JavaScript Group By Array

Question

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array_count_values for javascript instead

Let's say I have simple JavaScript array like the following:

var array = ['Car', 'Car', 'Truck', 'Boat', 'Truck'];

I want to group and count of each so I would expect a key/value map of:

{
  Car   : 2,
  Truck : 2,
  Boat  : 1
}

Rudolf Mühlbauer · Accepted Answer · 2012-10-14 11:41:06Z

59

var arr = [ 'Car', 'Car', 'Truck', 'Boat', 'Truck' ];
var hist = {};
arr.map( function (a) { if (a in hist) hist[a] ++; else hist[a] = 1; } );
console.log(hist);

results in

{ Car: 2, Truck: 2, Boat: 1 }

This works, too:

hist = arr.reduce( function (prev, item) { 
  if ( item in prev ) prev[item] ++; 
  else prev[item] = 1; 
  return prev; 
}, {} );

edited Oct 14, 2012 at 11:41

answered Oct 13, 2012 at 12:48

Rudolf Mühlbauer

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6 Comments

Kat Lim Ruiz Over a year ago

first solution doesn't work for values that are not compatible with variable names like GUID or integers, though it is great

ludo Over a year ago

Just a note, according to developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/… map creates a copy of the array. In your case you're not even using that copy, so it will be better to doe is like so: arr.forEach(function (a) { if (a in hist) hist[a] ++; else hist[a] = 1; }) because forEach will not create a new copy

user3116322 Over a year ago

If you are using Undescore, instead of arr.reduce you can also try _.countBy(list, iterator) underscorejs.org/#countBy

blazkovicz Over a year ago

thanks for exampla of creating dict out of array with reduce

Daniel Over a year ago

You should use Array.prototype.forEach() instead of Array.prototype.map() in your first solution.

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Shubhanshu Mishra · Accepted Answer · 2012-10-13 12:53:35Z

6

You can loop through each index and save it in a dictionary and increment it when every that key is found.

count = {};
for(a in array){
  if(count[array[a]])count[array[a]]++;
  else count[array[a]]=1;
}

Output will be:

Boat: 1
Car: 2
Truck: 2

answered Oct 13, 2012 at 12:53

Shubhanshu Mishra

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