32
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Challenge

Given a nonempty list of real numbers, compute its median.

Definitions

The median is computed as follows: First sort the list,

  • if the number of entries is odd, the median is the value in the center of the sorted list,
  • otherwise the median is the arithmetic mean of the two values closest to the center of the sorted list.

Examples

[1,2,3,4,5,6,7,8,9] -> 5
[1,4,3,2] -> 2.5
[1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,-5,100000,1.3,1.4] -> 1.5
[1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,-5,100000,1.3,1.4] -> 1.5
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  • \$\begingroup\$ Can we output as a fraction over 2 (e.g. 7/2 or 8/2) \$\endgroup\$ – Sriotchilism O'Zaic Jan 8 '17 at 23:34
  • \$\begingroup\$ According to this fractions are fine. \$\endgroup\$ – flawr Jan 8 '17 at 23:36
  • 15
    \$\begingroup\$ How is this not already a challenge? \$\endgroup\$ – orlp Jan 8 '17 at 23:53
  • 1
    \$\begingroup\$ @orlp This is a subset of this challenge. \$\endgroup\$ – AdmBorkBork Jan 9 '17 at 17:44
  • 3
    \$\begingroup\$ It's also makes a nice fastest code challenge as there are some interesting linear time algorithms. \$\endgroup\$ – user9206 Jan 10 '17 at 10:51

54 Answers 54

18
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Python 2, 48 bytes

An unnamed function which returns the result. -1 byte thanks to xnor.

lambda l:l.sort()or(l[len(l)/2]+l[~len(l)/2])/2.

The first step is obviously to sort the array, using l.sort(). However, we can only have one statement in a lambda, so we utilise the fact that the sort function returns None by adding an or - as None is falsy in Python, this tells it to evaluate and return the next part of the statement.

Now we have the sorted list, we need to find either the middle, or the middle two, values.

Using a conditional to check the length's parity would be too verbose, so instead we get the indexes len(l)/2 and ~len(l)/2:

  • The first is floor(length / 2), which gets the middle element if the length is odd, or the left item in the central pair if the length is even.
  • The second is the binary inversion of the list length's, evaluating to -1 - floor(length / 2). Due Python's negative indexing, this essentially does the same as the first index, but backwards from the end of the array.

If the list is of odd length, these indexes will point to the same value. If it is of even length, then they will point to the central two items.

Now that we have these two indexes, we find these values in the list, sum them, and divide them by 2. The trailing decimal place in /2. makes sure that it is float division rather than integer division.

The result is implicitly returned, as this is a lambda function.

Try it online!

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  • \$\begingroup\$ Looks like a lambda wins out despite the repetition: lambda l:l.sort()or(l[len(l)/2]+l[~len(l)/2])/2. \$\endgroup\$ – xnor Jan 9 '17 at 5:15
  • \$\begingroup\$ @xnor Thanks! When I tried that, I accidentally counted the f=, thinking it was 1 byte longer. \$\endgroup\$ – FlipTack Jan 9 '17 at 7:17
13
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Python3 - 31 30 bytes

Saved a byte thanks to @Dennis!

I wasn't planning on a builtin answer, but I found this module and thought it was really cool cuz I had no idea it existed.

from statistics import*;median

Try it online here.

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  • 6
    \$\begingroup\$ from statistics import*;median saves a byte. \$\endgroup\$ – Dennis Jan 9 '17 at 3:48
  • \$\begingroup\$ @Dennis oh cool. is that always shorter? \$\endgroup\$ – Maltysen Jan 9 '17 at 21:55
  • 2
    \$\begingroup\$ It always beats using __import__, but import math;math.log would beat from math import*;log. \$\endgroup\$ – Dennis Jan 9 '17 at 22:00
9
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Actually, 1 byte

Try it online!

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  • \$\begingroup\$ Right tool for the job. \$\endgroup\$ – Mego Jan 9 '17 at 20:17
9
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Jelly, 9 bytes

L‘HịṢµ÷LS

Try it online!

Explanation

I'm still getting the hang of Jelly... I wasn't able to find built-ins for either the median or the mean of a list, but it's very convenient for this challenge that Jelly allows non-integer indices into lists, in which case it will return a pair of the two closest values. That means we can work with half the input length as an index, and get a pair of values when we need to average it.

L          Get the length of the input.
 ‘         Increment it.
  H        Halve it. This gives us the index of the median for an odd-length list
           (Jelly uses 1-based indexing), and a half-integer between the two indices
           we need to average for even-length lists.
   ịṢ      Use this as an index into the sorted input. As explained above this will
           either give us the median (in case of an odd-length list) or a pair of
           values we'll now need to average.
     µ     Starts a monadic chain which is then applied to this median or pair...
      ÷L     Divide by the length. L treats atomic values like singleton lists.
        S    Sum. This also treats atomic values like singleton lists. Hence this
             monadic chain leaves a single value unchanged but will return the
             mean of a pair.
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  • \$\begingroup\$ Of course, Æṁ will work now \$\endgroup\$ – caird coinheringaahing Oct 15 '17 at 18:37
9
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Brain-Flak, 914 + 1 = 915 bytes

([]){({}[()]<(([])<{({}[()]<([([({}<(({})<>)<>>)<><({}<>)>]{}<(())>)](<>)){({}())<>}{}({}<><{}{}>){{}<>(<({}<({}<>)<>>)<>({}<>)>)}{}({}<>)<>>)}{}<>{}>[()]){({}[()]<({}<>)<>>)}{}<>>)}{}([]<(()())>(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<(())>){((<{}{}([[]]()){({}()()<{}>)}{}(({}){}<([]){{}{}([])}{}>)>))}{}{(<{}([[]]()()){({}()()<{}>)}{}({}{}<([]){{}{}([])}{}>)>)}{}([(({}<((((((()()()){}){}){}()){})[()()()])>)<(())>)](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}(({}<>)<{(<{}([{}])>)}{}{(({})<((()()()()()){})>)({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{})({}<({}<>)<>>((((()()()){}){}){}){})((()()()()()){})<>({}<>)(()()){({}[()]<([([({})](<()>))](<>())){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}(({})<>)<>{(<{}([{}])>)}{}({}<>)<>({}<><({}<>)>)>)}{}({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>)<>(({}{}[(())])){{}{}(((<{}>)))}{}{}{(<{}<>([{}])><>)}{}<>}{}>){(<{}(((((()()()()())){}{})){}{})>)}{}

Requires the -A flag to run.

Try it online!

Explanation

The backbone of this algorithm is a bubble sort I wrote a while ago.

([]){({}[()]<(([])<{({}[()]<([([({}<(({})<>)<>>)<><({}<>)>]{}<(())>)](<>)){({}())<>}{}({}<><{}{}>){{}<>(<({}<({}<>)<>>)<>({}<>)>)}{}({}<>)<>>)}{}<>{}>[()]){({}[()]<({}<>)<>>)}{}<>>)}{}

I don't remember how this works so don't ask me. But I do know it sorts the stack and even works for negatives

After everything has been sorted I find 2 times the median with the following chunk

([]<(()())>(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{}<(())>)  #Stack height modulo 2
{((<{}{}          #If odd
 ([[]]())         #Push negative stack height +1
 {                #Until zero 
  ({}()()<{}>)    #add 2 to the stack height and pop one
 }{}              #Get rid of garbage
 (({}){}<         #Pickup and double the top value
 ([]){{}{}([])}{} #Remove everything on the stack
 >)               #Put it back down
>))}{}            #End if
{(<{}                     #If even
  ([[]]()())              #Push -sh + 2
  {({}()()<{}>)}{}        #Remove one value for every 2 in that value
  ({}{}<([]){{}{}([])}{}>)#Add the top two and remove everything under them
>)}{}                     #End if

Now all that is left is to make convert to ASCII

([(({}<((((((()()()){}){}){}()){})[()()()])>)<(())>)](<>)){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}(({}<>)<
{(<{}([{}])>)}{}  #Absolute value (put "/2" beneath everything)

{                 #Until the residue is zero 
(({})<            #|Convert to base 10
((()()()()()){})  #|
>)                #|...
({}(<>))<>{(({})){({}[()])<>}{}}{}<>([{}()]{})
({}<({}<>)<>>((((()()()){}){}){}){})((()()()()()){})<>({}<>)
                  #|
(()()){({}[()]<([([({})](<()>))](<>())){({}())<>}{}<>{}{}<>(({})){{}{}<>(<(())>)}{}(({})<>)<>{(<{}([{}])>)}{}({}<>)<>({}<><({}<>)>)>)}{}({}(<>))<>([()]{()<(({})){({}[()])<>}{}>}{}<><{}{}>)<>(({}{}[(())])){{}{}(((<{}>)))}{}{}{(<{}<>([{}])><>)}{}<>
}{}               #|
>)
{(<{}(((((()()()()())){}{})){}{})>)}{}  #If it was negative put a minus sign
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7
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R, 6 bytes

median

Not surprising that R, a statistical programming language, has this built-in.

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  • 4
    \$\begingroup\$ R beating Jelly :D:D:D \$\endgroup\$ – JAD Jan 9 '17 at 13:24
5
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MATL, 4 bytes

.5Xq

This finds the 0.5-quantile, which is the median.

Try it online!

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  • \$\begingroup\$ I was just about to figure it out! \$\endgroup\$ – flawr Jan 8 '17 at 23:31
  • \$\begingroup\$ Ah no, I mean I was figuring out how to do it in MATL=) (But I had a 5 byte solution, so yeah...) \$\endgroup\$ – flawr Jan 8 '17 at 23:32
  • \$\begingroup\$ @flawr Post it then! It will surely be more interesting than mine \$\endgroup\$ – Luis Mendo Jan 9 '17 at 0:18
  • \$\begingroup\$ Nope, it was the same as yours just with an i in front :) \$\endgroup\$ – flawr Jan 9 '17 at 10:39
  • \$\begingroup\$ @flawr The same i that you suggested to make implicit? :-P \$\endgroup\$ – Luis Mendo Jan 9 '17 at 11:03
5
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Pyth - 11 bytes

Finds the average of the middle item taken both backwards and forwards.

.O@R/lQ2_BS

Test Suite.

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5
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Octave, 38 bytes

@(x)mean(([1;1]*sort(x))(end/2+[0 1]))

This defines an anonymous function. Input is a row vector.

Try it online!

Explanation

            sort(x)                 % Sort input x, of length k
      [1;1]*                        % Matrix-multiply by column vector of two ones
                                    % This vertically concatenates the sort(x) with 
                                    % itself. In column-major order, this effectively 
                                    % repeats each entry of sort(x)
     (             )(end/2+[0 1])   % Select the entry at position end/2 and the next.
                                    % Entries are indexed in column-major order. Since
                                    % the array has 2*k elements, this picks the k-th 
                                    % and (k+1)-th. Because entries were repeated, for
                                    % odd k this takes the original (k+1)/2-th entry
                                    % (1-based indexing) twice. For even k this takes
                                    % the original (k/2)-th and (k/2+1)-th entries
mean(                            )  % Mean of the two selected entries
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  • 1
    \$\begingroup\$ Ugh... clever use of "bsxfun" and mean :-) \$\endgroup\$ – Stewie Griffin Jan 9 '17 at 11:51
5
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JavaScript, 57 52 bytes

v=>(v.sort((a,b)=>a-b)[(x=v.length)>>1]+v[--x>>1])/2

Sort the array numerically. If the array is an even length, find the 2 middle numbers and average them. If the array is odd, find the middle number twice and divide by 2.

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  • 1
    \$\begingroup\$ I've found that Array.sort() doesn't work properly with decimals \$\endgroup\$ – TrojanByAccident Jan 9 '17 at 0:49
  • 3
    \$\begingroup\$ It does if you pass in a sorting function as I did. If you call Array.sort() with no parameters, it uses an alphabetic sort. \$\endgroup\$ – Grax32 Jan 9 '17 at 0:51
  • \$\begingroup\$ Interesting. Didn't know that \$\endgroup\$ – TrojanByAccident Jan 9 '17 at 0:52
  • \$\begingroup\$ You can save a few bytes by using the return value of sort() directly and getting rid of the t variable: v=>(v.sort((a,b)=>a-b)[(x=v.length)>>1]+v[--x>>1])/2 \$\endgroup\$ – Arnauld Jan 9 '17 at 7:29
  • 1
    \$\begingroup\$ Not that you should necessarily correct for this, but if x>=2**31, this would fail. >> is a sign-propagating right shift, meaning that when the number is interpreted as a 32 bit integer, if the msb is set, then it stays set, making the result negative for 2**32>x>=2**31. For x>=2**32, it just yields 0. \$\endgroup\$ – Patrick Roberts Jan 10 '17 at 12:40
5
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Matlab/Octave, 6 bytes

A boring built-in:

median

Try it online!

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  • \$\begingroup\$ I forget the rules for anonymous functions in MATLAB/Octave, should this be @median? \$\endgroup\$ – Giuseppe Oct 1 '18 at 16:18
  • \$\begingroup\$ @Giuseppe I don't know what the currently accepted way to score built-in functions is. \$\endgroup\$ – flawr Oct 1 '18 at 16:48
4
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Mathematica, 6 bytes

Median

As soon as I figure out Mthmtca, I'm posting a solution in it.

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  • \$\begingroup\$ In Mthmtca 0.1/10.1.0.0, the code would have the bytes CBC8 (ËÈ). However, until I apply another patch, the notion of function-calling might not meet PPCG's standards. \$\endgroup\$ – LegionMammal978 Jan 9 '17 at 13:32
4
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Perl 6, 31 bytes

*.sort[{($/=$_/2),$/-.5}].sum/2

Try it

Expanded:

*\     # WhateverCode lambda ( this is the parameter )

.sort\ # sort it

[{     # index into the sorted list using a code ref to calculate the positions

  (
    $/ = $_ / 2 # the count of elements divided by 2 stored in 「$/」
  ),            # that was the first index

  $/ - .5       # subtract 1/2 to get the second index

                # indexing operations round down to nearest Int
                # so both are effectively the same index if given
                # an odd length array

}]\

.sum / 2        # get the average of the two values
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  • 1
    \$\begingroup\$ 24 bytes \$\endgroup\$ – Jo King Sep 30 '18 at 13:19
4
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APL (Dyalog Unicode), 14 bytes

≢⊃2+/2/⊂∘⍋⌷÷∘2

Try it online!

This is a train. The original dfn was {(2+/2/⍵[⍋⍵])[≢⍵]÷2}.

The train is structured as follows

┌─┼───┐
≢ ⊃ ┌─┼───┐
    2 / ┌─┼───┐
    ┌─┘ 2 / ┌─┼─┐
    +       ∘ ⌷ ∘
           ┌┴┐ ┌┴┐
           ⊂ ⍋ ÷ 2

denotes the right argument.

index

  • ⊂∘⍋ the indices which indexed into results in being sorted

  • ÷∘2 into divided by 2

2/ replicate this twice, so 1 5 7 8 becomes 1 1 5 5 7 7 8 8

2+/ take the pairwise sum, this becomes (1+1)(1+5)(5+5)(5+7)(7+7)(7+8)(8+8)

from this pick

  • element with index equal to the length of

Previous solutions

{.5×+/(⍵[⍋⍵])[(⌈,⌊).5×1+≢⍵]}
{+/(2/⍵[⍋⍵]÷2)[0 1+≢⍵]}
{+/¯2↑(1-≢⍵)↓2/⍵[⍋⍵]÷2}
{(2+/2/⍵[⍋⍵])[≢⍵]÷2}
{(≢⍵)⊃2+/2/⍵[⍋⍵]÷2}
≢⊃2+/2/2÷⍨⊂∘⍋⌷⊢
≢⊃2+/2/⊂∘⍋⌷÷∘2
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3
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Common Lisp, 89

(lambda(s &aux(m(1-(length s)))(s(sort s'<)))(/(+(nth(floor m 2)s)(nth(ceiling m 2)s))2))

I compute the mean of elements at position (floor middle) and (ceiling middle), where middle is the zero-based index for the middle element of the sorted list. It is possible for middle to be a whole number, like 1 for an input list of size 3 such as (10 20 30), or a fraction for lists with an even numbers of elements, like 3/2 for (10 20 30 40). In both cases, we compute the expected median value.

(lambda (list &aux
             (m (1-(length list)))
             (list (sort list #'<)))
  (/ (+ (nth (floor m 2) list)
        (nth (ceiling m 2) list))
     2))
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3
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Vim, 62 bytes

I originally did this in V using only text manipulation until the end, but got frustrated with handling [X] and [X,Y], so here's the easy version. They're about the same length.

c$:let m=sort(")[(len(")-1)/2:len(")/2]
=(m[0]+m[-1])/2.0

Try it online!

Unprintables:

c$^O:let m=sort(^R")[(len(^R")-1)/2:len(^R")/2]
^R=(m[0]+m[-1])/2.0

Honorable mention:

  • ^O takes you out of insert mode for one command (the let command).
  • ^R" inserts the text that was yanked (in this case the list)
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3
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TI-Basic, 2 bytes

median(Ans

Very straightforward.

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  • 2
    \$\begingroup\$ Ans is not an allowed I/O method. \$\endgroup\$ – Mego Jan 9 '17 at 18:46
  • 1
    \$\begingroup\$ @Mego your link and comment confuses me... according to the vote, it is allowed. Am I missing something? \$\endgroup\$ – Patrick Roberts Jan 10 '17 at 12:55
  • \$\begingroup\$ @PatrickRoberts There's actually some debate currently about the threshold for acceptability. Several users (myself included) have been following the rule that a method needs at least +5 and at least twice as many upvotes as downvotes, which was the rule originally stated in that post (it's been removed since), and is the rule followed for standard loopholes. \$\endgroup\$ – Mego Jan 10 '17 at 12:56
  • \$\begingroup\$ Whoever removed my comment twice from my own post is annoying. Since there's no clearly accepted rule on acceptability, I don't see the problem here. You can see my answers on SO for how this is used as arguments to a program. \$\endgroup\$ – Timtech Jan 10 '17 at 22:38
  • \$\begingroup\$ @Mego +38 is more than twice -18 \$\endgroup\$ – Timtech Aug 12 '17 at 2:53
3
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C#, 126 bytes

using System.Linq;float m(float[] a){var x=a.Length;return a.OrderBy(g=>g).Skip(x/2-(x%2==0?1:0)).Take(x%2==0?2:1).Average();}

Pretty straightforward, here with LINQ to order the values, skip half the list, take one or two values depending on even/odd and average them.

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  • \$\begingroup\$ You need to include using System.Linq; into your byte count, however you can cancel this out by making some changes. Compile to a Func<float[], float> and assign the value of the modulo to a variable for 106 bytes: using System.Linq;a=>{int x=a.Length,m=x%2<1?1:0;return a.OrderBy(g=>g).Skip(x/2-m).Take(++m).Average();}; \$\endgroup\$ – TheLethalCoder Jan 10 '17 at 13:23
  • \$\begingroup\$ @TheLethalCoder I'm never quite sure what constitutes a complete program. You are right about the using. Concatenating the declarations of the modulus with the length is also a good idea. I experimented around a bit with that but couldn't get it to be shorter than putting it twice in there. I would venture to say that your optimizations are worth an answer by itself, as they are quite substantial and I would not have come up with them. \$\endgroup\$ – Jens Jan 10 '17 at 22:33
  • \$\begingroup\$ The challenge doesn't state that you need a full program so an anonymous method is fine. Beyond that I only stated some common golfing tips so no need for me to add an answer just golf your own! \$\endgroup\$ – TheLethalCoder Jan 10 '17 at 22:47
3
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C++ 112 Bytes

Thanks to @original.legin for helping me save bytes.

#include<vector>
#include<algorithm>
float a(float*b,int s){std::sort(b,b+s);return(b[s/2-(s&1^1)]+b[s/2])/2;}

Usage:

    int main()
    {
        int n = 4;
        float e[4] = {1,4,3,2};
        std::cout<<a(e,n); /// Prints 2.5

        n = 9;
        float e1[9] = {1,2,3,4,5,6,7,8,9};
        std::cout<<a(e1,n); /// Prints 5

        n = 13;
        float e2[13] = {1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,-5,100000,1.3,1.4};
        std::cout<<a(e2,n); /// Prints 1.5

        return 0;
    }
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  • 1
    \$\begingroup\$ You could use float instead of double to save two bytes. Also, on GCC, you can use #import<vector> and #import<algorithm> instead of #include. (Note that you don't need the space after either the #include or #import) \$\endgroup\$ – Steadybox Jan 10 '17 at 2:27
  • \$\begingroup\$ @Steadybox I didn't count the two includes in the score. Should I? Also, I mainly use Clang so I don't know much about GCC but thanks. \$\endgroup\$ – Wade Tyler Jan 10 '17 at 3:56
  • \$\begingroup\$ Yes, the includes should be included in the byte count if the code doesn't compile without them. \$\endgroup\$ – Steadybox Jan 11 '17 at 12:39
3
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J, 16 14 bytes

2%~#{2#/:~+\:~

Try it online!

In addition to BMO's array duplication trick, I found that we can add the whole array sorted in two directions. Then I realized that the two steps can be reversed, i.e. add the two arrays, then duplicate them and take the nth element.

How it works

2%~#{2#/:~+\:~
                Input: array of length n
       /:~      Sort ascending
           \:~  Sort descending
          +     Add the two element-wise
     2#         Duplicate each element
   #{           Take n-th element
2%~             Halve

Previous answers

J with stats addon, 18 bytes

load'stats'
median

Try it online!

Library function FTW.

median's implementation looks like this:

J, 31 bytes

-:@(+/)@((<.,>.)@(-:@<:@#){/:~)

Try it online!

How it works

-:@(+/)@((<.,>.)@(-:@<:@#){/:~)
         (<.,>.)@(-:@<:@#)       Find center indices:
                  -:@<:@#          Compute half of given array's length - 1
          <.,>.                    Form 2-element array of its floor and ceiling
                          {/:~   Extract elements at those indices from sorted array
-:@(+/)                          Sum and half

A bit of golfing gives this:

J, 28 bytes

2%~[:+/(<.,>.)@(-:@<:@#){/:~

Try it online!

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  • 1
    \$\begingroup\$ Nicely done, the J port of my APL answer would be #{0,2+/\2#-:/:] at a close 15 bytes (man I miss ⎕io). \$\endgroup\$ – Cows quack Oct 2 '18 at 17:52
2
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J, 19 bytes

<.@-:@#{(/:-:@+\:)~

Explanation:

        (        )~   apply monadic argument twice to dyadic function 
         /:           /:~ = sort the list upwards
               \:     \:~ = sort the list downwards
           -:@+       half of sum of both lists, element-wise
<.@-:@#               floor of half of length of list
       {              get that element from the list of sums
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  • \$\begingroup\$ You can save a byte by removing the parentheses and applying ~ directly to each <.@-:@#{/:~-:@+\:~ \$\endgroup\$ – miles Jan 9 '17 at 1:04
2
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JavaScript, 273 Bytes

function m(l){a=(function(){i=l;o=[];while(i.length){p1=i[0];p2=0;for(a=0;a<i.length;a++)if(i[a]<p1){p1=i[a];p2=a}o.push(p1);i[p2]=i[i.length-1];i.pop()}return o})();return a.length%2==1?l[Math.round(l.length/2)-1]:(l[Math.round(l.length/2)-1]+l[Math.round(l.length/2)])/2}
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2
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Java 7, 99 bytes

Golfed:

float m(Float[]a){java.util.Arrays.sort(a);int l=a.length;return l%2>0?a[l/2]:(a[l/2-1]+a[l/2])/2;}

Ungolfed:

float m(Float[] a)
{
    java.util.Arrays.sort(a);
    int l = a.length;
    return l % 2 > 0 ? a[l / 2] : (a[l / 2 - 1] + a[l / 2]) / 2;
}

Try it online

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  • \$\begingroup\$ I'm a bit disappointed even Java 7 has a short enough sorting syntax that en.wikipedia.org/wiki/… is suboptimal \$\endgroup\$ – JollyJoker Jan 9 '17 at 13:24
  • \$\begingroup\$ Dont you need to count the import for java.util.Arrays? \$\endgroup\$ – FlipTack Jan 9 '17 at 15:23
  • \$\begingroup\$ Whoops, thank you for noticiting. :) \$\endgroup\$ – peech Jan 9 '17 at 15:53
  • \$\begingroup\$ Hello from the future! You can save 14 bytes by using integer division truncation to handle length parity. See my Java 8 answer. \$\endgroup\$ – Jakob Aug 12 '17 at 2:40
2
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Pari/GP - 37 39 Bytes

Let a be a rowvector containing the values.

b=vecsort(a);n=#b+1;(b[n\2]+b[n-n\2])/2  \\ 39 byte              

n=1+#b=vecsort(a);(b[n\2]+b[n-n\2])/2    \\ obfuscated but only 37 byte

Since Pari/GP is interactive, no additional command is needed to display the result.


For the "try-it-online" link a line before and after is added. To get printed, the median-result in stored in variable w

a=vector(8,r,random(999))           
n=1+#b=vecsort(a);w=(b[n\2]+b[n-n\2])/2      
print(a);print(b);print(w)       

Try it online!

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2
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Japt, 20 bytes

n gV=0|½*Ul)+Ug~V)/2

Test it online! Japt really lacks any built-ins necessary to create a really short answer for this challenge...

Explanation

n gV=0|½*Ul)+Ug~V)/2  // Implicit: U = input list
n                     // Sort U.
   V=0|½*Ul)          // Set variable V to floor(U.length / 2).
  g                   // Get the item at index V in U.
            +Ug~V     // Add to that the item at index -V - 1 in U.
                 )/2  // Divide by 2 to give the median.
                      // Implicit: output result of last expression
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2
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Java 8, 71 bytes

Parity is fun! Here's a lambda from double[] to Double.

l->{java.util.Arrays.sort(l);int s=l.length;return(l[s/2]+l[--s/2])/2;}

Nothing too complex going on here. The array gets sorted, and then I take the mean of two numbers from the array. There are two cases:

  • If the length is even, then the first number is taken from just ahead of the middle of the array, and the second number is taken from the position before that by integer division. The mean of these numbers is the median of the input.
  • If the length is odd, s and s-1 both divide to the index of the middle element. The number is added to itself and the result divided by two, yielding the original value.

Try It Online

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2
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SmileBASIC, 45 bytes

DEF M A
L=LEN(A)/2SORT A?(A[L-.5]+A[L])/2
END

Gets the average of the elements at floor(length/2) and floor(length/2-0.5) Very simple, but I was able to save 1 byte by moving things around:

DEF M A
SORT A    <- extra line break
L=LEN(A)/2?(A[L-.5]+A[L])/2
END
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2
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Husk, 10 bytes

½ΣF~e→←½OD

Try it online!

Explanation

This function uses that the median of \$[a_1 \dots a_N]\$ is the same as the median of \$[a_1 \; a_1 \dots a_N \; a_N]\$ which avoids the ugly distinction of odd-/even-length lists.

½ΣF~e→←½OD  -- example input: [2,3,4,1]
         D  -- duplicate: [2,3,4,1,2,3,4,1]
        O   -- sort: [1,1,2,2,3,3,4,4]
       ½    -- halve: [[1,1,2,2],[3,3,4,4]]
  F         -- fold the following
   ~        -- | compose the arguments ..
     →      -- | | last element: 2
      ←     -- | | first element: 3
    e       -- | .. and create list: [2,3]
            -- : [2,3]
 Σ          -- sum: 5
½           -- halve: 5/2

Unfortunately ½ for lists has the type [a] -> [[a]] and not [a] -> ([a],[a]) which doesn't allow F~+→← since foldl1 needs a function of type a -> a -> a as first argument, forcing me to use e.

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2
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R without using the median builtin, 51 bytes

function(x,n=sum(x|1)+1)mean(sort(x)[n/2+0:1*n%%2])

Try it online!

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  • 2
    \$\begingroup\$ function(x)mean(x,.5) \$\endgroup\$ – ngm Oct 1 '18 at 14:30
2
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GolfScript, 27 25 20 17 bytes

~..+$\,(>2<~+"/2"

Takes input as an array of integers on stdin. Outputs as an unreduced fraction. Try it online!

Explanation

The median of the array, as BMO's Husk answer explains, is equal to the median of an array twice as long where each element is repeated twice. So we concatenate the array to itself, sort, and take the mean of the middle two elements. If the length of the original array is \$l\$, the middle two elements of the doubled array are at indices \$l-1\$ and \$l\$.

~                  Evaluate input (converting string -> array)
 ..                Duplicate twice
   +               Concatenate two of the copies
    $              Sort the doubled array
     \,            Swap with the non-doubled array and get its length: l
       (           Decrement: l-1
        >          Array slice: all elements at index (l-1) and greater
         2<        Array slice: first two elements (originally at indices l-1 and l)
           ~       Dump array elements to stack
            +      Add
             "/2"  Push that string
                   Output all items on stack without separator

The output will be something like 10/2.

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