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))