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Remove Duplicate Values from an Array using JS or jQuery

Working with arrays is a common task in JavaScript, but sometimes you may encounter arrays with duplicate values. Removing these duplicates can be essential for data manipulation and optimization.

In this post, we’ll explore how to remove duplicate values from an array using both vanilla JavaScript and jQuery, along with performance tips and handling edge cases.

Removing Duplicates with JavaScript (ES6)

One of the easiest and most efficient ways to remove duplicates from an array in modern JavaScript is by using the Set object combined with the spread operator.

A Set is a special type of object that only stores unique values, making it a simple way to filter out duplicates.

// Original array with duplicates
const arrayWithDuplicates = [1, 2, 3, 2, 4, 1, 5];

// Remove duplicates using Set and spread operator
const uniqueArray = [...new Set(arrayWithDuplicates)];

console.log(uniqueArray); // Output: [1, 2, 3, 4, 5]

In the example above, the Set automatically removes duplicates, and the [...new Set()] converts the set back into an array.

Removing Duplicates with JavaScript (Older Versions)

If you need to support older browsers or prefer not to use ES6 features, you can use a traditional loop or filter() method to remove duplicates.

// Original array with duplicates
var arrayWithDuplicates = [1, 2, 3, 2, 4, 1, 5];

// Remove duplicates using filter method
var uniqueArray = arrayWithDuplicates.filter(function(item, index) {
    return arrayWithDuplicates.indexOf(item) === index;
});

console.log(uniqueArray); // Output: [1, 2, 3, 4, 5]

This code uses the filter() method to create a new array by checking if the first occurrence of each item matches its current index. If it does, the item is included in the result.

Removing Duplicates with jQuery

Although jQuery does not have a built-in function for removing duplicates from arrays, you can easily combine jQuery with native JavaScript methods. Here’s how you can do it using jQuery.

// Original array with duplicates
var arrayWithDuplicates = [1, 2, 3, 2, 4, 1, 5];

// jQuery way of removing duplicates (using $.each)
var uniqueArray = [];

$.each(arrayWithDuplicates, function(i, el){
    if($.inArray(el, uniqueArray) === -1) uniqueArray.push(el);
});

console.log(uniqueArray); // Output: [1, 2, 3, 4, 5]

In this example, $.inArray() checks whether the element is already present in the uniqueArray. If not, it adds the element to the array.

 

Removing Duplicates from an Array of Objects

If you’re working with an array of objects, removing duplicates requires a different approach. Here’s how to do it using JavaScript’s reduce() method.

// Array of objects with duplicates
var arrayWithObjects = [
    { id: 1, name: 'John' },
    { id: 2, name: 'Jane' },
    { id: 1, name: 'John' },
    { id: 3, name: 'Bob' }
];

// Remove duplicates by comparing object IDs
var uniqueArray = arrayWithObjects.reduce((accumulator, current) => {
    if (!accumulator.some(item => item.id === current.id)) {
        accumulator.push(current);
    }
    return accumulator;
}, []);

console.log(uniqueArray);
// Output: [{ id: 1, name: 'John' }, { id: 2, name: 'Jane' }, { id: 3, name: 'Bob' }]

This code uses reduce() to iterate over the array and adds objects to the new array only if their id doesn’t already exist in the accumulator array.

Performance Considerations

When working with large arrays, it’s important to choose the most efficient method for removing duplicates. The ES6 Set method is generally the fastest and most efficient way for simple arrays with primitive values, while methods like filter() and reduce() might be slower, especially for arrays of objects.

// Performance tip: Using Set for large arrays
const largeArray = Array.from({length: 1000000}, (_, i) => i % 1000); // Array with duplicates
const uniqueLargeArray = [...new Set(largeArray)];
console.log(uniqueLargeArray.length); // Fast operation even with large datasets

If performance is critical, testing each method with your specific dataset is advisable to ensure you’re using the fastest approach.

Handling Edge Cases

It’s important to consider edge cases when removing duplicates, such as:

  • Empty arrays: Ensure your code handles empty arrays gracefully, returning an empty array.
  • Non-primitive values: Arrays that contain null, undefined, or complex objects might require special handling depending on your use case.
// Handling empty arrays
const emptyArray = [];
const uniqueEmptyArray = [...new Set(emptyArray)];
console.log(uniqueEmptyArray); // Output: []

// Handling non-primitive values
const arrayWithNulls = [1, null, 2, null, 3];
const uniqueArrayWithNulls = [...new Set(arrayWithNulls)];
console.log(uniqueArrayWithNulls); // Output: [1, null, 2, 3]

Practical Use Cases

Removing duplicates is a common task in many real-world scenarios, including:

  • Form input validation: Ensuring users do not enter duplicate data, such as email addresses or usernames.
  • Cleaning data from APIs: When working with data fetched from external APIs, duplicates can arise, and cleaning the data ensures that your application processes unique entries only.
  • Optimizing data storage: Removing duplicates before saving data to a database helps reduce redundancy and storage costs.

Conclusion

Removing duplicate values from arrays is a common task in JavaScript, and there are multiple ways to accomplish it depending on the version of JavaScript you’re using.

Whether you’re using the modern ES6 approach with Set or older methods like filter(), both are effective. Additionally, if you’re using jQuery, you can easily integrate JavaScript methods to achieve the same result.

Be mindful of performance when working with large arrays, and always consider edge cases such as empty arrays or non-primitive values. By applying these techniques, you can ensure your arrays remain clean and efficient in your applications!

Roee Yossef
Roee Yossef

I develop pixel-perfect custom WordPress themes, delivering high-performance, SEO-optimized websites. Have a project in mind or need assistance? Feel free to contact me!

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