[R] test if elements of a character vector contain letters

[Marc Schwartz]

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 <ruipbarradas at 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