Can someone help me find a better alternative to this Javascript for loop I've written?

Question

Thanks in advance to anyone who can think of a more efficient or a better way to do what my Javascript code below does:

var availableCharacters=Array("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9");

for (counter=0; counter<availableCharacters.length; counter++){
  if(availableCharacters[counter]=="i"||
     availableCharacters[counter]=="l"||
     availableCharacters[counter]=="I"||
     availableCharacters[counter]=="L"||
     availableCharacters[counter]=="1"||
     availableCharacters[counter]=="0"||
     availableCharacters[counter]=="O"){
      availableCharacters.splice(counter, 1);
    }
}

What I'm trying to do is run through an array and remove any elements in that array that are "i", "l", "I", "L", "1", "0" or "O". Whilst this does work it seems like it might be slow and a bit cumbersome. If there is a better way? If not then not a problem, but most of the time when I do something that doesnt seem right to me, it's not! So I thought I'd ask S.O.

Thanks :)

possible duplicate of [Check variable equality against a list of values](http://stackoverflow.com/questions/4728144/check-variable-equality-against-a-list-of-values) — pimvdb, Jun 19 '12 at 21:21

score 6 · Accepted Answer · edited Jun 19 '12 at 21:25

6

More recent browsers support Array.filter:

var availableCharacters = ........;
availableCharacters = availableCharacters.filter(function(a) {
  return !a.match(/[ilLI10O]/);
});

For older browsers, however, the for loop given by Mark Linus is good.

edited Jun 19 '12 at 21:25

Alexander

23,432
11
63
73

answered Jun 19 '12 at 21:24

Niet the Dark Absol

320,036
81
464
592

Thank you, I'm a fan of short and easy to read code. As it mentioned below that Regexes aren't taxing on modern day browsers, I think this fixes my problem very well. Can anyone see mobile Safari having a problem with looping though an 250-300 element array with this method? It's not critical but if this code would perform well on iOS devices aswell that would be a bonus! – GhostInTheSecureShell Jun 19 '12 at 21:33

Danilo Valente · Answer 2 · 2012-06-19T21:32:09.240

4

var availableCharacters=["a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"];
for (counter=0; counter<availableCharacters.length; counter++){
    if(/[ilIL10O]/.test(availableCharacters[counter])){
        availableCharacters.splice(counter, 1);
    }
}

edited Jun 19 '12 at 21:32

answered Jun 19 '12 at 21:21

Danilo Valente

11,270
8
53
67

Yes, a raw and simple regex :) – Danilo Valente Jun 19 '12 at 21:23
Brilliant, thanks :) Are Regexes demanding on browsers at all? – GhostInTheSecureShell Jun 19 '12 at 21:24
Yes, they are. Also, take a look on Kolink's answer – Danilo Valente Jun 19 '12 at 21:25
modern browser engines deal with regexes surprizingly fast `.split('a').join('b')` is slower then `.replace(/a/g,'b')`, for example. PS: instead of using an object constructor `Array` without the `new` keyword, use brackets... – Elias Van Ootegem Jun 19 '12 at 21:27
Even cleaner might be using one string and doing `.split("")`. – pimvdb Jun 19 '12 at 21:29

score 1 · Answer 3 · answered Jun 19 '12 at 21:23

1

Alternative solution with simple filter list:

var availableCharacters=Array("a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z", "0", "1", "2", "3",     "4", "5", "6", "7", "8", "9");
var filter = ['i', 'l', 'I', 'L', '1', '0', 'O'];
for (counter=0; counter<availableCharacters.length; counter++){
    if(filter.indexOf(availableCharacters[counter]) >= 0) {
        availableCharacters.splice(counter, 1);
    }
}

answered Jun 19 '12 at 21:23

jsalonen

29,593
15
91
109

No, `in` checks for keys which are `0, 1, 2, ...` in arrays. Use `indexOf` instead. – pimvdb Jun 19 '12 at 21:23
Very true. should be fixed now. – jsalonen Jun 19 '12 at 21:24

Cuadue · Answer 4 · 2012-06-20T01:08:40.160

0

If you're willing to use jQuery you can use grep() and write it like this:

Fiddle here

var availableChars = ["a", "b", "c", '1', 'i', 'o', "9"];
var result = $.grep(availableChars, function(c) {
    return !c.match(/[^il10o]/i)
})
document.write(result)

edited Jun 20 '12 at 01:08

answered Jun 19 '12 at 21:29

Cuadue

3,769
3
24
38

1

This essentially returns an array of the characters in the regexp. I think you meant to negate. – pimvdb Jun 19 '12 at 21:31

score 0 · Answer 5 · answered Jun 20 '12 at 01:48

filter is nice, but a simple string replace works without a net-

   var availableCharacters=["a", "b", "c", "d", "e", "f", "g", "h", "i", 
   "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", 
   "x", "y", "z", "0", "1", "2", "3", "4", "5", "6", "7", "8", "9"];

  var allowed= availableCharacters.join('').replace(/[iLl10O]/g,'').split('');

  returned value: (Array)
  a,b,c,d,e,f,g,h,j,k,m,n,o,p,q,r,s,t,u,v,w,x,y,z,2,3,4,5,6,7,8,9

Can someone help me find a better alternative to this Javascript for loop I've written?

5 Answers5