JavaScript Exercise: Find the Number of Unique Letters in a String

Discussion

1. Wes

   This looks like the beginning of a simple javascript word game…you get the letters and amount of times they occur and then have to figure out the string.
2. Jonathan Snook

   Here’s my attempt:

   var str = 'hellodavidthisisatestofobjectusage';
   var uniq = '';
   for (var i = 0; i< str.length; i++) {
    if(uniq.indexOf( str[i] ) == -1){
      uniq += str[i];
    }
   }
   console.log(uniq.length);
   

   This is probably self explanatory but it goes through and checks if the character is in the “uniq”ue string. If not, it adds it to the end. Then it just determines the length of the unique string, since all characters contained within are… unique. It’s basically the same as yours but it saves having to loop through the object keys just to get a count. I feel like there’s a regex way of doing this but I’m not sure how.
3. Rakesh Juyal [RJ]

   Walsh, u certainly know how to perform the easiest task in the toughest way.
4. David Walsh

   My intention Rakesh was to keep the number of times each appeared — which I now realize I didn’t state. Jonathan’s solution looks great.
5. James

   My attempt:

   function letterCount(str) {
       var ret = {},
           len = str.length;
       str = str.split('').reverse();
       while (len--) ret[str[len]] = ret[str[len]] + 1 || 1;
       return ret;
   }
   

   It iterates (in reverse) through the string and adds a new property of that character to the returned object. It iterates in reverse because that’s quicker than the alternative – i.e. having to check the length property on every iteration…
6. grigri

   function getCharCounts(s) {
     var letters = {};
     s.replace(/\S/g, function(l){letters[l] = (isNaN(letters[l]) ? 1 : letters[l] + 1);});
     return letters;
   }
   
   console.log(getCharCounts('hellodavidthisishowtodoitwithregularexpressions'));
   
7. James

   I forgot to mention, my function returns a very usable object, constructed like this: { h: 2, e: 4, l: 2, o: 3, etc…. }
8. David Walsh

   Oooh, I like that James.
9. Keeto

   This is fun. :)

   var str = “hellodavidthisisatestofobjectusage”;
   var letters = {};

   for (var x = 0, y = str.length; x < y; x++) {
   letters[str[x]] = letters[str[x]] + 1 || 1;
   }
10. Keeto

    hah! I just realized that the whole thing could be made smaller:

    var str = “hellodavidthisisatestofobjectusage”, letters = {};
    for (var x = 0, y = str.length; x < y; x++) letters[str[x]] = letters[str[x]] + 1 || 1;

    ^^b
11. grigri

    Nice one keeto, this makes mine smaller too:

    var str='thisisultrashortwithregexps', letters={};
    s.replace(/\S/g, function(l){letters[l]=letters[l]+1||1);});
12. Keeto

    Cute one, grigri. Although I forgot I could make mine even shorter:

    var str=’thisisultrashortwithregexps’, letters={};
    for (var x in str) letters[str[x]] = letters[str[x]] + 1 || 1;

    More David! More!
13. David Walsh

    Great solutions everyone! I added the Object.size() to be able to show a result so mine is pretty close to everyone’s.

    More in the future?
14. rakesh juyal

    @Gri: nice use of regex.
15. Guillermo Rauch

    I like @grigri, with the correction of \w instead of \S because the OP said ‘letters’ :)

    Taking that into account, that’s the solution that most elegantly lets you count letters alone.
16. Raphael Montanaro

    This might be a great read:

    http://dreaminginjavascript.wordpress.com/2008/08/22/eliminating-duplicates/

    Not the same thing, but brilliant as well.
17. Mark Sanborn

    I am currently learning python so I did:

    s = ‘thisissomestringwithsomechars’
    count = {}
    for c in s:
    count[c] = count.get(c,0) + 1
    print count

    Which outputs:
    {‘a’: 1, ‘c’: 1, ‘e’: 2, ‘g’: 1, ‘i’: 4, ‘h’: 3, ‘m’: 2, ‘o’: 2, ‘n’: 1, ‘s’: 6, ‘r’: 2, ‘t’: 3, ‘w’: 1}
18. Sucheta

    I m new to javascript.
    I want to find no of unique letters in a string using javascript.
    pls help me out
19. Chandra Sekhar

    Using reduce,

    function getLetterCount(arr){
        return arr.reduce(function(prev,next){
            prev[next] = (prev[next] + 1) || 1;
            return prev;
        },{});
    }
    
    console.log(getLetterCount("dfglksdhfglhjhkdjfhgfdg".split('')));