How to Return All Odd Numbers in a JavaScript Array?

Find all odd numbers in a JavaScript array

You can find all odd numbers in a JavaScript array by:

You should avoid checks for odd numbers with "n % 2 === 1" (where n is an integer, and % is the remainder operator) because, for negative numbers, this would return -1. Therefore, it may be easier to simply check if the number is not an even number (i.e. n % 2 !== 0) because it accounts for both, negative and positive numbers (since -0 === 0 in JavaScript). Otherwise, you would need an extra/explicit check for negative numbers.

Using Array.prototype.filter()

To get all odd numbers in an array of integers using the Array.prototype.filter() method, you can do the following:

// ES5+
const numbers = [-5, -2, -1, 0, 1, 3, 4, 7];
const oddNumbers = numbers.filter(function (number) {
    return number % 2 !== 0;
});

console.log(oddNumbers); // [-5, -1, 1, 3, 7]

This would return a new array containing all odd numbers. In ES6+, you can shorten this to a one-liner using the arrow function syntax:

// ES6+
const numbers = [-5, -2, -1, 0, 1, 3, 4, 7];
const oddNumbers = numbers.filter((number) => number % 2 !== 0);

console.log(oddNumbers); // [-5, -1, 1, 3, 7]

When no matches are found, an empty array is returned:

// ES6+
const numbers = [-2, 0, 4];
const oddNumbers = numbers.filter((number) => number % 2 !== 0);

console.log(oddNumbers); // []

Using a Loop

You can loop over an array of integers and create a new array to which you add all odd numbers to, for example, like so:

const numbers = [-5, -2, -1, 0, 1, 3, 4, 7];
const oddNumbers = [];

for (let i = 0; i < numbers.length; i++) {
    if (numbers[i] % 2 !== 0) {
        oddNumbers.push(numbers[i]);
    }
}

console.log(oddNumbers); // [-5, -1, 1, 3, 7]

This would return a new array containing all odd numbers, and when no match is found, an empty array is returned.


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