JavaScript function returning unique values sorted by count

Question

I am looking for a concise JavaScript function which takes some values and returns the unique values sorted by the number of occurrences, the most frequent first.

For example, if the input is the array [3,2,2,2,2,1,2,1] then the output should be [2,1,3].

What other collections do you mean? There are only objects ({}) and arrays ([]), and I'm not sure how you'd apply this to objects. — Matt Ball
– Matt Ball, Commented Feb 14, 2012 at 16:23
Your probably right, I didn't think so much about that, I have removed that stuff from the original question. — user1069609
– user1069609, Commented Feb 14, 2012 at 16:29

Jamund Ferguson · Accepted Answer · 2012-02-14 18:25:38Z

17

Here's my first stab. I bet we can make it simpler, but this seems okay.

 function fancyUnique(arr) {

   var counts = {}; // store counts for each value
   var fancy = []; // put final results in this array
   var count = 0; // initialize count

   // create counts object to store counts for each value of arr
   for (var i = 0; i < arr.length; i++) {
     count = counts[arr[i]] || 0;  // use existing count or start at 0
     count++; // increment count
     counts[arr[i]] = count; // update counts object with latest count
   }

   // take all keys from counts object and add to array
   // also: object keys are string, so must parseInt()
   for (var key in counts) {
     fancy.push(parseInt(key, 10)); 
   }

   // sort results array in highest to lowest order
   return fancy.sort(function(a, b) { 
     return counts[b] - counts[a]; 
   })
 }


fancyUnique([22,22,1,1,1,1]) // [ 1, 22 ]

edited Feb 14, 2012 at 18:25

answered Feb 14, 2012 at 16:33

Jamund Ferguson

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5 Comments

user1069609 Over a year ago

I like it. I can't see any way to improve it actually. Did you just write that from scratch or is it from your lib? :)

user1069609 Over a year ago

It's a little cryptic, e.g. what happens in the count = uniques[arr[i]] || 0; line, could you add a few comments please

Jamund Ferguson Over a year ago

I will add a comment. It just says basically if there is no count yet for a specific value then initialize it at 0.

nzifnab Over a year ago

Mm... stacksort took this: [[5,8,4,9], [12,32,6,5,8], [4,2]] and turned it into this: [5,12,4] using this answer. Little bit odd.

AJMansfield Over a year ago

Fails on the array [1, 2, 3, 4, 1, 4, 32, 1, 3, 1, 1.0004, 5.03274, 2.93456e27, 2.93456e27, 2.93456e39, 0, 1], output should be [1, 3, 4, 2.93456e27, 2, 3, 32, 1.0004, 5.03274, 2.93456e39, 0] but instead was [1,1,3,4,0,2,32,5,2,2].

Akila Galappatthy · Accepted Answer · 2013-08-30 09:33:35Z

0

var data=[3,2,2,2,2,1,2,1];
console.log(countUnique(data));

function countUnique(data){
  var count={};
  $(data).each(function(index){                      
       count[data[index]] = count[data[index]]+1 || 1;             
  });
  return count;
}

output= "Object {1: 2, 2: 5, 3: 1}"

This works for strings as well.

Reading the output can be done using the following loop

   for(var key in count)
    {
             console.log(key); // prints the index of the array
             console.log(count[key]); //prints the number of occurrences  

    }

answered Aug 30, 2013 at 9:33

Akila Galappatthy

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