Merge/flatten an array of arrays

Question

I have a JavaScript array like:

[["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]

How would I go about merging the separate inner arrays into one like:

["$6", "$12", "$25", ...]

Answer 1

You can use concat to merge arrays:

var arrays = [
  ["$6"],
  ["$12"],
  ["$25"],
  ["$25"],
  ["$18"],
  ["$22"],
  ["$10"]
];
var merged = [].concat.apply([], arrays);

console.log(merged);

Using the apply method of concat will just take the second parameter as an array, so the last line is identical to this:

var merged2 = [].concat(["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]);

---

There is also the Array.prototype.flat() method (introduced in ES2019) which you could use to flatten the arrays, although it is only available in Node.js starting with version 11, and not at all in Internet Explorer.

const arrays = [
      ["$6"],
      ["$12"],
      ["$25"],
      ["$25"],
      ["$18"],
      ["$22"],
      ["$10"]
    ];
const merge3 = arrays.flat(1); //The depth level specifying how deep a nested array structure should be flattened. Defaults to 1.
console.log(merge3);
    

Answer 2

Here's a short function that uses some of the newer JavaScript array methods to flatten an n-dimensional array.

function flatten(arr) {
  return arr.reduce(function (flat, toFlatten) {
    return flat.concat(Array.isArray(toFlatten) ? flatten(toFlatten) : toFlatten);
  }, []);
}

Usage:

flatten([[1, 2, 3], [4, 5]]); // [1, 2, 3, 4, 5]
flatten([[[1, [1.1]], 2, 3], [4, 5]]); // [1, 1.1, 2, 3, 4, 5]

Answer 3

There is a confusingly hidden method, which constructs a new array without mutating the original one:

var oldArray = [[1],[2,3],[4]];
var newArray = Array.prototype.concat.apply([], oldArray);
console.log(newArray); // [ 1, 2, 3, 4 ]

Answer 4

It can be best done by javascript reduce function.

var arrays = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]];

arrays = arrays.reduce(function(a, b){
     return a.concat(b);
}, []);

Or, with ES2015:

arrays = arrays.reduce((a, b) => a.concat(b), []);

js-fiddle

Mozilla docs

Answer 5

There's a new native method called flat to do this exactly.

(As of late 2019, flat is now published in the ECMA 2019 standard, and core-js@3 (babel's library) includes it in their polyfill library)

const arr1 = [1, 2, [3, 4]];
arr1.flat(); 
// [1, 2, 3, 4]

const arr2 = [1, 2, [3, 4, [5, 6]]];
arr2.flat();
// [1, 2, 3, 4, [5, 6]]

// Flatten 2 levels deep
const arr3 = [2, 2, 5, [5, [5, [6]], 7]];
arr3.flat(2);
// [2, 2, 5, 5, 5, [6], 7];

// Flatten all levels
const arr4 = [2, 2, 5, [5, [5, [6]], 7]];
arr4.flat(Infinity);
// [2, 2, 5, 5, 5, 6, 7];

Answer 6

Most of the answers here don't work on huge (e.g. 200 000 elements) arrays, and even if they do, they're slow. polkovnikov.ph's answer has the best performance, but it doesn't work for deep flattening.

Here is the fastest solution, which works also on arrays with multiple levels of nesting:

const flatten = function(arr, result = []) {
  for (let i = 0, length = arr.length; i < length; i++) {
    const value = arr[i];
    if (Array.isArray(value)) {
      flatten(value, result);
    } else {
      result.push(value);
    }
  }
  return result;
};

Examples

Huge arrays

flatten(Array(200000).fill([1]));

It handles huge arrays just fine. On my machine this code takes about 14 ms to execute.

Nested arrays

flatten(Array(2).fill(Array(2).fill(Array(2).fill([1]))));

It works with nested arrays. This code produces [1, 1, 1, 1, 1, 1, 1, 1].

Arrays with different levels of nesting

flatten([1, [1], [[1]]]);

It doesn't have any problems with flattening arrays like this one.

Answer 7

Update: it turned out that this solution doesn't work with large arrays. It you're looking for a better, faster solution, check out this answer.

---

function flatten(arr) {
  return [].concat(...arr)
}

Is simply expands arr and passes it as arguments to concat(), which merges all the arrays into one. It's equivalent to [].concat.apply([], arr).

You can also try this for deep flattening:

function deepFlatten(arr) {
  return flatten(           // return shalowly flattened array
    arr.map(x=>             // with each x in array
      Array.isArray(x)      // is x an array?
        ? deepFlatten(x)    // if yes, return deeply flattened x
        : x                 // if no, return just x
    )
  )
}

See demo on JSBin.

References for ECMAScript 6 elements used in this answer:

• Spread operator
• Arrow functions

---

Side note: methods like find() and arrow functions are not supported by all browsers, but it doesn't mean that you can't use these features right now. Just use Babel — it transforms ES6 code into ES5.

Answer 8

You can use Underscore:

var x = [[1], [2], [3, 4]];

_.flatten(x); // => [1, 2, 3, 4]

Answer 9

Generic procedures mean we don't have to rewrite complexity each time we need to utilize a specific behaviour.

concatMap (or flatMap) is exactly what we need in this situation.

// concat :: ([a],[a]) -> [a]
const concat = (xs,ys) =>
  xs.concat (ys)

// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = f => xs =>
  xs.map(f).reduce(concat, [])

// id :: a -> a
const id = x =>
  x

// flatten :: [[a]] -> [a]
const flatten =
  concatMap (id)

// your sample data
const data =
  [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]

console.log (flatten (data))

foresight

And yes, you guessed it correctly, it only flattens one level, which is exactly how it should work

Imagine some data set like this

// Player :: (String, Number) -> Player
const Player = (name,number) =>
  [ name, number ]

// team :: ( . Player) -> Team
const Team = (...players) =>
  players

// Game :: (Team, Team) -> Game
const Game = (teamA, teamB) =>
  [ teamA, teamB ]

// sample data
const teamA =
  Team (Player ('bob', 5), Player ('alice', 6))

const teamB =
  Team (Player ('ricky', 4), Player ('julian', 2))

const game =
  Game (teamA, teamB)

console.log (game)
// [ [ [ 'bob', 5 ], [ 'alice', 6 ] ],
//   [ [ 'ricky', 4 ], [ 'julian', 2 ] ] ]

Ok, now say we want to print a roster that shows all the players that will be participating in game …

const gamePlayers = game =>
  flatten (game)

gamePlayers (game)
// => [ [ 'bob', 5 ], [ 'alice', 6 ], [ 'ricky', 4 ], [ 'julian', 2 ] ]

If our flatten procedure flattened nested arrays too, we'd end up with this garbage result …

const gamePlayers = game =>
  badGenericFlatten(game)

gamePlayers (game)
// => [ 'bob', 5, 'alice', 6, 'ricky', 4, 'julian', 2 ]

---

rollin' deep, baby

That's not to say sometimes you don't want to flatten nested arrays, too – only that shouldn't be the default behaviour.

We can make a deepFlatten procedure with ease …

// concat :: ([a],[a]) -> [a]
const concat = (xs,ys) =>
  xs.concat (ys)

// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = f => xs =>
  xs.map(f).reduce(concat, [])

// id :: a -> a
const id = x =>
  x

// flatten :: [[a]] -> [a]
const flatten =
  concatMap (id)

// deepFlatten :: [[a]] -> [a]
const deepFlatten =
  concatMap (x =>
    Array.isArray (x) ? deepFlatten (x) : x)

// your sample data
const data =
  [0, [1, [2, [3, [4, 5], 6]]], [7, [8]], 9]

console.log (flatten (data))
// [ 0, 1, [ 2, [ 3, [ 4, 5 ], 6 ] ], 7, [ 8 ], 9 ]

console.log (deepFlatten (data))
// [ 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 ]

There. Now you have a tool for each job – one for squashing one level of nesting, flatten, and one for obliterating all nesting deepFlatten.

Maybe you can call it obliterate or nuke if you don't like the name deepFlatten.

---

Don't iterate twice !

Of course the above implementations are clever and concise, but using a .map followed by a call to .reduce means we're actually doing more iterations than necessary

Using a trusty combinator I'm calling mapReduce helps keep the iterations to a minium; it takes a mapping function m :: a -> b, a reducing function r :: (b,a) ->b and returns a new reducing function - this combinator is at the heart of transducers; if you're interested, I've written other answers about them

// mapReduce = (a -> b, (b,a) -> b, (b,a) -> b)
const mapReduce = (m,r) =>
  (acc,x) => r (acc, m (x))

// concatMap :: (a -> [b]) -> [a] -> [b]
const concatMap = f => xs =>
  xs.reduce (mapReduce (f, concat), [])

// concat :: ([a],[a]) -> [a]
const concat = (xs,ys) =>
  xs.concat (ys)

// id :: a -> a
const id = x =>
  x

// flatten :: [[a]] -> [a]
const flatten =
  concatMap (id)
  
// deepFlatten :: [[a]] -> [a]
const deepFlatten =
  concatMap (x =>
    Array.isArray (x) ? deepFlatten (x) : x)

// your sample data
const data =
  [ [ [ 1, 2 ],
      [ 3, 4 ] ],
    [ [ 5, 6 ],
      [ 7, 8 ] ] ]

console.log (flatten (data))
// [ [ 1. 2 ], [ 3, 4 ], [ 5, 6 ], [ 7, 8 ] ]

console.log (deepFlatten (data))
// [ 1, 2, 3, 4, 5, 6, 7, 8 ]

Answer 10

To flatten an array of single element arrays, you don't need to import a library, a simple loop is both the simplest and most efficient solution :

for (var i = 0; i < a.length; i++) {
  a[i] = a[i][0];
}

To downvoters: please read the question, don't downvote because it doesn't suit your very different problem. This solution is both the fastest and simplest for the asked question.

Answer 11

A solution for the more general case, when you may have some non-array elements in your array.

function flattenArrayOfArrays(a, r){
    if(!r){ r = []}
    for(var i=0; i<a.length; i++){
        if(a[i].constructor == Array){
            r.concat(flattenArrayOfArrays(a[i], r));
        }else{
            r.push(a[i]);
        }
    }
    return r;
}

Answer 12

What about using reduce(callback[, initialValue]) method of JavaScript 1.8

list.reduce((p,n) => p.concat(n),[]);

Would do the job.

Answer 13

Another ECMAScript 6 solution in functional style:

Declare a function:

const flatten = arr => arr.reduce(
  (a, b) => a.concat(Array.isArray(b) ? flatten(b) : b), []
);

and use it:

flatten( [1, [2,3], [4,[5,[6]]]] ) // -> [1,2,3,4,5,6]

 const flatten = arr => arr.reduce(
         (a, b) => a.concat(Array.isArray(b) ? flatten(b) : b), []
       );


console.log( flatten([1, [2,3], [4,[5],[6,[7,8,9],10],11],[12],13]) )

Consider also a native function Array.prototype.flat() (proposal for ES6) available in last releases of modern browsers. Thanks to @(Константин Ван) and @(Mark Amery) mentioned it in the comments.

The flat function has one parameter, specifying the expected depth of array nesting, which equals 1 by default.

[1, 2, [3, 4]].flat();                  // -> [1, 2, 3, 4]

[1, 2, [3, 4, [5, 6]]].flat();          // -> [1, 2, 3, 4, [5, 6]]

[1, 2, [3, 4, [5, 6]]].flat(2);         // -> [1, 2, 3, 4, 5, 6]

[1, 2, [3, 4, [5, 6]]].flat(Infinity);  // -> [1, 2, 3, 4, 5, 6]

let arr = [1, 2, [3, 4]];

console.log( arr.flat() );

arr =  [1, 2, [3, 4, [5, 6]]];

console.log( arr.flat() );
console.log( arr.flat(1) );
console.log( arr.flat(2) );
console.log( arr.flat(Infinity) );

Answer 14

const common = arr.reduce((a, b) => [...a, ...b], [])

Answer 15

You can also try the new Array.flat() method. It works in the following manner:

let arr = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]].flat()

console.log(arr);

The flat() method creates a new array with all sub-array elements concatenated into it recursively up to the 1 layer of depth (i.e. arrays inside arrays)

If you want to also flatten out 3 dimensional or even higher dimensional arrays you simply call the flat method multiple times. For example (3 dimensions):

let arr = [1,2,[3,4,[5,6]]].flat().flat().flat();

console.log(arr);

Be careful!

Array.flat() method is relatively new. Older browsers like ie might not have implemented the method. If you want you code to work on all browsers you might have to transpile your JS to an older version. Check for MDN web docs for current browser compatibility.

Answer 16

You can use Array.flat() with Infinity for any depth of nested array.

var arr = [ [1,2,3,4], [1,2,[1,2,3]], [1,2,3,4,5,[1,2,3,4,[1,2,3,4]]], [[1,2,3,4], [1,2,[1,2,3]], [1,2,3,4,5,[1,2,3,4,[1,2,3,4]]]] ];

let flatten = arr.flat(Infinity)

console.log(flatten)

check here for browser compatibility

Answer 17

Please note: When Function.prototype.apply ([].concat.apply([], arrays)) or the spread operator ([].concat(...arrays)) is used in order to flatten an array, both can cause stack overflows for large arrays, because every argument of a function is stored on the stack.

Here is a stack-safe implementation in functional style that weighs up the most important requirements against one another:

• reusability
• readability
• conciseness
• performance

// small, reusable auxiliary functions:

const foldl = f => acc => xs => xs.reduce(uncurry(f), acc); // aka reduce

const uncurry = f => (a, b) => f(a) (b);

const concat = xs => y => xs.concat(y);


// the actual function to flatten an array - a self-explanatory one-line:

const flatten = xs => foldl(concat) ([]) (xs);

// arbitrary array sizes (until the heap blows up :D)

const xs = [[1,2,3],[4,5,6],[7,8,9]];

console.log(flatten(xs));


// Deriving a recursive solution for deeply nested arrays is trivially now


// yet more small, reusable auxiliary functions:

const map = f => xs => xs.map(apply(f));

const apply = f => a => f(a);

const isArray = Array.isArray;


// the derived recursive function:

const flattenr = xs => flatten(map(x => isArray(x) ? flattenr(x) : x) (xs));

const ys = [1,[2,[3,[4,[5],6,],7],8],9];

console.log(flattenr(ys));

As soon as you get used to small arrow functions in curried form, function composition and higher order functions, this code reads like prose. Programming then merely consists of putting together small building blocks that always work as expected, because they don't contain any side effects.

Answer 18

ES6 One Line Flatten

See lodash flatten, underscore flatten (shallow true)

function flatten(arr) {
  return arr.reduce((acc, e) => acc.concat(e), []);
}

or

function flatten(arr) {
  return [].concat.apply([], arr);
}

Tested with

test('already flatted', () => {
  expect(flatten([1, 2, 3, 4, 5])).toEqual([1, 2, 3, 4, 5]);
});

test('flats first level', () => {
  expect(flatten([1, [2, [3, [4]], 5]])).toEqual([1, 2, [3, [4]], 5]);
});

ES6 One Line Deep Flatten

See lodash flattenDeep, underscore flatten

function flattenDeep(arr) {
  return arr.reduce((acc, e) => Array.isArray(e) ? acc.concat(flattenDeep(e)) : acc.concat(e), []);
}

Tested with

test('already flatted', () => {
  expect(flattenDeep([1, 2, 3, 4, 5])).toEqual([1, 2, 3, 4, 5]);
});

test('flats', () => {
  expect(flattenDeep([1, [2, [3, [4]], 5]])).toEqual([1, 2, 3, 4, 5]);
});

Answer 19

Using the spread operator:

const input = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]];
const output = [].concat(...input);
console.log(output); // --> ["$6", "$12", "$25", "$25", "$18", "$22", "$10"]

Answer 20

A Haskellesque approach

function flatArray([x,...xs]){
  return x ? [...Array.isArray(x) ? flatArray(x) : [x], ...flatArray(xs)] : [];
}

var na = [[1,2],[3,[4,5]],[6,7,[[[8],9]]],10];
    fa = flatArray(na);
console.log(fa);

Answer 21

If you only have arrays with 1 string element:

[["$6"], ["$12"], ["$25"], ["$25"]].join(',').split(',');

will do the job. Bt that specifically matches your code example.

Answer 22

ES6 way:

const flatten = arr => arr.reduce((acc, next) => acc.concat(Array.isArray(next) ? flatten(next) : next), [])

const a = [1, [2, [3, [4, [5]]]]]
console.log(flatten(a))

ES5 way for flatten function with ES3 fallback for N-times nested arrays:

var flatten = (function() {
  if (!!Array.prototype.reduce && !!Array.isArray) {
    return function(array) {
      return array.reduce(function(prev, next) {
        return prev.concat(Array.isArray(next) ? flatten(next) : next);
      }, []);
    };
  } else {
    return function(array) {
      var arr = [];
      var i = 0;
      var len = array.length;
      var target;

      for (; i < len; i++) {
        target = array[i];
        arr = arr.concat(
          (Object.prototype.toString.call(target) === '[object Array]') ? flatten(target) : target
        );
      }

      return arr;
    };
  }
}());

var a = [1, [2, [3, [4, [5]]]]];
console.log(flatten(a));

Answer 23

var arrays = [["a"], ["b", "c"]];
Array.prototype.concat.apply([], arrays);

// gives ["a", "b", "c"]

(I'm just writing this as a separate answer, based on comment of @danhbear.)

Answer 24

I recommend a space-efficient generator function:

function* flatten(arr) {
  if (!Array.isArray(arr)) yield arr;
  else for (let el of arr) yield* flatten(el);
}

// Example:
console.log(...flatten([1,[2,[3,[4]]]])); // 1 2 3 4

If desired, create an array of flattened values as follows:

let flattened = [...flatten([1,[2,[3,[4]]]])]; // [1, 2, 3, 4]

Answer 25

I would rather transform the whole array, as-is, to a string, but unlike other answers, would do that using JSON.stringify and not use the toString() method, which produce an unwanted result.

With that JSON.stringify output, all that's left is to remove all brackets, wrap the result with start & ending brackets yet again, and serve the result with JSON.parse which brings the string back to "life".

• Can handle infinite nested arrays without any speed costs.
• Can rightly handle Array items which are strings containing commas.

var arr = ["abc",[[[6]]],["3,4"],"2"];

var s = "[" + JSON.stringify(arr).replace(/\[|]/g,'') +"]";
var flattened = JSON.parse(s);

console.log(flattened)

• Only for multidimensional Array of Strings/Numbers (not Objects)

Answer 26

That's not hard, just iterate over the arrays and merge them:

var result = [], input = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"]];

for (var i = 0; i < input.length; ++i) {
    result = result.concat(input[i]);
}

Answer 27

It looks like this looks like a job for RECURSION!

• Handles multiple levels of nesting
• Handles empty arrays and non array parameters
• Has no mutation
• Doesn't rely on modern browser features

Code:

var flatten = function(toFlatten) {
  var isArray = Object.prototype.toString.call(toFlatten) === '[object Array]';

  if (isArray && toFlatten.length > 0) {
    var head = toFlatten[0];
    var tail = toFlatten.slice(1);

    return flatten(head).concat(flatten(tail));
  } else {
    return [].concat(toFlatten);
  }
};

Usage:

flatten([1,[2,3],4,[[5,6],7]]);
// Result: [1, 2, 3, 4, 5, 6, 7] 

Answer 28

I have done it using recursion and closures

function flatten(arr) {

  var temp = [];

  function recursiveFlatten(arr) { 
    for(var i = 0; i < arr.length; i++) {
      if(Array.isArray(arr[i])) {
        recursiveFlatten(arr[i]);
      } else {
        temp.push(arr[i]);
      }
    }
  }
  recursiveFlatten(arr);
  return temp;
}

Answer 29

I was goofing with ES6 Generators the other day and wrote this gist. Which contains...

function flatten(arrayOfArrays=[]){
  function* flatgen() {
    for( let item of arrayOfArrays ) {
      if ( Array.isArray( item )) {
        yield* flatten(item)
      } else {
        yield item
      }
    }
  }

  return [...flatgen()];
}

var flatArray = flatten([[1, [4]],[2],[3]]);
console.log(flatArray);

Basically I'm creating a generator that loops over the original input array, if it finds an array it uses the yield* operator in combination with recursion to continually flatten the internal arrays. If the item is not an array it just yields the single item. Then using the ES6 Spread operator (aka splat operator) I flatten out the generator into a new array instance.

I haven't tested the performance of this, but I figure it is a nice simple example of using generators and the yield* operator.

But again, I was just goofing so I'm sure there are more performant ways to do this.

Answer 30

just the best solution without lodash

let flatten = arr => [].concat.apply([], arr.map(item => Array.isArray(item) ? flatten(item) : item))

Answer 31

Here is the recursive way...

function flatten(arr){
    let newArray = [];
    for(let i=0; i< arr.length; i++){
        if(Array.isArray(arr[i])){
          newArray =  newArray.concat(flatten(arr[i]))
        }else{
          newArray.push(arr[i])
        }
    }
  return newArray; 
}

console.log(flatten([1, 2, 3, [4, 5] ])); // [1, 2, 3, 4, 5]
console.log(flatten([[[[1], [[[2]]], [[[[[[[3]]]]]]]]]]))  // [1,2,3]
console.log(flatten([[1],[2],[3]])) // [1,2,3]

Answer 32

I propose two short solutions without recursion. They are not optimal from a computational complexity point of view, but work fine in average cases:

let a = [1, [2, 3], [[4], 5, 6], 7, 8, [9, [[10]]]];

// Solution #1
while (a.find(x => Array.isArray(x)))
    a = a.reduce((x, y) => x.concat(y), []);

// Solution #2
let i = a.findIndex(x => Array.isArray(x));
while (i > -1)
{
    a.splice(i, 1, ...a[i]);
    i = a.findIndex(x => Array.isArray(x));
}

Answer 33

The logic here is to convert input array to string and remove all brackets([]) and parse output to array. I'm using ES6 template feature for this.

var x=[1, 2, [3, 4, [5, 6,[7], 9],12, [12, 14]]];

var y=JSON.parse(`[${JSON.stringify(x).replace(/\[|]/g,'')}]`);

console.log(y)

Answer 34

if you use lodash, you can just use its flatten method: https://lodash.com/docs/4.17.14#flatten

The nice thing about lodash is that it also has methods to flatten the arrays:

i) recursively: https://lodash.com/docs/4.17.14#flattenDeep

ii) upto n levels of nesting: https://lodash.com/docs/4.17.14#flattenDepth

For example

const _ = require("lodash");
const pancake =  _.flatten(array)

Answer 35

Nowadays the best and easy way to do this is joining and spliting the array like this.

var multipleArrays = [["$6","$Demo"], ["$12",["Multi","Deep"]], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]]

var flattened = multipleArrays.join().split(",")

This solution works with multiple levels and is also oneliner.

DEMO

EDIT for ECMAScript 6

Since ECMAScript 6 has been standardized, you can change the operation [].concat.apply([], arrays); for [].concat(...arrays);

var flattened = [].concat(...input);

DEMO

EDIT Most Efficient solution

The most efficient way to solve the problem is using a loop. You can compare the "ops/sec" velocity here

var flattened=[];
for (var i=0; i<input.length; ++i) {
    var current = input[i];
    for (var j=0; j<current.length; ++j)
        flattened.push(current[j]);
} 

DEMO

Hope It Helps

Answer 36

const flatten = array => array.reduce((a, b) => a.concat(Array.isArray(b) ? flatten(b) : b), []); 

Per request, Breaking down the one line is basically having this.

function flatten(array) {
  // reduce traverses the array and we return the result
  return array.reduce(function(acc, b) {
     // if is an array we use recursion to perform the same operations over the array we found 
     // else we just concat the element to the accumulator
     return acc.concat( Array.isArray(b) ? flatten(b) : b);
  }, []); // we initialize the accumulator on an empty array to collect all the elements
}

Answer 37

Here is a version in Typescript based on the answer by artif3x, with a bonus implementation of flatMap for Scala fans.

function flatten<T>(items: T[][]): T[] {
  return items.reduce((prev, next) => prev.concat(next), []);
}

function flatMap<T, U>(items: T[], f: (t: T) => U[]): U[] {
  return items.reduce((prev, next) => prev.concat(f(next)), new Array<U>());
}

Answer 38

I think array.flat(Infinity) is a perfect solution. But flat function is a relatively new function and may not run in older versions of browsers. We can use recursive function for solving this.

const arr = ["A", ["B", [["B11", "B12", ["B131", "B132"]], "B2"]], "C", ["D", "E", "F", ["G", "H", "I"]]]
const flatArray = (arr) => {
    const res = []
    for (const item of arr) {
        if (Array.isArray(item)) {
            const subRes = flatArray(item)
            res.push(...subRes)
        } else {
            res.push(item)
        }
    }

    return res
}

console.log(flatArray(arr))

Answer 39

function flatten(input) {
  let result = [];
  
  function extractArrayElements(input) {
    for(let i = 0; i < input.length; i++){
      if(Array.isArray(input[i])){
        extractArrayElements(input[i]);
      }else{
        result.push(input[i]);
      }
    }
  }
  
  extractArrayElements(input);
  
  return result;
}


// let input = [1,2,3,[4,5,[44,7,8,9]]];
// console.log(flatten(input));

// output [1,2,3,4,5,6,7,8,9]

Answer 40

In Node environment, if your version is not support Array.prototype.flat but you use es6 or ts, then you can do like this

const arrays = [
  ["$6"],
  ["$12"],
  ["$25"],
  ["$25"],
  ["$18"],
  ["$22"],
  ["$10"]
];
const merged = [].concat(...arrays);

console.log(merged);

Answer 41

There is 2 way to flat your array

*1-First One You Can Use Array.prototype.flat Method

const veryDeep = [[1, [2, 2, [3, [4, [5, [6]]]]], 1]];

veryDeep.flat(Infinity)
 // [1, 2, 2, 3, 4, 5, 6, 1]
 /**  If you don't know the depth of the array, 
simply pass Infinity **\

Or You Can Put Argument If You Know How Much Nested Your Array Is

const twoLevelsDeep = [[1, [2, 2], 1]];
//depth = 1
twoLevelsDeep.flat();
[1, [2, 2], 1]

//depth = 2
twoLevelsDeep.flat(2);
// [1, 2, 2, 1]

*2-Second Way You Can Flat Your Array By Using Spread Operator In ES6 And Concat It Will Flat Your Array Regardless How Many Nested In It

 const multidimension = [, [, , ], , [, ]];

   [].concat(...multidimension); // This will return: [, , , , , , 
   ]

   const multidimension = [1, [2, 3, 4], 5, [6, 7]];

   [].concat(...multidimension); // This will return: [1, 2, 3, 4, 5, 6, 7]

Answer 42

Here's another deep flatten for modern browsers:

function flatten(xs) {
  xs = Array.prototype.concat.apply([], xs);
  return xs.some(Array.isArray) ? flatten(xs) : xs;
};

Answer 43

if your array only consists out of integers or strings you can use this dirty hack:

var arr = [345,2,[34],2,[524,[5456]],[5456]];
var flat = arr.toString().split(',');

Works, in FF, IE and Chrome didn't test the other browsers yet.

Answer 44

To flatten a two-dimensional array in one line:

[[1, 2], [3, 4, 5]].reduce(Function.prototype.apply.bind(Array.prototype.concat))
// => [ 1, 2, 3, 4, 5 ]

Answer 45

I'm aware that this is hacky, but the must succinct way I know of to flatten an array(of any depth!) of strings(without commas!) is to turn the array into a string and then split the string on commas:

var myArray =[["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]];
var myFlatArray = myArray.toString().split(',');

myFlatArray;
// ["$6", "$12", "$25", "$25", "$18", "$22", "$10", "$0", "$15", "$3", "$75", "$5", "$100", "$7", "$3", "$75", "$5"]

This should work on any depth of nested arrays containing only strings and numbers(integers and floating points) with the caveat that numbers will be converted to strings in the process. This can be solved with a little mapping:

var myArray =[[[1,2],[3,4]],[[5,6],[7,8]],[[9,0]]];
var myFlatArray = myArray.toString().split(',').map(function(e) { return parseInt(e); });
myFlatArray;
// [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

Answer 46

/**
* flatten an array first level
* @method flatten
* @param array {Array}
* @return {Array} flatten array
*/
function flatten(array) {
  return array.reduce((acc, current) => acc.concat(current), []);
}


/**
* flatten an array recursively
* @method flattenDeep
* @param array {Array}
* @return {Array} flatten array
*/
function flattenDeep(array) {
  return array.reduce((acc, current) => {
    return Array.isArray(current) ? acc.concat(flattenDeep(current)) : acc.concat([current]);
  }, []);
}

/**
* flatten an array recursively limited by depth
* @method flattenDepth
* @param array {Array}
* @return {Array} flatten array
*/
function flattenDepth(array, depth) {
  if (depth === 0) {
    return array;
  }
  return array.reduce((acc, current) => {
    return Array.isArray(current) ? acc.concat(flattenDepth(current, --depth)) : acc.concat([current]);
  }, []);
}

Answer 47

Just to add to the great solutions. I used recursion to solve this.

            const flattenArray = () => {
                let result = [];
                return function flatten(arr) {
                    for (let i = 0; i < arr.length; i++) {
                        if (!Array.isArray(arr[i])) {
                            result.push(arr[i]);
                        } else {
                            flatten(arr[i])
                        }
                    }
                    return result;
                }
            }

Test results: https://codepen.io/ashermike/pen/mKZrWK

Answer 48

It's better to do it in a recursive way, so if still another array inside the other array, can be filtered easily...

const flattenArray = arr =>
  arr.reduce(
    (res, cur) =>
       !Array.isArray(cur) 
       ? res.concat(cur)
       : res.concat(flattenArray(cur)), []);

And you can call it like:

flattenArray([[["Alireza"], "Dezfoolian"], ["is a"], ["developer"], [[1, [2, 3], ["!"]]]);

and the result isas below:

["Alireza", "Dezfoolian", "is a", "developer", 1, 2, 3, "!"]

Answer 49

Ways for making flatten array

• using Es6 flat()
• using Es6 reduce()
• using recursion
• using string manipulation

[1,[2,[3,[4,[5,[6,7],8],9],10]]] - [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]

// using Es6 flat() 
let arr = [1,[2,[3,[4,[5,[6,7],8],9],10]]]
console.log(arr.flat(Infinity))

// using Es6 reduce()
let flatIt = (array) => array.reduce(
  (x, y) => x.concat(Array.isArray(y) ? flatIt(y) : y), []
)
console.log(flatIt(arr))

// using recursion
function myFlat(array) {
  let flat = [].concat(...array);
  return flat.some(Array.isArray) ? myFlat(flat) : flat;
}
console.log(myFlat(arr));

// using string manipulation
let strArr = arr.toString().split(','); 
for(let i=0;i<strArr.length;i++)
  strArr[i]=parseInt(strArr[i]);

console.log(strArr)

Answer 50

This solution will work for any depth level (specifying how deep a nested array structure should be) of array.

function flatten(obj) {
    var out = [];
    function cleanElements(input) {
        for (var i  in input){
            if (input[i]instanceof Array){
                cleanElements(input[i]);
            }
            else {
                out.push(input[i]);
            }
        }
    }
    cleanElements(obj);
    return (out);
}

Answer 51

Here is the fastest solution in Typescript, which works also on arrays with multiple levels of nesting:

export function flatten<T>(input: Array<any>, output: Array<T> = []): Array<T> {
    for (const value of input) {
        Array.isArray(value) ? flatten(value, output) : output.push(value);
    }
    return output;
}

and than:

const result = flatten<MyModel>(await Promise.all(promises));

Answer 52

Simply using spread operator we can flatten in the following way.

var OriginalArray = [[5, 1], [6], [2], [8]];
var newArray = [];

for (let i = 0; i < OriginalArray.length; i++) {
  newArray.push(...OriginalArray[i]);
}

console.log(newArray)

Answer 53

There's a much faster way of doing this than using the merge.concat.apply() method listed in the top answer, and by faster I mean more than several orders of magnitude faster. This assumes your environment has access to the ES5 Array methods.

var array2d = [
  ["foo", "bar"],
  ["baz", "biz"]
];
merged = array2d.reduce(function(prev, next) {
    return prev.concat(next);
});

Here's the jsperf link: http://jsperf.com/2-dimensional-array-merge

Answer 54

You can flatten an array of arrays using Array.prototype.reduce() and Array.prototype.concat()

var data = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]].reduce(function(a, b) {
  return a.concat(b);
}, []);
console.log(data);

Related docs: https://developer.mozilla.org/en/docs/Web/JavaScript/Reference/Global_Objects/Array/concat

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/Reduce

Answer 55

I originally thought to use the .reduce method and recursively call a function to flatten inner arrays, however this can lead to stack overflows when you are working with a deeply nested array of deeply nested arrays. Using concat is also not the best way to go, because each iteration will create a new shallow copy of the array. What we can do instead is this:

const flatten = arr => {
    for(let i = 0; i < arr.length;) {
        const val = arr[i];
        if(Array.isArray(val)) {
            arr.splice(i, 1, ...val);
        } else {
            i ++;
        }
    }
    return arr;
}

We are not creating new arrays via concat and we are not recursively calling any functions.

http://jsbin.com/firiru/4/edit?js,console

Answer 56

Recursive version that works on all datatypes

 /*jshint esversion: 6 */

// nested array for testing
let nestedArray = ["firstlevel", 32, "alsofirst", ["secondlevel", 456,"thirdlevel", ["theinnerinner", 345, {firstName: "Donald", lastName: "Duck"}, "lastinner"]]];

// wrapper function to protect inner variable tempArray from global scope;
function flattenArray(arr) {

  let tempArray = [];

  function flatten(arr) {
    arr.forEach(function(element) {
      Array.isArray(element) ? flatten(element) : tempArray.push(element);     // ternary check that calls flatten() again if element is an array, hereby making flatten() recursive.
    });
  }

  // calling the inner flatten function, and then returning the temporary array
  flatten(arr);
  return tempArray;
}

// example usage:
let flatArray = flattenArray(nestedArray);

Answer 57

The following code will flatten deeply nested arrays:

/**
 * [Function to flatten deeply nested array]
 * @param  {[type]} arr          [The array to be flattened]
 * @param  {[type]} flattenedArr [The flattened array]
 * @return {[type]}              [The flattened array]
 */
function flattenDeepArray(arr, flattenedArr) {
  let length = arr.length;

  for(let i = 0; i < length; i++) {
    if(Array.isArray(arr[i])) {
      flattenDeepArray(arr[i], flattenedArr);
    } else {
      flattenedArr.push(arr[i]);
    }
  }

  return flattenedArr;
}

let arr = [1, 2, [3, 4, 5], [6, 7]];

console.log(arr, '=>', flattenDeepArray(arr, [])); // [ 1, 2, [ 3, 4, 5 ], [ 6, 7 ] ] '=>' [ 1, 2, 3, 4, 5, 6, 7 ]

arr = [1, 2, [3, 4], [5, 6, [7, 8, [9, 10]]]];

console.log(arr, '=>', flattenDeepArray(arr, [])); // [ 1, 2, [ 3, 4 ], [ 5, 6, [ 7, 8, [Object] ] ] ] '=>' [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]

Answer 58

You can use Ramda JS flatten

var arr = [[1,2], [3], [4,5]];
var flattenedArray = R.flatten(arr); 

console.log(flattenedArray)

<script src="https://cdnjs.cloudflare.com/ajax/libs/ramda/0.25.0/ramda.js"></script>

Answer 59

let arr = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]];
arr = arr.reduce((a, b) => a.concat(b)); // flattened

Answer 60

Simple flatten util i've written

const flatten = (arr, result = []) => {
    if (!Array.isArray(arr)){
        return [...result, arr];
    }
     arr.forEach((a) => {
         result = flatten(a, result)
    })

    return result
}

console.log(flatten([1,[2,3], [4,[5,6,[7,8]]]])) // [ 1, 2, 3, 4, 5, 6, 7, 8 ]

Answer 61

Define an array of arrays in javascript called foo and flatten that array of arrays into a single array using javascript's array concat builtin method:

const foo = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]] 
console.log({foo}); 

const bar = [].concat(...foo) 
console.log({bar});

Should print:

{ foo: 
   [ [ '$6' ],
     [ '$12' ],
     [ '$25' ],
     [ '$25' ],
     [ '$18' ],
     [ '$22' ],
     [ '$10' ] ] }
{ bar: [ '$6', '$12', '$25', '$25', '$18', '$22', '$10' ] }

Answer 62

Since ES2019 has introduced flat and flatMap then flat any array can be done like this:

const myArr = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]
const myArrFlat= myArr.flat(Infinity);
console.log(myArrFlat); // ["$6", "$12", "$25", ...]

Ref. https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/flat

Answer 63

Recursively calling the deepFlatten function so we can spread the inner array without using any external helper method is the way to go.

const innerArr = ['a', 'b'];
const multiDimArr = [[1, 2], 3, 4, [5, 6, innerArr], 9];

const deepFlatten = (arr) => {
  const flatList = [];
  arr.forEach(item => {
    Array.isArray(item)
     ? flatList.push(...deepFlatten(item)) // recursive call
     : flatList.push(item)
  });
  return flatList;
}

Answer 64

solve this problem using single line of code..

function unflatten(arr){
    return arr.reduce((prevProps, nextProps) => {
        return prevProps.concat(Array.isArray(nextProps) ? unflatten(nextProps) : nextProps)
    }, [])
}

https://codepen.io/jyotishman/pen/BayVeoQ

Answer 65

Use this method to flat two arrays

arr1.concat(...arr2);

Answer 66

[1,[2,3],[4,[5,6]]].reduce(function(p, c) {
    return p.concat(c instanceof Array ? 
                    c.reduce(arguments.callee, []) : 
                    [c]); 
}, []);

Answer 67

Here is my version of it. It allows you to flatten a complicated object which could be used in more scenarios:

Input

var input = {
   a: 'asdf',
   b: [1,2,3],
   c: [[1,2],[3,4]],
   d: {subA: [1,2]}
}

Code

The function is like this:

function flatten (input, output) {

  if (isArray(input)) {
    for(var index = 0, length = input.length; index < length; index++){
      flatten(input[index], output);
    }
  }
  else if (isObject(input)) {
    for(var item in input){
      if(input.hasOwnProperty(item)){
        flatten(input[item], output);
      }
    }
  }
  else {
    return output.push(input);
  }
};

function isArray(obj) {
  return Array.isArray(obj) || obj.toString() === '[object Array]';
}

function isObject(obj) {
  return obj === Object(obj);
}

Usage

var output = []

flatten(input, output);

Output

["asdf", 1, 2, 3, 1, 2, 3, 4, 1, 2]

Answer 68

If you need to support IE8 and, therefore, can't use methods such as reduce or isArray, here is a possible solution. It is a verbose approach to help you to understand the recursive algorithm.

function flattenArray(a){

    var aFinal = [];

    (function recursiveArray(a){

        var i,
            iCount = a.length;

        if (Object.prototype.toString.call(a) === '[object Array]') {
            for (i = 0; i < iCount; i += 1){
                recursiveArray(a[i]);
            }
        } else {
            aFinal.push(a);
        }

    })(a);

    return aFinal;

}

var aMyArray = [6,3,4,[12,14,15,[23,24,25,[34,35],27,28],56],3,4];

var result = flattenArray(aMyArray);

console.log(result);

Answer 69

var arrays = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"], ["$0"], ["$15"],["$3"], ["$75"], ["$5"], ["$100"], ["$7"], ["$3"], ["$75"], ["$5"]];
var merged = [].concat.apply([], arrays);
alert(merged);

Answer 70

Array.prototype.flatten = Array.prototype.flatten || function() {
    return [].reduce.call(this, function(flat, toFlatten) {
        return flat.concat(Array.isArray(toFlatten) ? toFlatten.flatten() : toFlatten);
    },[])
};

Answer 71

I didn't find here solution for large arrays when flattening is not deep. So, my version:

function flatten(arr){
    result = []
    for (e of arr)
        result.push(...e)
    return result
}

Answer 72

You can just keep on using Array.flat() method to achieve this even when array is a lot more nested.

[1,2,3,[2]].flat() 

is equivalent to

[1,2,3,[2]].flat(1)

so as your nesting increases you can keep on increasing number.

eg:

[1,[2,[3,[4]]]].flat(3) // [1, 2, 3, 4]

And if you are not sure about level of nesting you can just pass Infinity as parameter

[1,2,3,[2,[3,[3,[34],43],[34]]]].flat(Infinity) //[1, 2, 3, 2, 3, 3, 34, 43, 34]

Answer 73

try this method,

arr = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]]
concat_arr = arr.concat.apply([], arr)
console.log(concat_arr)

Answer 74

For Scala users looking for a way to replicate Seq.flatten in Javascript, here is a pimp of Array:

Array.prototype.flatten = function() {
  return [].concat.apply([], this);
};

which can be used this way:

[[12, 3, 5], [1], [], [3, 4]].flatten() // [12, 3, 5, 1, 3, 4]

Answer 75

I answer this question just with ES6, assume the deep array is:

const deepArray = ['1',[['a'],['b']],[2],[[[['4',[3,'c']]]],[5]]];

If you know or guess the depth of your arrays is not more than a number like 7, use below code:

const flatArray = deepArray.flat(7);

But if you don't know the depth of your deep arrays or your JavaScript engine doesn't support flat like react-native JavaScriptCore, use below function that is used JavaScript reduce function:

 const deepFlatten = arr =>
         arr.reduce(
           (acc, val) =>
             Array.isArray(val) 
               ? acc.concat(deepFlatten(val)) 
               : acc.concat(val),
             []
         );

Both of methods return below result:

["1", "a", "b", 2, "4", 3, "c", 5]

Answer 76

I have a simple solution without using in a special js function. (like reduce etc)

const input = [[0, 1], [2, 3], [4, 5]]
let flattened=[];

for (let i=0; i<input.length; ++i) {
    let current = input[i];
    for (let j=0; j<current.length; ++j)
        flattened.push(current[j]);
}

Answer 77

Much simpler and straight-forward one; with option to deep flatten;

const flatReduce = (arr, deep) => {
    return arr.reduce((acc, cur) => {
        return acc.concat(Array.isArray(cur) && deep ? flatReduce(cur, deep) : cur);
    }, []);
};

console.log(flatReduce([1, 2, [3], [4, [5]]], false)); // =>  1,2,3,4,[5]
console.log(flatReduce([1, 2, [3], [4, [5, [6, 7, 8]]]], true)); // => 1,2,3,4,5,6,7,8

Answer 78

ES-Next solutions

1. Array.from / Set / Array.flat / Array.sort

// 1. remove duplicate item
// 2. flat array
let arr= [["$6", ["$18"]], ["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]];

let result = Array.from(new Set(arr.flat(Infinity)));

console.log(`flat array with unique filter`, result);
// 

// 1. remove duplicate item
// 2. sort by Asc order (munber only)
// 3. flat array

// let arr= [["$6", ["$18"]], ["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]];

let arr= [["6", ["18"]], ["6"], ["12"], ["25"], ["25"], ["18"], ["22"], ["10"]];

let result = Array.from(new Set(arr.flat(Infinity))).map(Number).sort((a,b)=> a > b ? 1: -1);

console.log(`flat array with unique filter & Asc sort`, result);
//

2. ...spread / Set / Array.flat / Array.sort

// 1. remove duplicate item
// 2. flat array
let arr= [["$6", ["$18"]], ["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]];

let result = [...new Set(arr.flat(Infinity))];

console.log(`flat array with unique filter`, result);
// 

// 1. remove duplicate item
// 2. sort by Asc order(number only)
// 3. flat array

//let arr= [["$6", ["$18"]], ["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]];

let arr= [["6", ["18"]], ["6"], ["12"], ["25"], ["25"], ["18"], ["22"], ["10"]];

let result = [...new Set(arr.flat(Infinity))].map(Number).sort((a,b)=> a > b ? 1: -1);

console.log(`flat array with unique filter & Asc sort`, result);
// 

refs

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Set

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Spread_syntax

Answer 79

Yet another approach using jQuery $.map() function. From jQuery documentation:

The function can return an array of values, which will be flattened into the full array.

var source = [["$6"], ["$12"], ["$25"], ["$25"], ["$18"], ["$22"], ["$10"]];
var target = $.map(source, function(value) { return value; }); // ["$6", "$12", "$25", "$25", "$18", "$22", "$10"]

Answer 80

I have just try to solve the problem without using any inbuild function.

var arr = [1, 3, 4, 65, [3, 5, 6, 9, [354, 5, 43, 54, 54, 6, [232, 323, 323]]]];
var result = [];

function getSingleArray(inArr) {
  for (var i = 0; i < inArr.length; i++) {
    if (typeof inArr[i] == "object") {
      getSingleArray(inArr[i]); // Calling Recursively
    } else {
      result.push(inArr[i]);
    }
  }
}

getSingleArray(arr);
console.log(result); // [1, 3, 4, 65, 3, 5, 6, 9, 354, 5, 43, 54, 54, 6, 232, 323, 323]

Answer 81

What about deep flatten & Object Oriented ?

[23, [34, 454], 12, 34].flatten();
// -->   [23,34, 454, 12, 34]

---

[23, [34, 454,[66,55]], 12, 34].flatten();

// -->  [23, 34, 454, [66,55], 12, 34]

DEEP Flatten :

[23, [34, 454,[66,55]], 12, 34].flatten(true);

// --> [23, 34, 454, 66, 55, 12, 34]

DEMO

CDN

---

If all array elements are Integer,Float,... or/and String , So just , do this trick :

var myarr=[1,[7,[9.2]],[3],90];
eval('myarr=['+myarr.toString()+']');
print(myarr);
// [1, 7, 9.2, 3, 90]

DEMO

Answer 82

I believe that the best way to do this would be something like this:

var flatten = function () {
  return [].slice.call(arguments).toString().split(',');
};

Answer 83

Sheer Magic of ES6

const flat = A => A.reduce((A, a) => Array.isArray(a) ? [...A, ...flat(a)] : [...A, a], []);

Answer 84

Using code from there.

I would write:

myArray.enumerable().selectMany(function(x) { return x; }).array()

Answer 85

I was just looking for the faster and easy solution to do this, why? because I get this as a one interview question and I got curious, so I made this:

function flattenArrayOfArrays(a, r){
    if(!r){ r = []}
    for(var i=0; i<a.length; i++){
        if(a[i].constructor == Array){
            flattenArrayOfArrays(a[i], r);
        }else{
            r.push(a[i]);
        }
    }
    return r;
}

var i = [[1,2,[3]],4,[2,3,4,[4,[5]]]], output;

// Start timing now
console.time("flatten");
output = new Array(JSON.stringify(i).replace(/[^\w\s,]/g,"")); 
output
// ... and stop.
console.timeEnd("flatten");

// Start timing now
console.time("flatten2");
output = [].concat.apply([], i)
output
// ... and stop.
console.timeEnd("flatten2");

// Start timing now
console.time("flatten3");
output = flattenArrayOfArrays(i)
output
// ... and stop.
console.timeEnd("flatten3");

I used the most popular answers here and my solution. I guess somebody will find this interesting. Cheers!