I need to remove the even numbers for this array
function removeEvens(numbers) {
}
/* Do not modify code below this line */
const oddNumbers = removeEvens([1, 2, 3, 4, 5]);
console.log(oddNumbers, `<-- should equal [1, 3, 5]`);
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Does this answer your question? Javascript - Filtering even numbersTomerikoo– Tomerikoo2023-02-22 08:03:35 +00:00Commented Feb 22, 2023 at 8:03
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3 Answers
The first step is how you check for evens - you can do this using the modulus operator (%) to see if the remainder when divided by 2 is 0, meaning that the number is even. Next you can filter the array to include only the numbers which pass this test:
function removeEvens(numbers) {
return numbers.filter(n => n % 2 !== 0); // if a number is even, remove it
}
const oddNumbers = removeEvens([1, 2, 3, 4, 5]);
console.log(oddNumbers);2 Comments
Joundill
Is 3.1 even? :)
awarrier99
@Joundill well it wasn't clearly specified whether there can be floats in the Array, but I updated my answer to not make this assumption. Although, it doesn't even make sense to be checking for the parity of an irrational number so I'm sure this won't have an effect
You can use .filter to do this with one line.
function removeEvens(numbers) {
return numbers.filter(n => n % 2 !== 0);
}
const oddNumbers = removeEvens([1, 2, 3, 4, 5]);
console.log(oddNumbers, '<-- should equal [1, 3, 5]');
Comments
Here is the code:
function removeEvens(numbers) {
return numbers.filter(n => n % 2 == 1);
}
/* Do not modify code below this line */
const oddNumbers = removeEvens([1, 2, 3, 4, 5]);
console.log(oddNumbers);