## Map-Reduce : How to count in a collection

Question

Important edit : I can't use `filter` - the purpose is pedagogic.

I have an array in which I would want to count the number of its elements that verify a boolean, using only `map` and `reduce`.

# Count of the array's size

I already wrote something that counts the array's size (ie. : the number of all of its elements), using `reduce` :

``````const array_numbers = [12, 15, 1, 1]; // Size : 4
console.log(
array_numbers.reduce((acc) => {
return acc + 1;
}, 0)
);
``````

# Count of the array's elements checking a boolean condition

Now I would want to count only the elements that verify a boolean condition. Thus, I must use `map` before `reduce` and the elements of the `map`'s returned array will be only the good elements.

So I wrote this code but it doesn't work... Indeed, I put `null` when I encounter a not-good element (and `null` is counted as en element unfortunately).

NB : here, the boolean condition is "is the element even ? (%2 == 0)".

``````const array_numbers = [12, 15, 1, 1]; // Size : 4
console.log(
array_numbers.map((current_value) => {
if(current_value % 2 == 0) {
return current_value;
}
return null;

}).reduce((acc) => {
return acc + 1;

}, 0)
);
``````

Show source
2017-01-03 17:01 4 Answers

## Answers to Map-Reduce : How to count in a collection ( 4 )

1. `Array#filter` the array and check the length:

``````const array_numbers = [12, 15, 1, 1];
const result = array_numbers.filter((n) => n % 2 === 0).length;

console.log(result);``````

Or count using `Array#reduce`:

``````const array_numbers = [12, 15, 1, 1, 4];
const result = array_numbers.reduce((r, n) => n % 2 ? r : r + 1, 0);

console.log(result);``````

Or if you must, you can use `Array#map` with `Array#reduce`:

``````const array_numbers = [12, 15, 1, 1, 4];
const result = array_numbers
.map((n) => n % 2 === 0 ? 1 : 0) // map to 1 or 0 according to the condition
.reduce((r, n) => r + n); // sum everything

console.log(result);``````

2. You can use `Array.prototype.filter` to filter the even numbers - and you don't need the `reduce()` function - you can use `length` of the array returned by the `filter()` function.

Or you can use `reduce()` method alone like below:

See demos below:

``````const array_numbers = [12, 15, 1, 1]; // Size : 4

// using filter
console.log(
array_numbers.filter((current_value) => {
return current_value % 2 == 0;
}).length
);

// using reduce
console.log(
array_numbers.reduce((prev, curr) => {
return curr % 2 == 0 ? prev + 1 : prev;
}, 0)
);``````

3. Since per your comment you must use reduce for some reason:

``````arr.reduce((acc, n) => { n % 2 ? acc + n : acc }, 0);
``````

The map is unecessary.

4. As Jared Smith mentioned, you don't need to use `map` for this task. `Array.reduce()` gets the current element as a second argument in the callback function which you can use to check if it satisfies your given condition.

So, again assuming that you must use either `map` and/or `reduce`:

``````const myArray = [1,2,3,4,5,6];
const condition = function(a) {
// let's say
return a %2 == 0;
}

let result = myArray.reduce((acc, val) => {
return condition(val) ? acc + 1 : acc
}, 0);

console.log(result);``````