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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)
);

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)
);

• javascript
• arrays
• mapreduce
• reduce

Share Follow edited Jan 3, 2017 at 16:52 JarsOfJam-Scheduler asked Jan 3, 2017 at 16:46 JarsOfJam-SchedulerJarsOfJam-Scheduler 3,17955 gold badges3535 silver badges9090 bronze badges 3

• you're looking for filter.... – Jared Smith Commented Jan 3, 2017 at 16:49
• I know, but I must recreate this function - it's pedagogic in fact. – JarsOfJam-Scheduler Commented Jan 3, 2017 at 16:50
• See my answer then. – Jared Smith Commented Jan 3, 2017 at 16:54

Add a ment  | 

4 Answers 4

Sorted by: Reset to default 5

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);

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)
);

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);

Since per your ment you must use reduce for some reason:

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

The map is unecessary.

本文标签： javascriptMapReduceHow to count in a collectionStack Overflow