Array traversal in JavaScript to fill two maps

Question

var numerList = [1, 3, 7, 2, 4, 16, 22, 23];
var evenNoLst = numerList.map(function(no) {
  return ((no % 2) === 0);
});
console.log(evenNoLst)

Above code for me is creating a map of even numbers, now I also want to have odd number list. Do I need to traverse through number list again? or is there a way to have two maps using single traversal of an array.

I am using Javascript.

Answer 1

Here's a way to split it in one go with reduce:

var numberList = [1, 3, 7, 2, 4, 16, 22, 23];

var grouped = numberList.reduce(function (acc, x){  
  acc[x%2].push(x);
  return acc;
}, [[], []]);

console.log(grouped);

The result is an array with two arrays inside: the first one has the even numbers and the second one the odd ones.

Answer 2

You could take the logical NOT operator and map all boolean values.

var numerList = [1, 3, 7, 2, 4, 16, 22, 23],
    evenNoLst = numerList.map(no => no % 2 === 0),
    oddNoLst = evenNoLst.map(b => !b);

console.log(evenNoLst);
console.log(oddNoLst);

.as-console-wrapper { max-height: 100% !important; top: 0; }

With a single loop approach

var numerList = [1, 3, 7, 2, 4, 16, 22, 23],
    oddNoLst = [],
    evenNoLst = [];

numerList.forEach(function (no) {
    var even = no % 2 === 0;
    evenNoLst.push(even);
    oddNoLst.push(!even);
});

console.log(evenNoLst);
console.log(oddNoLst);

.as-console-wrapper { max-height: 100% !important; top: 0; }

With for ... of loop

var numerList = [1, 3, 7, 2, 4, 16, 22, 23],
    oddNoLst = [],
    evenNoLst = [],
    no, even;

for (no of numerList) {
    even = no % 2 === 0;
    evenNoLst.push(even);
    oddNoLst.push(!even);
}

console.log(evenNoLst);
console.log(oddNoLst);

.as-console-wrapper { max-height: 100% !important; top: 0; }