How can I sum all unique numbers of an array?

Question

I've the following data:

var array = [2, 4, 12, 4, 3, 2, 2, 5, 0];

I want to sum 2 + 4 + 12 + 3 + 5 and want to show the result using JavaScript without any library, by simply using a for loop and if / else statements.

Answer 1

You can make use of ES6 Sets and .reduce() method of arrays:

 let array = [2, 4, 12, 4, 3, 2, 2, 5, 0]; let sum = [...new Set(array)].reduce((a, c) => (a + c), 0); console.log(sum); 

Answer 2

you can try following using Set and Array.reduce

 var array = [2,4,12,4,3,2,2,5,0]; let set = new Set(); let sum = array.reduce((a,c) => { if(!set.has(c)) {set.add(c); a += c; } return a; }, 0); console.log(sum); 

Answer 3

 const distinct = (array) => array ? array.reduce((arr, item) => (arr.find(i => i === item) ? [...arr] : [...arr, item]), []) : array; const sum = distinct([2,4,12,4,3,2,2,5,0]).reduce((a,b) => a + b); console.log(sum); 

Answer 4

Here is a simple solution using a loop:

 const array = [2, 4, 12, 4, 3, 2, 2, 5, 0]; function sumNotCommon(array) { const sortedArray = array.slice().sort(); let sum = 0; for (let i = 0; i < sortedArray.length; ++i) { if (i === 0 || sortedArray[i] !== sortedArray[i-1]) sum += sortedArray[i]; } return sum; } console.log(sumNotCommon(array)); 

First it sorts the array and then iterates through it ignoring equal numbers that follow each other.