list of integers and strings and returns a new list with only a numbers

Question

I know howe to do it using filter but I wondering if therei s a posibility to run this code with loop "for". Thank you for help.

function filter_list(l) {
let result = [];
  for (i = 0; i < l.length; i++) {
    if (typeof(l) === "number") {
   return l[i];
  }
 }  

}

console.log(filter_list(["abc", 1, 2, "def", 3, 4, "ghi"]));

Does this answer your question? [Remove Strings, Keep Numbers In Array With JavaScript](https://stackoverflow.com/questions/55403168/remove-strings-keep-numbers-in-array-with-javascript) — Ivar, Mar 09 '21 at 12:47

score 0 · Answer 1 · answered Mar 09 '21 at 12:47

0

You could filter by check the finiteness of the value.

function filter_list(l) {
    return l.filter(Number.isFinite);
}

console.log(filter_list(["abc", 1, 2, "def", 3, 4, "ghi"]));

answered Mar 09 '21 at 12:47

Nina Scholz

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I know howe to do it using filter but I wondering if therei s a posibility to run this code with loop "for". Thank you for help. – Jakub Mar 09 '21 at 12:50
@Jakub you can push the value to result when type is number and return that result list in the last. – gorak Mar 09 '21 at 12:51

score 0 · Accepted Answer · answered Mar 09 '21 at 12:51

You have a number of errors in your code that are causing it not to work as intended.

In your condition you are testing if the typeof the passed array is 'number' rather than each element.
next, you don't want to return on true matches, instead you want to push the element to result.
last, you need to declare i

function filter_list(l) {
  let result = [];
  for (let i = 0; i < l.length; i++) {
    let element = l[i];
    if (typeof element === "number") {
      result.push(element);
    }
  }
  return result;
}

console.log(filter_list(["abc", 1, 2, "def", 3, 4, "ghi"]));

list of integers and strings and returns a new list with only a numbers

2 Answers2