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" "b10" "c8" "d3" "e6" "f1" "g5" ...
> system.time(is.letter(x))
user system elapsed
0.011 0.000 0.010
> system.time(is.number(x))
user system elapsed
0.010 0.000 0.011
Regards,
Marc Schwartz
On Aug 6, 2012, at 11:51 AM, Rui Barradas <ruipbarra...@sapo.pt> wrote:
> 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 loop.
> has_letter <- function(x){
> sapply(x, function(y){
> y <- as.integer(charToRaw(y))
> any((65 <= y & y <= 90) | (97 <= y & y <= 122))
> })
> }
>
> x <- c(letters, 1:26)
> x[1:10] <- paste(x[1:10], sample(1:10, 10), sep='')
> x <- rep(x, 1e3)
>
> t1 <- system.time(is_letter(x))
> t2 <- system.time(has_letter(x))
> rbind(t1, t2, t1/t2)
> user.self sys.self elapsed user.child sys.child
> t1 15.69 0 15.74 NA NA
> t2 0.50 0 0.50 NA NA
> 31.38 NaN 31.48 NA NA
>
>
> Em 06-08-2012 17:25, Liviu Andronic escreveu:
>> 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" "c2" "d3" "e6" "f1" "g5" "h8" "i9" "j4"
>>> x
>> [1] "a10" "b7" "c2" "d3" "e6" "f1" "g5" "h8" "i9" "j4" "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" "13"
>> "14" "15" "16"
>> [43] "17" "18" "19" "20" "21" "22" "23" "24" "25" "26"
>>
>>
>> How do you test whether the elements of the vector contain at least
>> one letter (or at least one digit) and obtain a logical vector of the
>> same dimension? I came up with the following awkward function:
>> is_letter <- function(x, pattern=c(letters, LETTERS)){
>> sapply(x, function(y){
>> any(sapply(pattern, function(z) grepl(z, y, fixed=T)))
>> })
>> }
>>
>>> is_letter(x)
>> a10 b7 c2 d3 e6 f1 g5 h8 i9 j4 k
>> l m n o
>> TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>> TRUE TRUE TRUE TRUE
>> p q r s t u v w x y z
>> 1 2 3 4
>> TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>> FALSE FALSE FALSE FALSE
>> 5 6 7 8 9 10 11 12 13 14 15
>> 16 17 18 19
>> FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
>> FALSE FALSE FALSE FALSE
>> 20 21 22 23 24 25 26
>> FALSE FALSE FALSE FALSE FALSE FALSE FALSE
>>> is_letter(x, 0:9) ##function slightly misnamed
>> a10 b7 c2 d3 e6 f1 g5 h8 i9 j4 k
>> l m n o
>> TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE FALSE
>> FALSE FALSE FALSE FALSE
>> p q r s t u v w x y z
>> 1 2 3 4
>> FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
>> TRUE TRUE TRUE TRUE
>> 5 6 7 8 9 10 11 12 13 14 15
>> 16 17 18 19
>> TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>> TRUE TRUE TRUE TRUE
>> 20 21 22 23 24 25 26
>> TRUE TRUE TRUE TRUE TRUE TRUE TRUE
>>
>>
>> Is there a nicer way to do this? Regards
>> Liviu
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