JavaScript - Join arrays of an array with concat method instead of push method

Question

I have a problem that asks me to join arrays of an array and return a single array in the form of [ array[0][0], array[0][1], array[1][0], array[1][1], etc. ]. I solved it using the push method in nested for-loops, but the prompt says that I should be familiar with the concat method. I know the concat method syntax and how it works, but I can't figure out how to use it to do what the prompt asks for.

Here's my solution using the push method:

function joinArrayOfArrays(arr) {
  var joined = [];
  for (var i = 0; i < arr.length; i++) {
    for (var k = 0; k < arr[i].length; k++) {
      joined.push(arr[i][k]);
    }
  } 
  return joined;
}

joinArrayOfArrays([[1, 4], [true, false], ['x', 'y']]);

// => [ 1, 4, true, false, 'x', 'y' ]

How would I return the same output using the concat method?

guijob · Accepted Answer · 2017-07-15 03:01:59Z

4

Try using reduce:

arr.reduce((a, e) => a.concat(e))

edited Jul 15, 2017 at 3:01

answered Jul 15, 2017 at 2:38

guijob

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2 Comments

Ateş Göral Over a year ago

No need for the second [] argument. When you omit it, it means you want to use the first item in the array as the starting value. Also, a little white space doesn't hurt anyone: arr.reduce((a, e) => a.concat(e))

guijob Over a year ago

thank you for your advices! i've just made an edit to implement this tips.

guest271314 · Accepted Answer · 2017-07-15 02:43:33Z

2

You can use spread element preceding array as parameter to .concat()

let res = [].concat(...[[1, 4], [true, false], ['x', 'y']]);

console.log(res);

Using a function

const joinArrayOfArrays = (arrays = []) => [].concat(...arrays);

let res = joinArrayOfArrays([[1, 4], [true, false], ['x', 'y']]);

console.log(res);

answered Jul 15, 2017 at 2:43

guest271314

1

Comments

nem035 · Accepted Answer · 2017-07-15 02:38:33Z

1

If you want to use concatenation, you wouldn't need the second loop, you can just concatenate each sub array within a single loop. One thing to remember is that concat doesn't modify exist array but returns a new one:

function joinArrayOfArrays(arr) {
  var joined = [];
  for (var i = 0; i < arr.length; i++) {
    joined = joined.concat(arr[i]) // concat sub array
  } 
  return joined;
}

console.log(
  joinArrayOfArrays([[1, 4], [true, false], ['x', 'y']])
);

You can also use a spread in a similar manner:

function joinArrayOfArrays(arr) {
  var joined = [];
  for (var i = 0; i < arr.length; i++) {
    joined = [...joined, ...arr[i]] // spread
  } 
  return joined;
}

console.log(
  joinArrayOfArrays([[1, 4], [true, false], ['x', 'y']])
);

Using reduce, as suggested by J. Guilherme, is a more elegant way to do this.

answered Jul 15, 2017 at 2:38

nem035

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4 Comments

Michael Over a year ago

Thank you! I had everything except for the the assignment of joined to the concat method.

nem035 Over a year ago

@Michael glad to help mate :)

RobG Over a year ago

With rest and spread: var joinArrayOfArrays = (...args)=>[].concat(...args);. ;-)

Michael Over a year ago

Thanks, I'm new. Now I see that the reason I couldn't get the solution using array concatenation is that I was assuming concat modified the existing array. So thanks also for that key piece of information.

Andrea Cookson · Accepted Answer · 2017-07-15 05:55:48Z

0

If you are trying to practice your array methods, I suggest looking up the reduce method on MDN and adapt this solution to this problem as an alternative to a for-loop:

function joinArrayOfArrays(arr) {
      return arr.reduce(function(a, b) {
        return a.concat(b);
      }); 
    }

    var output = joinArrayOfArrays([[1, 4], [true, false], ['x', 'y']]);
    console.log(output); 
    // [1, 4, true, false, "x", "y"]

answered Jul 15, 2017 at 5:55

Andrea Cookson

1

1 Comment

Michael Over a year ago

Thanks! This particular problem was asking specifically for the concat method but I'll definitely do that.

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