# Averages/Median

Averages/Median
You are encouraged to solve this task according to the task description, using any language you may know.

Write a program to find the   median   value of a vector of floating-point numbers.

The program need not handle the case where the vector is empty, but must handle the case where there are an even number of elements.   In that case, return the average of the two middle values.

There are several approaches to this.   One is to sort the elements, and then pick the element(s) in the middle.

Sorting would take at least   O(n logn).   Another approach would be to build a priority queue from the elements, and then extract half of the elements to get to the middle element(s).   This would also take   O(n logn).   The best solution is to use the   selection algorithm   to find the median in   O(n)   time.

procedure FindMedian is

f: array(1..10) of float := ( 4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5 );
min_idx: integer;
min_val, median_val, swap: float;

begin
for i in f'range loop
min_idx := i;
min_val := f(i);
for j in i+1 .. f'last loop
if f(j) < min_val then
min_idx := j;
min_val := f(j);
end if;
end loop;
swap := f(i); f(i) := f(min_idx); f(min_idx) := swap;
end loop;

if f'length mod 2 /= 0 then
median_val := f( f'length/2+1 );
else
median_val := ( f(f'length/2) + f(f'length/2+1) ) / 2.0;
end if;

end FindMedian;

## ALGOL 68

Translation of: C
INT max_elements = 1000000;

# Return the k-th smallest item in array x of length len #
PROC quick_select = (INT k, REF[]REAL x) REAL:
BEGIN

PROC swap = (INT a, b) VOID:
BEGIN
REAL t = x[a];
x[a] := x[b]; x[b] := t
END;

INT left := 1, right := UPB x;
INT pos, i;
REAL pivot;

WHILE left < right DO
pivot := x[k];
swap (k, right);
pos := left;
FOR i FROM left TO right DO
IF x[i] < pivot THEN
swap (i, pos);
pos +:= 1
FI
OD;
swap (right, pos);
IF pos = k THEN break FI;
IF pos < k THEN left := pos + 1
ELSE right := pos - 1
FI
OD;
break:
SKIP;
x[k]
END;

# Initialize random length REAL array with random doubles #
INT length = ENTIER (next random * max_elements);
[length]REAL x;
FOR i TO length DO
x[i] := (next random * 1e6 - 0.5e6)
OD;

REAL median :=
IF NOT ODD length THEN
# Even number of elements, median is average of middle two #
(quick_select (length % 2, x) + quick_select(length % 2 - 1, x)) / 2
ELSE
# select middle element #
quick_select(length % 2, x)
FI;

# Sanity testing of median #
INT less := 0, more := 0, eq := 0;
FOR i TO length DO
IF x[i] < median THEN less +:= 1
ELIF x[i] > median THEN more +:= 1
ELSE eq +:= 1
FI
OD;
print (("length: ", whole (length,0), new line, "median: ", median, new line,
"<: ", whole (less,0), new line,
">: ", whole (more, 0), new line,
"=: ", whole (eq, 0), new line))

Sample output:

length: 97738
median: -2.52550126608709e  +3
<: 48868
>: 48870
=: 0

## AntLang

AntLang has a built-in median function.

median[list]

## APL

median←{v←⍵[⍋⍵]⋄.5×v[⌈¯1+.5×⍴v]+v[⌊.5×⍴v]} ⍝ Assumes ⎕IO←0

First, the input vector ⍵ is sorted with ⍵[⍋⍵] and the result placed in v. If the dimension ⍴v of v is odd, then both ⌈¯1+.5×⍴v and ⌊.5×⍴v give the index of the middle element. If ⍴v is even, ⌈¯1+.5×⍴v and ⌊.5×⍴v give the indices of the two middle-most elements. In either case, the average of the elements at these indices gives the median.

Note that the index origin ⎕IO is assumed zero. To set it to zero use:
⎕IO←0

If you prefer an index origin of 1, use this code instead:

⎕IO←1
median←{v←⍵[⍋⍵] ⋄ 0.5×v[⌈0.5×⍴v]+v[⌊1+0.5×⍴v]}

This code was tested with ngn/apl and Dyalog 12.1. You can try this function online with ngn/apl. Note that ngn/apl currently only supports index origin 0. Examples:

median 1 5 3 6 4 2
3.5

median 1 5 3 2 4
3

median 4.4 2.3 ¯1.7 7.5 6.6 0.0 1.9 8.2 9.3 4.5
4.45

median 4.1 4 1.2 6.235 7868.33
4.1

median 4.1 5.6 7.2 1.7 9.3 4.4 3.2
4.4

median 4.1 7.2 1.7 9.3 4.4 3.2
4.25

Caveats: To keep it simple, no input validation is done. If you input a vector with zero elements (e.g., ⍳0), you get an INDEX ERROR. If you input a vector with 1 element, you get a RANK ERROR. Only (rank 1) numeric vectors of dimension 2 or more are supported. If you input a (rank 2 or more) matrix, you get a RANK ERROR. If you input a string (vector of chars), you get a DOMAIN ERROR:

median ⍳0
INDEX ERROR

median 66.6
RANK ERROR

median (2 2)⍴⍳4 ⍝ 2x2 matrix
RANK ERROR

median 'HELLO'
DOMAIN ERROR

## AppleScript

### By iteration

set alist to {1, 2, 3, 4, 5, 6, 7, 8}
set med to medi(alist)

on medi(alist)

set temp to {}
set lcount to count every item of alist
if lcount is equal to 2 then
return (item (random number from 1 to 2) of alist)
else if lcount is less than 2 then
return item 1 of alist
else --if lcount is greater than 2
set min to findmin(alist)
set max to findmax(alist)
repeat with x from 1 to lcount
if x is not equal to min and x is not equal to max then set end of temp to item x of alist
end repeat
set med to medi(temp)
end if
return med

end medi

on findmin(alist)

set min to 1
set alength to count every item of alist
repeat with x from 1 to alength
if item x of alist is less than item min of alist then set min to x
end repeat
return min

end findmin

on findmax(alist)

set max to 1
set alength to count every item of alist
repeat with x from 1 to alength
if item x of alist is greater than item max of alist then set max to x
end repeat
return max

end findmax

### Composing functionally

Using a quick select algorithm:

Translation of: JavaScript
-- MEDIAN ---------------------------------------------------------------------

-- median :: [Num] -> Num
on median(xs)
-- nth :: [Num] -> Int -> Maybe Num
script nth
on |λ|(xxs, n)
if length of xxs > 0 then
set {x, xs} to uncons(xxs)

script belowX
on |λ|(y)
y < x
end |λ|
end script

set {ys, zs} to partition(belowX, xs)
set k to length of ys
if k = n then
x
else
if k > n then
|λ|(ys, n)
else
|λ|(zs, n - k - 1)
end if
end if
else
missing value
end if
end |λ|
end script

set n to length of xs
if n > 0 then
tell nth
if n mod 2 = 0 then
(|λ|(xs, n div 2) + |λ|(xs, (n div 2) - 1)) / 2
else
|λ|(xs, n div 2)
end if
end tell
else
missing value
end if
end median

-- TEST -----------------------------------------------------------------------
on run

map(median, [¬
[], ¬
[5, 3, 4], ¬
[5, 4, 2, 3], ¬
[3, 4, 1, -8.4, 7.2, 4, 1, 1.2]])

--> {missing value, 4, 3.5, 2.1}
end run

-- GENERIC FUNCTIONS ----------------------------------------------------------

-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
tell mReturn(f)
set lng to length of xs
set lst to {}
repeat with i from 1 to lng
set end of lst to |λ|(item i of xs, i, xs)
end repeat
return lst
end tell
end map

-- Lift 2nd class handler function into 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
if class of f is script then
f
else
script
property |λ| : f
end script
end if
end mReturn

-- partition :: predicate -> List -> (Matches, nonMatches)
-- partition :: (a -> Bool) -> [a] -> ([a], [a])
on partition(f, xs)
tell mReturn(f)
set lst to {{}, {}}
repeat with x in xs
set v to contents of x
set end of item ((|λ|(v) as integer) + 1) of lst to v
end repeat
end tell
{item 2 of lst, item 1 of lst}
end partition

-- uncons :: [a] -> Maybe (a, [a])
on uncons(xs)
if length of xs > 0 then
{item 1 of xs, rest of xs}
else
missing value
end if
end uncons
Output:
{missing value, 4, 3.5, 2.1}

## Applesoft BASIC

100 REMMEDIAN
110 K = INT(L/2) : GOSUB 150
120 R = X(K)
130 IF L - 2 * INT (L / 2) THEN R = (R + X(K + 1)) / 2
140 RETURN

150 REMQUICK SELECT
160 LT = 0:RT = L - 1
170 FOR J = LT TO RT STEP 0
180 PT = X(K)
190 P1 = K:P2 = RT: GOSUB 300
200 P = LT
210 FOR I = P TO RT - 1
220 IF X(I) < PT THEN P1 = I:P2 = P: GOSUB 300:P = P + 1
230 NEXT I
240 P1 = RT:P2 = P: GOSUB 300
250 IF P = K THEN RETURN
260 IF P < K THEN LT = P + 1
270 IF P > = K THEN RT = P - 1
280 NEXT J
290 RETURN

300 REMSWAP
310 H = X(P1):X(P1) = X(P2)
320 X(P2) = H: RETURN
Example:
X(0)=4.4 : X(1)=2.3 : X(2)=-1.7 : X(3)=7.5 : X(4)=6.6 : X(5)=0.0 : X(6)=1.9 : X(7)=8.2 : X(8)=9.3 : X(9)=4.5 : X(10)=-11.7
L = 11 : GOSUB 100MEDIAN
? R
Output:
5.95

## AutoHotkey

Takes the lower of the middle two if length is even

seq = 4.1, 7.2, 1.7, 9.3, 4.4, 3.2, 5
MsgBox % median(seq, "`,") ; 4.1

median(seq, delimiter)
{
Sort, seq, ND%delimiter%
StringSplit, seq, seq, % delimiter
median := Floor(seq0 / 2)
Return seq%median%
}

## AWK

AWK arrays can be passed as parameters, but not returned, so they are usually global.

#!/usr/bin/awk -f

BEGIN {
d[1] = 3.0
d[2] = 4.0
d[3] = 1.0
d[4] = -8.4
d[5] = 7.2
d[6] = 4.0
d[7] = 1.0
d[8] = 1.2
showD("Before: ")
gnomeSortD()
showD("Sorted: ")
printf "Median: %f\n", medianD()
exit
}

function medianD( len, mid) {
len = length(d)
mid = int(len/2) + 1
if (len % 2) return d[mid]
else return (d[mid] + d[mid-1]) / 2.0
}

function gnomeSortD( i) {
for (i = 2; i <= length(d); i++) {
if (d[i] < d[i-1]) gnomeSortBackD(i)
}
}

function gnomeSortBackD(i, t) {
for (; i > 1 && d[i] < d[i-1]; i--) {
t = d[i]
d[i] = d[i-1]
d[i-1] = t
}
}

function showD(p, i) {
printf p
for (i = 1; i <= length(d); i++) {
printf d[i] " "
}
print ""
}

Example output:

Before: 3 4 1 -8.4 7.2 4 1 1.2
Sorted: -8.4 1 1 1.2 3 4 4 7.2
Median: 2.100000

## BaCon

DECLARE a[] = { 4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2 } TYPE FLOATING
DECLARE b[] = { 4.1, 7.2, 1.7, 9.3, 4.4, 3.2 } TYPE FLOATING

DEF FN Dim(x) = SIZEOF(x) / SIZEOF(double)

DEF FN Median(x) = IIF(ODD(Dim(x)), x[(Dim(x)-1)/2], (x[Dim(x)/2-1]+x[Dim(x)/2])/2 )

SORT a
PRINT "Median of a: ", Median(a)

SORT b
PRINT "Median of b: ", Median(b)
Output:
Median of a: 4.4
Median of b: 4.25

## BASIC

Works with: FreeBASIC
Works with: PowerBASIC
Works with: QB64
Works with: QBasic
Works with: Visual Basic

This uses the Quicksort function described at Quicksort#BASIC, with arr()'s type changed to SINGLE.

Note that in order to truly work with the Windows versions of PowerBASIC, the module-level code must be contained inside FUNCTION PBMAIN. Similarly, in order to work under Visual Basic, the same module-level code must be contained with Sub Main.

DECLARE FUNCTION median! (vector() AS SINGLE)

DIM vec1(10) AS SINGLE, vec2(11) AS SINGLE, n AS INTEGER

RANDOMIZE TIMER

FOR n = 0 TO 10
vec1(n) = RND * 100
vec2(n) = RND * 100
NEXT
vec2(11) = RND * 100

PRINT median(vec1())
PRINT median(vec2())

FUNCTION median! (vector() AS SINGLE)
DIM lb AS INTEGER, ub AS INTEGER, L0 AS INTEGER
lb = LBOUND(vector)
ub = UBOUND(vector)
REDIM v(lb TO ub) AS SINGLE
FOR L0 = lb TO ub
v(L0) = vector(L0)
NEXT
quicksort v(), lb, ub
IF ((ub - lb + 1) MOD 2) THEN
median = v((ub + lb) / 2)
ELSE
median = (v(INT((ub + lb) / 2)) + v(INT((ub + lb) / 2) + 1)) / 2
END IF
END FUNCTION

## BBC BASIC

INSTALL @lib\$+"SORTLIB"
Sort% = FN_sortinit(0,0)

DIM a(6), b(5)
a() = 4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2
b() = 4.1, 7.2, 1.7, 9.3, 4.4, 3.2

PRINT "Median of a() is " ; FNmedian(a())
PRINT "Median of b() is " ; FNmedian(b())
END

DEF FNmedian(a())
LOCAL C%
C% = DIM(a(),1) + 1
CALL Sort%, a(0)
= (a(C% DIV 2) + a((C%-1) DIV 2)) / 2

Output:

Median of a() is 4.4
Median of b() is 4.25

## Bracmat

Bracmat has no floating point numbers, so we have to parse floating point numbers as strings and convert them to rational numbers. Each number is packaged in a little list and these lists are accumulated in a sum. Bracmat keeps sums sorted, so the median is the term in the middle of the list, or the average of the two terms in the middle of the list.

(median=
begin decimals end int list med med1 med2 num number
. 0:?list
& whl
' ( @( !arg
:  ?
((%@:~" ":~",") ?:?number)
((" "|",") ?arg|:?arg)
)
& @( !number
: ( #?int "." [?begin #?decimals [?end
& !int+!decimals*10^(!begin+-1*!end):?num
| ?num
)
)
& (!num.)+!list:?list
)
& !list:?+[?end
& (  !end*1/2:~/
& !list:?+[!(=1/2*!end+-1)+(?med1.)+(?med2.)+?
& !med1*1/2+!med2*1/2:?med
| !list:?+[(div\$(1/2*!end,1))+(?med.)+?
)
& !med
);

median\$" 4.1 4 1.2 6.235 7868.33"
41/10

median\$"4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5"
89/20

median\$"1, 5, 3, 2, 4"
3

median\$"1, 5, 3, 6, 4, 2"
7/2

## C

#include <stdio.h>
#include <stdlib.h>

typedef struct floatList {
float *list;
int size;
} *FloatList;

int floatcmp( const void *a, const void *b) {
if (*(const float *)a < *(const float *)b) return -1;
else return *(const float *)a > *(const float *)b;
}

float median( FloatList fl )
{
qsort( fl->list, fl->size, sizeof(float), floatcmp);
return 0.5 * ( fl->list[fl->size/2] + fl->list[(fl->size-1)/2]);
}

int main()
{
static float floats1[] = { 5.1, 2.6, 6.2, 8.8, 4.6, 4.1 };
static struct floatList flist1 = { floats1, sizeof(floats1)/sizeof(float) };

static float floats2[] = { 5.1, 2.6, 8.8, 4.6, 4.1 };
static struct floatList flist2 = { floats2, sizeof(floats2)/sizeof(float) };

printf("flist1 median is %7.2f\n", median(&flist1)); /* 4.85 */
printf("flist2 median is %7.2f\n", median(&flist2)); /* 4.60 */
return 0;
}

### Quickselect algorithm

Average O(n) time:

#include <stdio.h>
#include <stdlib.h>
#include <time.h>

#define MAX_ELEMENTS 1000000

/* Return the k-th smallest item in array x of length len */
double quick_select(int k, double *x, int len)
{
inline void swap(int a, int b)
{
double t = x[a];
x[a] = x[b], x[b] = t;
}

int left = 0, right = len - 1;
int pos, i;
double pivot;

while (left < right)
{
pivot = x[k];
swap(k, right);
for (i = pos = left; i < right; i++)
{
if (x[i] < pivot)
{
swap(i, pos);
pos++;
}
}
swap(right, pos);
if (pos == k) break;
if (pos < k) left = pos + 1;
else right = pos - 1;
}
return x[k];
}

int main(void)
{
int i, length;
double *x, median;

/* Initialize random length double array with random doubles */
srandom(time(0));
length = random() % MAX_ELEMENTS;
x = malloc(sizeof(double) * length);
for (i = 0; i < length; i++)
{
// shifted by RAND_MAX for negative values
// divide by a random number for floating point
x[i] = (double)(random() - RAND_MAX / 2) / (random() + 1); // + 1 to not divide by 0
}

if (length % 2 == 0) // Even number of elements, median is average of middle two
{
median = (quick_select(length / 2, x, length) + quick_select(length / 2 - 1, x, length / 2)) / 2;
}
else // select middle element
{
median = quick_select(length / 2, x, length);
}

/* Sanity testing of median */
int less = 0, more = 0, eq = 0;
for (i = 0; i < length; i++)
{
if (x[i] < median) less ++;
else if (x[i] > median) more ++;
else eq ++;
}
printf("length: %d\nmedian: %lf\n<: %d\n>: %d\n=: %d\n", length, median, less, more, eq);

free(x);
return 0;
}

Output:

length: 992021
median: 0.000473
<: 496010
>: 496010
=: 1

## C++

This function runs in linear time on average.

#include <algorithm>

// inputs must be random-access iterators of doubles
// Note: this function modifies the input range
template <typename Iterator>
double median(Iterator begin, Iterator end) {
// this is middle for odd-length, and "upper-middle" for even length
Iterator middle = begin + (end - begin) / 2;

// This function runs in O(n) on average, according to the standard
std::nth_element(begin, middle, end);

if ((end - begin) % 2 != 0) { // odd length
return *middle;
} else { // even length
// the "lower middle" is the max of the lower half
Iterator lower_middle = std::max_element(begin, middle);
return (*middle + *lower_middle) / 2.0;
}
}

#include <iostream>

int main() {
double a[] = {4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2};
double b[] = {4.1, 7.2, 1.7, 9.3, 4.4, 3.2};

std::cout << median(a+0, a + sizeof(a)/sizeof(a[0])) << std::endl; // 4.4
std::cout << median(b+0, b + sizeof(b)/sizeof(b[0])) << std::endl; // 4.25

return 0;
}

## C#

using System;
using System.Linq;

namespace Test
{
class Program
{
static void Main()
{
double[] myArr = new double[] { 1, 5, 3, 6, 4, 2 };

myArr = myArr.OrderBy(i => i).ToArray();
// or Array.Sort(myArr) for in-place sort

int mid = myArr.Length / 2;
double median;

if (myArr.Length % 2 == 0)
{
//we know its even
median = (myArr[mid] + myArr[mid - 1]) / 2.0;
}
else
{
//we know its odd
median = myArr[mid];
}

Console.WriteLine(median);
}
}
}

## Clojure

Simple:

(defn median [ns]
(let [ns (sort ns)
cnt (count ns)
mid (bit-shift-right cnt 1)]
(if (odd? cnt)
(nth ns mid)
(/ (+ (nth ns mid) (nth ns (dec mid))) 2))))

## COBOL

Intrinsic function:

FUNCTION MEDIAN(some-table (ALL))

## Common Lisp

The recursive partitioning solution, without the median of medians optimization.

((defun select-nth (n list predicate)
"Select nth element in list, ordered by predicate, modifying list."
(do ((pivot (pop list))
(ln 0) (left '())
(rn 0) (right '()))
((endp list)
(cond
((< n ln) (select-nth n left predicate))
((eql n ln) pivot)
((< n (+ ln rn 1)) (select-nth (- n ln 1) right predicate))
(t (error "n out of range."))))
(if (funcall predicate (first list) pivot)
(psetf list (cdr list)
(cdr list) left
left list
ln (1+ ln))
(psetf list (cdr list)
(cdr list) right
right list
rn (1+ rn)))))

(defun median (list predicate)
(select-nth (floor (length list) 2) list predicate))

## D

import std.stdio, std.algorithm;

T median(T)(T[] nums) pure nothrow {
nums.sort();
if (nums.length & 1)
return nums[\$ / 2];
else
return (nums[\$ / 2 - 1] + nums[\$ / 2]) / 2.0;
}

void main() {
auto a1 = [5.1, 2.6, 6.2, 8.8, 4.6, 4.1];
writeln("Even median: ", a1.median);

auto a2 = [5.1, 2.6, 8.8, 4.6, 4.1];
writeln("Odd median: ", a2.median);
}
Output:
Even median: 4.85
Odd median:  4.6

## Delphi

program AveragesMedian;

{\$APPTYPE CONSOLE}

uses Generics.Collections, Types;

function Median(aArray: TDoubleDynArray): Double;
var
lMiddleIndex: Integer;
begin
TArray.Sort<Double>(aArray);

lMiddleIndex := Length(aArray) div 2;
if Odd(Length(aArray)) then
Result := aArray[lMiddleIndex]
else
Result := (aArray[lMiddleIndex - 1] + aArray[lMiddleIndex]) / 2;
end;

begin
Writeln(Median(TDoubleDynArray.Create(4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2)));
Writeln(Median(TDoubleDynArray.Create(4.1, 7.2, 1.7, 9.3, 4.4, 3.2)));
end.

## E

TODO: Use the selection algorithm, whatever that is

def median(list) {
def sorted := list.sort()
def count := sorted.size()
def mid1 := count // 2
def mid2 := (count - 1) // 2
if (mid1 == mid2) { # avoid inexact division
return sorted[mid1]
} else {
return (sorted[mid1] + sorted[mid2]) / 2
}
}
? median([1,9,2])
# value: 2

? median([1,9,2,4])
# value: 3.0

## EchoLisp

(define (median L) ;; O(n log(n))
(set! L (vector-sort! < (list->vector L)))
(define dim (// (vector-length L) 2))
(if (integer? dim)
(// (+ [L dim] [L (1- dim)]) 2)
[L (floor dim)]))

(median '( 3 4 5))
4
(median '(6 5 4 3))
4.5
(median (iota 10000))
4999.5
(median (iota 10001))
5000

## Elena

ELENA 3.4 :

import system'routines.
import system'math.
import extensions.

extension op
{
median
[
var aSorted := self ascendant.

var aLen := aSorted length.
if (aLen == 0)
[ ^ nil ];
[
var aMiddleIndex := aLen / 2.
if (aLen mod:2 == 0)
[ ^ (aSorted[aMiddleIndex - 1] + aSorted[aMiddleIndex]) / 2 ];
[ ^ aSorted[aMiddleIndex] ]
]
]
}

public program
[
var a1 := (4.1r, 5.6r, 7.2r, 1.7r, 9.3r, 4.4r, 3.2r).
var a2 := (4.1r, 7.2r, 1.7r, 9.3r, 4.4r, 3.2r).

console printLine("median of (",a1,") is ",a1 median).
console printLine("median of (",a2,") is ",a2 median).

]
Output:
median of (4.1,5.6,7.2,1.7,9.3,4.4,3.2) is 4.4
median of (4.1,7.2,1.7,9.3,4.4,3.2) is 4.25

## Elixir

Translation of: Erlang
defmodule Average do
def median([]), do: nil
def median(list) do
len = length(list)
sorted = Enum.sort(list)
mid = div(len, 2)
if rem(len,2) == 0, do: (Enum.at(sorted, mid-1) + Enum.at(sorted, mid)) / 2,
else: Enum.at(sorted, mid)
end
end

median = fn list -> IO.puts "#{inspect list} => #{inspect Average.median(list)}" end
median.([])
Enum.each(1..6, fn i ->
(for _ <- 1..i, do: :rand.uniform(6)) |> median.()
end)
Output:
[] => nil
[4] => 4
[1, 6] => 3.5
[5, 2, 4] => 4
[2, 3, 5, 1] => 2.5
[3, 2, 6, 3, 2] => 3
[6, 4, 2, 3, 1, 3] => 3.0

## Erlang

-module(median).
-import(lists, [nth/2, sort/1]).
-compile(export_all).

median(Unsorted) ->
Sorted = sort(Unsorted),
Length = length(Sorted),
Mid = Length div 2,
Rem = Length rem 2,
(nth(Mid+Rem, Sorted) + nth(Mid+1, Sorted)) / 2.

## ERRE

PROGRAM MEDIAN

DIM X[10]

PROCEDURE QUICK_SELECT
LT=0 RT=L-1
J=LT
REPEAT
PT=X[K]
SWAP(X[K],X[RT])
P=LT
FOR I=P TO RT-1 DO
IF X[I]<PT THEN SWAP(X[I],X[P]) P=P+1 END IF
END FOR
SWAP(X[RT],X[P])
IF P=K THEN EXIT PROCEDURE END IF
IF P<K THEN LT=P+1 END IF
IF P>=K THEN RT=P-1 END IF
UNTIL J>RT
END PROCEDURE

PROCEDURE MEDIAN
K=INT(L/2)
QUICK_SELECT
R=X[K]
IF L-2*INT(L/2)<>0 THEN R=(R+X[K+1])/2 END IF
END PROCEDURE

BEGIN
PRINT(CHR\$(12);) !CLS
X[0]=4.4 X[1]=2.3 X[2]=-1.7 X[3]=7.5 X[4]=6.6 X[5]=0
X[6]=1.9 X[7]=8.2 X[8]=9.3 X[9]=4.5 X[10]=-11.7
L=11
MEDIAN
PRINT(R)
END PROGRAM

Ouput is 5.95

## Euler Math Toolbox

The following function does much more than computing the median. It can handle a matrix of x values row by row. Then it can handle multiplicities in the vector v. Moreover it can search for the p median, not only the p=0.5 median.

>type median
function median (x, v: none, p)

## Default for v : none
## Default for p : 0.5

m=rows(x);
if m>1 then
y=zeros(m,1);
loop 1 to m;
y[#]=median(x[#],v,p);
end;
return y;
else
if v<>none then
{xs,i}=sort(x); vsh=v[i];
n=cols(xs);
ns=sum(vsh);
i=1+p*(ns-1); i0=floor(i);
vs=cumsum(vsh);
loop 1 to n
if vs[#]>i0 then
return xs[#];
elseif vs[#]+1>i0 then
k=#+1;
repeat;
if vsh[k]>0 or k>n then break; endif;
k=k+1;
end;
return (1-(i-i0))*xs[#]+(i-i0)*xs[k]+0;
endif;
end;
return xs[n];
else
xs=sort(x);
n=cols(x);
i=1+p*(n-1); i0=floor(i);
if i0==n then return xs[n]; endif;
return (i-i0)*xs[i+1]+(1-(i-i0))*xs[i];
endif;
endif;
endfunction
>median(1:10)
5.5
>median(1:9)
5
>median(1:10,p=0.2)
2.8
>0.2*10+0.8*1
2.8

## Euphoria

function median(sequence s)
atom min,k
-- Selection sort of half+1
for i = 1 to length(s)/2+1 do
min = s[i]
k = 0
for j = i+1 to length(s) do
if s[j] < min then
min = s[j]
k = j
end if
end for
if k then
s[k] = s[i]
s[i] = min
end if
end for
if remainder(length(s),2) = 0 then
return (s[\$/2]+s[\$/2+1])/2
else
return s[\$/2+1]
end if
end function

? median({ 4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5 })

Output:

4.45

## Excel

Assuming the values are entered in the A column, type into any cell which will not be part of the list :

=MEDIAN(A1:A10)

Assuming 10 values will be entered, alternatively, you can just type

=MEDIAN(

and then select the start and end cells, not necessarily in the same row or column.

The output for the first expression, for any 10 numbers is

23 11,5
21
12
3
19
7
23
11
9
0

## F#

Median of Medians algorithm implementation

let rec splitToFives list =
match list with
| a::b::c::d::e::tail ->
([a;b;c;d;e])::(splitToFives tail)
| [] -> []
| _ ->
let left = 5 - List.length (list)
let last = List.append list (List.init left (fun _ -> System.Double.PositiveInfinity) )
in [last]

let medianFromFives =
List.map ( fun (i:float list) ->
List.nth (List.sort i) 2 )

let start l =
let rec magicFives list k =
if List.length(list) <= 10 then
List.nth (List.sort list) (k-1)
else
let s = splitToFives list
let M = medianFromFives s
let m = magicFives M (int(System.Math.Ceiling((float(List.length M))/2.)))
let (ll,lg) = List.partition ( fun i -> i < m ) list
let (le,lg) = List.partition ( fun i -> i = m ) lg
in
if (List.length ll >= k) then
magicFives ll k
else if (List.length ll + List.length le >= k ) then m
else
magicFives lg (k-(List.length ll)-(List.length le))
in
let len = List.length l in
if (len % 2 = 1) then
magicFives l ((len+1)/2)
else
let a = magicFives l (len/2)
let b = magicFives l ((len/2)+1)
in (a+b)/2.

let z = [1.;5.;2.;8.;7.;2.]
start z
let z' = [1.;5.;2.;8.;7.]
start z'

## Factor

The quicksort-style solution, with random pivoting. Takes the lesser of the two medians for even sequences.

USING: arrays kernel locals math math.functions random sequences ;
IN: median

: pivot ( seq -- pivot ) random ;

: split ( seq pivot -- {lt,eq,gt} )
[ [ < ] curry partition ] keep
[ = ] curry partition
3array ;

DEFER: nth-in-order
:: nth-in-order-recur ( seq ind -- elt )
seq dup pivot split
dup [ length ] map 0 [ + ] accumulate nip
dup [ ind <= [ 1 ] [ 0 ] if ] map sum 1 -
[ swap nth ] curry [email protected]
ind swap -
nth-in-order ;

: nth-in-order ( seq ind -- elt )
dup 0 =
[ drop first ]
[ nth-in-order-recur ]
if ;

: median ( seq -- median )
dup length 1 - 2 / floor nth-in-order ;

Usage:

( scratchpad ) 11 iota median .
5
( scratchpad ) 10 iota median .
4

## Forth

This uses the O(n) algorithm derived from quicksort.

-1 cells constant -cell
: cell- -cell + ;

defer lessthan ( [email protected] [email protected] -- ? ) ' < is lessthan

: mid ( l r -- mid ) over - 2/ -cell and + ;

: exch ( addr1 addr2 -- ) dup @ >r over @ swap ! r> swap ! ;

: part ( l r -- l r r2 l2 )
2dup mid @ >r ( r: pivot )
2dup begin
swap begin dup @ [email protected] lessthan while cell+ repeat
swap begin [email protected] over @ lessthan while cell- repeat
2dup <= if 2dup exch >r cell+ r> cell- then
2dup > until r> drop ;

0 value midpoint

: select ( l r -- )
begin 2dup < while
part
dup midpoint >= if nip nip ( l l2 ) else
over midpoint <= if drop rot drop swap ( r2 r ) else
2drop 2drop exit then then
repeat 2drop ;

: median ( array len -- m )
1- cells over + 2dup mid to midpoint
select midpoint @ ;
create test 4 , 2 , 1 , 3 , 5 ,

test 4 median . \ 2
test 5 median . \ 3

## Fortran

Works with: Fortran version 90 and later
program Median_Test

real :: a(7) = (/ 4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2 /), &
b(6) = (/ 4.1, 7.2, 1.7, 9.3, 4.4, 3.2 /)

print *, median(a)
print *, median(b)

contains

function median(a, found)
real, dimension(:), intent(in) :: a
! the optional found argument can be used to check
! if the function returned a valid value; we need this
! just if we suspect our "vector" can be "empty"
logical, optional, intent(out) :: found
real :: median

integer :: l
real, dimension(size(a,1)) :: ac

if ( size(a,1) < 1 ) then
if ( present(found) ) found = .false.
else
ac = a
! this is not an intrinsic: peek a sort algo from
! Category:Sorting, fixing it to work with real if
call sort(ac)

l = size(a,1)
if ( mod(l, 2) == 0 ) then
median = (ac(l/2+1) + ac(l/2))/2.0
else
median = ac(l/2+1)
end if

if ( present(found) ) found = .true.
end if

end function median

end program Median_Test
If one refers to Quickselect_algorithm#Fortran which offers function FINDELEMENT(K,A,N) that returns the value of A(K) when the array of N elements has been rearranged if necessary so that A(K) is the K'th in order, then, supposing that a version is devised using the appropriate type for array A,
K = N/2
MEDIAN = FINDELEMENT(K + 1,A,N)
IF (MOD(N,2).EQ.0) MEDIAN = (FINDELEMENT(K,A,N) + MEDIAN)/2

As well as returning a result, the function possibly re-arranges the elements of the array, which is not "pure" behaviour. Not to the degree of fully sorting them, merely that all elements before K are not larger than A(K) as it now is, and all elements after K are not smaller than A(K).

## FreeBASIC

' FB 1.05.0 Win64

Sub quicksort(a() As Double, first As Integer, last As Integer)
Dim As Integer length = last - first + 1
If length < 2 Then Return
Dim pivot As Double = a(first + length\ 2)
Dim lft As Integer = first
Dim rgt As Integer = last
While lft <= rgt
While a(lft) < pivot
lft +=1
Wend
While a(rgt) > pivot
rgt -= 1
Wend
If lft <= rgt Then
Swap a(lft), a(rgt)
lft += 1
rgt -= 1
End If
Wend
quicksort(a(), first, rgt)
quicksort(a(), lft, last)
End Sub

Function median(a() As Double) As Double
Dim lb As Integer = LBound(a)
Dim ub As Integer = UBound(a)
Dim length As Integer = ub - lb + 1
If length = 0 Then Return 0.0/0.0 '' NaN
If length = 1 Then Return a(ub)
Dim mb As Integer = (lb + ub) \2
If length Mod 2 = 1 Then Return a(mb)
Return (a(mb) + a(mb + 1))/2.0
End Function

Dim a(0 To 9) As Double = {4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5}
quicksort(a(), 0, 9)
Print "Median for all 10 elements  : "; median(a())
' now get rid of final element
Dim b(0 To 8) As Double = {4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3}
quicksort(b(), 0, 8)
Print "Median for first 9 elements : "; median(b())
Print
Print "Press any key to quit"
Sleep
Output:
Median for all 10 elements  :  4.45
Median for first 9 elements :  4.4

## GAP

Median := function(v)
local n, w;
w := SortedList(v);
n := Length(v);
return (w[QuoInt(n + 1, 2)] + w[QuoInt(n, 2) + 1]) / 2;
end;

a := [41, 56, 72, 17, 93, 44, 32];
b := [41, 72, 17, 93, 44, 32];

Median(a);
# 44
Median(b);
# 85/2

## Go

### Sort

Go built-in sort. O(n log n).

package main

import (
"fmt"
"sort"
)

func main() {
fmt.Println(median([]float64{3, 1, 4, 1})) // prints 2
fmt.Println(median([]float64{3, 1, 4, 1, 5})) // prints 3
}

func median(a []float64) float64 {
sort.Float64s(a)
half := len(a) / 2
m := a[half]
if len(a)%2 == 0 {
m = (m + a[half-1]) / 2
}
return m
}

### Partial selection sort

The task description references the WP entry for "selection algorithm" which (as of this writing) gives just one pseudocode example, which is implemented here. As the WP article notes, it is O(kn). Unfortunately in the case of median, k is n/2 so the algorithm is O(n^2). Still, it gives the idea of median by selection. Note that the partial selection sort does leave the k smallest values sorted, so in the case of an even number of elements, the two elements to average are available after a single call to sel().

package main

import "fmt"

func main() {
fmt.Println(median([]float64{3, 1, 4, 1})) // prints 2
fmt.Println(median([]float64{3, 1, 4, 1, 5})) // prints 3
}

func median(a []float64) float64 {
half := len(a) / 2
med := sel(a, half)
if len(a)%2 == 0 {
return (med + a[half-1]) / 2
}
return med
}

func sel(list []float64, k int) float64 {
for i, minValue := range list[:k+1] {
minIndex := i
for j := i + 1; j < len(list); j++ {
if list[j] < minValue {
minIndex = j
minValue = list[j]
list[i], list[minIndex] = minValue, list[i]
}
}
}
return list[k]
}

### Quickselect

It doesn't take too much more code to implement a quickselect with random pivoting, which should run in expected time O(n). The qsel function here permutes elements of its parameter "a" in place. It leaves the slice somewhat more ordered, but unlike the sort and partial sort examples above, does not guarantee that element k-1 is in place. For the case of an even number of elements then, median must make two separate qsel() calls.

package main

import (
"fmt"
"math/rand"
)

func main() {
fmt.Println(median([]float64{3, 1, 4, 1})) // prints 2
fmt.Println(median([]float64{3, 1, 4, 1, 5})) // prints 3
}

func median(list []float64) float64 {
half := len(list) / 2
med := qsel(list, half)
if len(list)%2 == 0 {
return (med + qsel(list, half-1)) / 2
}
return med
}

func qsel(a []float64, k int) float64 {
for len(a) > 1 {
px := rand.Intn(len(a))
pv := a[px]
last := len(a) - 1
a[px], a[last] = a[last], pv
px = 0
for i, v := range a[:last] {
if v < pv {
a[px], a[i] = v, a[px]
px++
}
}
if px == k {
return pv
}
if k < px {
a = a[:px]
} else {
// swap elements. simply assigning a[last] would be enough to
// allow qsel to return the correct result but it would leave slice
// "a" unusable for subsequent use. we want this full swap so that
// we can make two successive qsel calls in the case of median
// of an even number of elements.
a[px], a[last] = pv, a[px]
a = a[px+1:]
k -= px + 1
}
}
return a[0]
}

## Groovy

Solution (brute force sorting, with arithmetic averaging of dual midpoints (even sizes)):

def median(Iterable col) {
def s = col as SortedSet
if (s == null) return null
if (s.empty) return 0
def n = s.size()
def m = n.intdiv(2)
def l = s.collect { it }
n%2 == 1 ? l[m] : (l[m] + l[m-1])/2
}

Test:

def a = [4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5]
def sz = a.size()

(0..sz).each {
println """\${median(a[0..<(sz-it)])} == median(\${a[0..<(sz-it)]})
\${median(a[it..<sz])} == median(\${a[it..<sz]})"""

}

Output:

4.45 == median([4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5])
4.45 == median([4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5])
4.4 == median([4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3])
4.5 == median([2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5])
3.35 == median([4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2])
5.55 == median([-1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5])
2.3 == median([4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9])
6.6 == median([7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5])
3.35 == median([4.4, 2.3, -1.7, 7.5, 6.6, 0.0])
5.55 == median([6.6, 0.0, 1.9, 8.2, 9.3, 4.5])
4.4 == median([4.4, 2.3, -1.7, 7.5, 6.6])
4.5 == median([0.0, 1.9, 8.2, 9.3, 4.5])
3.35 == median([4.4, 2.3, -1.7, 7.5])
6.35 == median([1.9, 8.2, 9.3, 4.5])
2.3 == median([4.4, 2.3, -1.7])
8.2 == median([8.2, 9.3, 4.5])
3.35 == median([4.4, 2.3])
6.9 == median([9.3, 4.5])
4.4 == median([4.4])
4.5 == median([4.5])
0 == median([])
0 == median([])

This uses a quick select algorithm and runs in expected O(n) time.

import Data.List (partition)

nth :: Ord t => [t] -> Int -> t
nth (x:xs) n
| k == n = x
| k > n = nth ys n
| otherwise = nth zs \$ n - k - 1
where
(ys, zs) = partition (< x) xs
k = length ys

medianMay :: (Fractional a, Ord a) => [a] -> Maybe a
medianMay xs
| n < 1 = Nothing
| even n = Just ((nth xs (div n 2) + nth xs (div n 2 - 1)) / 2.0)
| otherwise = Just (nth xs (div n 2))
where
n = length xs

main :: IO ()
main =
mapM_
(printMay . medianMay)
[[], [7], [5, 3, 4], [5, 4, 2, 3], [3, 4, 1, -8.4, 7.2, 4, 1, 1.2]]
where
printMay = maybe (putStrLn "(not defined)") print
Output:
(not defined)
7.0
4.0
3.5
2.1
Or
Library: hstats
> Math.Statistics.median [1,9,2,4]
3.0

## HicEst

If the input has an even number of elements, median is the mean of the middle two values:

REAL :: n=10, vec(n)

vec = RAN(1)
SORT(Vector=vec, Sorted=vec) ! in-place Merge-Sort

IF( MOD(n,2) ) THEN ! odd n
median = vec( CEILING(n/2) )
ELSE
median = ( vec(n/2) + vec(n/2 + 1) ) / 2
ENDIF

## Icon and Unicon

A quick and dirty solution:

procedure main(args)
write(median(args))
end

procedure median(A)
A := sort(A)
n := *A
return if n % 2 = 1 then A[n/2+1]
else (A[n/2]+A[n/2+1])/2.0 | 0 # 0 if empty list
end

Sample outputs:

->am 3 1 4 1 5 9 7 6 3
4
->am 3 1 4 1 5 9 7 6
4.5
->

## J

The verb median is available from the stats/base addon and returns the mean of the two middle values for an even number of elements:

require 'stats/base'
median 1 9 2 4
3

The definition given in the addon script is:

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

If, for an even number of elements, both values were desired when those two values are distinct, then the following implementation would suffice:

median=: [email protected](<. , >.)@midpt { /:~
median 1 9 2 4
2 4

## Java

Works with: Java version 1.5+

Sorting:

// Note: this function modifies the input list
public static double median(List<Double> list) {
Collections.sort(list);
return (list.get(list.size() / 2) + list.get((list.size() - 1) / 2)) / 2;
}
Works with: Java version 1.5+

Using priority queue (which sorts under the hood):

public static double median2(List<Double> list) {
PriorityQueue<Double> pq = new PriorityQueue<Double>(list);
int n = list.size();
for (int i = 0; i < (n - 1) / 2; i++)
if (n % 2 != 0) // odd length
return pq.poll();
else
return (pq.poll() + pq.poll()) / 2.0;
}

## JavaScript

### ES5

function median(ary) {
if (ary.length == 0)
return null;
ary.sort(function (a,b){return a - b})
var mid = Math.floor(ary.length / 2);
if ((ary.length % 2) == 1) // length is odd
return ary[mid];
else
return (ary[mid - 1] + ary[mid]) / 2;
}

median([]); // null
median([5,3,4]); // 4
median([5,4,2,3]); // 3.5
median([3,4,1,-8.4,7.2,4,1,1.2]); // 2.1

### ES6

Using a quick select algorithm

(() => {
'use strict';

// median :: [Num] -> Num
function median(xs) {
// nth :: [Num] -> Int -> Maybe Num
let nth = (xxs, n) => {
if (xxs.length > 0) {
let [x, xs] = uncons(xxs),
[ys, zs] = partition(y => y < x, xs),
k = ys.length;

return k === n ? x : (
k > n ? nth(ys, n) : nth(zs, n - k - 1)
);
} else return undefined;
},
n = xs.length;

return even(n) ? (
(nth(xs, div(n, 2)) + nth(xs, div(n, 2) - 1)) / 2
) : nth(xs, div(n, 2));
}

// GENERIC

// partition :: (a -> Bool) -> [a] -> ([a], [a])
let partition = (p, xs) =>
xs.reduce((a, x) =>
p(x) ? [a[0].concat(x), a[1]] : [a[0], a[1].concat(x)], [
[],
[]
]),

// uncons :: [a] -> Maybe (a, [a])
uncons = xs => xs.length ? [xs[0], xs.slice(1)] : undefined,

// even :: Integral a => a -> Bool
even = n => n % 2 === 0,

// div :: Num -> Num -> Int
div = (x, y) => Math.floor(x / y);

return [
[],
[5, 3, 4],
[5, 4, 2, 3],
[3, 4, 1, -8.4, 7.2, 4, 1, 1.2]
].map(median);
})();
Output:
[
null,
4,
3.5,
2.1
]

## jq

def median:
length as \$length
| sort as \$s
| if \$length == 0 then null
else (\$length / 2 | floor) as \$l2
| if (\$length % 2) == 0 then
(\$s[\$l2 - 1] + \$s[\$l2]) / 2
else \$s[\$l2]
end
end ;
This definition can be used in a jq program, but to illustrate how it can be used as a command line filter, suppose the definition and the program median are in a file named median.jq, and that the file in.dat contains a sequence of arrays, such as
[4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2]
[4.1, 7.2, 1.7, 9.3, 4.4, 3.2]
Then invoking the jq program yields a stream of values:
\$ jq -f median.jq in.dat
4.4
4.25

## Julia

Julia has a built-in median() function

# Version 5.2
function median2(n)
s = sort(n)
len = length(n)
if len % 2 == 0
return (s[floor(Int, len / 2) + 1] + s[floor(Int, len / 2)]) / 2
else
return s[floor(Int, len / 2) + 1]
end
end

a = [4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2]
b = [4.1, 7.2, 1.7, 9.3, 4.4, 3.2]

@show a b median2(a) median(a) median2(b) median(b)
Output:
a = [4.1,5.6,7.2,1.7,9.3,4.4,3.2]
b = [4.1,7.2,1.7,9.3,4.4,3.2]
median2(a) = 4.4
median(a) = 4.4
median2(b) = 4.25
median(b) = 4.25

## K

med:{a:[email protected]<x; i:(#a)%2; :[(#a)!2; [email protected]; {(+/x)%#x} [email protected],i-1]}
v:10*6 _draw 0
v
5.961475 2.025856 7.262835 1.814272 2.281911 4.854716
med[v]
3.568313
med[1_ v]
2.281911

An alternate solution which works in the oK implementation using the same dataset v from above and shows both numbers around the median point on even length datasets would be:

med:{a:[email protected]<x; i:_(#a)%2
\$[2!#a; [email protected]; |[email protected],i-1]}
med[v]
2.2819 4.8547

## Kotlin

Works with: Kotlin version 1.0+
fun median(l: List<Double>) = l.sorted().let { (it[it.size / 2] + it[(it.size - 1) / 2]) / 2 }

fun main(args: Array<String>) {
median(listOf(5.0, 3.0, 4.0)).let { println(it) } // 4
median(listOf(5.0, 4.0, 2.0, 3.0)).let { println(it) } // 3.5
median(listOf(3.0, 4.0, 1.0, -8.4, 7.2, 4.0, 1.0, 1.2)).let { println(it) } // 2.1
}

## Lasso

can't use Lasso's built in median method because that takes 3 values, not an array of indeterminate length

Lasso's built in function is "median( value_1, value_2, value_3 )"

define median_ext(a::array) => {
#a->sort

if(#a->size % 2) => {
// odd numbered element array, pick middle
return #a->get(#a->size / 2 + 1)

else
// even number elements in array
return (#a->get(#a->size / 2) + #a->get(#a->size / 2 + 1)) / 2.0
}
}

median_ext(array(3,2,7,6)) // 4.5
median_ext(array(3,2,9,7,6)) // 6

## Liberty BASIC

dim a( 100), b( 100) ' assumes we will not have vectors of more terms...

a\$ ="4.1,5.6,7.2,1.7,9.3,4.4,3.2"
print "Median is "; median( a\$) ' 4.4 7 terms
print
a\$ ="4.1,7.2,1.7,9.3,4.4,3.2"
print "Median is "; median( a\$) ' 4.25 6 terms
print
a\$ ="4.1,4,1.2,6.235,7868.33" ' 4.1
print "Median of "; a\$; " is "; median( a\$)
print
a\$ ="1,5,3,2,4" ' 3
print "Median of "; a\$; " is "; median( a\$)
print
a\$ ="1,5,3,6,4,2" ' 3.5
print "Median of "; a\$; " is "; median( a\$)
print
a\$ ="4.4,2.3,-1.7,7.5,6.6,0.0,1.9,8.2,9.3,4.5" ' 4.45
print "Median of "; a\$; " is "; median( a\$)

end

function median( a\$)
i =1
do
v\$ =word\$( a\$, i, ",")
if v\$ ="" then exit do
print v\$,
a( i) =val( v\$)
i =i +1
loop until 0
print

sort a(), 1, i -1

for j =1 to i -1
print a( j),
next j
print

middle =( i -1) /2
intmiddle =int( middle)
if middle <>intmiddle then median= a( 1 +intmiddle) else median =( a( intmiddle) +a( intmiddle +1)) /2
end function

4.1 5.6 7.2 1.7 9.3 4.4 3.2
Median is 4.4

4.1 7.2 1.7 9.3 4.4 3.2
Median is 4.25

4.1 4 1.2 6.235 7868.33
Median of 4.1,4,1.2,6.235,7868.33 is 4.1

1 5 3 2 4
Median of 1,5,3,2,4 is 3

1 5 3 6 4 2
Median of 1,5,3,6,4,2 is 3.5

4.4 2.3 -1.7 7.5 6.6 0.0 1.9 8.2 9.3 4.5
Median of 4.4,2.3,-1.7,7.5,6.6,0.0,1.9,8.2,9.3,4.5 is 4.45

## Lingo

on median (numlist)
-- numlist = numlist.duplicate() -- if input list should not be altered
numlist.sort()
if numlist.count mod 2 then
return numlist[numlist.count/2+1]
else
return (numlist[numlist.count/2]+numlist[numlist.count/2+1])/2.0
end if
end

## LiveCode

LC has median as a built-in function

put median("4.1,5.6,7.2,1.7,9.3,4.4,3.2") & "," & median("4.1,7.2,1.7,9.3,4.4,3.2")
returns 4.4, 4.25

To make our own, we need own own floor function first

function floor n
if n < 0 then
return (trunc(n) - 1)
else
return trunc(n)
end if
end floor

function median2 x
local n, m
set itemdelimiter to comma
sort items of x ascending numeric
put the number of items of x into n
put floor(n / 2) into m
if n mod 2 is 0 then
return (item m of x + item (m + 1) of x) / 2
else
return item (m + 1) of x
end if
end median2

returns the same as the built-in median, viz.
put median2("4.1,5.6,7.2,1.7,9.3,4.4,3.2") & "," & median2("4.1,7.2,1.7,9.3,4.4,3.2")
4.4,4.25

## LSL

integer MAX_ELEMENTS = 10;
integer MAX_VALUE = 100;
default {
state_entry() {
list lst = [];
integer x = 0;
for(x=0 ; x<MAX_ELEMENTS ; x++) {
lst += llFrand(MAX_VALUE);
}
llOwnerSay("lst=["+llList2CSV(lst)+"]");
llOwnerSay("Geometric Mean: "+(string)llListStatistics(LIST_STAT_GEOMETRIC_MEAN, lst));
llOwnerSay(" Max: "+(string)llListStatistics(LIST_STAT_MAX, lst));
llOwnerSay(" Mean: "+(string)llListStatistics(LIST_STAT_MEAN, lst));
llOwnerSay(" Median: "+(string)llListStatistics(LIST_STAT_MEDIAN, lst));
llOwnerSay(" Min: "+(string)llListStatistics(LIST_STAT_MIN, lst));
llOwnerSay(" Num Count: "+(string)llListStatistics(LIST_STAT_NUM_COUNT, lst));
llOwnerSay(" Range: "+(string)llListStatistics(LIST_STAT_RANGE, lst));
llOwnerSay(" Std Dev: "+(string)llListStatistics(LIST_STAT_STD_DEV, lst));
llOwnerSay(" Sum: "+(string)llListStatistics(LIST_STAT_SUM, lst));
llOwnerSay(" Sum Squares: "+(string)llListStatistics(LIST_STAT_SUM_SQUARES, lst));
}
}

Output:

lst=[23.815209, 85.890704, 10.811144, 31.522696, 54.619416, 12.211729, 42.964463, 87.367889, 7.106129, 18.711078]
Geometric Mean:    27.325070
Max:    87.367889
Mean:    37.502046
Median:    27.668953
Min:     7.106129
Num Count:    10.000000
Range:    80.261761
Std Dev:    29.819840
Sum:   375.020458
Sum Squares: 22067.040048

## Lua

function median (numlist)
if type(numlist) ~= 'table' then return numlist end
table.sort(numlist)
if #numlist %2 == 0 then return (numlist[#numlist/2] + numlist[#numlist/2+1]) / 2 end
return numlist[math.ceil(#numlist/2)]
end

print(median({4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2}))
print(median({4.1, 7.2, 1.7, 9.3, 4.4, 3.2}))

## Maple

### Builtin

This works for numeric lists or arrays, and is designed for large data sets.

> Statistics:-Median( [ 1, 5, 3, 2, 4 ] );
3.

> Statistics:-Median( [ 1, 5, 3, 6, 2, 4 ] );
3.50000000000000

### Using a sort

This solution can handle exact numeric inputs. Instead of inputting a container of some kind, it simply finds the median of its arguments.

median1 := proc()
local L := sort( [ args ] );
( L[ iquo( 1 + nargs, 2 ) ] + L[ 1 + iquo( nargs, 2 ) ] ) / 2
end proc:

For example:

> median1( 1, 5, 3, 2, 4 ); # 3
3

> median1( 1, 5, 3, 6, 4, 2 ); # 7/2
7/2

## Mathematica / Wolfram Language

Built-in function:

Median[{1, 5, 3, 2, 4}]
Median[{1, 5, 3, 6, 4, 2}]
Output:
3
7/2

Custom function:

mymedian[x_List]:=Module[{t=Sort[x],L=Length[x]},
If[Mod[L,2]==0,
(t[[L/2]]+t[[L/2+1]])/2
,
t[[(L+1)/2]]
]
]

Example of custom function:

mymedian[{1, 5, 3, 2, 4}]
mymedian[{1, 5, 3, 6, 4, 2}]
Output:
3
7/2

## MATLAB

If the input has an even number of elements, function returns the mean of the middle two values:

function medianValue = findmedian(setOfValues)
medianValue = median(setOfValues);
end

## Maxima

/* built-in */
median([41, 56, 72, 17, 93, 44, 32]); /* 44 */
median([41, 72, 17, 93, 44, 32]); /* 85/2 */

## MUMPS

MEDIAN(X)
;X is assumed to be a list of numbers separated by "^"
;I is a loop index
;L is the length of X
;Y is a new array
QUIT:'\$DATA(X) "No data"
QUIT:X="" "Empty Set"
NEW I,ODD,L,Y
SET L=\$LENGTH(X,"^"),ODD=L#2,I=1
;The values in the vector are used as indices for a new array Y, which sorts them
FOR QUIT:I>L SET Y(\$PIECE(X,"^",I))=1,I=I+1
;Go to the median index, or the lesser of the middle if there is an even number of elements
SET J="" FOR I=1:1:\$SELECT(ODD:L\2+1,'ODD:L/2) SET J=\$ORDER(Y(J))
QUIT \$SELECT(ODD:J,'ODD:(J+\$ORDER(Y(J)))/2)

USER>W \$\$MEDIAN^ROSETTA("-1.3^2.43^3.14^17^2E-3")
3.14
USER>W \$\$MEDIAN^ROSETTA("-1.3^2.43^3.14^17^2E-3^4")
3.57
USER>W \$\$MEDIAN^ROSETTA("")
Empty Set
USER>W \$\$MEDIAN^ROSETTA
No data

## NetRexx

Translation of: Java
/* NetRexx */
options replace format comments java crossref symbols nobinary

class RAvgMedian00 public

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method median(lvector = java.util.List) public static returns Rexx
cvector = ArrayList(lvector) -- make a copy of input to ensure it's contents are preserved
Collections.sort(cvector, RAvgMedian00.RexxComparator())
kVal = ((Rexx cvector.get(cvector.size() % 2)) + (Rexx cvector.get((cvector.size() - 1) % 2))) / 2
return kVal

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method median(rvector = Rexx[]) public static returns Rexx
return median(ArrayList(Arrays.asList(rvector)))

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method show_median(lvector = java.util.List) public static returns Rexx
mVal = median(lvector)
say 'Meadian:' mVal.format(10, 6, 3, 6, 's')', Vector:' (Rexx lvector).space(0)
return mVal

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method show_median(rvector = Rexx[]) public static returns Rexx
return show_median(ArrayList(Arrays.asList(rvector)))

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method run_samples() public static
show_median([Rexx 10.0]) -- 10.0
show_median([Rexx 10.0, 9.0, 8.0, 7.0, 6.0, 5.0, 4.0, 3.0, 2.0, 1.0]) -- 5.5
show_median([Rexx 9, 8, 7, 6, 5, 4, 3, 2, 1]) -- 5.0
show_median([Rexx 1.0, 9, 2.0, 4.0]) -- 3.0
show_median([Rexx 3.0, 1, 4, 1.0, 5.0, 9, 7.0, 6.0]) -- 4.5
show_median([Rexx 3, 4, 1, -8.4, 7.2, 4, 1, 1.2]) -- 2.1
show_median([Rexx -1.2345678e+99, 2.3e+700]) -- 1.15e+700
show_median([Rexx 4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2]) -- 4.4
show_median([Rexx 4.1, 7.2, 1.7, 9.3, 4.4, 3.2]) -- 4.25
show_median([Rexx 28.207, 74.916, 51.695, 72.486, 51.118, 3.241, 73.807]) -- 51.695
show_median([Rexx 27.984, 89.172, 0.250, 66.316, 41.805, 60.043]) -- 50.924
show_median([Rexx 5.1, 2.6, 6.2, 8.8, 4.6, 4.1]) -- 4.85
show_median([Rexx 5.1, 2.6, 8.8, 4.6, 4.1]) -- 4.6
show_median([Rexx 4.4, 2.3, -1.7, 7.5, 6.6, 0.0, 1.9, 8.2, 9.3, 4.5]) -- 4.45
show_median([Rexx 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 0, 0, 0, 0.11]) -- 3.0
show_median([Rexx 10, 20, 30, 40, 50, -100, 4.7, -11e+2]) -- 15.0
show_median([Rexx 9.3, -2.0, 4.0, 7.3, 8.1, 4.1, -6.3, 4.2, -1.0, -8.4]) -- 4.05
show_median([Rexx 8.3, -3.6, 5.7, 2.3, 9.3, 5.4, -2.3, 6.3, 9.9]) -- 5.7
return

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method main(args = String[]) public static
run_samples()
return

-- =============================================================================
class RAvgMedian00.RexxComparator implements Comparator

-- ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
method compare(i1=Object, i2=Object) public returns int
i = Rexx i1
j = Rexx i2

if i < j then return -1
if i > j then return +1
else return 0

Output:

## NewLISP

; median.lsp
; oofoe 2012-01-25

(define (median lst)
(sort lst) ; Sorts in place.
(if (empty? lst)
nil
(letn ((n (length lst))
(h (/ (- n 1) 2)))
(if (zero? (mod n 2))
(div (add (lst h) (lst (+ h 1))) 2)
(lst h))
)))

(define (test lst) (println lst " -> " (median lst)))

(test '())
(test '(5 3 4))
(test '(5 4 2 3))
(test '(3 4 1 -8.4 7.2 4 1 1.2))

(exit)

Sample output:

() -> nil
(5 3 4) -> 4
(5 4 2 3) -> 3.5
(3 4 1 -8.4 7.2 4 1 1.2) -> 2.1

## Nim

Translation of: Python
import algorithm, strutils

proc median(xs: seq[float]): float =
var ys = xs
sort(ys, system.cmp[float])
0.5 * (ys[ys.high div 2] + ys[ys.len div 2])

var a = @[4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2]
echo formatFloat(median(a), precision = 0)
a = @[4.1, 7.2, 1.7, 9.3, 4.4, 3.2]
echo formatFloat(median(a), precision = 0)

Example Output:

4.4
4.25

## Oberon-2

Oxford Oberon-2

MODULE Median;
IMPORT Out;
CONST
MAXSIZE = 100;

PROCEDURE Partition(VAR a: ARRAY OF REAL; left, right: INTEGER): INTEGER;
VAR
pValue,aux: REAL;
store,i,pivot: INTEGER;
BEGIN
pivot := right;
pValue := a[pivot];
aux := a[right];a[right] := a[pivot];a[pivot] := aux; (* a[pivot] <-> a[right] *)
store := left;
FOR i := left TO right -1 DO
IF a[i] <= pValue THEN
aux := a[store];a[store] := a[i];a[i]:=aux;
INC(store)
END
END;
aux := a[right];a[right] := a[store]; a[store] := aux;
RETURN store
END Partition;

(* QuickSelect algorithm *)
PROCEDURE Select(a: ARRAY OF REAL; left,right,k: INTEGER;VAR r: REAL);
VAR
pIndex, pDist : INTEGER;
BEGIN
IF left = right THEN r := a[left]; RETURN END;
pIndex := Partition(a,left,right);
pDist := pIndex - left + 1;
IF pDist = k THEN
r := a[pIndex];RETURN
ELSIF k < pDist THEN
Select(a,left, pIndex - 1, k, r)
ELSE
Select(a,pIndex + 1, right, k - pDist, r)
END
END Select;

PROCEDURE Median(a: ARRAY OF REAL;left,right: INTEGER): REAL;
VAR
idx,len : INTEGER;
r1,r2 : REAL;
BEGIN
len := right - left + 1;
idx := len DIV 2 + 1;
r1 := 0.0;r2 := 0.0;
Select(a,left,right,idx,r1);
IF ODD(len) THEN RETURN r1 END;
Select(a,left,right,idx - 1,r2);
RETURN (r1 + r2) / 2;
END Median;

VAR
ary: ARRAY MAXSIZE OF REAL;
r: REAL;
BEGIN
r := 0.0;
Out.Fixed(Median(ary,0,0),4,2);Out.Ln; (* empty *)
ary[0] := 5;
ary[1] := 3;
ary[2] := 4;
Out.Fixed(Median(ary,0,2),4,2);Out.Ln;
ary[0] := 5;
ary[1] := 4;
ary[2] := 2;
ary[3] := 3;
Out.Fixed(Median(ary,0,3),4,2);Out.Ln;
ary[0] := 3;
ary[1] := 4;
ary[2] := 1;
ary[3] := -8.4;
ary[4] := 7.2;
ary[5] := 4;
ary[6] := 1;
ary[7] := 1.2;
Out.Fixed(Median(ary,0,7),4,2);Out.Ln;
END Median.

Output:

0.00
4.00
3.50
2.10

## Objeck

use Structure;

bundle Default {
class Median {
function : Main(args : String[]) ~ Nil {
numbers := FloatVector->New([4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2]);
DoMedian(numbers)->PrintLine();

numbers := FloatVector->New([4.1, 7.2, 1.7, 9.3, 4.4, 3.2]);
DoMedian(numbers)->PrintLine();
}

function : native : DoMedian(numbers : FloatVector) ~ Float {
if(numbers->Size() = 0) {
return 0.0;
}
else if(numbers->Size() = 1) {
return numbers->Get(0);
};

numbers->Sort();

i := numbers->Size() / 2;
if(numbers->Size() % 2 = 0) {
return (numbers->Get(i - 1) + numbers->Get(i)) / 2.0;
};

return numbers->Get(i);
}
}
}

## OCaml

(* note: this modifies the input array *)
let median array =
let len = Array.length array in
Array.sort compare array;
(array.((len-1)/2) +. array.(len/2)) /. 2.0;;

let a = [|4.1; 5.6; 7.2; 1.7; 9.3; 4.4; 3.2|];;
median a;;
let a = [|4.1; 7.2; 1.7; 9.3; 4.4; 3.2|];;
median a;;

## Octave

Of course Octave has its own median function we can use to check our implementation. The Octave's median function, however, does not handle the case you pass in a void vector.

function y = median2(v)
if (numel(v) < 1)
y = NA;
else
sv = sort(v);
l = numel(v);
if ( mod(l, 2) == 0 )
y = (sv(floor(l/2)+1) + sv(floor(l/2)))/2;
else
y = sv(floor(l/2)+1);
endif
endif
endfunction

a = [4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2];
b = [4.1, 7.2, 1.7, 9.3, 4.4, 3.2];

disp(median2(a)) % 4.4
disp(median(a))
disp(median2(b)) % 4.25
disp(median(b))

## ooRexx

call testMedian .array~of(10, 9, 8, 7, 6, 5, 4, 3, 2, 1)
call testMedian .array~of(10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 0, 0, 0, .11)
call testMedian .array~of(10, 20, 30, 40, 50, -100, 4.7, -11e2)
call testMedian .array~new

::routine testMedian
use arg numbers
say "numbers =" numbers~toString("l", ", ")
say "median =" median(numbers)
say

::routine median
use arg numbers

if numbers~isempty then return 0
-- make a copy so the sort does not alter the
-- original set. This also means this will
-- work with lists and queues as well
numbers = numbers~makearray

-- sort and return the middle element
numbers~sortWith(.numbercomparator~new)
size = numbers~items
-- this handles the odd value too
return numbers[size%2 + size//2]

-- a custom comparator that sorts strings as numeric values rather than
-- strings
::class numberComparator subclass comparator
::method compare
use strict arg left, right
-- perform the comparison on the names. By subtracting
-- the two and returning the sign, we give the expected
-- results for the compares
return (left - right)~sign

## Oz

declare
fun {Median Xs}
Len = {Length Xs}
Mid = Len div 2 + 1 %% 1-based index
Sorted = {Sort Xs Value.'<'}
in
if {IsOdd Len} then {Nth Sorted Mid}
else ({Nth Sorted Mid} + {Nth Sorted Mid-1}) / 2.0
end
end
in
{Show {Median [4.1 5.6 7.2 1.7 9.3 4.4 3.2]}}
{Show {Median [4.1 7.2 1.7 9.3 4.4 3.2]}}

## PARI/GP

Sorting solution.

median(v)={
vecsort(v)[#v\2]
};

Linear-time solution, mostly proof-of-concept but perhaps suitable for large lists.

BFPRT(v,k=#v\2)={
if(#v<15, return(vecsort(v)[k]));
my(u=List(),pivot,left=List(),right=List());
forstep(i=1,#v-4,5,
listput(u,BFPRT([v[i],v[i+1],v[i+2],v[i+3],v[i+4]]))
);
pivot=BFPRT(Vec(u));
u=0;
for(i=1,#v,
if(v[i]<pivot,
listput(left,v[i])
,
listput(right,v[i])
)
);
if(k>#left,
BFPRT(right, k-#left)
,
BFPRT(left, k)
)
};

## Pascal

Works with: Free_Pascal
Program AveragesMedian(output);

type
TDoubleArray = array of double;

procedure bubbleSort(var list: TDoubleArray);
var
i, j, n: integer;
t: double;
begin
n := length(list);
for i := n downto 2 do
for j := 0 to i - 1 do
if list[j] > list[j + 1] then
begin
t := list[j];
list[j] := list[j + 1];
list[j + 1] := t;
end;
end;

function Median(aArray: TDoubleArray): double;
var
lMiddleIndex: integer;
begin
bubbleSort(aArray);
lMiddleIndex := (high(aArray) - low(aArray)) div 2;
if Odd(Length(aArray)) then
Median := aArray[lMiddleIndex + 1]
else
Median := (aArray[lMiddleIndex + 1] + aArray[lMiddleIndex]) / 2;
end;

var
A: TDoubleArray;
i: integer;

begin
randomize;
setlength(A, 7);
for i := low(A) to high(A) do
begin
A[i] := 100 * random;
write (A[i]:7:3, ' ');
end;
writeln;
writeln('Median: ', Median(A):7:3);

setlength(A, 6);
for i := low(A) to high(A) do
begin
A[i] := 100 * random;
write (A[i]:7:3, ' ');
end;
writeln;
writeln('Median: ', Median(A):7:3);
end.

Output:

% ./Median
28.207  74.916  51.695  72.486  51.118   3.241  73.807
Median:  51.695
27.984  89.172   0.250  66.316  41.805  60.043
Median:  50.924

## Perl

Translation of: Python
sub median {
my @a = sort {\$a <=> \$b} @_;
return (\$a[\$#a/2] + \$a[@a/2]) / 2;
}

## Perl 6

Works with: Rakudo version 2016.08
sub median {
my @a = sort @_;
return (@a[(*-1) div 2] + @a[* div 2]) / 2;
}

Notes:

• The div operator does integer division. The / operator (rational number division) would work too, since the array subscript automatically coerces to Int, but using div is more explicit (i.e. clearer to readers) as well as faster, and thus recommended in cases like this.
• The * inside the subscript stands for the array's length (see documentation).

In a slightly more compact way:

sub median { @_.sort[(*-1)/2, */2].sum / 2 }

## Phix

The obvious simple way:

function median(sequence s)
atom res=0
integer l = length(s), k = floor((l+1)/2)
if l then
s = sort(s)
res = s[k]
if remainder(l,2)=0 then
res = (res+s[k+1])/2
end if
end if
return res
end function

It is also possible to use the quick_select routine for a small (20%) performance improvement, which as suggested below may with luck be magnified by retaining any partially sorted results.

function medianq(sequence s)
atom res=0, tmp
integer l = length(s), k = floor((l+1)/2)
if l then
{s,res} = quick_select(s,k)
if remainder(l,2)=0 then
{s,tmp} = quick_select(s,k+1)
res = (res+tmp)/2
end if
end if
return res -- (or perhaps return {s,res})
end function

## PHP

This solution uses the sorting method of finding the median.

function median(\$arr)
{
sort(\$arr);
\$count = count(\$arr); //count the number of values in array
\$middleval = floor((\$count-1)/2); // find the middle value, or the lowest middle value
if (\$count % 2) { // odd number, middle is the median
\$median = \$arr[\$middleval];
} else { // even number, calculate avg of 2 medians
\$low = \$arr[\$middleval];
\$high = \$arr[\$middleval+1];
\$median = ((\$low+\$high)/2);
}
return \$median;
}

echo median(array(4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2)) . "\n"; // 4.4
echo median(array(4.1, 7.2, 1.7, 9.3, 4.4, 3.2)) . "\n"; // 4.25

## PicoLisp

(de median (Lst)
(let N (length Lst)
(if (bit? 1 N)
(get (sort Lst) (/ (inc N) 2))
(setq Lst (nth (sort Lst) (/ N 2)))
(/ (+ (car Lst) (cadr Lst)) 2) ) ) )

(scl 2)
(prinl (round (median (1.0 2.0 3.0))))
(prinl (round (median (1.0 2.0 3.0 4.0))))
(prinl (round (median (5.1 2.6 6.2 8.8 4.6 4.1))))
(prinl (round (median (5.1 2.6 8.8 4.6 4.1))))

Output:

2.00
2.50
4.85
4.60

## PL/I

call sort(A);
n = dimension(A,1);
if iand(n,1) = 1 then /* an odd number of elements */
median = A(n/2);
else /* an even number of elements */
median = (a(n/2) + a(trunc(n/2)+1) )/2;

## PowerShell

This function returns an object containing the minimal amount of statistical data, including Median, and could be modified to take input directly from the pipeline.

All statistical properties could easily be added to the output object.

function Measure-Data
{
[CmdletBinding()]
[OutputType([PSCustomObject])]
Param
(
[Parameter(Mandatory=\$true,
Position=0)]
[double[]]
\$Data
)

Begin
{
function Get-Mode ([double[]]\$Data)
{
if (\$Data.Count -gt (\$Data | Select-Object -Unique).Count)
{
\$groups = \$Data | Group-Object | Sort-Object -Property Count -Descending

return (\$groups | Where-Object {[double]\$_.Count -eq [double]\$groups[0].Count}).Name | ForEach-Object {[double]\$_}
}
else
{
return \$null
}
}

function Get-StandardDeviation ([double[]]\$Data)
{
\$variance = 0
\$average = \$Data | Measure-Object -Average | Select-Object -Property Count, Average

foreach (\$number in \$Data)
{
\$variance += [Math]::Pow((\$number - \$average.Average),2)
}

return [Math]::Sqrt(\$variance / (\$average.Count-1))
}

function Get-Median ([double[]]\$Data)
{
if (\$Data.Count % 2)
{
return \$Data[[Math]::Floor(\$Data.Count/2)]
}
else
{
return (\$Data[\$Data.Count/2], \$Data[\$Data.Count/2-1] | Measure-Object -Average).Average
}
}
}
Process
{
\$Data = \$Data | Sort-Object

\$Data | Measure-Object -Maximum -Minimum -Sum -Average |
Select-Object -Property Count,
Sum,
Minimum,
Maximum,
@{Name='Range'; Expression={\$_.Maximum - \$_.Minimum}},
@{Name='Mean' ; Expression={\$_.Average}} |
Add-Member -MemberType NoteProperty -Name Median -Value (Get-Median \$Data) -PassThru |
Add-Member -MemberType NoteProperty -Name StandardDeviation -Value (Get-StandardDeviation \$Data) -PassThru |
Add-Member -MemberType NoteProperty -Name Mode -Value (Get-Mode \$Data) -PassThru
}
}

\$statistics = Measure-Data 4, 5, 6, 7, 7, 7, 8, 1, 1, 1, 2, 3
\$statistics

Output:
Count             : 12
Sum               : 52
Minimum           : 1
Maximum           : 8
Range             : 7
Mean              : 4.33333333333333
Median            : 4.5
StandardDeviation : 2.67423169368609
Mode              : {1, 7}

Median only:

\$statistics.Median

Output:
4.5

## Prolog

median(L, Z) :-
length(L, Length),
I is Length div 2,
Rem is Length rem 2,
msort(L, S),
maplist(sumlist, [[I, Rem], [I, 1]], Mid),
maplist(nth1, Mid, [S, S], X),
sumlist(X, Y),
Z is Y/2.

## Pure

median x = (/(2-rem)) \$ foldl1 (+) \$ take (2-rem) \$ drop (mid-(1-rem)) \$ sort (<=) x
when len = # x;
mid = len div 2;
rem = len mod 2;
end;
Output:
> median [1, 3, 5];
3.0
> median [1, 2, 3, 4];
2.5

## PureBasic

Procedure.d median(Array values.d(1), length.i)
If length = 0 : ProcedureReturn 0.0 : EndIf
SortArray(values(), #PB_Sort_Ascending)
If length % 2
ProcedureReturn values(length / 2)
EndIf
ProcedureReturn 0.5 * (values(length / 2 - 1) + values(length / 2))
EndProcedure

Protected length.i, i.i
ReDim values(length - 1)
For i = 0 To length - 1
Next
ProcedureReturn i
EndProcedure

Dim floats.d(0)
Restore array1
Debug median(floats(), length)
Restore array2
Debug median(floats(), length)

DataSection
array1:
Data.i 7
Data.d 4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2
array2:
Data.i 6
Data.d 4.1, 7.2, 1.7, 9.3, 4.4, 3.2
EndDataSection

## Python

def median(aray):
srtd = sorted(aray)
alen = len(srtd)
return 0.5*( srtd[(alen-1)//2] + srtd[alen//2])

a = (4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2)
print a, median(a)
a = (4.1, 7.2, 1.7, 9.3, 4.4, 3.2)
print a, median(a)

## R

R has its built-in median function.

Translation of: Octave
omedian <- function(v) {
if ( length(v) < 1 )
NA
else {
sv <- sort(v)
l <- length(sv)
if ( l %% 2 == 0 )
(sv[floor(l/2)+1] + sv[floor(l/2)])/2
else
sv[floor(l/2)+1]
}
}

a <- c(4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2)
b <- c(4.1, 7.2, 1.7, 9.3, 4.4, 3.2)

print(median(a)) # 4.4
print(omedian(a))
print(median(b)) # 4.25
print(omedian(b))

## Racket

#lang racket
(define (median numbers)
(define sorted (list->vector (sort (vector->list numbers) <)))
(define count (vector-length numbers))
(if (zero? count)
#f
(/ (+ (vector-ref sorted (floor (/ (sub1 count) 2)))
(vector-ref sorted (floor (/ count 2))))
2)))

(median '#(5 3 4)) ;; 4
(median '#()) ;; #f
(median '#(5 4 2 3)) ;; 7/2
(median '#(3 4 1 -8.4 7.2 4 1 1.2)) ;; 2.1

## REBOL

median: func [
"Returns the midpoint value in a series of numbers; half the values are above, half are below."
block [any-block!]
/local len mid
][
if empty? block [return none]
block: sort copy block
len: length? block
mid: to integer! len / 2
either odd? len [
][
(block/:mid) + (pick block add 1 mid) / 2
]
]

## REXX

/*REXX program finds the  median  of a  vector  (and displays the  vector  and  median).*/
/* ══════════vector════════════ ══show vector═══ ════════show result═══════════ */
v= 1 9 2 4  ; say "vector" v; say 'median──────►' median(v); say
v= 3 1 4 1 5 9 7 6  ; say "vector" v; say 'median──────►' median(v); say
v= '3 4 1 -8.4 7.2 4 1 1.2'; say "vector" v; say 'median──────►' median(v); say
v= -1.2345678e99 2.3e700 ; say "vector" v; say 'median──────►' median(v); say
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
eSORT: procedure expose @. #; parse arg \$; #= words(\$) /*\$: is the vector. */
do g=1 for #; @.g= word(\$, g); end /*g*/ /*convert list──►array*/
h=# /*#: number elements.*/
do while h>1; h= h % 2 /*cut entries by half.*/
do i=1 for #-h; j= i; k= h + i /*sort lower section. */
do while @.k<@.j; parse value @.j @.k with @.k @.j /*swap.*/
if h>=j then leave; j= j - h; k= k - h /*diminish J and K.*/
end /*while @.k<@.j*/
end /*i*/
end /*while h>1*/ /*end of exchange sort*/
return
/*──────────────────────────────────────────────────────────────────────────────────────*/
median: procedure; call eSORT arg(1); m= # % 2 /*  % is REXX's integer division.*/
n= m + 1 /*N: the next element after M. */
if # // 2 then return @.n /*[odd?] // ◄───REXX's ÷ remainder*/
return (@.m + @.n) / 2 /*process an even─element vector. */
output:
vector: 1 9 2 4
median──────► 3

vector: 3 1 4 1 5 9 7 6
median──────► 4.5

vector: 3 4 1 -8.4 7.2 4 1 1.2
median──────► 2.1

vector: -1.2345678e99  2.3e700
median──────► 1.15000000E+700

## Ring

aList = [5,4,2,3]
see "medium : " + median(aList) + nl

func median aray
srtd = sort(aray)
alen = len(srtd)
if alen % 2 = 0
return (srtd[alen/2] + srtd[alen/2 + 1]) / 2.0
else return srtd[ceil(alen/2)] ok

## Ruby

def median(ary)
return nil if ary.empty?
mid, rem = ary.length.divmod(2)
if rem == 0
ary.sort[mid-1,2].inject(:+) / 2.0
else
ary.sort[mid]
end
end

p median([]) # => nil
p median([5,3,4]) # => 4
p median([5,4,2,3]) # => 3.5
p median([3,4,1,-8.4,7.2,4,1,1.2]) # => 2.1

Alternately:

def median(aray)
srtd = aray.sort
alen = srtd.length
(srtd[(alen-1)/2] + srtd[alen/2]) / 2.0
end

## Run BASIC

sqliteconnect #mem, ":memory:"
mem\$ = "CREATE TABLE med (x float)"
#mem execute(mem\$)

a\$ ="4.1,5.6,7.2,1.7,9.3,4.4,3.2" :gosub [median]
a\$ ="4.1,7.2,1.7,9.3,4.4,3.2" :gosub [median]
a\$ ="4.1,4,1.2,6.235,7868.33" :gosub [median]
a\$ ="1,5,3,2,4" :gosub [median]
a\$ ="1,5,3,6,4,2" :gosub [median]
a\$ ="4.4,2.3,-1.7,7.5,6.6,0.0,1.9,8.2,9.3,4.5"  :gosub [median]'
end
[median]
#mem execute("DELETE FROM med")
for i = 1 to 100
v\$ = word\$( a\$, i, ",")
if v\$ = "" then exit for
mem\$ = "INSERT INTO med values(";v\$;")"
#mem execute(mem\$)
next i
mem\$ = "SELECT AVG(x) as median FROM (SELECT x FROM med
ORDER BY x LIMIT 2 - (SELECT COUNT(*) FROM med) % 2
OFFSET (SELECT (COUNT(*) - 1) / 2
FROM med))"

#mem execute(mem\$)
#row = #mem #nextrow()
median = #row median()
print " Median :";median;chr\$(9);" Values:";a\$

RETURN
Output:
Median :4.4	 Values:4.1,5.6,7.2,1.7,9.3,4.4,3.2
Median :4.25	 Values:4.1,7.2,1.7,9.3,4.4,3.2
Median :4.1	 Values:4.1,4,1.2,6.235,7868.33
Median :3.0	 Values:1,5,3,2,4
Median :3.5	 Values:1,5,3,6,4,2
Median :4.45	 Values:4.4,2.3,-1.7,7.5,6.6,0.0,1.9,8.2,9.3,4.5

## Rust

Sorting, then obtaining the median element:

fn median(mut xs: Vec<f64>) -> f64 {
// sort in ascending order, panic on f64::NaN
xs.sort_by(|x,y| x.partial_cmp(y).unwrap() );
let n = xs.len();
if n % 2 == 0 {
(xs[n/2] + xs[n/2 - 1]) / 2.0
} else {
xs[n/2]
}
}

fn main() {
let nums = vec![2.,3.,5.,0.,9.,82.,353.,32.,12.];
println!("{:?}", median(nums))
}
Output:
9

## Scala

Works with: Scala version 2.8
def median[T](s: Seq[T])(implicit n: Fractional[T]) = {
import n._
val (lower, upper) = s.sortWith(_<_).splitAt(s.size / 2)
if (s.size % 2 == 0) (lower.last + upper.head) / fromInt(2) else upper.head
}

This isn't really optimal. The methods splitAt and last are O(n/2) on many sequences, and then there's the lower bound imposed by the sort. Finally, we call size two times, and it can be O(n).

## Scheme

Translation of: Python

Using Rosetta Code's bubble-sort function

(define (median l)
(* (+ (list-ref (bubble-sort l >) (round (/ (- (length l) 1) 2)))
(list-ref (bubble-sort l >) (round (/ (length l) 2)))) 0.5))

Using SRFI-95:

(define (median l)
(* (+ (list-ref (sort l less?) (round (/ (- (length l) 1) 2)))
(list-ref (sort l less?) (round (/ (length l) 2)))) 0.5))

## Seed7

\$ include "seed7_05.s7i";
include "float.s7i";

const type: floatList is array float;

const func float: median (in floatList: floats) is func
result
var float: median is 0.0;
local
var floatList: sortedFloats is 0 times 0.0;
begin
sortedFloats := sort(floats);
if odd(length(sortedFloats)) then
median := sortedFloats[succ(length(sortedFloats)) div 2];
else
median := 0.5 * (sortedFloats[length(sortedFloats) div 2] +
sortedFloats[succ(length(sortedFloats) div 2)]);
end if;
end func;

const proc: main is func
local
const floatList: flist1 is [] (5.1, 2.6, 6.2, 8.8, 4.6, 4.1);
const floatList: flist2 is [] (5.1, 2.6, 8.8, 4.6, 4.1);
begin
writeln("flist1 median is " <& median(flist1) digits 2 lpad 7); # 4.85
writeln("flist2 median is " <& median(flist2) digits 2 lpad 7); # 4.60
end func;

## Sidef

func median(arry) {
var srtd = arry.sort;
var alen = srtd.length;
srtd[(alen-1)/2]+srtd[alen/2] / 2;
}

## Slate

[email protected](Sequence traits) median
[
s isEmpty
ifTrue: [Nil]
ifFalse:
[| sorted |
sorted: s sort.
sorted length `cache isEven
ifTrue: [(sorted middle + (sorted at: sorted indexMiddle - 1)) / 2]
ifFalse: [sorted middle]]
].
inform: { 4.1 . 5.6 . 7.2 . 1.7 . 9.3 . 4.4 . 3.2 } median.
inform: { 4.1 . 7.2 . 1.7 . 9.3 . 4.4 . 3.2 } median.

## Smalltalk

Works with: GNU Smalltalk
OrderedCollection extend [
median [
self size = 0
ifFalse: [ |s l|
l := self size.
s := self asSortedCollection.
(l rem: 2) = 0
ifTrue: [ ^ ((s at: (l//2 + 1)) + (s at: (l//2))) / 2 ]
ifFalse: [ ^ s at: (l//2 + 1) ]
]
ifTrue: [ ^nil ]
]
].
{ 4.1 . 5.6 . 7.2 . 1.7 . 9.3 . 4.4 . 3.2 } asOrderedCollection
median displayNl.
{ 4.1 . 7.2 . 1.7 . 9.3 . 4.4 . 3.2 } asOrderedCollection
median displayNl.

## Stata

Use summarize to compute the median of a variable (as well as other basic statistics).

set obs 100000
gen x=rbeta(0.2,1.3)
quietly summarize x, detail
display r(p50)

Here is a straightforward implementation using sort.

program calcmedian, rclass sortpreserve
sort `1'
if mod(_N,2)==0 {
return scalar p50=(`1'[_N/2]+`1'[_N/2+1])/2
}
else {
return scalar p50=`1'[(_N-1)/2]
}
end

calcmedian x
display r(p50)

## Tcl

proc median args {
set list [lsort -real \$args]
set len [llength \$list]
# Odd number of elements
if {\$len & 1} {
return [lindex \$list [expr {(\$len-1)/2}]]
}
# Even number of elements
set idx2 [expr {\$len/2}]
set idx1 [expr {\$idx2-1}]
return [expr {
([lindex \$list \$idx1] + [lindex \$list \$idx2])/2.0
}]
}

puts [median 3.0 4.0 1.0 -8.4 7.2 4.0 1.0 1.2]; # --> 2.1

## TI-83 BASIC

Using the built-in function:

median({1.1, 2.5, 0.3241})

## TI-89 BASIC

median({3, 4, 1, -8.4, 7.2, 4, 1, 1})

## Ursala

the simple way (sort first and then look in the middle)

#import std
#import flo

median = fleq-<; @K30K31X eql?\~&rh div\2.+ [email protected]

test program, once with an odd length and once with an even length vector

#cast %eW

examples =

median~~ (
<9.3,-2.0,4.0,7.3,8.1,4.1,-6.3,4.2,-1.0,-8.4>,
<8.3,-3.6,5.7,2.3,9.3,5.4,-2.3,6.3,9.9>)

output:

(4.050000e+00,5.700000e+00)

## Vala

Requires --pkg posix -X -lm compilation flags in order to use POSIX qsort, and to have access to math library.

int compare_numbers(void* a_ref, void* b_ref) {
double a = *(double*) a_ref;
double b = *(double*) b_ref;
return a > b ? 1 : a < b ? -1 : 0;
}

double median(double[] elements) {
double[] clone = elements;
Posix.qsort(clone, clone.length, sizeof(double), compare_numbers);
double middle = clone.length / 2.0;
int first = (int) Math.floor(middle);
int second = (int) Math.ceil(middle);
return (clone[first] + clone[second]) / 2;
}
void main() {
double[] array1 = {2, 4, 6, 1, 7, 3, 5};
double[] array2 = {2, 4, 6, 1, 7, 3, 5, 8};
print(@"\$(median(array1)) \$(median(array2))\n");
}

## Vedit macro language

This is a simple implementation for positive integers using sorting. The data is stored in current edit buffer in ascii representation. The values must be right justified.

The result is returned in text register @10. In case of even number of items, the lower middle value is returned.

Sort(0, File_Size, NOCOLLATE+NORESTORE)
EOF Goto_Line(Cur_Line/2)
Reg_Copy(10, 1)

## Wortel

@let {
; iterative
med1 &l @let {a @sort l s #a i @/s 2 ?{%%s 2 ~/ 2 +`-i 1 a `i a `i a}}

; tacit
med2 ^(\~/2 @sum @(^(\&![#~f #~c] \~/2 \~-1 #) @` @id) @sort)

[[
!med1 [4 2 5 2 1]
!med1 [4 5 2 1]
!med2 [4 2 5 2 1]
!med2 [4 5 2 1]
]]
}
Returns:
[2 3 2 3]

## Yabasic

Translation of: Lua
sub floor(x)
return int(x + .05)
end sub

sub ceil(x)
if x > int(x) x = x + 1
return x
end sub

SUB ASort\$(matriz\$())
local last, gap, first, tempi\$, tempj\$, i, j

last = arraysize(matriz\$(), 1)

gap = floor(last / 10) + 1
while(TRUE)
first = gap + 1
for i = first to last
tempi\$ = matriz\$(i)
j = i - gap
while(TRUE)
tempj\$ = matriz\$(j)
if (tempi\$ >= tempj\$) then
j = j + gap
break
end if
matriz\$(j+gap) = tempj\$
if j <= gap then
break
end if
j = j - gap
wend
matriz\$(j) = tempi\$
next i
if gap = 1 then
return
else
gap = floor(gap / 3.5) + 1
end if
wend
END SUB

sub median(numlist\$)
local numlist\$(1), n

n = token(numlist\$, numlist\$(), ", ")

ASort\$(numlist\$())

if mod(n, 2) = 0 then return (val(numlist\$(n / 2)) + val(numlist\$(n / 2 + 1))) / 2 end if
return val(numlist\$(ceil(n / 2)))
end sub

print median("4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2") // 4.4
print median("4.1, 7.2, 1.7, 9.3, 4.4, 3.2") // 4.25

## zonnon

module Averages;

type
Vector = array {math} * of real;

procedure Partition(var a: Vector; left, right: integer): integer;
var
pValue,aux: real;
store,i,pivot: integer;
begin
pivot := right;
pValue := a[pivot];
aux := a[right];a[right] := a[pivot];a[pivot] := aux; (* a[pivot] <-> a[right] *)
store := left;
for i := left to right -1 do
if a[i] <= pValue then
aux := a[store];a[store] := a[i];a[i]:=aux;
inc(store)
end
end;
aux := a[right];a[right] := a[store]; a[store] := aux;
return store
end Partition;

(* QuickSelect algorithm *)
procedure Select(a: Vector; left,right,k: integer;var r: real);
var
pIndex, pDist : integer;
begin
if left = right then r := a[left]; return end;
pIndex := Partition(a,left,right);
pDist := pIndex - left + 1;
if pDist = k then
r := a[pIndex];return
elsif k < pDist then
Select(a,left, pIndex - 1, k, r)
else
Select(a,pIndex + 1, right, k - pDist, r)
end
end Select;

procedure Median(a: Vector): real;
var
idx: integer;
r1,r2 : real;
begin
idx := len(a) div 2 + 1;
r1 := 0.0;r2 := 0.0;
Select(a,0,len(a) - 1,idx,r1);
if odd(len(a)) then return r1 end;
Select(a,0,len(a) - 1,idx - 1,r2);
return (r1 + r2) / 2;
end Median;

var
ary: Vector;
r: real;

begin
ary := new Vector(3);
ary := [5.0,3.0,4.0];
writeln(Median(ary):10:2);
ary := new Vector(4);
ary := [5.0,4.0,2.0,3.0];
writeln(Median(ary):10:2);
ary := new Vector(8);
ary := [3.0,4.0,1.0,-8.4,7.2,4.0,1.0,1.2];
writeln(Median(ary):10:2)
end Averages.

4
3,5
2,1

## zkl

Using the Quickselect algorithm#zkl for O(n) time:

var quickSelect=Import("quickSelect").qselect;

fcn median(xs){
n:=xs.len();
if (n.isOdd) return(quickSelect(xs,n/2));
( quickSelect(xs,n/2-1) + quickSelect(xs,n/2) )/2;
}
median(T( 5.1, 2.6, 6.2, 8.8, 4.6, 4.1 )); //-->4.85
median(T( 5.1, 2.6, 8.8, 4.6, 4.1 )); //-->4.6