Find average of each array within an array

Question

I'm trying to write a map/reduce to get the average of each array within an array.

For example.

[[1][2,3][4,5,6,7]] => [1, 2.5, 5.5]

Right now this is my code where result is the array of arrays:

result.map(array => {
  return array.reduce((a, b) => (a + b)) / array.length;
})

const result = [
  [1],
  [2, 3],
  [4, 5, 6, 7]
]

console.log(result.map(array => {
  return array.reduce((a, b) => (a + b)) / array.length;
}))

Any help to get the desired output is much appreciated. As it stands, my output is reducing to an array of NaN's instead of the averages.

With the syntax corrected your code works. The problem is somewhere else. For example the result array contains non numbers. — Yury Tarabanko
– Yury Tarabanko, Commented Oct 30, 2019 at 17:56

Nina Scholz · Accepted Answer · 2019-10-30 17:54:59Z

5

You need a closing parentesis.

By using Array#reduce with arrays with unknown length, you need to take a start value, which is in this case of a length of zero the result.

var result = [[1], [2, 3], [4, 5, 6, 7]],
    avg = result.map(array => array.reduce((a, b) => a + b, 0) / array.length);
    //                                                    ^^^                ^
    //                                                    optional           required

console.log(avg);

edited Oct 30, 2019 at 17:54

answered Oct 30, 2019 at 17:48

Nina Scholz

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1 Comment

Yury Tarabanko Over a year ago

Not really [1].reduce((a, b) => a + b) The problem would be for empty arrays.

Avocado · Accepted Answer · 2019-10-30 17:50:42Z

1

you must provide a second argument to the reduce function, the initial value of a. So:

result.map(array => {
  return array.reduce((a, b) => a + b, 0) / array.length;
});

You may also want to ensure that array.length > 0 before you divide by it

answered Oct 30, 2019 at 17:50

Avocado

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1 Comment

adiga Over a year ago

OP's code works fine without the the initialValue argument as well

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