JavaScript: Sort an array of String numbers

Question

I have an array of school grades that looks like the following.

(Note 'N' stands for no grades and 'K' stands for kindergarten)

const toSort = ['1','3','4','5','6','7','9','10','11','12','K','2','N','8'];

Using the JavaScript sort() method, I would like to arrange the array so it will look like:

const sorted = ['K','1','2','3','4','5','6','7','8','9','10','11','12','N'];

Here is my attempt at it:

const toSort = ['1', '3', '4', '5', '6', '7', '9', '10', '11', '12', 'K', '2', 'N', '8'];

toSort.sort();
// Produces: ["1", "10", "11", "12", "2", "3", "4", "5", "6", "7", "8", "9", "K", "N"]

const test = toSort.sort((a, b) => {
  if (a === 'K') {
    return -1;
  }

  return Number(a) < Number(b) ? -1 : Number(a) > Number(b) ? 1 : 0;
});

console.log(test)

https://jsbin.com/pocajayala/1/edit?html,js,console,output

How I can resolve this?

What's the sorting logic that needs to be implemented? Why is K < 1 while N > 12? — Tom O.
– Tom O., Commented Jul 27, 2018 at 14:52
These are school grades. 'N' stands for no grade, 'K' stands for kindergarten — RyeGuy
– RyeGuy, Commented Jul 27, 2018 at 14:52
Just like 'K', make a special case for 'N'. Then convert everything else to a Number. No need for any ternary operators. — Code-Apprentice
– Code-Apprentice, Commented Jul 27, 2018 at 14:54
Your issue is that Number("K") and Number("N") are both NaN. You only remove the case where the parameter a is "K", not where b is "K" or where either is "N". — ASDFGerte
– ASDFGerte, Commented Jul 27, 2018 at 14:55

Peter Mortensen · Accepted Answer · 2023-01-24 18:25:07Z

22

You can use the string's native prototype localeCompare() function like so:

['1', '3', '4', '5', '6', '7', '9', '10', '11', '12', 'K', '2', 'N', '8']
    .sort((a, b) => a.localeCompare(b, undefined, { numeric: true }))

//  ["1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "K", "N"]

It works well with numbers among other characters too.

edited Jan 24, 2023 at 18:25

Peter Mortensen

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answered Jun 3, 2021 at 8:33

Liviu

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2 Comments

Yosi Leibman Over a year ago

Learned a new native function, thanks! that's a clean solution

devnik Over a year ago

What a smart solution. I really like this.

ponury-kostek · Accepted Answer · 2018-07-27 14:59:14Z

7

const toSort = ['1', '3', '4', '5', '6', '7', '9', '10', '11', '12', 'K', '2', 'N', '8'];

const test = toSort.sort((a, b) => {
	// console.log('a ' + a + ' b ' + b);
	if (a === "K" || b === "N") {
		return -1;
	}
	if (a === "N" || b === "K") {
		return 1;
	}
	return +a - +b;
});

console.log(test)

answered Jul 27, 2018 at 14:59

ponury-kostek

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Comments

Aramil Rey · Accepted Answer · 2018-07-27 15:05:56Z

6

I would do it like this:

const toSort = ['1', '3', '4', '5', '6', '7', '9', '10', '11', '12', 'K', '2', 'N', '8'];

const transform = k => {
    if (k === 'K') return 0;
    else if (k === 'N') return 13;
    else return +k;
}

const test = toSort.sort((a, b) => transform(a) - transform(b));

console.log(test);

In case your letters don't have correlation with those specific numbers, and instead are always the biggest and the smallest, you can use Infinity and -Infinity on the transform function.

const transform = k => {
    if (k === 'K') return -Infinity;
    else if (k === 'N') return Infinity;
    else return +k;
}

edited Jul 27, 2018 at 15:05

answered Jul 27, 2018 at 15:00

Aramil Rey

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1 Comment

Andrew Bone Over a year ago

return k == 'K' ? 0 : k == 'N' ? 13 : +k if you felt like minifying it.

Saeed · Accepted Answer · 2018-07-27 15:00:16Z

0

I think this helps you

const toSort = ['1','3','4','5','6','7','9','10','11','12','K','2','N','8'];

toSort.sort(function(a, b) {
  if(a == 'K') {
    return -1; // always K is smaller
  }
  if(a == 'N')  {
    return 1 // always N is bigger
  }
  if(b == 'K') {
    return 1;
  }
  if(b == 'N')  {
    return -1
  }
  return Number(a) - Number(b);
});

console.log(toSort);

answered Jul 27, 2018 at 15:00

Saeed

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