calculate the median of an array with javascript

median.js

function median(values) {

    values.sort( function(a,b) {return a - b;} );

    var half = Math.floor(values.length/2);

    if(values.length % 2)
        return values[half];
    else
        return (values[half-1] + values[half]) / 2.0;
}

var list1 = [3, 8, 9, 1, 5, 7, 9, 21];
median(list1);

@ MichalPaszkiewicz Hello what is the use of the function(a,b) {return a - b;} it will sort it anyhow even if list1.sort().

@impari Unfortunately in javascript this isn't the case. You can test this using the following:

[10, 5].sort()

@alireza-saberi's fix along with a second function for already sorted arrays. The median should be just as fast to calculate for massive arrays as little ones, provided the input is already sorted.

function median(values) {
    values = values.slice(0).sort( function(a, b) {return a - b; } );

    return middle(values);
}

function middle(values) {
    var len = values.length;
    var half = Math.floor(len / 2);

    if(len % 2)
        return (values[half - 1] + values[half]) / 2.0;
    else
        return values[half];
}

var list1 = [3, 8, 9, 1, 5, 7, 9, 21];
median(list1);
list1.sort(function(a, b) {return a - b; });
middle(list1);

@willstott101 I believe your if-statement is wrong, when len is even then len % 2 will return 0, and the code will jump to the else. OP has it the other way around and it is correct. Better yet have len % 2 === 0 so that it is easier to understand at a glance, and not have to think about len % 2 returning 1 or 0 which then evaluates to true or false, etc..

For reference, here's the median() function of lodash.math:

math.median = function(arr) {
  arr = arr.slice(0); // create copy
  var middle = (arr.length + 1) / 2,
    sorted = math.sort(arr);
  return (sorted.length % 2) ? sorted[middle - 1] : (sorted[middle - 1.5] + sorted[middle - 0.5]) / 2;
};

nice !

Similar to lodash.math but without the custom sort method and a little more readable (in my opinion):

function median(numbers) {
  const middle = (numbers.length + 1) / 2;
  const sorted = [...numbers].sort((a, b) => a - b); // avoid mutating when sorting
  const isEven = sorted.length % 2 === 0;
  return isEven ? (sorted[middle - 1.5] + sorted[middle - 0.5]) / 2 : sorted[middle - 1];
}

EDIT: fixed as per @henrikra's comment.

@sqren, Thanks, worked like a charm.

@sqren Otherwise you answer was good but the number sorting didn't work

This version will work!

function median(numbers: number[]) {
  const middle = (numbers.length + 1) / 2;
  const sorted = [...numbers].sort((a, b) => a - b); // you have to add sorting function for numbers
  const isEven = sorted.length % 2 === 0;
  return isEven ? (sorted[middle - 1.5] + sorted[middle - 0.5]) / 2 : sorted[middle - 1];
}

@henrikra + @sqren

I think [...numbers] is pointless, because numbers is already an array. Please correct me if I am wrong.

@danielbayerlein sort will mutate the array. I used the spread operator to instead return a new shallow copy. Before the spread operator concat and slice did the trick:

arr.concat().sort()
# or
arr.slice(0).sort()

Some more discussion on this topic https://stackoverflow.com/questions/9592740/how-can-you-sort-an-array-without-mutating-the-original-array

@sqren Thank you for the explanation.

@danielbayerlein You are welcome.
Btw. Noticed your project https://github.com/danielbayerlein/git-pick. I've create a tool called backport: https://github.com/sqren/backport

Looks like we are doing something similar :D

I would also check the length first. if it is zero return. no need to proceed further.