Array average with parameter (...rest)

Question

I'm trying to write a function to calculate the elements average in an array using the parameter (...rest)

Here what I've tried:

function average(...nums) {
  let total = 0;
  
  for (const num of nums) {
    total += num;
  }
  
  return total / nums.length;
}

console.log(average(2, 6));
console.log(average(2, 3, 3, 5, 7, 10));
console.log(average(7, 1432, 12, 13, 100));
console.log(average());

But the last test returns NaN and I don't know why.

possible duplicate of https://stackoverflow.com/questions/36441258/numbers-system-in-javascripts — Bergi, Sep 26 '19 at 19:57
[How do I create a runnable stack snippet?](https://meta.stackoverflow.com/questions/358992) — adiga, Sep 26 '19 at 19:59

score 2 · Answer 1 · edited Oct 01 '20 at 19:33

2

function average(...nums)
{
    let total = 0;  
    for(const num of nums)
    {
        total +=num;
        n= nums.length;
    }
    return total/n;
}

console.log(average(2, 6));
console.log(average(2, 3, 3, 5, 7, 10));
console.log(average(7, 1432, 12, 13, 100));
console.log(average());

edited Oct 01 '20 at 19:33

derloopkat

5,980
15
38
42

answered Oct 01 '20 at 18:22

Hari Swapna Puli

19
3

MrGeek · Answer 2 · 2019-09-26T20:00:20.153

1

Because you're trying to divide 0 (total) by 0 (nums.length where nums is []), which is NaN in JavaScript.

You can have a check at the top of your function that returns a default value (say, 0), if the list is empty:

function average(...nums) {
    if (!nums.length) return 0;
    let total = 0;
    // rest
}

edited Sep 26 '19 at 20:00

answered Sep 26 '19 at 19:54

MrGeek

21,097
4
28
52

score 0 · Answer 3 · answered Sep 26 '19 at 19:56

0

You are returning 0/0 (which is NaN) when you pass nothing in, as nums becomes an empty array. and its length becomes 0.

You should return nums.length ? total/nums.length : 0;

answered Sep 26 '19 at 19:56

Abishek Aditya

747
3
11

Eugen Sunic · Answer 4 · 2019-09-26T20:19:03.523

Check it the argument which is an array exists, if not return 0 or some value stating that average cannot be calculated.

function average(...nums) {
  if (!nums.length) {
    return 'Cannot find average'
  }
  let total = 0;
  for (const num of nums) {

    total += num;
  }
  return total / nums.length;
}

console.log(average(2, 6));
console.log(average(2, 3, 3, 5, 7, 10));
console.log(average(7, 1432, 12, 13, 100));
console.log(average());

score 0 · Answer 5 · answered Mar 08 '22 at 11:27

function average(...inputs) {
    let avg = 0;
    let sum = 0;
    for(const input of inputs){
        sum += input;
        count= inputs.length;
    }
    avg = sum / count;
    return avg;
}

console.log(average(2, 6));
console.log(average(2, 3, 3, 5, 7, 10));
console.log(average(7, 1432, 12, 13, 100));
console.log(average());

Array average with parameter (...rest)

5 Answers5