R help - Nov 2016 - Frequency of a character in a string

?nchar in the base R should also help...

On Mon, Nov 14, 2016 at 2:26 PM, Ismail SEZEN <sezenismail at gmail.com>
wrote:
>
> > On 14 Nov 2016, at 11:44, Ferri Leberl <ferri.leberl at gmx.at>
wrote:
> >
> >
> > Dear All,
> > Is there a function to count the occurences of a certain character in
a
> string resp. in a vector of strings?
> > Thank you in advance!
> > Yours, Ferri
> >
>
> library(stringr)
> ?str_count
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.
>
	[[alternative HTML version deleted]]

Yes, but it need some help, since nchar gives the length of the
*entire* string; e.g.

## to count "a" 's  :
> x <-(c("abbababba","bbabbabbaaaba"))
> nchar(gsub("[^a]","",x))
[1] 4 6

This is one of about 8 zillion ways to do this in base R if you don't
want to use a specialized package.

Just for curiosity: Can anyone comment on what is the most efficient
way to do this using base R pattern matching?

Cheers,
Bert
Bert Gunter

"The trouble with having an open mind is that people keep coming along
and sticking things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )


On Mon, Nov 14, 2016 at 5:52 AM, Brijesh Mishra
<brijeshkmishra at gmail.com> wrote:
> ?nchar in the base R should also help...
>
> On Mon, Nov 14, 2016 at 2:26 PM, Ismail SEZEN <sezenismail at
gmail.com> wrote:
>
>>
>> > On 14 Nov 2016, at 11:44, Ferri Leberl <ferri.leberl at
gmx.at> wrote:
>> >
>> >
>> > Dear All,
>> > Is there a function to count the occurences of a certain character
in a
>> string resp. in a vector of strings?
>> > Thank you in advance!
>> > Yours, Ferri
>> >
>>
>> library(stringr)
>> ?str_count
>>
>> ______________________________________________
>> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
>> 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.
>>
>
>         [[alternative HTML version deleted]]
>
> ______________________________________________
> R-help at r-project.org mailing list -- To UNSUBSCRIBE and more, see
> 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.

On Mon, 14 Nov 2016, Bert Gunter wrote:
> Yes, but it need some help, since nchar gives the length of the
> *entire* string; e.g.
>
> ## to count "a" 's  :
>
>> x <-(c("abbababba","bbabbabbaaaba"))
>> nchar(gsub("[^a]","",x))
> [1] 4 6
>
> This is one of about 8 zillion ways to do this in base R if you don't
> want to use a specialized package.
>
> Just for curiosity: Can anyone comment on what is the most efficient
> way to do this using base R pattern matching?
>
Most efficient? There probably is no uniformly most efficient way to do 
this as the timing will depend on the distribution of "a" in the atoms
of
any vector as well as the length of the vector.

But here is one way to avoid the regular expression matching:

lengths(strsplit(paste0("X", x,
"X"),"a",fixed=TRUE)) - 1


Chuck