Divide array into two arrays odd and even

Question

This is the code but it returns in unidentified . It should return two arrays where the first one is a set of odd numbers in the original array and the second is odd

function evenAndOdd(array) {
    let odd = [];
    let even = [];
    for (i = 0; i < array.length; i++) {
        if (array[i] % 2 === 0) {
            even.push(array[i]);
        } else {
            odd.push(array[i]);
        }
    }
    return even,odd;
}

What do you mean by "it returns in unidentified"? A JS function can only return one value. — Valentin
– Valentin, Commented Oct 21, 2022 at 16:25
Probably best to understand about scope [in this case, of the arrays you define in the function] and to decide how you want the information returned - for example perhaps as an array of two arrays, or using arrays that already have been created outside the function itself or....? — A Haworth
– A Haworth, Commented Oct 21, 2022 at 16:26
Note you cannot return two separate values, you can return them in object or an array but not separately, eg return [even,odd]; — Patrick Evans
– Patrick Evans, Commented Oct 21, 2022 at 16:28
I think it should return odd. JavaScript is not Python, and , is an operator. What you could return though is [even,odd], and then destructure it later. — tevemadar
– tevemadar, Commented Oct 21, 2022 at 16:28

Nestoro · Accepted Answer · 2022-10-21 16:31:27Z

1

adding curly braces around your return values converts them into accessible objects

function evenAndOdd(array) {
    let odd = [];
    let even = [];
    for (i = 0; i < array.length; i++) {
        if (array[i] % 2 === 0) {
            even.push(array[i]);
        } else {
            odd.push(array[i]);
        }
    }
    return {even,odd};
}


console.log(evenAndOdd([1,2,3]))

answered Oct 21, 2022 at 16:31

Nestoro

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Comments

insertcreativename · Accepted Answer · 2022-10-21 16:27:59Z

0

You cant return two separate values from the same function. Instead, you can return a new array containing both of the other arrays.

function evenAndOdd(array){
  let odd = [];
  let even = [];
  for (i=0; i<array.length; i++){
    if (array[i] %2===0){
      even.push(array[i]);
    }else{
      odd.push(array[i]);
    }
  }
  return [even,odd];
}

answered Oct 21, 2022 at 16:27

insertcreativename

1

Comments

Sarah · Accepted Answer · 2022-10-21 16:30:43Z

0

If you're returning two or more variables, you can return them inside an array such as:

function evenAndOdd(array) {
  let odd = [];
  let even = [];
  for (i = 0; i < array.length; i++) {
    if (array[i] % 2 === 0) {
      even.push(array[i]);
    } else {
      odd.push(array[i]);
    }
  }
  return [even, odd];
}

console.log(evenAndOdd([1,2,3,4,5,6]));

answered Oct 21, 2022 at 16:30

Sarah

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Comments

shivam dwivedi · Accepted Answer · 2022-10-21 16:33:46Z

0

function evenAndOdd(array) {
    let odd = [];
    let even = [];
    for (i = 0; i < array.length; i++) {
        if (array[i] % 2 === 0) {
            even.push(array[i]);
        } else {
            odd.push(array[i]);
        }
    }

    return [even, odd];
}

var [even, odd] = evenAndOdd([0,1,2,3,4,5,6,7,8,9])
console.log("Even " + even)
console.log("Odd " + odd)

answered Oct 21, 2022 at 16:33

shivam dwivedi

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Comments

Nixit Ahir · Accepted Answer · 2022-10-21 16:34:49Z

0

You have to pass a valid number array.
and all I have done is just created an object with odd and even values array and return it because your code was only returning odd numbers.

function evenAndOdd(array) {
  let odd = [];
  let even = [];
  let length = array.length
  for (i = 0; i < length; i++) {
    if (array[i] % 2 === 0) {
      even.push(array[i]);
    } else {
      odd.push(array[i]);
    }
  }
  return {
    odd,
    even
  };
}

console.log(evenAndOdd([1, 2, 3, 4, 5, 6, 7, 8, 9]))

edited Oct 21, 2022 at 16:34

answered Oct 21, 2022 at 16:27

Nixit Ahir

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1 Comment

tevemadar Over a year ago

You can shortcut that into return {even, odd};.

trincot · Accepted Answer · 2025-09-26 11:54:20Z

0

As of ECMAScript 2024 there is a native function Object.groupBy for this:

const { 0: even, 1: odd } = Object.groupBy(arr, x => x % 2);

Example:

const arr = [10, 3, 9, 6, 12, 7];
const { 0: even, 1: odd } = Object.groupBy(arr, x => x % 2);
console.log("odd:", ...odd);
console.log("even:", ...even);

answered Sep 26 at 11:54

trincot

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Comments

doctorjay · Accepted Answer · 2022-10-21 17:07:12Z

-1

arr[i] & 1 === 0 is more elegant and should run more quickly than arr[i] % 2 === 0 if you're working with a large array

answered Oct 21, 2022 at 17:07

doctorjay

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1 Comment

trincot Sep 26 at 12:02

arr[i] & 1 === 0 is never true, because === has precedence over &. Apparently you didn't check your answer before posting.

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