Javascript recursive array flattening

Question

I'm exercising and trying to write a recursive array flattening function. The code goes here:

function flatten() {
 var flat = [];
 for (var i = 0; i < arguments.length; i++) {
 if (arguments[i] instanceof Array) {
 flat.push(flatten(arguments[i]));
 }
 flat.push(arguments[i]);
 }
 return flat;
}

The problem is that if I pass there an array or nested arrays I get the "maximum call stack size exceeded" error. What am I doing wrong?

• Etheryte: Just as a side-note: stackoverflow.com/questions/10865025/…
• basilikum: flatten.apply(this, arguments[i]); but it's not the only issue.
• Penny Liu: Does this answer your question? Merge/flatten an array of arrays

Answer 1

The problem is how you are passing the processing of array, if the value is an array then you are keep calling it causing an infinite loop

function flatten() {
 var flat = [];
 for (var i = 0; i < arguments.length; i++) {
 if (arguments[i] instanceof Array) {
 flat.push.apply(flat, flatten.apply(this, arguments[i]));
 } else {
 flat.push(arguments[i]);
 }
 }
 return flat;
}

Demo: Fiddle

Here's a more modern version:

function flatten(items) {
 const flat = [];
 items.forEach(item => {
 if (Array.isArray(item)) {
 flat.push(...flatten(item));
 } else {
 flat.push(item);
 }
 });
 return flat;
}

• dotnetCarpenter: Here is another modern one: function flatten(array, accu = []) { array.forEach(a => { if(Array.isArray(a)) flatten(a, accu) else accu.push(a) }) return accu }
• mrwnt10: Why don't the declarations of flat in the recursive calls of flatten() mask the outermost flat variable?
• Nicholas Gentile: @mrwnt10 I don't mean to sound sarcastic, but its because it is in fact, the outermost flat variable. The underlying flats in each successive frame are ultimately returned and then pushed into the top level flat where it is finally returned.
• Gy Tis: why are you using forEach() instead of for...of?

Answer 2

The clean way to flatten an Array in 2019 with ES6 is flat()

Short Answer:

array.flat(Infinity)

Detailed Answer:

const array = [1, 1, [2, 2], [[3, [4], 3], 2]]
// All layers
array.flat(Infinity) // [1, 1, 2, 2, 3, 4, 3, 2]
// Varying depths
array.flat() // [1, 1, 2, 2, Array(3), 2]
array.flat(2) // [1, 1, 2, 2, 3, Array(1), 3, 2]
array.flat().flat() // [1, 1, 2, 2, 3, Array(1), 3, 2]
array.flat(3) // [1, 1, 2, 2, 3, 4, 3, 2]
array.flat().flat().flat() // [1, 1, 2, 2, 3, 4, 3, 2]

Mozilla Docs

Can I Use - 96% Oct '23

Answer 3

If the item is array, we simply add all the remaining items to this array

function flatten(array, result) {
 if (array.length === 0) {
 return result
 }
 var head = array[0]
 var rest = array.slice(1)
 if (Array.isArray(head)) {
 return flatten(head.concat(rest), result)
 }
 result.push(head)
 return flatten(rest, result)
}
console.log(flatten([], []))
console.log(flatten([1], []))
console.log(flatten([1,2,3], []))
console.log(flatten([1,2,[3,4]], []))
console.log(flatten([1,2,[3,[4,5,6]]], []))
console.log(flatten([[1,2,3],[4,5,6]], []))
console.log(flatten([[1,2,3],[[4,5],6,7]], []))
console.log(flatten([[1,2,3],[[4,5],6,[7,8,9]]], []))

• jasonseminara: This should be the chosen answer. It does not use any iterative loops.

Answer 4

[...arr.toString().split(",")]

Use the toString() method of the Object. Use a spread operator (...) to make an array of string and split it by ",".

Example:

let arr =[["1","2"],[[[3]]]]; // output : ["1", "2", "3"]

• Scott Sauyet: This will turn numbers into strings and will break apart strings that include commas.

Answer 5

A Haskellesque approach...

function flatArray([x,...xs]){
 return x !== undefined ? [...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);

So i think the above code snippet could be made easier to understand with proper indenting;

function flatArray([x,...xs]){
 return x !== undefined ? [ ...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 6

If you assume your first argument is an array, you can make this pretty simple.

function flatten(a) {
 return a.reduce((flat, i) => {
 if (Array.isArray(i)) {
 return flat.concat(flatten(i));
 }
 return flat.concat(i);
 }, []);
 }

If you did want to flatten multiple arrays just concat them before passing.

Answer 7

If someone looking for flatten array of objects (e.g. tree) so here is a code:

function flatten(items) {
 const flat = [];
 items.forEach(item => {
 flat.push(item)
 if (Array.isArray(item.children) && item.children.length > 0) {
 flat.push(...flatten(item.children));
 delete item.children
 }
 delete item.children
 });
 return flat;
}
var test = [
	{children: [
 {children: [], title: '2'}
 ],
 title: '1'},
	{children: [
 {children: [], title: '4'},
 {children: [], title: '5'}
 ],
 title: '3'}
]
console.log(flatten(test))

Answer 8

Your code is missing an else statement and the recursive call is incorrect (you pass the same array over and over instead of passing its items).

Your function could be written like this:

function flatten() {
 // variable number of arguments, each argument could be:
 // - array
 // array items are passed to flatten function as arguments and result is appended to flat array
 // - anything else
 // pushed to the flat array as-is
 var flat = [],
 i;
 for (i = 0; i < arguments.length; i++) {
 if (arguments[i] instanceof Array) {
 flat = flat.concat(flatten.apply(null, arguments[i]));
 } else {
 flat.push(arguments[i]);
 }
 }
 return flat;
}
// flatten([[[[0, 1, 2], [0, 1, 2]], [[0, 1, 2], [0, 1, 2]]], [[[0, 1, 2], [0, 1, 2]], [[0, 1, 2], [0, 1, 2]]]]);
// [0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2]

• mrwnt10: Why don't the declarations of flat in the recursive calls of flatten() mask the outer flat variable?
• Salman Arshad: @mrwnt10 (if I understand your question correctly) var flat creates a local variable so it won't overwrite the outer flat variable.

Answer 9

Modern but not crossbrowser

function flatten(arr) {
 return arr.flatMap(el => {
 if(Array.isArray(el)) {
 return flatten(el);
 } else {
 return el;
 }
 });
}

Answer 10

This is a Vanilla JavaScript solution to this problem

var _items = {'keyOne': 'valueOne', 'keyTwo': 'valueTwo', 'keyThree': ['valueTree', {'keyFour': ['valueFour', 'valueFive']}]};
// another example
// _items = ['valueOne', 'valueTwo', {'keyThree': ['valueTree', {'keyFour': ['valueFour', 'valueFive']}]}];
// another example
/*_items = {"data": [{
 "rating": "0",
 "title": "The Killing Kind",
 "author": "John Connolly",
 "type": "Book",
 "asin": "0340771224",
 "tags": "",
 "review": "i still haven't had time to read this one..."
}, {
 "rating": "0",
 "title": "The Third Secret",
 "author": "Steve Berry",
 "type": "Book",
 "asin": "0340899263",
 "tags": "",
 "review": "need to find time to read this book"
}]};*/
function flatten() {
 var results = [],
 arrayFlatten;
 arrayFlatten = function arrayFlattenClosure(items) {
 var key;
 
 for (key in items) {
 if ('object' === typeof items[key]) {
 arrayFlatten(items[key]);
 } else {
 results.push(items[key]);
 }
 }
 };
 arrayFlatten(_items);
 
 return results;
}
console.log(flatten());

Answer 11

Here's a recursive reduce implementation taken from absurdum that mimics lodash's _.concat()

It can take any number of array or non-array arguments. The arrays can be any level of depth. The resulting output will be a single array of flattened values.

export const concat = (...arrays) => {
 return flatten(arrays, []);
}
function flatten(array, initial = []) {
 return array.reduce((acc, curr) => {
 if(Array.isArray(curr)) {
 acc = flatten(curr, acc);
 } else {
 acc.push(curr);
 }
 return acc;
 }, initial);
}

It can take any number of arrays or non-array values as input.

Source: I'm the author of absurdum

Answer 12

Here you are my functional approach:

const deepFlatten = (array => (array, start = []) => array.reduce((acc, curr) => {
 return Array.isArray(curr) ? deepFlatten(curr, acc) : [...acc, curr];
}, start))();
console.log(deepFlatten([[1,2,[3, 4, [5, [6]]]],7]));

• Scott Sauyet: A late response, but I'm curious why you would choose the extra IIFE wrapper and your start parameter. Why not just const deepFlatten = (xs) => xs .reduce ((a, x) => [...a, ... (Array .isArray (x) ? deepFlatten (x) : [x])], []) ?

Answer 13

A recursive approach to flatten an array in JavaScript is as follows.

function flatten(array) {
 let flatArray = [];
 for (let i = 0; i < array.length; i++) {
 if (Array.isArray(array[i])) {
 flatArray.push(...flatten(array[i]));
 } else {
 flatArray.push(array[i]);
 }
 }
 return flatArray;
}
let array = [[1, 2, 3], [[4, 5], 6, [7, 8, 9]]];
console.log(flatten(array)); 
// Output = [ 1, 2, 3, 4, 5, 6, 7, 8, 9 ]
let array2 = [1, 2, [3, [4, 5, 6]]];
console.log(flatten(array2)); 
// Output = [ 1, 2, 3, 4, 5, 6 ]

Answer 14

The function below flat the array and mantains the type of every item not changing them to a string. It is usefull if you need to flat arrays that not contains only numbers like items. It flat any kind of array with free of side effect.

function flatten(arr) {
 for (let i = 0; i < arr.length; i++) {
 arr = arr.reduce((a, b) => a.concat(b),[])
 }
 return arr
}
console.log(flatten([1, 2, [3, [[4]]]]));
console.log(flatten([[], {}, ['A', [[4]]]]));

Answer 15

Solution using forEach

function flatten(arr) {
 const flat = [];
 arr.forEach((item) => {
 Array.isArray(item) ? flat.push(...flatten(item)) : flat.push(item);
 });
 return flat; 
}

Solution using reduce

function flatten(arr) {
 return arr.reduce((acc, curr) => {
 if (Array.isArray(curr)) {
 return [...acc, ...flatten(curr)];
 } else {
 return [...acc, curr];
 }
 }, []);
}

Answer 16

Another answer in the list of answers, flattening an array with recursion:

let arr = [1, 2, [3, 4, 5, [6, 7, [[8], 9, [10]], [11, 13]], 15], [16, [17]]];
let newArr = [];
function steamRollAnArray(list) {
 for (let i = 0; i < list.length; i++) {
 if (Array.isArray(list[i])) {
 steamRollAnArray(list[i]);
 } else {
 newArr.push(list[i]);
 }
 }
}
steamRollAnArray(arr);
console.log(newArr);

To simplify, check whether the element at an index is an array itself and if so, pass it to the same function. If its not an array, push it to the new array.

Answer 17

This should work

function flatten() {
 var flat = [
 ];
 for (var i = 0; i < arguments.length; i++) {
 flat = flat.concat(arguments[i]);
 }
 var removeIndex = [
 ];
 for (var i = flat.length - 1; i >= 0; i--) {
 if (flat[i] instanceof Array) {
 flat = flat.concat(flatten(flat[i]));
 removeIndex.push(i);
 }
 }
 for (var i = 0; i < removeIndex.length; i++) {
 flat.splice(removeIndex - i, 1);
 }
 return flat;
}

Answer 18

The other answers already did point to the source of the OP's code malfunction. Writing more descriptive code, the problem literally boils down to an "array-detection/-reduce/-concat-recursion" ...

(function (Array, Object) {
//"use strict";
 var
 array_prototype = Array.prototype,
 array_prototype_slice = array_prototype.slice,
 expose_internal_class = Object.prototype.toString,
 isArguments = function (type) {
 return !!type && (/^\[object\s+Arguments\]$/).test(expose_internal_class.call(type));
 },
 isArray = function (type) {
 return !!type && (/^\[object\s+Array\]$/).test(expose_internal_class.call(type));
 },
 array_from = ((typeof Array.from == "function") && Array.from) || function (listAlike) {
 return array_prototype_slice.call(listAlike);
 },
 array_flatten = function flatten (list) {
 list = (isArguments(list) && array_from(list)) || list;
 if (isArray(list)) {
 list = list.reduce(function (collector, elm) {
 return collector.concat(flatten(elm));
 }, []);
 }
 return list;
 }
 ;
 array_prototype.flatten = function () {
 return array_flatten(this);
 };
}(Array, Object));

borrowing code from one of the other answers as proof of concept ...

console.log([
 [[[0, 1, 2], [0, 1, 2]], [[0, 1, 2], [0, 1, 2]]],
 [[[0, 1, 2], [0, 1, 2]], [[0, 1, 2], [0, 1, 2]]]
].flatten());
//[0, 1, 2, 0, 1, 2, 0, 1, 2, 0, 1, 2, ..., ..., ..., 0, 1, 2]

Answer 19

I hope you got all kind of different. One with a combination of recursive and "for loop"/high-order function. I wanted to answer without for loop or high order function.

Check the first element of the array is an array again. If yes, do recursive till you reach the inner-most array. Then push it to the result. I hope I approached it in a pure recursive way.

function flatten(arr, result = []) {
 if(!arr.length) return result;
 (Array.isArray(arr[0])) ? flatten(arr[0], result): result.push(arr[0]);
 return flatten(arr.slice(1),result)
}

Answer 20

I think the problem is the way you are using arguments.

since you said when you pass a nested array, it causes "maximum call stack size exceeded" Error.

because arguments[0] is a reference pointed to the first param you passed to the flatten function. for example:

 flatten([1,[2,[3]]]) // arguments[0] will always represents `[1,[2,[3]]]`

so, you code ends up calling flatten with the same param again and again.

to solve this problem, i think it's better to use named arguments, rather than using arguments, which essentially not a "real array".

Answer 21

There are few ways to do this:

1. using the flat method and Infinity keyword:

   const flattened = arr.flat(Infinity);

2. You can flatten any array using the methods reduce and concat like this:

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

Read more at: https://www.techiedelight.com/recursively-flatten-nested-array-javascript/

Answer 22

const nums = [1,2,[3,4,[5]]];
const chars = ['a',['b','c',['d',['e','f']]]];
const mixed = ['a',[3,6],'c',[1,5,['b',[2,'e']]]]; 
const flatten = (arr,res=[]) => res.concat(...arr.map((el) => (Array.isArray(el)) ? flatten(el) : el));
console.log(flatten(nums)); // [ 1, 2, 3, 4, 5 ]
console.log(flatten(chars)); // [ 'a', 'b', 'c', 'd', 'e', 'f' ]
console.log(flatten(mixed)); // [ 'a', 3, 6, 'c', 1, 5, 'b', 2, 'e' ]

Here is the breakdown:

1. loop over "arr" with "map"

arr.map((el) => ...)

2. on each iteration we'll use a ternary to check whether each "el" is an array or not

(Array.isArray(el))

3. if "el" is an array, then invoke "flatten" recursively and pass in "el" as its argument

flatten(el)

4. if "el" is not an array, then simply return "el"

: el

5. lastly, concatenate the outcome of the ternary with "res"

res.concat(...arr.map((el) => (Array.isArray(el)) ? flatten(el) : el));

--> the spread operator will copy all the element(s) instead of the array itself while concatenating with "res"

Answer 23

var nestedArr = [1, 2, 3, [4, 5, [6, 7, [8, [9]]]], 10];
let finalArray = [];
const getFlattenArray = (array) => {
 array.forEach(element => {
 if (Array.isArray(element)) {
 getFlattenArray(element)
 } else {
 finalArray.push(element)
 }
 });
}
getFlattenArray(nestedArr);

In the finalArray you will get the flattened array

• Scott Sauyet: Please don't add duplicate answers to several existing questions.

Answer 24

I think you are very close. One of the problems are that you call the flatten function with the same arguments. We can make use of the spread operator (...) to make sure we are calling flatten on the array inside of arguments[i], and not repeating the same arguments.

We also need to make a few more adjustments so we're not pushing more items into our array than we should

function flatten() {
 var flat = [];
 for (var i = 0; i < arguments.length; i++) {
 if (arguments[i] instanceof Array) {
 flat.push(...flatten(...arguments[i]));
 } else {
 flat.push(arguments[i]);
 }
 }
 return flat;
}
console.log(flatten([1,2,3,[4,5,6,[7,8,9]]],[10,11,12]));

Answer 25

function flatArray(input) {
 if (input[0] === undefined) return [];
 if (Array.isArray(input[0]))
 return [...flatArray(input[0]), ...flatArray(input.slice(1))];
 return [input[0], ...flatArray(input.slice(1))];
}

• Tyler2P: Your answer could be improved by adding more information on what the code does and how it helps the OP.

Answer 26

Here is my exampl

let arra = [1, [3,4], [5,6,4], 8]
let flatten2 = (oneArray) => {
var finalArray = [] 
for(var i=0;i<oneArray.length; i++){
if(Array.isArray(oneArray[i])){
finalArray.push(...oneArray[i])
} else{
finalArray.push(oneArray[i])
}
}
return finalArray
}
console.log(flatten2(arra))

e without recursion. Its simply using ES6

let flatten2 = (oneArray) => {
 var finalArray = [] 
 for(var i=0;i<oneArray.length; i++){
 if(Array.isArray(oneArray[i])){
 finalArray.push(...oneArray[i])
 } else {
 finalArray.push(oneArray[i])
}
}
return finalArray
 }
 console.log(flatten2(arra)

Answer 27

I found this very short syntaxed and functional.

function flattenArray(arr) {
 return arr.reduce((acc, val) => 
 Array.isArray(val) ? acc.concat(flattenArray(val)) : acc.concat(val), []);
}
const nestedArray = [1, [2, [3, [4, 5]], 6], 7];
console.log(flattenArray(nestedArray)); // Output: [1, 2, 3, 4, 5, 6, 7]

Answer 28

you should add stop condition for the recursion .

as an example if len (arguments[i]) ==0 return

• yxfxmx: Why doesn't it just stop when i==arguments.length is true?
• yxfxmx: so, what's the problem?(

Answer 29

Here is a solution without any iteration or helper variables:

function flatten(arr) {
 if (arr.length === 0) {
 return [];
 } else if (Array.isArray(arr[0])) {
 return flatten(arr[0]).concat(flatten(arr.slice(1)));
 } else {
 return [arr[0]].concat(flatten(arr.slice(1)));
 }
}