Javascript sum of arrays and average

If you wanted to do it a bit more functionally, you could do something like this:

function sumAverage(arrays) {
  const average = arrays.reduce((acc, arr) => {
    const total = arr.reduce((total, num) => total += num, 0);
    return acc += total / arr.length;
  }, 0);
  
  return Math.round(average);
}

console.log('sum average:', sumAverage([[1,2,3], [4,5,6]]));

Just try this method..this kind of issues sometimes occured for me.

For example

 var total = 0;
 total = total + sum / a.length;

And every concat use this method..

Because you are assigning the value [] with the same name as the argument? This works, see jFiddle

function sumAverage(arr) {
  var result = 0;

  //arr = [];
  //a = [];
  //b = [];
  a = arr[0];
  b = arr[1];

  var sum = 0;

  // compute sum of elements in the array
  for(var j = 0; j < a.length; j++ ){
    sum += a[j] ;
  }

// get the  average of elements in the array
var total = 0;
total += sum / a.length;

var add = 0;

for(var i = 0; i < b.length; i++)
    add += b[i];

  var math = 0;
  math += add / b.length;
  result += math + total;

Math.round(result);
  return result;
}
document.write(sumAverage([[2,3,4,5], [6,7,8,9]]));

As said in comments, you reset your arguments...

Use the variable "arguments" for dynamic function parameters.

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Functions/arguments

I suggest to use two nested loops, one for the outer array and one for the inner arrays. Then sum values, calculate the average and add averages.

function sumAverage(array) {
    var result = 0,
        sum,
        i, j;

    for (i = 0; i < array.length; i++) {
        sum = 0;
        for (j = 0; j < array[i].length; j++) {
            sum += array[i][j];
        }
        result += Math.round(sum / array[i].length);
    }
    return result;
}

console.log(sumAverage([[2, 3, 4, 5], [6, 7, 8, 9]])); // 12

The problem is that you are emptying arr by saying arr = [].

Later, you are iterating over a which is empty too. Again when you say total += sum / a.length;, sum is 0 and a.length is 0 so 0/0 becomes NaN. Similarly for math. Adding Nan to NaN is again NaN and that's what you get.

Solution is to not empty passed arr and modify your code like below:

function sumAverage(arr) {
  var result = 0;
  // Your code here
  // set an array
  //arr = [];
  //a = [];
  //b = [];
  a = arr[0];
  b = arr[1];

  var sum = 0;

  // compute sum of elements in the array
  for (var j = 0; j < a.length; j++) {
    sum += a[j];
  }

  // get the  average of elements in the array
  var total = 0;
  total = sum / a.length;


  var add = 0;

  for (var i = 0; i < b.length; i++)
    add += b[i];

  var math = 0;
  math = add / b.length;

  result = math + total;

  result = Math.round(result);

  return result;
}
console.log(sumAverage([
  [2, 3, 4, 5],
  [6, 7, 8, 9]
]));

Basically I see a mistake here: arr[0] = a; arr[1] = b;

That should be a= arr[0]; b= arr[1];

and then remove: arr = [];

I suggest you write your function like this:

function sum(arr) {
var arr1 = arr[0]
var sum1 = 0;
arr1.map(function(e){sum1+=e});
var arr2 = arr[1]
var sum2 = 0;
arr2.map(function(e){sum2+=e});
return Math.round(sum1/arr1.length + sum2/arr2.length);
}