On Tue, Aug 7, 2012 at 10:26 PM, Marc Schwartz marc_schwa...@me.com wrote:
since there are alpha-numerics present, whereas the first option will:
grepl([^[:alnum:]], ab%)
[1] TRUE
So, use the first option.
And I should start reading more carefully. The above works fine for me.
I ended up
On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
is.letter - function(x) grepl([[:alpha:]], x)
is.number - function(x) grepl([[:digit:]], x)
Quick follow-up question.
I'm always reluctant to create functions that would resemble the
method of a function (here, is() ),
On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
is.letter - function(x) grepl([[:alpha:]], x)
is.number - function(x) grepl([[:digit:]], x)
Another follow-up. To test for (non-)alphanumeric one would do the following:
x - c(letters, 1:26, '+', '-', '%^')
x[1:10] -
On Tue, Aug 7, 2012 at 4:28 AM, Liviu Andronic landronim...@gmail.com wrote:
On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
is.letter - function(x) grepl([[:alpha:]], x)
is.number - function(x) grepl([[:digit:]], x)
Quick follow-up question.
I'm always reluctant
On Aug 7, 2012, at 3:02 PM, Liviu Andronic landronim...@gmail.com wrote:
On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
is.letter - function(x) grepl([[:alpha:]], x)
is.number - function(x) grepl([[:digit:]], x)
Another follow-up. To test for (non-)alphanumeric
On Aug 7, 2012, at 3:18 PM, Marc Schwartz marc_schwa...@me.com wrote:
On Aug 7, 2012, at 3:02 PM, Liviu Andronic landronim...@gmail.com wrote:
On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
is.letter - function(x) grepl([[:alpha:]], x)
is.number - function(x)
On Tue, Aug 7, 2012 at 10:18 PM, Marc Schwartz marc_schwa...@me.com wrote:
That will get you values where punctuation characters are used, but there may
be other non-alphanumeric characters in the vector. There may be ASCII
control codes, tabs, newlines, CR, LF, spaces, etc. which would not
nzchar(x) !is.na(x)
No?
-- Bert
On Mon, Aug 6, 2012 at 9:25 AM, Liviu Andronic landronim...@gmail.com wrote:
Dear all
I'm pretty sure that I'm approaching the problem in a wrong way.
Suppose the following character vector:
(x[1:10] - paste(x[1:10], sample(1:10, 10), sep=''))
[1] a10 b7
Hello,
Fun as an exercise in vectorization. 30 times faster. Don't look, guess.
Gave it up? Ok, here it is.
is_letter - function(x, pattern=c(letters, LETTERS)){
sapply(x, function(y){
any(sapply(pattern, function(z) grepl(z, y, fixed=T)))
})
}
# test ascii codes, just one
On 08/06/2012 09:51 AM, Rui Barradas wrote:
Hello,
Fun as an exercise in vectorization. 30 times faster. Don't look, guess.
system.time(res0 - grepl([[:alpha:]], x))
user system elapsed
0.060 0.000 0.061
system.time(res1 - has_letter(x))
user system elapsed
3.728 0.008
Perhaps I am missing something, but why use sapply() when grepl() is already
vectorized?
is.letter - function(x) grepl([:alpha:], x)
is.number - function(x) grepl([:digit:], x)
x - c(letters, 1:26)
x[1:10] - paste(x[1:10], sample(1:10, 10), sep='')
x - rep(x, 1e3)
str(x)
chr [1:52000] a2
Hi,
Not sure whether this is you wanted.
x-letters
(x[1:10] - paste(x[1:10], sample(1:10, 10), sep=''))
x1-c(x,1:26)
x1
[1] a4 b3 c5 d2 e9 f6 g1 h8 i10 j7 k l
[13] m n o p q r s t u v w x
[25] y z 1 2 3 4 5 6 7 8 9 10
[37] 11 12
On Aug 6, 2012, at 12:06 PM, Marc Schwartz marc_schwa...@me.com wrote:
Perhaps I am missing something, but why use sapply() when grepl() is already
vectorized?
is.letter - function(x) grepl([:alpha:], x)
is.number - function(x) grepl([:digit:], x)
Sorry, typos in the above from my CP.
: Monday, August 06, 2012 12:07 PM
To: Rui Barradas
Cc: r-help
Subject: Re: [R] test if elements of a character vector contain letters
Perhaps I am missing something, but why use sapply() when grepl() is
already vectorized?
is.letter - function(x) grepl([:alpha:], x)
is.number - function(x
On Mon, Aug 6, 2012 at 6:42 PM, Bert Gunter gunter.ber...@gene.com wrote:
nzchar(x) !is.na(x)
No?
It doesn't work for what I need:
x
[1] a10 b8 c9 d2 e3 f4 g1 h7 i6 j5 k
l m n
[15] o p q r s t u v w x y
z 1 2
[29] 3 4 5 6 7 8 9 10 11 12
You probably mean grepl('[a-zA-Z]', x)
Regards,
Yihui
--
Yihui Xie xieyi...@gmail.com
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Mon, Aug 6, 2012 at 3:29 PM, Liviu Andronic landronim...@gmail.com wrote:
On Mon, Aug
On Mon, Aug 6, 2012 at 7:35 PM, Marc Schwartz marc_schwa...@me.com wrote:
is.letter - function(x) grepl([[:alpha:]], x)
is.number - function(x) grepl([[:digit:]], x)
This does exactly what I wanted:
x
[1] a10 b8 c9 d2 e3 f4 g1 h7 i6 j5 k
l m n
[15] o p q r s t u v
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