# Compute the Median

### 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

• Can we output as a fraction over 2 (e.g. 7/2 or 8/2) Commented Jan 8, 2017 at 23:34
• According to this fractions are fine. Commented Jan 8, 2017 at 23:36
• How is this not already a challenge?
– orlp
Commented Jan 8, 2017 at 23:53
• @orlp This is a subset of this challenge. Commented Jan 9, 2017 at 17:44
• It's also makes a nice fastest code challenge as there are some interesting linear time algorithms.
– user9206
Commented Jan 10, 2017 at 10:51

# 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


# Rwithout 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!

• function(x)mean(x,.5)
– ngm
Commented Oct 1, 2018 at 14:30

# GolfScript, 272520 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
"/2"  Push that string
Output all items on stack without separator


The output will be something like 10/2.

# Python 3, 59 bytes

f=lambda l:l.sort()or(len(l)<3)*(l[0]+l[-1])/2or f(l[1:-1])


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This is a recursive version:

• the list is sorted
• if there are 1 or 2 elements left, we output the median since 0 and -1 are both first and last with a single or atwo element list
• if not, we remove first and last elements and call f.

# Vyxal, 6 bytes

s∆ṁwfṁ


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sort the input, get the ∆ṁiddle item(s), wrap in a list, flatten and get the ṁean of that. (wf is needed to handle cases where the list is of odd length)

Or, without a trivial built-in

## Vyxal, 11 bytes

L‹½₍⌈⌊$s$İṁ


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## Explained

L‹½₍⌈⌊$s$İṁ
L‹         # Length of the input - 1 (this accounts for 0 indexing)
½        # halved
₍⌈⌊      # a list of the ceiling and floor of that number
$s # the input list sorted$İ  # indexed at the positions in the ceil,floor list
ṁ # take the average of that


## Racket 113 bytes

(let*((L(sort L >))(n(length L))(r list-ref))(if(odd? n)(r L(floor(/ n 2)))(/(+(r L(-(/ n 2)1))(r L(/ n 2)))2)))


Ungolfed:

(define (median L)
(let* ((L (sort L >))
(n (length L))
(lr list-ref))
(if (odd? n)
(lr L (floor (/ n 2)))
(/(+ (lr L (sub1(/ n 2)))
(lr L (/ n 2)))
2))))


Testing:

(median '(1 2 3))
(median '(1 2 3 4))


Output:

2
2 1/2


## Clojure, 65 bytes

#(/(apply +(take 2(drop(-(count %)1)(sort(for[c % i[0 1]]c)))))2)


An other approach I tried:

#(apply +(map *(for[i(range)](get{-2 0.5 -1 1 0 0.5}(-(* i 2)(count %))0))(sort %)))


# Haxe, 104 bytes

Not amazing, what with the function keywords and a mandatory sorting function …

function f(l,?a)return(l[(a={l[0]+=.0;l.sort(function(x,y)return x>y?1:-1);l.length;})>>1]+l[a-1>>1])/2;


With some whitespace:

function f(l, ?a)
return (
l[(a = {
l[0] += .0;
l.sort(
function(x, y) return x > y ? 1 : -1
);
l.length;
}) >> 1]
+ l[a - 1 >> 1]
) / 2;


I used l[0]+=.0; to let Haxe know the type of l. The alternative would be l:Array<Float> in the arguments. Then l is sorted, its length is stored in a, and then we basically do (l[a / 2] + l[(a - 1) / 2]) / 2.

• So much abuse of Haxe's "everything is an expression" paradigm going on here, I love it. Commented Aug 11, 2017 at 18:55

# Swift, 93 bytes

let m:([Double])->Double={{c,s in c%2==0 ?(s[c/2-1]+s[c/2])/2:s[c/2]}($0.count,$0.sorted())}


This takes about 10 seconds to compile on my machine but it works. It declares the constant m of type [Double] -> Double.

# T-SQL, 101 67

DECLARE @ table(i real)
INSERT @ values(1),(3),(20),(4)

SELECT top 1PERCENTILE_CONT(.5)WITHIN GROUP(ORDER BY i)OVER()FROM @


Try it out

# C#, 75 bytes

a=>{Array.Sort(a);int m=a.Length;return m%2>0?a[m/2]:(a[m/2-1]+a[m/2])/2;};


An anonymous function which computes the median.

Full program with ungolfed method and test cases:

using System;

public class Program
{
public static void Main()
{
Func<double[],double> f =
a =>
{
Array.Sort(a);  // built-in sort function for arrays
int m = a.Length;   // stores the number of elements from the array
return m % 2 > 0 ? a[m/2] : ( a[m/2-1] + a[m/2] ) / 2;
// if the array has an odd number of elements, the central number will be returned
// otherwise, the average of the two central elements
};

// test cases:
Console.WriteLine(f(new double[]{1,2,3,4,5,6,7,8,9}));  // 5
Console.WriteLine(f(new double[]{1,4,3,2}));    // 2.5
Console.WriteLine(f(new double[]{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
}
}

• It should be System.Array not just Array Commented Jan 10, 2017 at 13:13

## CJAM - 21

q~]$__,2/=\_,(2/=+2d/  • q~] reads input to array • $__ sorts it and makes 2 copies
• , gets length of array
• 2/ divides that by 2 rounded down
• = finds the number at that index
• /_ puts original array at top of stack and copies it
• ,( gets length of array - 1
• 2/ divides that by 2 rounded down
• = finds the number at that index
• + adds the two array elements extracted 2d/ divides them by 2 as a double (so no rounding)

If the number of array elements N is odd, floor(N/2) = floor((N-1)/2). If N is even the two center elements are selected and the mean is found.

Longer but working alternative strategies:

q~]$__,2/)<_,@W%<&_:+\,d/ q~]$:A,2/_(A,2%$A=@A=+2d/\; q~]$_Vf*_,2/.5t_W%.+.*:+


Ruby 50 48 Bytes

-2 bytes thanks to @Conor O'Brien

->(l){l.sort!;e=l.length;(l[~-e/2]+l[e/2])/2.0}

• You could save two bytes by removing the parentheses around the first l, and say ~-e instead of (e-1). Commented Jan 10, 2017 at 22:01

# Racket, 95 bytes

Using the trusty old match syntax. The pattern (list _ m ... _) matches the middle of a list (that is, it omits the first and last element).

(λ(l)(let f([l(sort l <)])(match l[(list x)x][(list x y)(/(+ x y)2)][(list _ m ... _)(f m)])))


## Ungolfed

(λ (l)
(let f ([l (sort l <)])
(match l
[(list x) x]
[(list x y) (/ (+ x y) 2)]
[(list _ m ... _) (f m)])))


# 8th, 108105 93 bytes

: m ' n:cmp a:sort a:len 2 n:/mod swap not if n:1- 2 a:slice a:open n:+ 2 n:/ else a:@ then ;


SED (Stack Effect Diagram) is a -- a n

Test

[1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,1.5,-5,100000,1.3,1.4] m .


Output

1.50000


\ Median
: m \ a -- a n
' n:cmp a:sort \ Sort array
a:len          \ Get array length
2 n:/mod       \ Remainder and quotient
swap           \ Remainder on TOS
not if
\ Array contains an even number of items
\ Get arithmetic mean of the two values closest to the center of the sorted list
n:1- 2 a:slice a:open n:+ 2 n:/
else
\ Array contains an odd number of items
\ Get the central value
a:@
then ;


# Perl 5, 58 + 2 (-ap) = 60 bytes

$_=(@F=sort{$a<=>$b}@F)%2?@F[@F/2]:@F[@F/2]/2+@F[@F/2-1]/2  Try it online! Input is split into the @F array by the '-a' flag. @F gets sorted. Then, its length is checked to see if it is odd or even. If odd, result is the middle element. If even, result is half of the element to the left of middle plus half of the element to the right of the middle. ## Julia 0.6.0 (9 bytes) (6 bytes) median(a) median where a is an array. It's not a very exciting answer but it's cool that Julia has a built in function for the median. edit: I didn't know I could just write median! • You can use median and count it as 6 bytes! Commented Aug 11, 2017 at 18:34 • I didn't know thanks a lot! Commented Aug 11, 2017 at 18:42 # APL, 26 bytes {3>≢⍵:(+/÷≢)⍵⋄∇1↓¯1↓⍵[⍋⍵]}  Try it online! ### How? • 3>≢⍵:(+/÷≢)⍵ - if the length of the array is less then 3, return the average • ⋄ - otherwise • ∇1↓¯1↓⍵[⍋⍵] - return the sorted array with the first and last elements removed. • How is this the first APL answer? Commented Aug 11, 2017 at 20:14 # GolfScript, 44 bytes ~$.,:l;~l 2%{l 2/$}{{l 2/$}2*+'/2'+}if{\;}l*


Try it online!

Explanation

~$.,:l;~l 2%{l.2/-}{{l 2/$}2*+'/2'+}if{\;}l*   |
~                                              Create array from input string
$Sort array . Duplicate array , Pop and count the top array :l Assign variable l ; Pop ~ Convert array into individual integers l Push variable l onto stack 2% Push 2 and perform mod {l 2/$}                            If block
{l 2/                              push variable l and divide by 2
$} Copy/push value at index (push(stack[pop()])) {{l 2/$}2*+'/2'+}           Else block
{l 2/                      Push l/2
$} Copy 2* Perform block {} twice + Add top two of stack (result of copies) '/2'+}if Push and add '/2'. End if {\;} New block. Swap top two elements then pop l* Perform previous block l times  # k, 23 bytes Basically a slightly golfed version of q's canonical med in k. {avg x(<x)@_.5*-1 0+#x}  # PHP, 70 77 bytes Not exactly optimal but works. Requires that the values are passed over GET. <?sort($_GET);die(($C=count($G=$_GET))&1?$G[~-$C/2]:($G[$C/2]+$G[$C/2-1])/2);  The result will be displayed in the browser and as the return code. Thanks to Titus for fixing it, at the cost of 7 bytes. • 1) You have a typo: ~~ should be ~-. 2) You compute the arithmetic mean of the whole array, but should only "of the two values closest to the center"; i.e. ?$G[~-$C/2]:($G[$C/2]+$G[$C/2-1])/2. Commented Sep 30, 2018 at 5:20 • Thank you for the fix. It is sad that it got longer :/ Commented Sep 30, 2018 at 12:09 # Haskell 1.2 (Gofer), 44 bytes f[x]=x f[x,y]=(x+y)/2.0 f(x:y)=f.init$sort y


Try it online!

# 05AB1E, 2 bytes

Åm


No need for an explanation, since Åm is a builtin which will:

Median. Sorts the list, then returns either the middle element or the average of the middle elements depending on the parity of the length of the list.

# Lua, 64 63 bytes

function f(t)table.sort(t)h=#t//2return(t[h+1]+t[h+#t%2])/2 end


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Sort table, get the position halfway through the table by integer-dividing table length by two, return average of element at half position plus one and element at half position if table length is even, else at half position plus one.

This solution is only valid in Lua 5.3 and onwards where there is integer division, // (and where integers can be squished right next to the keyword return). In Lua 5.1, the equivalent is math.floor(a/b), which would add several bytes.

• You can remove the space between 2 and return
– Jo King
Commented Dec 22, 2018 at 11:12
• Thanks! I didn't try that, though I noticed before that integer plus keyword (with no intervening space) sometimes works in Lua 5.3; for instance if 0then end is valid. Commented Dec 22, 2018 at 11:17
• I think the rule is that the keyword can't start with a letter that might make the number look like a hexadecimal. For example, you can't remove spaces before end, do, else etc.
– Jo King
Commented Dec 22, 2018 at 11:18
• Oh yeah, the lexer segments 0end as 0e, nd because it starts by simply reading a series of hexadecimal digits, decimal dots, or exponent markers into the buffer. So that part of the process accepts things like 1e+10e-10 or 1.1.1 before trying to get their numerical value and throwing a "malformed number" error. Commented Dec 22, 2018 at 11:30

# Tcl, 100 bytes

proc M L {expr ([lindex [set S [lsort -r $L]] [set h [expr [llength$L]/2]]]+[lindex $S end-$h])/2.}


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# Tcl, 97 bytes

proc M L {expr ([lindex [set S [lsort $L]] [set h [expr [llength$L]/2]]]+[lindex $S end-$h])/2.}


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# Tcl, 123 bytes

proc M L {set I [lindex [set S [lsort $L]] [expr [set n [llength$L]]/2]]
expr {$n%2?$I:($I+[lindex$S [expr $n/2-1]])/2.}}  Try it online! # Tcl, 124 bytes proc M L {set n [llength [set S [lsort$L]]]
set I [lindex $S [expr$n/2]]
expr {$n%2?$I:($I+[lindex$S [expr $n/2-1]])/2.}}  Try it online! # Tcl, 133 bytes proc M L {proc G L\ i {lindex$L [expr $i]} expr {[set n [llength [set S [lsort$L]]]]%2?[G $S$n/2]:([G $S$n/2]+[G $S$n/2-1])/2.}}


Try it online!

proc M L {expr {[set n [llength [set S [lsort $L]]]]%2?[lindex$S [expr $n/2]]:([lindex$S [expr $n/2]]+[lindex$S [expr $n/2-1]])/2.}}  Try it online! Still very ungolfed, my first minimum viable product! • Save some bytes with recursive version (translate from my python3 version) -6 bytes Commented Dec 22, 2018 at 15:34 • @david Meanwhile you posted, I was outgolfing myself. I tried to golf your suggestion a little more as tio.run/##jU/LCoMwELz7FXPw0FJqffYB/… , but I've already shortened my code more than it. Thanks anyway Commented Dec 22, 2018 at 16:05 • OK right! Your code is great, I propose you to save still some : 118 bytes Commented Dec 22, 2018 at 16:18 • and even more (-2) in removing an unuseful pair of final braces! Commented Dec 22, 2018 at 16:20 • Could not grasp how do you distinguish the odd from the even case. Are you doing the average with self in the last line for the odd case? Commented Dec 22, 2018 at 16:47 # Arn, 13 12 bytes Version: 1.1.5 Iõå)n├┼U■¨Mõ  # Explained Unpacked: z&:}+\$.:<:_&:-

z&               Zip array with itself, bind to the following
:_             Flatten
:<           Sort ascending
$. Split in half :}+\ Fold with :}+. &:- Bind the half symbol to the expression  How :}+\ works: Takes the last element of the first array and adds to the second array, which is casted to the first element of that array automatically (finally, that feature actually is useful!). Where a symbol needs to take a value but none is provided, _ is inserted, the implied variable. Ungolfed this looks like ( _ z _ ) & ( :}+\ ($. ( :< ( _ :_ ) ) ) ) & ( :- _ )


# Arn, the boring way, 3 bytes

med


Built in function, can't be compressed.

# Pip, 16 bytes

SN:g(CHg+Rgoi)/2


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### Explanation

SN:g(CHg+Rgoi)/2
g              List of command-line arguments
SN:               Sort it (using numeric comparison) in-place
g          Take that sorted list
Rg       Its reverse
CH           Chop into two halves (if odd length, second half is longer)
(      oi)    Select the item at index 0 of the sublist at index 1 (o=1, i=0)
/2  Divide it by 2


# Thunno 2, 1 byte

ṁ


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Built-in.

## Thunno 2, 11 bytes

l⁻½çṃN$Ṡsim  Attempt This Online! Non built-in. #### Explanation l⁻½çṃN$Ṡsim  # Implicit input
l⁻           # Length of input, decremented
½          # Halve this number
ç         # Parallelly apply:
ṃ        #  Ceiling
N       #  Floor
\$Ṡ     # Input list sorted
si   # Index into this list
m  # Mean of this list
# Implicit output


# Python 3.8 (pre-release), 49 44 bytes

-5 Inspired by Luis Mendos Octave answer.

lambda x:sum(sorted(x+x)[(n:=~len(x)):-n])/2


Try it online!

• You can remove the space after the or and the parentheses around n:=len(x)//2 Commented Jun 17, 2023 at 11:51
• Thanks! Removing the parentheses gives syntax error. Commented Jun 17, 2023 at 14:04
• @Hunaphu it works in newer versions of Python: Attempt This Online (with Python 3.10) Commented Jun 17, 2023 at 14:11

# APL(NARS), 31 chars, 62 bytes

{2÷⍨x[⌊k]+(⌈k←2÷⍨1+≢⍵)⌷x←⍵[⍋⍵]}


test

  t←{2÷⍨x[⌊k]+(⌈k←2÷⍨1+≢⍵)⌷x←⍵[⍋⍵]}
t 5 4 3 2 1
3
t 4 3 2 1
2.5
t 5 40 30 2 1
5
t 5 40 30 2
17.5
35÷2
17.5
t ,80
80
t 9 3 4 8 7 6
6.5