Sort an array which contains number and strings

Question

I am trying to sort an array which contains strings, numbers, and numbers as strings (ex. '1','2'). I want to sort this array so that the sorted array contains numbers first and then strings that contain a number and then finally strings.

var arr = [9,5,'2','ab','3',-1 ] // to be sorted
arr.sort()
// arr = [-1, 5, 9, "2", "3","ab"] // expected result
//arr = [-1, "2", 5, 9, "ab"] // actual result

I have also tried

var number =[];
var char =[];
arr.forEach(a=>{
 if(typeof a == 'number') number.push(a);
 else char.push(a);
})
arr = (number.sort((a,b)=> a>b)).concat(char.sort((a,b)=> a>b))
// arr = [-1, 5, 9, "2", "3","ab"] // expected result
//  arr = [-1, 5, 9, "2", "ab", "3"]// actual result

Answer 1

The shortest is probably:

 arr.sort((a, b) => ((typeof b === "number") - (typeof a === "number")) || (a > b ? 1 : -1));

Answer 2

You can sort the numbers first and then the non-numbers by using .filter() to separate both data-types.

See working example below (read code comments for explanation):

const arr = [9,5,'2','ab','3',-1];

const nums = arr.filter(n => typeof n == "number").sort(); // If the data type of a given element is a number store it in this array (and then sort)
const non_nums = arr.filter(x => typeof x != "number").sort(); // Store everything that is not a number in an array (and then sort)

const res = [...nums, ...non_nums]; // combine the two arrays
console.log(res); // [-1, 5, 9, "2", "3", "ab"]