CODEWITHSUNDEEP
javascriptsortingnumbers
user1432532
user1432532

Reputation: 349

Why does ".sort()" in JavaScript sort incredibly small numbers as incredibly large numbers?

If I were to run "[6,3,8,7,5,2,1,9,4,0].sort()" I get "[0,1,2,3,4,5,6,7,8,9]" as an output like you would expect. It sorts the numbers from smallest to largest. But if I were to run "[2,3,1,4e-20].sort()" I get "[1,2,3,4e-20]" as an output. Why does the ".sort()" function treat "4e-20" as larger than "3"? Even if you run "3>4e-20" you get "true" as an output, showing that JavaScript does realize that "4e-20" is an incredibly tiny number. But the sort function treats it as an incredibly larger number. Why is this? and is there some way I can change it, or do I just have to write my own function?

Upvotes: 1

Views: 1059

Answers (3)

reem
reem

Reputation: 7246

Array.sort() in javascript does not sort by < and >, it instead does a lexical sort, which means it treats the numbers as strings and sorts them that way. This is what causes the longer 4e-20 to be sorted higher than the shorter and (lexically) lower 3.

While a - b works, this is a theoretically faster comparison because it is branchless: (a > b) - (a < b)

Upvotes: 4

Kodlee Yin
Kodlee Yin

Reputation: 1089

From MDN Array.sort:

"The default sort order is according to string Unicode code points."

It then goes on "[to] compare numbers instead of strings, the compare function can simply subtract b from a. The following function will sort the array ascending":

function compareNumbers(a, b) {
  return a - b;
}

Upvotes: 6

Danilo Valente
Danilo Valente

Reputation: 11352

You might want to do this:

[2, 3, 1, 4e-20].sort(function (a, b) {
    return a - b;
});

That happens because (according to MDN) when no sorter function is provided, it uses an unicode string comparator.

Upvotes: 4

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