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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