CODEWITHSUNDEEP
javascriptarraysalgorithmcartesian-product
viebel
viebel

Reputation: 20670

Cartesian product of multiple arrays in JavaScript

How would you implement the Cartesian product of multiple arrays in JavaScript?

As an example,

cartesian([1, 2], [10, 20], [100, 200, 300])

should return

[
  [1, 10, 100],
  [1, 10, 200],
  [1, 10, 300],
  [2, 10, 100],
  [2, 10, 200]
  ...
]

Upvotes: 186

Views: 93034

Answers (30)

Trevor Dixon
Trevor Dixon

Reputation: 24352

Similar in spirit to others, but highly readable imo.

const productOfTwo = (a, b) => a.flatMap(c => b.map(d => [c, d].flat()));
[['a', 'b', 'c'], ['+', '-'], [1, 2, 3]].reduce(productOfTwo);

Upvotes: 3

Mulan
Mulan

Reputation: 135217

functional programming

This question is tagged functional-programming so let's take a look at the List monad:

One application for this monadic list is representing nondeterministic computation. List can hold results for all execution paths in an algorithm...

Well that sounds like a perfect fit for cartesian. JavaScript gives us Array and the monadic binding function is Array.prototype.flatMap, so let's put them to use -

const cartesian = (...all) => {
  const loop = (t, a, ...more) =>
    a === undefined
      ? [ t ]
      : a.flatMap(x => loop([ ...t, x ], ...more))
  return loop([], ...all)
}

console.log(cartesian([1,2], [10,20], [100,200,300]))

[1,10,100]
[1,10,200]
[1,10,300]
[1,20,100]
[1,20,200]
[1,20,300]
[2,10,100]
[2,10,200]
[2,10,300]
[2,20,100]
[2,20,200]
[2,20,300]

more recursion

Other recursive implementations include -

const cartesian = (a, ...more) =>
  a == null
    ? [[]]
    : cartesian(...more).flatMap(c => a.map(v => [v,...c]))

for (const p of cartesian([1,2], [10,20], [100,200,300]))
  console.log(JSON.stringify(p))
.as-console-wrapper { min-height: 100%; top: 0; }

[1,10,100]
[2,10,100]
[1,20,100]
[2,20,100]
[1,10,200]
[2,10,200]
[1,20,200]
[2,20,200]
[1,10,300]
[2,10,300]
[1,20,300]
[2,20,300]

Note the different order above. You can get lexicographic order by inverting the two loops. Be careful to avoid duplicating work by calling cartesian inside the loop like Nick's answer -

const bind = (x, f) => f(x)

const cartesian = (a, ...more) =>
  a == null
    ? [[]]
    : bind(cartesian(...more), r => a.flatMap(v => r.map(c => [v,...c])))

for (const p of cartesian([1,2], [10,20], [100,200,300]))
  console.log(JSON.stringify(p))
.as-console-wrapper { min-height: 100%; top: 0; }

[1,10,100]
[1,10,200]
[1,10,300]
[1,20,100]
[1,20,200]
[1,20,300]
[2,10,100]
[2,10,200]
[2,10,300]
[2,20,100]
[2,20,200]
[2,20,300]

generators

Another option is to use generators. A generator is a good fit for combinatorics because the solution space can become very large. Generators offer lazy evaluation so they can be paused/resumed/canceled at any time -

function* cartesian(a, ...more) {
  if (a == null) return yield []
  for (const v of a)
    for (const c of cartesian(...more)) // ⚠️
      yield [v, ...c]
}
  
for (const p of cartesian([1,2], [10,20], [100,200,300]))
  console.log(JSON.stringify(p))
.as-console-wrapper { min-height: 100%; top: 0; }

[1,10,100]
[1,10,200]
[1,10,300]
[1,20,100]
[1,20,200]
[1,20,300]
[2,10,100]
[2,10,200]
[2,10,300]
[2,20,100]
[2,20,200]
[2,20,300]

Maybe you saw that we called cartesian in a loop in the generator. If you suspect that can be optimized, it can! Here we use a generic tee function that forks any iterator n times -

function* cartesian(a, ...more) {
  if (a == null) return yield []
  for (const t of tee(cartesian(...more), a.length)) // ✅
    for (const v of a)
      for (const c of t) // ✅
        yield [v, ...c]
}

Where tee is implemented as -

function tee(g, n = 2) {
  const memo = []
  function* iter(i) {
    while (true) {
      if (i >= memo.length) {
        const w = g.next()
        if (w.done) return
        memo.push(w.value)
      }
      else yield memo[i++]
    }
  }
  return Array.from(Array(n), _ => iter(0))
}

Even in small tests cartesian generator implemented with tee performs twice as fast.

Upvotes: 14

N S Niko
N S Niko

Reputation: 105

I like this for readability purposes..

const _getPotentialArrayCombos = (arrayOfArrays)=>{
    let potentialArrayCombos
    for(const array of arrayOfArrays){
        if(!potentialArrayCombos){
            potentialArrayCombos=[]
            for(const value of array){
                potentialArrayCombos.push([value])
            }
        }else{
            const newPotentialArrayCombos = []
            for(const combo of potentialArrayCombos){
                for(const value of array){
                    newPotentialArrayCombos.push([...combo,value])
                }
            }
            potentialArrayCombos = newPotentialArrayCombos 
        }
    }
    return potentialArrayCombos
}

Upvotes: -1

AnyWhichWay
AnyWhichWay

Reputation: 774

A non-recursive approach that adds the ability to filter and modify the products before actually adding them to the result set.

Note: the use of .map rather than .forEach. In some browsers, .map runs faster.

function crossproduct(arrays, rowtest, rowaction) {
  // Calculate the number of elements needed in the result
  var result_elems = 1,
    row_size = arrays.length;
  arrays.map(function(array) {
    result_elems *= array.length;
  });
  var temp = new Array(result_elems),
    result = [];

  // Go through each array and add the appropriate
  // element to each element of the temp
  var scale_factor = result_elems;
  arrays.map(function(array) {
    var set_elems = array.length;
    scale_factor /= set_elems;
    for (var i = result_elems - 1; i >= 0; i--) {
      temp[i] = (temp[i] ? temp[i] : []);
      var pos = i / scale_factor % set_elems;
      // deal with floating point results for indexes,
      // this took a little experimenting
      if (pos < 1 || pos % 1 <= .5) {
        pos = Math.floor(pos);
      } else {
        pos = Math.min(array.length - 1, Math.ceil(pos));
      }
      temp[i].push(array[pos]);
      if (temp[i].length === row_size) {
        var pass = (rowtest ? rowtest(temp[i]) : true);
        if (pass) {
          if (rowaction) {
            result.push(rowaction(temp[i]));
          } else {
            result.push(temp[i]);
          }
        }
      }
    }
  });
  return result;
}

console.log(
crossproduct([[1, 2], [10, 20], [100, 200, 300]],null,null)
)

Upvotes: 1

ckozl
ckozl

Reputation: 6761

It seems the community thinks this to be trivial and/or easy to find a reference implementation. However, upon brief inspection I couldn't find one, … either that or maybe it's just that I like re-inventing the wheel or solving classroom-like programming problems. Either way its your lucky day:

function cartProd(paramArray) {
 
  function addTo(curr, args) {
    
    var i, copy, 
        rest = args.slice(1),
        last = !rest.length,
        result = [];
    
    for (i = 0; i < args[0].length; i++) {
      
      copy = curr.slice();
      copy.push(args[0][i]);
      
      if (last) {
        result.push(copy);
      
      } else {
        result = result.concat(addTo(copy, rest));
      }
    }
    
    return result;
  }
  
  
  return addTo([], Array.prototype.slice.call(arguments));
}


>> console.log(cartProd([1,2], [10,20], [100,200,300]));
>> [
     [1, 10, 100], [1, 10, 200], [1, 10, 300], [1, 20, 100], 
     [1, 20, 200], [1, 20, 300], [2, 10, 100], [2, 10, 200], 
     [2, 10, 300], [2, 20, 100], [2, 20, 200], [2, 20, 300]
   ]

Full reference implementation that's relatively efficient… 😁

On efficiency: You could gain some by taking the if out of the loop and having 2 separate loops since it is technically constant and you'd be helping with branch prediction and all that mess, but that point is kind of moot in JavaScript.

Upvotes: 31

sebnukem
sebnukem

Reputation: 8323

Here's a non-fancy, straightforward recursive solution:

function cartesianProduct(a) { // a = array of array
  var i, j, l, m, a1, o = [];
  if (!a || a.length == 0) return a;

  a1 = a.splice(0, 1)[0]; // the first array of a
  a = cartesianProduct(a);
  for (i = 0, l = a1.length; i < l; i++) {
    if (a && a.length)
      for (j = 0, m = a.length; j < m; j++)
        o.push([a1[i]].concat(a[j]));
    else
      o.push([a1[i]]);
  }
  return o;
}

console.log(cartesianProduct([[1, 2], [10, 20], [100, 200, 300]]));
// [
//   [1,10,100],[1,10,200],[1,10,300],
//   [1,20,100],[1,20,200],[1,20,300],
//   [2,10,100],[2,10,200],[2,10,300],
//   [2,20,100],[2,20,200],[2,20,300]
// ]

Upvotes: 25

Nick
Nick

Reputation: 147156

Here's a recursive one-liner that works using only flatMap and map:

const inp = [
  [1, 2],
  [10, 20],
  [100, 200, 300]
];

const cartesian = (first, ...rest) => 
  rest.length ? first.flatMap(v => cartesian(...rest).map(c => [v].concat(c))) 
              : first;

console.log(cartesian(...inp));

Upvotes: 4

Brandon McConnell
Brandon McConnell

Reputation: 6119

Another, even more simplified, 2021-style answer using only reduce, map, and concat methods:

const cartesian = (...arr) => arr.reduce((a,c) => a.map(e => c.map(f => e.concat([f]))).reduce((a,c) => a.concat(c), []), [[]]);

console.log(cartesian([1, 2], [10, 20], [100, 200, 300]));

Upvotes: 5

Georgiy Bukharov
Georgiy Bukharov

Reputation: 396

2021 version of David Tang's great answer
Also inspired with Neil Mountford's answer

  • fast (faster than accepted answer by 95%)
  • works with any data types

const getAllCombinations = (arraysToCombine) => {
  const divisors = [];
  let combinationsCount = 1;
  for (let i = arraysToCombine.length - 1; i >= 0; i--) {
      divisors[i] = divisors[i + 1] ? divisors[i + 1] * arraysToCombine[i + 1].length : 1;
      combinationsCount *= (arraysToCombine[i].length || 1);
  }

  const getCombination = (n, arrays, divisors) => arrays.reduce((acc, arr, i) => {
      acc.push(arr[Math.floor(n / divisors[i]) % arr.length]);
      return acc;
  }, []);

  const combinations = [];
  for (let i = 0; i < combinationsCount; i++) {
      combinations.push(getCombination(i, arraysToCombine, divisors));
  }
  return combinations;
};

console.log(getAllCombinations([['a', 'b'], ['c', 'z'], ['d', 'e', 'f']]));

Benchmarks: https://jsbench.me/gdkmxhm36d/1

Upvotes: 0

cviejo
cviejo

Reputation: 4398

For those happy with a ramda solution:

import { xprod, flatten } from 'ramda';

const cartessian = (...xs) => xs.reduce(xprod).map(flatten)

Or the same without dependencies and two lego blocks for free (xprod and flatten):

const flatten = xs => xs.flat();

const xprod = (xs, ys) => xs.flatMap(x => ys.map(y => [x, y]));

const cartessian = (...xs) => xs.reduce(xprod).map(flatten);

Upvotes: 2

rsp
rsp

Reputation: 111316

2020 Update: 1-line (!) answer with vanilla JS

Original 2017 Answer: 2-line answer with vanilla JS: (see updates below)

All of the answers here are overly complicated, most of them take 20 lines of code or even more.

This example uses just two lines of vanilla JavaScript, no lodash, underscore or other libraries:

let f = (a, b) => [].concat(...a.map(a => b.map(b => [].concat(a, b))));
let cartesian = (a, b, ...c) => b ? cartesian(f(a, b), ...c) : a;

Update:

This is the same as above but improved to strictly follow the Airbnb JavaScript Style Guide - validated using ESLint with eslint-config-airbnb-base:

const f = (a, b) => [].concat(...a.map(d => b.map(e => [].concat(d, e))));
const cartesian = (a, b, ...c) => (b ? cartesian(f(a, b), ...c) : a);

Special thanks to ZuBB for letting me know about linter problems with the original code.

Update 2020:

Since I wrote this answer we got even better builtins, that can finally let us reduce (no pun intended) the code to just 1 line!

const cartesian =
  (...a) => a.reduce((a, b) => a.flatMap(d => b.map(e => [d, e].flat())));

Special thanks to inker for suggesting the use of reduce.

Special thanks to Bergi for suggesting the use of the newly added flatMap.

Special thanks to ECMAScript 2019 for adding flat and flatMap to the language!

Example

This is the exact example from your question:

let output = cartesian([1,2],[10,20],[100,200,300]);

Output

This is the output of that command:

[ [ 1, 10, 100 ],
  [ 1, 10, 200 ],
  [ 1, 10, 300 ],
  [ 1, 20, 100 ],
  [ 1, 20, 200 ],
  [ 1, 20, 300 ],
  [ 2, 10, 100 ],
  [ 2, 10, 200 ],
  [ 2, 10, 300 ],
  [ 2, 20, 100 ],
  [ 2, 20, 200 ],
  [ 2, 20, 300 ] ]

Demo

See demos on:

Syntax

The syntax that I used here is nothing new. My example uses the spread operator and the rest parameters - features of JavaScript defined in the 6th edition of the ECMA-262 standard published on June 2015 and developed much earlier, better known as ES6 or ES2015. See:

The new methods from the Update 2020 example was added in ES2019:

It makes code like this so simple that it's a sin not to use it. For old platforms that don't support it natively you can always use Babel or other tools to transpile it to older syntax - and in fact my example transpiled by Babel is still shorter and simpler than most of the examples here, but it doesn't really matter because the output of transpilation is not something that you need to understand or maintain, it's just a fact that I found interesting.

Conclusion

There's no need to write hundred of lines of code that is hard to maintain and there is no need to use entire libraries for such a simple thing, when two lines of vanilla JavaScript can easily get the job done. As you can see it really pays off to use modern features of the language and in cases where you need to support archaic platforms with no native support of the modern features you can always use Babel, TypeScript or other tools to transpile the new syntax to the old one.

Don't code like it's 1995

JavaScript evolves and it does so for a reason. TC39 does an amazing job of the language design with adding new features and the browser vendors do an amazing job of implementing those features.

To see the current state of native support of any given feature in the browsers, see:

To see the support in Node versions, see:

To use modern syntax on platforms that don't support it natively, use Babel or TypeScript:

Upvotes: 239

iAmOren
iAmOren

Reputation: 2804

f=(a,b,c)=>a.flatMap(ai=>b.flatMap(bi=>c.map(ci=>[ai,bi,ci])))

This is for 3 arrays.
Some answers gave a way for any number of arrays.
This can easily contract or expand to less or more arrays.
I needed combinations of one set with repetitions, so I could have used: 

f(a,a,a)

but used: 

f=(a,b,c)=>a.flatMap(a1=>a.flatMap(a2=>a.map(a3=>[a1,a2,a3])))

Upvotes: 1

Danny
Danny

Reputation: 664

Here's a modified version of @viebel's code in plain Javascript, without using any library:

function cartesianProduct(arr) {
    return arr.reduce(function(a,b){
        return a.map(function(x){
            return b.map(function(y){
                return x.concat([y]);
            })
        }).reduce(function(a,b){ return a.concat(b) },[])
    }, [[]])
}

var a = cartesianProduct([[1, 2,3], [4, 5,6], [7, 8], [9,10]]);
console.log(JSON.stringify(a));

Upvotes: 61

Fred Kleuver
Fred Kleuver

Reputation: 8037

Here is a one-liner using the native ES2019 flatMap. No libraries needed, just a modern browser (or transpiler):

data.reduce((a, b) => a.flatMap(x => b.map(y => [...x, y])), [[]]);

It's essentially a modern version of viebel's answer, without lodash.

Upvotes: 22

adiga
adiga

Reputation: 35222

You could reduce the 2D array. Use flatMap on the accumulator array to get acc.length x curr.length number of combinations in each loop. [].concat(c, n) is used because c is a number in the first iteration and an array afterwards. 

const data = [ [1, 2], [10, 20], [100, 200, 300] ];

const output = data.reduce((acc, curr) =>
  acc.flatMap(c => curr.map(n => [].concat(c, n)))
)

console.log(JSON.stringify(output))

(This is based on Nina Scholz's answer)

Upvotes: 4

Andrei
Andrei

Reputation: 43

A more readable implementation

function productOfTwo(one, two) {
  return one.flatMap(x => two.map(y => [].concat(x, y)));
}

function product(head = [], ...tail) {
  if (tail.length === 0) return head;
  return productOfTwo(head, product(...tail));
}

const test = product(
  [1, 2, 3],
  ['a', 'b']
);

console.log(JSON.stringify(test));

Upvotes: 3

Miles Elam
Miles Elam

Reputation: 1778

Modern JavaScript in just a few lines. No external libraries or dependencies like Lodash.

function cartesian(...arrays) {
  return arrays.reduce((a, b) => a.flatMap(x => b.map(y => x.concat([y]))), [ [] ]);
}

console.log(
  cartesian([1, 2], [10, 20], [100, 200, 300])
    .map(arr => JSON.stringify(arr))
    .join('\n')
);

Upvotes: 4

viebel
viebel

Reputation: 20670

Here is a functional solution to the problem (without any mutable variable!) using reduce and flatten, provided by underscore.js:

function cartesianProductOf() {
    return _.reduce(arguments, function(a, b) {
        return _.flatten(_.map(a, function(x) {
            return _.map(b, function(y) {
                return x.concat([y]);
            });
        }), true);
    }, [ [] ]);
}

// [[1,3,"a"],[1,3,"b"],[1,4,"a"],[1,4,"b"],[2,3,"a"],[2,3,"b"],[2,4,"a"],[2,4,"b"]]
console.log(cartesianProductOf([1, 2], [3, 4], ['a']));  
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.9.1/underscore.js"></script>

Remark: This solution was inspired by http://cwestblog.com/2011/05/02/cartesian-product-of-multiple-arrays/

Upvotes: 93

LeandroP
LeandroP

Reputation: 377

For the record

Here it goes my version of it. I made it using the simplest javascript iterator "for()", so it's compatible on every case and has the best performance.

function cartesian(arrays){
    var quant = 1, counters = [], retArr = [];

    // Counts total possibilities and build the counters Array;
    for(var i=0;i<arrays.length;i++){
        counters[i] = 0;
        quant *= arrays[i].length;
    }

    // iterate all possibilities
    for(var i=0,nRow;i<quant;i++){
        nRow = [];
        for(var j=0;j<counters.length;j++){
            if(counters[j] < arrays[j].length){
                nRow.push(arrays[j][counters[j]]);
            } else { // in case there is no such an element it restarts the current counter
                counters[j] = 0;
                nRow.push(arrays[j][counters[j]]);
            }
            counters[j]++;
        }
        retArr.push(nRow);
    }
    return retArr;
}

Best regards.

Upvotes: 0

Chris Vouga
Chris Vouga

Reputation: 554

No libraries needed! :)

Needs arrow functions though and probably not that efficient. :/

const flatten = (xs) => 
    xs.flat(Infinity)

const binaryCartesianProduct = (xs, ys) =>
    xs.map((xi) => ys.map((yi) => [xi, yi])).flat()

const cartesianProduct = (...xss) =>
    xss.reduce(binaryCartesianProduct, [[]]).map(flatten)
      
console.log(cartesianProduct([1,2,3], [1,2,3], [1,2,3]))

Upvotes: 3

artnikpro
artnikpro

Reputation: 5869

For those who needs TypeScript (reimplemented @Danny's answer)

/**
 * Calculates "Cartesian Product" sets.
 * @example
 *   cartesianProduct([[1,2], [4,8], [16,32]])
 *   Returns:
 *   [
 *     [1, 4, 16],
 *     [1, 4, 32],
 *     [1, 8, 16],
 *     [1, 8, 32],
 *     [2, 4, 16],
 *     [2, 4, 32],
 *     [2, 8, 16],
 *     [2, 8, 32]
 *   ]
 * @see https://stackoverflow.com/a/36234242/1955709
 * @see https://en.wikipedia.org/wiki/Cartesian_product
 * @param arr {T[][]}
 * @returns {T[][]}
 */
function cartesianProduct<T> (arr: T[][]): T[][] {
  return arr.reduce((a, b) => {
    return a.map(x => {
      return b.map(y => {
        return x.concat(y)
      })
    }).reduce((c, d) => c.concat(d), [])
  }, [[]] as T[][])
}

Upvotes: 4

Juan
Juan

Reputation: 632

A simple, modified version of @viebel's code in plain Javascript:

function cartesianProduct(...arrays) {
  return arrays.reduce((a, b) => {
    return [].concat(...a.map(x => {
      const next = Array.isArray(x) ? x : [x];
      return [].concat(b.map(y => next.concat(...[y])));
    }));
  });
}

const product = cartesianProduct([1, 2], [10, 20], [100, 200, 300]);

console.log(product);
/*
[ [ 1, 10, 100 ],
  [ 1, 10, 200 ],
  [ 1, 10, 300 ],
  [ 1, 20, 100 ],
  [ 1, 20, 200 ],
  [ 1, 20, 300 ],
  [ 2, 10, 100 ],
  [ 2, 10, 200 ],
  [ 2, 10, 300 ],
  [ 2, 20, 100 ],
  [ 2, 20, 200 ],
  [ 2, 20, 300 ] ];
*/

Upvotes: 1

flare256
flare256

Reputation: 11

Yet another implementation. Not the shortest or fancy, but fast:

function cartesianProduct() {
    var arr = [].slice.call(arguments),
        intLength = arr.length,
        arrHelper = [1],
        arrToReturn = [];

    for (var i = arr.length - 1; i >= 0; i--) {
        arrHelper.unshift(arrHelper[0] * arr[i].length);
    }

    for (var i = 0, l = arrHelper[0]; i < l; i++) {
        arrToReturn.push([]);
        for (var j = 0; j < intLength; j++) {
            arrToReturn[i].push(arr[j][(i / arrHelper[j + 1] | 0) % arr[j].length]);
        }
    }

    return arrToReturn;
}

Upvotes: 1

le_m
le_m

Reputation: 20228

The following efficient generator function returns the cartesian product of all given iterables:

// Generate cartesian product of given iterables:
function* cartesian(head, ...tail) {
  const remainder = tail.length > 0 ? cartesian(...tail) : [[]];
  for (let r of remainder) for (let h of head) yield [h, ...r];
}

// Example:
console.log(...cartesian([1, 2], [10, 20], [100, 200, 300]));

It accepts arrays, strings, sets and all other objects implementing the iterable protocol.

Following the specification of the n-ary cartesian product it yields

  • [] if one or more given iterables are empty, e.g. [] or ''
  • [[a]] if a single iterable containing a single value a is given.

All other cases are handled as expected as demonstrated by the following test cases:

// Generate cartesian product of given iterables:
function* cartesian(head, ...tail) {
  const remainder = tail.length > 0 ? cartesian(...tail) : [[]];
  for (let r of remainder) for (let h of head) yield [h, ...r];
}

// Test cases:
console.log([...cartesian([])]);              // []
console.log([...cartesian([1])]);             // [[1]]
console.log([...cartesian([1, 2])]);          // [[1], [2]]

console.log([...cartesian([1], [])]);         // []
console.log([...cartesian([1, 2], [])]);      // []

console.log([...cartesian([1], [2])]);        // [[1, 2]]
console.log([...cartesian([1], [2], [3])]);   // [[1, 2, 3]]
console.log([...cartesian([1, 2], [3, 4])]);  // [[1, 3], [2, 3], [1, 4], [2, 4]]

console.log([...cartesian('')]);              // []
console.log([...cartesian('ab', 'c')]);       // [['a','c'], ['b', 'c']]
console.log([...cartesian([1, 2], 'ab')]);    // [[1, 'a'], [2, 'a'], [1, 'b'], [2, 'b']]

console.log([...cartesian(new Set())]);       // []
console.log([...cartesian(new Set([1]))]);    // [[1]]
console.log([...cartesian(new Set([1, 1]))]); // [[1]]

Upvotes: 48

Nina Scholz
Nina Scholz

Reputation: 386560

A single line approach, for better reading with indentations.

result = data.reduce(
    (a, b) => a.reduce(
        (r, v) => r.concat(b.map(w => [].concat(v, w))),
        []
    )
);

It takes a single array with arrays of wanted cartesian items.

var data = [[1, 2], [10, 20], [100, 200, 300]],
    result = data.reduce((a, b) => a.reduce((r, v) => r.concat(b.map(w => [].concat(v, w))), []));

console.log(result.map(a => a.join(' ')));
.as-console-wrapper { max-height: 100% !important; top: 0; }

Upvotes: 8

guneysus
guneysus

Reputation: 6502

var chars = ['A', 'B', 'C']
var nums = [1, 2, 3]

var cartesianProduct = function() {
  return _.reduce(arguments, function(a, b) {
    return _.flatten(_.map(a, function(x) {
      return _.map(b, function(y) {
        return x.concat(y);
      });
    }), true);
  }, [
    []
  ]);
};

console.log(cartesianProduct(chars, nums))
<script src="https://cdnjs.cloudflare.com/ajax/libs/underscore.js/1.8.3/underscore-min.js"></script>

Just converted @dummersl's answer from CoffeScript to JavaScript. It just works.

var chars = ['A', 'B', 'C']
var nums = [1, 2, 3]

var cartesianProduct = function() {
  return _.reduce(arguments, function(a, b) {
    return _.flatten(_.map(a, function(x) {
      return _.map(b, function(y) {
        return x.concat(y);
      });
    }), true);
  }, [[]]);
};

console.log( cartesianProduct(chars, nums) )

Upvotes: 0

aliep
aliep

Reputation: 1772

A simple "mind and visually friendly" solution.

enter image description here

// t = [i, length]

const moveThreadForwardAt = (t, tCursor) => {
  if (tCursor < 0)
    return true; // reached end of first array

  const newIndex = (t[tCursor][0] + 1) % t[tCursor][1];
  t[tCursor][0] = newIndex;

  if (newIndex == 0)
    return moveThreadForwardAt(t, tCursor - 1);

  return false;
}

const cartesianMult = (...args) => {
  let result = [];
  const t = Array.from(Array(args.length)).map((x, i) => [0, args[i].length]);
  let reachedEndOfFirstArray = false;

  while (false == reachedEndOfFirstArray) {
    result.push(t.map((v, i) => args[i][v[0]]));

    reachedEndOfFirstArray = moveThreadForwardAt(t, args.length - 1);
  }

  return result;
}

// cartesianMult(
//   ['a1', 'b1', 'c1'],
//   ['a2', 'b2'],
//   ['a3', 'b3', 'c3'],
//   ['a4', 'b4']
// );

console.log(cartesianMult(
  ['a1'],
  ['a2', 'b2'],
  ['a3', 'b3']
));

Upvotes: 1

Samuel Ventura
Samuel Ventura

Reputation: 19

Plain JS brute force approach that takes an array of arrays as input.

var cartesian = function(arrays) {
    var product = [];
    var precals = [];
    var length = arrays.reduce(function(acc, curr) {
        return acc * curr.length
    }, 1);
    for (var i = 0; i < arrays.length; i++) {
        var array = arrays[i];
        var mod = array.length;
        var div = i > 0 ? precals[i - 1].div * precals[i - 1].mod : 1;
        precals.push({
            div: div,
            mod: mod
        });
    }
    for (var j = 0; j < length; j++) {
        var item = [];
        for (var i = 0; i < arrays.length; i++) {
            var array = arrays[i];
            var precal = precals[i];
            var k = (~~(j / precal.div)) % precal.mod;
            item.push(array[k]);
        }
        product.push(item);
    }
    return product;
};

cartesian([
    [1],
    [2, 3]
]);

cartesian([
    [1],
    [2, 3],
    [4, 5, 6]
]);

Upvotes: -1

Simple.Js
Simple.Js

Reputation: 91

Just for a choice a real simple implementation using array's reduce:

const array1 = ["day", "month", "year", "time"];
const array2 = ["from", "to"];
const process = (one, two) => [one, two].join(" ");

const product = array1.reduce((result, one) => result.concat(array2.map(two => process(one, two))), []);

Upvotes: 1

ewcz
ewcz

Reputation: 13087

In my particular setting, the "old-fashioned" approach seemed to be more efficient than the methods based on more modern features. Below is the code (including a small comparison with other solutions posted in this thread by @rsp and @sebnukem) should it prove useful to someone else as well.

The idea is following. Let's say we are constructing the outer product of N arrays, a_1,...,a_N each of which has m_i components. The outer product of these arrays has M=m_1*m_2*...*m_N elements and we can identify each of them with a N-dimensional vector the components of which are positive integers and i-th component is strictly bounded from above by m_i. For example, the vector (0, 0, ..., 0) would correspond to the particular combination within which one takes the first element from each array, while (m_1-1, m_2-1, ..., m_N-1) is identified with the combination where one takes the last element from each array. Thus in order to construct all M combinations, the function below consecutively constructs all such vectors and for each of them identifies the corresponding combination of the elements of the input arrays.

function cartesianProduct(){
    const N = arguments.length;

    var arr_lengths = Array(N);
    var digits = Array(N);
    var num_tot = 1;
    for(var i = 0; i < N; ++i){
        const len = arguments[i].length;
        if(!len){
            num_tot = 0;
            break;
        }
        digits[i] = 0;
        num_tot *= (arr_lengths[i] = len);
    }

    var ret = Array(num_tot);
    for(var num = 0; num < num_tot; ++num){

        var item = Array(N);
        for(var j = 0; j < N; ++j){ item[j] = arguments[j][digits[j]]; }
        ret[num] = item;

        for(var idx = 0; idx < N; ++idx){
            if(digits[idx] == arr_lengths[idx]-1){
                digits[idx] = 0;
            }else{
                digits[idx] += 1;
                break;
            }
        }
    }
    return ret;
}
//------------------------------------------------------------------------------
let _f = (a, b) => [].concat(...a.map(a => b.map(b => [].concat(a, b))));
let cartesianProduct_rsp = (a, b, ...c) => b ? cartesianProduct_rsp(_f(a, b), ...c) : a;
//------------------------------------------------------------------------------
function cartesianProduct_sebnukem(a) {
    var i, j, l, m, a1, o = [];
    if (!a || a.length == 0) return a;

    a1 = a.splice(0, 1)[0];
    a = cartesianProduct_sebnukem(a);
    for (i = 0, l = a1.length; i < l; i++) {
        if (a && a.length) for (j = 0, m = a.length; j < m; j++)
            o.push([a1[i]].concat(a[j]));
        else
            o.push([a1[i]]);
    }
    return o;
}
//------------------------------------------------------------------------------
const L = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9];
const args = [L, L, L, L, L, L];

let fns = {
    'cartesianProduct': function(args){ return cartesianProduct(...args); },
    'cartesianProduct_rsp': function(args){ return cartesianProduct_rsp(...args); },
    'cartesianProduct_sebnukem': function(args){ return cartesianProduct_sebnukem(args); }
};

Object.keys(fns).forEach(fname => {
    console.time(fname);
    const ret = fns[fname](args);
    console.timeEnd(fname);
});

with node v6.12.2, I get following timings:

cartesianProduct: 427.378ms
cartesianProduct_rsp: 1710.829ms
cartesianProduct_sebnukem: 593.351ms

Upvotes: 4

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