Filter out array to have only unique values [duplicate]

Answer 1

You can use Map and Spread Operator:

var rawData = ["X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11"];

var unique = new Map();
rawData.forEach(d => unique.set(d, d));
var uniqueItems = [...unique.keys()];

console.log(uniqueItems);

Answer 2

This is for es2015 and above as far as I know. There are 'cleaner' options with ES6 but this a great way to do it (with TypeScript).

let values: any[] = [];
const distinct = (value: any, index: any, self: any) => {
    return self.indexOf(value) === index;
};
values = values.filter(distinct);

Answer 3

arr = ["I", "do", "love", "JavaScript", "and", "I", "also", "do", "love", "Java"];

uniqueArr = [... new Set(arr)];

// or

reallyUniqueArr = arr.filter((item, pos, ar) => ar.indexOf(item) === pos)

console.log(`${uniqueArr}\n${reallyUniqueArr}`)

Answer 4

You could use a hash table for look up and filter all not included values.

var data = ["X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11"],
    unique = data.filter(function (a) {
        return !this[a] && (this[a] = true);
    }, Object.create(null));

console.log(unique);

Answer 5

You can use Array.filter function to filter out elements of an array based on the return value of a callback function. The callback function runs for every element of the original array.

The logic for the callback function here is that if the indexOf value for current item is same as the index, it means the element has been encountered first time, so it can be considered unique. If not, it means the element has been encountered already, so should be discarded now.

var arr = ["X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11", "X_row7", "X_row4", "X_row6", "X_row10", "X_row8", "X_row9", "X_row11"];

var filteredArray = arr.filter(function(item, pos){
  return arr.indexOf(item)== pos; 
});

console.log( filteredArray );

Caveat: As pointed out by rob in the comments, this method should be avoided with very large arrays as it runs in O(N^2).

UPDATE (16 Nov 2017)

If you can rely on ES6 features, then you can use Set object and Spread operator to create a unique array from a given array, as already specified in @Travis Heeter's answer below:

var uniqueArray = [...new Set(array)]

Answer 6

Array.prototype.unique = function () {
  return [...new Set(this)]
}

then we can write:

const arr = [1, 5, 2, 2, 2, 3, 4, 3, 2, 1, 5]
const uniqueArr = arr.unique()