How to use reduce to return an object in Javascript?

Question

I'm trying to count the number of odd and even numbers in an array by using the Array.reduce() method. When I run the below code, I get the error "odd is not defined." How/where do I define odd to get this code to work?

var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2];
var oddEvenCounts = numbers.reduce(function(counts, number) {
   if (number % 2 === 1) {
     counts[odd]++
   } else {
     counts[even]++;
   }
   return counts;
 }, {});

Answer 1

Well, odd isn't defined. What you should do is either put odd/even in quotes ( counts['odd'] ) or use dot notation ( counts.odd ).

Also, since odd and even aren't defined, incrementing them would result into NaN . The initial value should instead be { odd: 0, even: 0 } .

 var numbers = [5, 3, 8, 6, 9, 1, 0, 2, 2]; var oddEvenCounts = numbers.reduce(function(counts, number) { if (number % 2 === 1) { counts['odd']++; } else { counts['even']++; } return counts; }, { odd: 0, even: 0 }); console.log(oddEvenCounts); 

Answer 2

This is a function that can do it for you.

 function oddEvenCounts(arr) { const counts = { even: 0, odd: 0 }; arr.forEach(n => { if(n % 2 === 0) { counts.even++; } else { counts.odd++ } }); return counts; } const array = [5, 3, 8, 6, 9, 1, 0, 2, 2]; console.log(oddEvenCounts(array)); 

Answer 3

Answer using ES6+

const sumEvenOdd = (numbersArray) => {
    return numbersArray.reduce((acc, current) => current % 2 === 0 ? {...acc,'even':acc['even'] + current} : {...acc, 'odd':acc['odd'] + current}, {"even":0, "odd":0})
}'

console.log(sumEvenOdd([1, 6, 8, 5, 3]));
// Expected results: {even: 14, odd: 9}