ordering array of numbers returns invalid result

Question

For some reason, sort doesnt sort correctly. Why is that?

console.log(
  [0,0,0,0,1,0,0,0,3,0,0,0,0,0,1,1,0,2,1,1,0,0,0,0,0,0].sort((a,b)=>a<b)
)

But the inverse works....

console.log(
  [0,0,0,0,1,0,0,0,3,0,0,0,0,0,1,1,0,2,1,1,0,0,0,0,0,0].sort((a,b)=>a>b)
)

The comparator function is wrong. [Check the documentation.](https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort) — Pointy, Oct 04 '18 at 12:44
https://stackoverflow.com/questions/1063007/how-to-sort-an-array-of-integers-correctly — Stefan Blamberg, Oct 04 '18 at 12:45
That's not how `sort` works, see the linked question's answers. You don't return a boolean, you return a *number*: negative, 0, or positive. More on MDN: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort So (if you want an ascending sort): `.sort((a,b) => a -b)` — T.J. Crowder, Oct 04 '18 at 12:45

Mamun · Answer 1 · 2018-10-04T12:51:34.063

1

Array.prototype.sort()

To compare numbers instead of strings, the compare function can simply subtract b from a.

The following function will sort the array ascending (if it doesn't contain Infinity and NaN):

console.log(
 [0,0,0,0,1,0,0,0,3,0,0,0,0,0,1,1,0,2,1,1,0,0,0,0,0,0].sort((a,b)=>a - b)
)

edited Oct 04 '18 at 12:51

answered Oct 04 '18 at 12:44

Mamun

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Is that the equivalent in C# of implementing `IComparable`? – Cid Oct 04 '18 at 12:46

score 0 · Answer 2 · answered Oct 04 '18 at 12:53

The function should return either a-b for ascending or b-a for descending

console.log([0,0,0,0,1,0,0,0,3,0,0,0,0,0,1,1,0,2,1,1,0,0,0,0,0,0,5].sort((a,b) => a - b)) 

console.log([0,0,0,0,1,0,0,0,3,0,0,0,0,0,1,1,0,2,1,1,0,0,0,0,0,0,5].sort((a,b) => b - a))

ordering array of numbers returns invalid result

2 Answers2