Map-Reduce：如何在集合中计数

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)
);

Answer 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); 

Answer 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) ); 

Answer 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.

Answer 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);