Using Javascript .sort() on an array to put all the zeros at the end

Question

I am trying to sort an array so that all the zeros are at the end. However, I don't want the list to be numerically sorted, all the numbers above zero should stay in the same order. Here's what I've got so far:

function placeZerosAtEnd(arr) {
    return arr.sort(compareForSort);
}
function compareForSort(first, second) {
    return first == 0 ? 1 : 0;
}
placeZerosAtEnd([9,0,9,1,0,2,0,1,1,0,3,0,1,9,9,0,0,0,0,0]);

This should return [9,9,1,2,1,1,3,1,9,9,0,0,0,0,0,0,0,0,0,0], but actually returns [3,9,9,1,9,2,9,1,1,1,0,0,0,0,0,0,0,0,0,0]. The zeros are correct, but the other numbers are in a strange order. What is going on here?

http://jsfiddle.net/v67fx4zk/

i guess its because of quicksort ("divide and conquer") so its a sort algorithms behavior. — Cracker0dks
– Cracker0dks, Commented Sep 23, 2014 at 23:17
You could do xs.filter(isntZero).concat(xs.filter(isZero)) — elclanrs
– elclanrs, Commented Sep 23, 2014 at 23:18
I get the expected result with your fiddle. Did you try also handling the case where second is 0 ? — Sebastien C.
– Sebastien C., Commented Sep 24, 2014 at 10:49

Stormdamage · Accepted Answer · 2014-09-24 10:43:06Z

11

As other have said, using .sort() was wrong in this instance. This is the code I used in the end, as suggested by elclanrs in the comments below my initial post.

function placeZerosAtEnd(arr) {
    return arr.filter(isntZero).concat(arr.filter(isZero));
}
function isntZero(element) {
    return element > 0;
}
function isZero(element) {
    return element == 0;
}

answered Sep 24, 2014 at 10:43

Stormdamage

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Comments

gazdagergo · Accepted Answer · 2019-01-11 15:16:28Z

3

The trivial solution with sort: "bubble down" zeros against non-zeros:

[9,0,9,1,0,2,0,1,1,0,3,0,1,9,9,0,0,0,0,0].sort((a, b) => {
  if (a !== 0 && b === 0) return -1;
  if (a === 0 && b !== 0) return 1;
  return 0;
})

// [9, 9, 1, 2, 1, 1, 3, 1, 9, 9, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]

edited Jan 11, 2019 at 15:16

answered Jan 11, 2019 at 13:24

gazdagergo

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Comments

Rick Hitchcock · Accepted Answer · 2014-09-23 23:19:44Z

2

I don't think you can accomplish this with the sort function. Depending on the sort method the browser uses (quicksort, merge sort, etc.), the sort function works by swapping items into place. So there's no telling where the non-zeroes will end up ... your algorithm simply ensures that they'll appear before the 0s.

Here's a function that will accomplish what you're trying:

function placeZerosAtEnd(arr) {
  for(var i = 0 ; i < arr.length ; i++) {
    if(arr[i]===0) arr.splice(arr.length-1, 0, arr.splice(i, 1)[0]);
  }
  return arr;
}

answered Sep 23, 2014 at 23:19

Rick Hitchcock

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Comments

adiga · Accepted Answer · 2020-07-10 16:37:50Z

2

you can use this some way too

let listSort = [9, 0, 9, 1, 0, 2, 0, 1, 1, 0, 3, 0, 1, 9, 9, 0, 0, 0, 0, 0],
    isntZero = [],
    isZero = []

for (let i = 0; i < listSort.length; i++) {
  if (listSort[i] == 0)
    isZero.push(listSort[i])
  else
    isntZero.push(listSort[i])
}

let output = isntZero.concat(isZero)

console.log(JSON.stringify(output))

edited Jul 10, 2020 at 16:37

adiga

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answered Jan 9, 2018 at 7:10

Davod Aslani Fakor

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Comments

Community · Accepted Answer · 2017-05-23 10:33:23Z

0

In your example the sort function is doing exactly what it's meant to.

You're expecting a stable sort, but a stable sort is not guaranteed by the javascript sort algorithm.

See this question for a discussion on stable sorting in javascript.

edited May 23, 2017 at 10:33

CommunityBot

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answered Sep 23, 2014 at 23:15

Andrew Shepherd

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Comments

Vijay Palaskar · Accepted Answer · 2020-07-10 17:01:40Z

0

You can use arr.sort to solve this problem

reference: Array.sort

   function placeZerosAtEnd(arr) {

     arr.sort(function(key1,key2) {

      if (key1 == 0 ) return 1;
      return -1;
    });    
    return arr;
  }

answered Jul 10, 2020 at 17:01

Vijay Palaskar

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Maksim Gerasimenko · Accepted Answer · 2021-10-21 14:23:10Z

-2

arr.sort((a,b) => a - b).reverse()

edited Oct 21, 2021 at 14:23

answered Oct 21, 2021 at 14:18

Maksim Gerasimenko

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1 Comment

76484 Over a year ago

Why not just do b - a and so remove the need for the .reverse()?

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