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How to Find Even or Odd Numbers in an Array in JavaScript

Filtering an array to separate its even and odd numbers is a classic programming exercise and a common data manipulation task. The most efficient and readable way to accomplish this in modern JavaScript is by using the Array.prototype.filter() method combined with the modulo operator (%).

This guide will teach you how to use this powerful combination to extract even or odd numbers from any array.

The Core Logic: The Modulo Operator (%)​

The key to determining if a number is even or odd is the modulo operator (%), which returns the remainder of a division operation.

  • A number is even if it can be divided by 2 with no remainder (number % 2 === 0).
  • A number is odd if it has a remainder of 1 when divided by 2 (number % 2 !== 0).

This simple check is the foundation of our filtering logic.

The filter() method is the perfect tool for this job. It creates a new array containing only the elements that pass a specific test.

For example, we have an array of numbers and we need to create two new arrays: one with only the even numbers, and one with only the odd numbers.

// Problem: How to separate the even and odd numbers from this array?
const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9];

Solution: By providing a simple callback function to filter(), we can create our new arrays in a single, readable line.

const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9];

// --- Find the EVEN numbers ---
const evenNumbers = numbers.filter(number => {
  return number % 2 === 0;
});
console.log('Even numbers:', evenNumbers); // Output: [2, 4, 6, 8]

// --- Find the ODD numbers ---
const oddNumbers = numbers.filter(number => {
  return number % 2 !== 0;
});
console.log('Odd numbers:', oddNumbers); // Output: [1, 3, 5, 7, 9]

This functional approach is concise, declarative, and the recommended best practice in modern JavaScript.

An Alternative: The Traditional for...of Loop​

While filter() is more concise, a traditional for...of loop is also a perfectly valid and readable way to accomplish the same goal. This imperative approach can be easier to understand for those new to functional programming.

The logic:

  1. Create a new, empty array to store the results.
  2. Loop through each number of the original array.
  3. Inside the loop, use an if statement with the modulo operator to check if the number is even or odd.
  4. If it meets the condition, push() it into the new array.

Solution:

const numbers = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const evenNumbers = [];

for (const number of numbers) {
  if (number % 2 === 0) {
    evenNumbers.push(number);
  }
}

console.log('Even numbers:', evenNumbers); // Output: [2, 4, 6, 8]
note

This method is slightly more verbose but achieves the exact same result and is just as performant for this task.

Conclusion​

Separating even and odd numbers from an array is a simple task once you understand the modulo operator.

  • The core logic for checking if a number is even is number % 2 === 0.
  • The Array.prototype.filter() method provides the most concise and modern solution for creating a new array based on this condition.
  • A traditional for...of loop is a perfectly valid and readable alternative that achieves the same result.

For most modern JavaScript projects, the filter() method is the preferred choice for its functional and declarative style.

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