R help - Apr 2005 - is there a function like %in% for characters?

like:

"a" %in% "abcd"
TRUE

Thanks.

I suppose here's one way:
> hasChar <- function(x, y) { length(grep(x, y)) > 0 }
> hasChar("a", "abcd")
[1] TRUE
> hasChar("e", "abcd")
[1] FALSE

Andy
> From: Terry Mu
> 
> like:
> 
> "a" %in% "abcd"
> TRUE
> 
> Thanks.
> 
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>

Terry Mu wrote on 4/2/2005 9:38 PM:
> like:
> 
> "a" %in% "abcd"
> TRUE
> 
> Thanks.
> 

See ?regexpr.

regexpr("a", "abcd") > 0

However, the first argument is not vectorized so you may also need 
something like:

 > sapply(c("a", "b", "e"), regexpr,
c("abcd", "bcde")) > 0
          a    b     e
[1,]  TRUE TRUE FALSE
[2,] FALSE TRUE  TRUE

Be sure to read up on regular expressions if pursuing this option.

HTH,

--sundar

Sundar Dorai-Raj wrote:
> 
> 
> Terry Mu wrote on 4/2/2005 9:38 PM:
> 
>> like:
>>
>> "a" %in% "abcd"
>> TRUE
>>
>> Thanks.
>>
> 
> 
> See ?regexpr.
> 
> regexpr("a", "abcd") > 0
> 
> However, the first argument is not vectorized so you may also need 
> something like:
> 
>  > sapply(c("a", "b", "e"), regexpr,
c("abcd", "bcde")) > 0
>          a    b     e
> [1,]  TRUE TRUE FALSE
> [2,] FALSE TRUE  TRUE
Alternatively you could use strsplit to split the original string into 
individual characters then fall back on %in%.  The only complication is 
that strsplit will return a list of one character vector and you must 
unlist it to get the character vector itself.

 > strsplit("abcf", "")
[[1]]
[1] "a" "b" "c" "f"

 > "a" %in% unlist(strsplit("abcf", ""))
[1] TRUE
 > "a" %in% unlist(strsplit("bcdf", ""))
[1] FALSE