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I have 2 arrays that I want a Cartesian product of. As an example:

Customer Array:

[10,A]
[11,B]

Debtor Array:

[88,W]
[99,X]

I want to produce a new customerDebtor array with:

[10,A,88,W]
[10,A,99,X]
[11,B,88,W]
[11,B,99,X]

I am attempting it with this code:

for (var i = 0; i < customerArray.length; i++) {
    for (var l = 0; l < debtorArray.length; l++) {
        $.each(customerArray[i], function (ndx, val) {
            //???? push values into customerDebtorMatrix array
        });
    }
}

I have 2 arrays that I want a Cartesian product of. As an example:

Customer Array:

[10,A]
[11,B]

Debtor Array:

[88,W]
[99,X]

I want to produce a new customerDebtor array with:

[10,A,88,W]
[10,A,99,X]
[11,B,88,W]
[11,B,99,X]

I am attempting it with this code:

for (var i = 0; i < customerArray.length; i++) {
    for (var l = 0; l < debtorArray.length; l++) {
        $.each(customerArray[i], function (ndx, val) {
            //???? push values into customerDebtorMatrix array
        });
    }
}

• javascript
• jquery

Share Improve this question Follow edited Sep 29, 2013 at 17:25 Ry-♦ 225k5656 gold badges492492 silver badges498498 bronze badges asked Sep 29, 2013 at 17:15 GregGreg 2,69033 gold badges3838 silver badges6060 bronze badges

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4 Answers 4

Sorted by: Reset to default 3

You don't need jquery for this:

var customerArray = [[10,'A'],[11,'B']];
var debtorArray = [[88,'W'],[99,'X']];

var customerDebtorMatrix = [];
for (var i = 0; i < customerArray.length; i++) {
    for (var l = 0; l < debtorArray.length; l++) {
        customerDebtorMatrix.push(customerArray[i].concat(debtorArray[l]));
    }
}

customerDebtorMatrix will be

[ [ 10, 'A', 88, 'W' ],
  [ 10, 'A', 99, 'X' ],
  [ 11, 'B', 88, 'W' ],
  [ 11, 'B', 99, 'X' ] ]

concatMap works well here:

(function() {
    'use strict';

    // cartProd :: [a] -> [b] -> [[a, b]]
    function cartProd(xs, ys) {
        return [].concat.apply([], xs.map(function (x) {
            return [].concat.apply([], ys.map(function (y) {
                return [[x, y]];
            }));
        }));
    }

    return cartProd(["alpha", "beta", "gamma"], [1, 2, 3]);

})(); 

Returning:

[["alpha", 1], ["alpha", 2], ["alpha", 3], ["beta", 1], ["beta", 2], ["beta", 3], ["gamma", 1], ["gamma", 2], ["gamma", 3]]

2017 version

Using you data:

// Customer Array:
let c = [[10, 'A'], [11, 'B']];
// Debtor Array:
let d = [[88, 'W'], [99, 'X']];

all you need is one line:

let r = [].concat(...c.map(c => (d.map(d => c.concat(d)))));

or this for patibility with older browsers that don't support the spread syntax:

let r = [].concat.apply([], c.map(c => (d.map(d => c.concat(d)))));

You can see that console.log(r); prints:

[ [ 10, 'A', 88, 'W' ],
  [ 10, 'A', 99, 'X' ],
  [ 11, 'B', 88, 'W' ],
  [ 11, 'B', 99, 'X' ] ]

which is exactly what you wanted.

No loops, no pushes to arrays, no jQuery, just a few simple function calls.

For more than two arrays, see this answer:

• Cartesian product of multiple arrays in JavaScript

Actually this would be the Kronecker (or Tensor) Product

Javascript code for the kronecker product of 2 arrays

function kronecker_product(a, b) 
{
    var i, j, k, al = a.length, bl = b.length, abl = al*bl, ab = new Array(abl);
    i = 0; j = 0;
    for (k=0; k<abl; k++)
    {
        if ( j>=bl) {j=0; i++;}
        ab[k] = [].concat(a[i],b[j]);
        j++;
    }
    return ab;
}

For disambiguation let me give the code for the cartesian product as well:

function cartesian_product(a, b) 
{
    var al = a.length, bl = b.length;
    return [a.concat(bl>al ? new Array(bl-al) : []), b.concat(al>bl ? new Array(al-bl) : [])];
}

The main difference is the number of ponents (length) of the final result, in kronecker the length should be the product of the lengths while in cartesian it should be the sum.

Example Kronecker

var a = ["a","b"], b = ["c","d"];
console.log(kronecker_product(a,b));

Output

[["a", "c"], ["a", "d"], ["b", "c"], ["b", "d"]]

Example Cartesian

var a = ["a","b"], b = ["c","d"];
console.log(cartesian_product(a,b));

Output

 [["a", "b"], ["c", "d"]]

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