Partitioning in JavaScript [duplicate]

Question

Please consider an array such as :

arrayAll = [1,2,3,4,5,6,7,8,9]

Is there a package that enable to do partitioning to obtain :

arrayALLPartionned = [[1,2,3],[4,5,6],[7,8,9]]

I can see how to do this with a for loop but would appreciate a "pre-made" function if existing.

Why do people prefer to call an external plugin function instead of writing 2 lines loop ? — Denys Séguret
– Denys Séguret, Commented Jul 5, 2012 at 13:28
Unfortunately, there is no pre-made function. Go with the for loop, but make it into a function too :) — Ry-
– Ry- ♦, Commented Jul 5, 2012 at 13:28
@dystroy Because if you're already using the library anyway, a function call is clearer. — millimoose
– millimoose, Commented Jul 5, 2012 at 13:29
@dystroy: a) It's neater b) It may not even be a plugin; it could be built-in. For example, PHP's array_chunk. — Ry-
– Ry- ♦, Commented Jul 5, 2012 at 13:29

georg · Accepted Answer · 2013-09-11 12:29:10Z

28

I think you will have to use a for loop, don't know of any inbuilt functions...

Try this function:

function splitarray(input, spacing)
{
    var output = [];

    for (var i = 0; i < input.length; i += spacing)
    {
        output[output.length] = input.slice(i, i + spacing);
    }

    return output;
}

edited Sep 11, 2013 at 12:29

georg

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answered Jul 5, 2012 at 13:45

starbeamrainbowlabs

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3 Comments

Denys Séguret Over a year ago

Using slice can avoid one loop.

starbeamrainbowlabs Over a year ago

Removed the extra for loop - Thank for pointing that out

Jay Sullivan Over a year ago

This is just as simple as the accepted answer, and doesn't depend on any external libraries. +1

fat · Accepted Answer · 2021-03-25 11:18:59Z

21

Here's a recursive solution:

function partition(array, n) {
  return array.length ? [array.splice(0, n)].concat(partition(array, n)) : [];
}

This takes advantage of the fact that Array#splice destructively remove the specified items, and returns them as the function value. Note that this will destroy the input array, leaving it empty.

edited Mar 25, 2021 at 11:18

fat

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answered Oct 7, 2014 at 7:08

user663031

Comments

millimoose · Accepted Answer · 2015-06-11 22:18:18Z

13

If using Underscore.js, you can implement this with groupBy() and values()

function partition(items, size) {
    var result = _.groupBy(items, function(item, i) {
        return Math.floor(i/size);
    });
    return _.values(result);
}

(This is less ugly in CoffeeScript.)

jsFiddle: http://jsfiddle.net/MW3BS/

edited Jun 11, 2015 at 22:18

answered Jul 5, 2012 at 13:40

millimoose

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2 Comments

Ron E Over a year ago

Note that Underscore.js now has a partition function: underscorejs.org/#partition

rrauenza Over a year ago

That partition function only splits the array in two.

Ross · Accepted Answer · 2015-09-09 09:02:18Z

I've added this solution to @dystroy's jspref here and it appears to run twice as fast as the other solutions. Edit: in Safari & Chrome but not Firefox

Here is functional style solution to add to the mix of answers here.

It is a higher order function called toPartitions which returns a callback for underscore's reduce method or the native array reduce method.

Example usage:

[1,2,3,4,5,6,7,8,9].reduce( toPartitions( 3 ), [] );

The function:

function toPartitions ( size ) {
    var partition = [];
    return function ( acc, v ) {
        partition.push( v );
        if ( partition.length === size ) {
            acc.push( partition );
            partition = [];
        }
        return acc;
    };
}

Like Clojure's partition it will not include a tail partition when there are not enough elements.

In your example you could do:

arrayALLPartionned = arrayAll.reduce( toPartitions( 3 ), [] ) );

If you don't want to use this with reduce, but just have a function which takes an array and partition size you could do:

function partition ( arr, size ) {
    return arr.reduce( toPartitions( size ), [] );
}

Therefore the solution would just be:

arrayALLPartionned = partition( arrayAll, 3 );

This is a cool solution. My only qualm is that this won't effectively handle scenarios wherein the input array is not evenly divisible by the partition size. For example, calling partition(arrayAll, 4) will return two partitions while the last value, 9, is left absent. The fix for that would be inside the if conditional: if (partition.length === size || index === list.length - 1) { /* ... */ } Of course, a more bespoke solution then would be to perhaps include an additional parameter (e.g., withRemainder) to indicate if the user wants a remainder partition in the output.
Here is a fixed version: function toPartitions(size) { var partition = []; return function (acc, val) { partition.push(val); if (partition.length === 1) { acc.push(partition); } if (partition.length === size) { partition = []; } return acc; }; }

Denys Séguret · Accepted Answer · 2012-07-05 14:08:27Z

4

One more solution, with no external library :

function partition(items, size) {
    var p = [];
    for (var i=Math.floor(items.length/size); i-->0; ) {
        p[i]=items.slice(i*size, (i+1)*size);
    }
    return p;
}

Demonstration : http://jsfiddle.net/dystroy/xtHXZ/

edited Jul 5, 2012 at 14:08

answered Jul 5, 2012 at 13:50

Denys Séguret

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5 Comments

500 Over a year ago

thank you for your attention ! Would you mind explaining the use of "-->" ? I had never seen that syntax before.

Denys Séguret Over a year ago

That simply means "decrement i and compare the value after decrement to 0". It's just like doing i-- and comparing afterwards as i>0. The intersting point in looping that way is that you don't have to evaluate the maximal range value more than once as >0 can be computed in one operation. I don't say you really have to do it to make a fast code, it's not so important, but it's a habit I have.

500 Over a year ago

excellent, i think it somehow correspond to my use of "reap and "sow" in Mathematica. thank you again.

Jay Sullivan Over a year ago

This answer will truncate the results if it doesn't divide evenly into size

tonnoz Over a year ago

this solution will not consider the remainder: in your fiddle, you will miss the last element if you run it with 4 as size

Peter Olson · Accepted Answer · 2012-07-05 13:39:50Z

You can write your own prototype method to do this

Array.prototype.partition = function(length) {
  var result = [];
  for(var i = 0; i < this.length; i++) {
    if(i % length === 0) result.push([]);
    result[result.length - 1].push(this[i]);
  }
  return result;
};

If you prefer not to add to the native prototype, you can write a simple function:

var partition = function(arr, length) {
  var result = [];
  for(var i = 0; i < arr.length; i++) {
    if(i % length === 0) result.push([]);
    result[result.length - 1].push(arr[i]);
  }
  return result;
};

You can see it in action on this jsFiddle demo.

Never modify the prototype of a native object, this can cause compatibility issues in the future.

Andy · Accepted Answer · 2012-07-05 13:33:12Z

1

Prototype has an array.partition function as well as an eachSlice() function. Sounds like eachSlice() is what you're looking for. If you're using jquery, there's a plug in to be able to use prototype functions. Here's a link to it... http://www.learningjquery.com/2009/02/implementing-prototypes-array-methods-in-jquery

answered Jul 5, 2012 at 13:33

Andy

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