I have an array and I want to double each odd number as well as return a new array with the newly doubled numbers. For some reason, the values won't double; here is my code.
function doubleOddNumbers(arr) {
return arr.filter(function(value) {
if(value%2 !== 0) {
return value * 2
}
})
}
When I look at my results in the console I continue to get the same odd numbers instead of them being doubled e.g. [1,2,3,4] will return [1,3] instead of [2, 6].
It can work this way
function doubleOddNumbers(arr) {
let newArr = [];
arr.forEach(function(value) {
if (value % 2 !== 0) {
newArr.push(value * 2);
}
});
return newArr;
}
console.log(doubleOddNumbers([1, 2, 3, 4, 5, 6, 7, 8]));
<!-- begin snippet: js hide: false console: true babel: false -->You can simply do it using Array.reduce.
function doubleOddNumbers(input) {
return input.reduce((acc, cur) => {
if (cur % 2 !== 0) {
acc.push(cur * 2);
}
return acc;
}, []);
}
console.log(doubleOddNumbers([1,2,3,4]));
[2, 6], I could be wrong though> var arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
undefined
> var x = arr.filter( ele => (ele % 2 != 0) ).map (ele => ele * 2)
undefined
> x
[ 2, 6, 10, 14, 18 ]
First of all, filter method's return statement always of boolean type. It should never be used assuming that it would return something. If you wanna return doubled odd values from array, do something like this -
arr.filter((value) => value % 2 !== 0).map((value) => value * 2);
.map()to transform/map those numbers. You should returntruefrom the filter method when you want to keep an item, andfalsewhen you want to remove the item.