "defabc", "qwerty")
>>
>> What I would like to find is the following: all elements where the word
>> 'abc' does not appear (i.e. 3 in this case of 'x').
>>
>> Since I am not really experienced with regular expressions, I started
>> slowly and thought I find all word were 'abc' actually does appear:
>>
>>> grep(pattern="abc", x=x)
>> [1] 1 2
>>
>> So far, so good. Now I read that ^ is the negation operator. But it can
>> also denote the beginning of a string as in:
>>
>>> grep(pattern="^abc", x=x)
>> [1] 1
>>
>> Of course, we need to put it inside square brackets to negate the
>> expression [1]
>>> grep(pattern="[^abc]", x=x)
>> [1] 1 2 3
>>
>> But this is not what I want either.
>>
>> I'd appreciate any help. I assume this is rather easy and
>> straightforward.
>>
>> Thanks,
>> Roland
>>
>>
>> [1] http://www.zytrax.com/tech/web/regex.htm: The ^ (circumflex or
>> caret) inside square brackets negates the expression....
>>
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>
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