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

[arun]

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"  "13"  "14"  "15"  "16"  "17"  "18"  "19"  "20"  "21"  "22" 
[49] "23"  "24"  "25"  "26" 


 grepl("^[[:alpha:]][[:digit:]]",x1)
 [1]  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE  TRUE FALSE FALSE
[13] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[25] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[37] FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE
[49] FALSE FALSE FALSE FALSE

A.K.



----- Original Message -----
From: Liviu Andronic <landronimirc at gmail.com>
To: "r-help at r-project.org Help" <r-help at r-project.org>
Cc: 
Sent: Monday, August 6, 2012 12:25 PM
Subject: [R] test if elements of a character vector contain letters

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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