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.

not common means, *for me*, `12 + 5 + 0` ... what you are looking for, is the sum of all unique numbers. — Nina Scholz, Sep 15 '18 at 14:53

score 4 · Answer 1 · answered Sep 15 '18 at 12:54

4

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);

answered Sep 15 '18 at 12:54

Mohammad Usman

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wow! Good to work with this method,But I want to use only for loop and if else statement. – Hasanuzzaman Sep 15 '18 at 13:14
@Hasanuzzaman Why would you not use the sleeker method? What browsers are you targeting? – Luca Kiebel Sep 15 '18 at 13:15
@Hasanuzzaman See this [post](https://stackoverflow.com/questions/1960473/get-all-unique-values-in-a-javascript-array-remove-duplicates) for more detailed answers. – Mohammad Usman Sep 15 '18 at 13:38
@Hasanuzzaman And [this](https://stackoverflow.com/questions/1230233/how-to-find-the-sum-of-an-array-of-numbers) one too. – Mohammad Usman Sep 15 '18 at 13:38

score 2 · Answer 2 · answered Sep 15 '18 at 12:55

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);

answered Sep 15 '18 at 12:55

Nikhil Aggarwal

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1

function houseNumbersSum(inputArray) { return inputArray.slice(0, inputArray.indexOf(0)).reduce((s, v) => s+v, 0); };houseNumbersSum(inputArray) //It works good also. – Hasanuzzaman Sep 15 '18 at 14:04
@Hasanuzzaman - There are many ways to achieve it. However, the above will not remove duplicates. – Nikhil Aggarwal Sep 15 '18 at 14:07

score 0 · Answer 3 · answered Sep 15 '18 at 13:03

0

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);

answered Sep 15 '18 at 13:03

Adrian Brand

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Krisztián Balla · Answer 4 · 2018-09-15T13:23:14.317

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.

How can I sum all unique numbers of an array?

4 Answers4