how to get sum of odd, even using reduce method, i have done as show in below code but returning undefined , @js-beginner
//code below
nums= [1,2,3,4,5,6,7,8,9]
function getOddEvenSum(numbers){
let{even,odd} = numbers.reduce((acc, cuu) => cuu%2 === 0?acc.even + cuu:acc.odd+cuu,{even:0, odd:0})
return {even, odd}
}
console.log(getOddEvenSum(nums)
//output i am getting below
{even:undefined, odd:undefined}
You can use Array.prototype.reduce like this:
const nums = [1, 2, 3, 4, 5, 6, 7, 8, 9];
const [odds, evens] = nums.reduce(
([odds, evens], cur) =>
cur % 2 === 0 ? [odds, evens + cur] : [odds + cur, evens],
[0, 0]
);
console.log(odds);
console.log(evens);The value that you return from your reduce callback will be the value of acc upon the next invocation/iteration of your array of numbers. Currently, your acc starts off as an object, but as you're only returning a number from your first iteration, all subsequent iterations will use a number as acc, which don't have .even or .odd properties. You could instead return a new object with updated even/odd properties so that acc remains as an object through all iterations:
const nums = [1,2,3,4,5,6,7,8,9];
function getOddEven(numbers){
return numbers.reduce((acc, cuu) => cuu % 2 === 0
? {odd: acc.odd, even: acc.even + cuu}
: {even: acc.even, odd: acc.odd+cuu},
{even:0, odd:0});
}
console.log(getOddEven(nums));This is not how the syntax of reduce works. One possible implementation:
function getOddEven(nums) {
return nums.reduce(
({odd, even}, num) => num % 2 === 0 ?
{odd, even: even + num} :
{odd: odd + num, even},
{odd: 0, even: 0},
);
}
I would argue that this is not very clear. Since performance is probably not critical, a clearer alternative would be:
function getOddEven(nums) {
return {
odd: nums.filter(num => num % 2 == 1).reduce((acc, num) => acc + num),
even: nums.filter(num => num % 2 == 0).reduce((acc, num) => acc + num),
};
}