Averages/Median: Difference between revisions
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end program Median_Test</lang> |
end program Median_Test</lang> |
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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, <lang Fortran> K = N/2 |
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MEDIAN = FINDELEMENT(K + 1,A,N) |
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IF (MOD(N,2).EQ.0) MEDIAN = (FINDELEMENT(K,A,N) + MEDIAN)/2 </lang> |
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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). |
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=={{header|FreeBASIC}}== |
=={{header|FreeBASIC}}== |
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Revision as of 04:34, 7 June 2017
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.
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Ada
<lang ada>with Ada.Text_IO, Ada.Float_Text_IO;
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;
Ada.Text_IO.Put( "Median value: " );
Ada.Float_Text_IO.Put( median_val );
Ada.Text_IO.New_line;
end FindMedian;</lang>
ALGOL 68
<lang algol68>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))</lang> Sample output:
length: 97738 median: -2.52550126608709e +3 <: 48868 >: 48870 =: 0
AntLang
AntLang has a built-in median function. <lang AntLang>median[list]</lang>
APL
<lang APL>median←{v←⍵[⍋⍵]⋄.5×v[⌈¯1+.5×⍴v]+v[⌊.5×⍴v]} ⍝ Assumes ⎕IO←0</lang>
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: <lang APL>⎕IO←0</lang>
If you prefer an index origin of 1, use this code instead: <lang APL> ⎕IO←1 median←{v←⍵[⍋⍵] ⋄ 0.5×v[⌈0.5×⍴v]+v[⌊1+0.5×⍴v]} </lang>
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
<lang AppleScript>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</lang>
Composing functionally
Using a quick select algorithm:
<lang AppleScript>-- 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</lang>
- Output:
<lang AppleScript>{missing value, 4, 3.5, 2.1}</lang>
Applesoft BASIC
<lang 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</lang>Example:<lang ApplesoftBASIC>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</lang>Output:
5.95
AutoHotkey
Takes the lower of the middle two if length is even <lang AutoHotkey>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%
}</lang>
AWK
AWK arrays can be passed as parameters, but not returned, so they are usually global.
<lang awk>#!/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 ""
} </lang>
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
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.
<lang qbasic>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</lang>
See also: BBC BASIC, Liberty BASIC, PureBasic, TI-83 BASIC, TI-89 BASIC.
BBC BASIC
<lang bbcbasic> INSTALL @lib$+"SORTLIB"
Sort% = FN_sortinit(0,0)
DIM a(6), b(5)
a() = 4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2
b() = 4.1, 7.2, 1.7, 9.3, 4.4, 3.2
PRINT "Median of a() is " ; FNmedian(a())
PRINT "Median of b() is " ; FNmedian(b())
END
DEF FNmedian(a())
LOCAL C%
C% = DIM(a(),1) + 1
CALL Sort%, a(0)
= (a(C% DIV 2) + a((C%-1) DIV 2)) / 2
</lang> Output:
Median of a() is 4.4 Median of b() is 4.25
Bracmat
Bracmat has no floating point numbers, so we have to parse floating point numbers as strings and convert them to rational numbers. Each number is packaged in a little list and these lists are accumulated in a sum. Bracmat keeps sums sorted, so the median is the term in the middle of the list, or the average of the two terms in the middle of the list.
<lang bracmat>(median=
begin decimals end int list med med1 med2 num number
. 0:?list
& whl
' ( @( !arg
: ?
((%@:~" ":~",") ?:?number)
((" "|",") ?arg|:?arg)
)
& @( !number
: ( #?int "." [?begin #?decimals [?end
& !int+!decimals*10^(!begin+-1*!end):?num
| ?num
)
)
& (!num.)+!list:?list
)
& !list:?+[?end
& ( !end*1/2:~/
& !list:?+[!(=1/2*!end+-1)+(?med1.)+(?med2.)+?
& !med1*1/2+!med2*1/2:?med
| !list:?+[(div$(1/2*!end,1))+(?med.)+?
)
& !med
);</lang>
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
<lang 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;
}</lang>
Quickselect algorithm
Average O(n) time: <lang c>#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;
} </lang>
Output: <lang c>length: 992021 median: 0.000473 <: 496010 >: 496010 =: 1</lang>
C++
This function runs in linear time on average. <lang cpp>#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;
}</lang>
C#
<lang csharp>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);
Console.ReadLine();
}
}
} </lang>
Clojure
Simple: <lang lisp>(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))))</lang>
COBOL
Intrinsic function: <lang cobol>FUNCTION MEDIAN(some-table (ALL))</lang>
Common Lisp
The recursive partitioning solution, without the median of medians optimization.
<lang lisp>((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))</lang>
D
<lang 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);
}</lang>
- Output:
Even median: 4.85 Odd median: 4.6
Delphi
<lang 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.</lang>
E
TODO: Use the selection algorithm, whatever that is
<lang e>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
}
}</lang>
<lang e>? median([1,9,2])
- value: 2
? median([1,9,2,4])
- value: 3.0</lang>
EchoLisp
<lang scheme> (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
</lang>
Elena
ELENA 3.x : <lang elena>#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. ].
].
]
}
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 writeLine:"median of (":a1:") is ":(a1 median).
console writeLine:"median of (":a2:") is ":(a2 median).
console readChar.
].</lang>
- 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
<lang elixir>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)</lang>
- 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
<lang 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.</lang>
ERRE
<lang> 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 </lang> 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.
<lang Euler Math Toolbox> >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
</lang>
Euphoria
<lang 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 })</lang>
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 :
<lang excel> =MEDIAN(A1:A10) </lang>
Assuming 10 values will be entered, alternatively, you can just type
<lang excel> =MEDIAN( </lang> 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
<lang> 23 11,5 21 12 3 19 7 23 11 9 0 </lang>
F#
Median of Medians algorithm implementation <lang fsharp> 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'
</lang>
Factor
The quicksort-style solution, with random pivoting. Takes the lesser of the two medians for even sequences. <lang factor>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 bi@ 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 ;</lang>
Usage: <lang factor>( scratchpad ) 11 iota median . 5 ( scratchpad ) 10 iota median . 4</lang>
Forth
This uses the O(n) algorithm derived from quicksort. <lang forth>-1 cells constant -cell
- cell- -cell + ;
defer lessthan ( a@ b@ -- ? ) ' < 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 @ r@ lessthan while cell+ repeat swap begin r@ 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 @ ;</lang>
<lang forth>create test 4 , 2 , 1 , 3 , 5 ,
test 4 median . \ 2 test 5 median . \ 3</lang>
Fortran
<lang fortran>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
! it uses integer instead.
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</lang>
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, <lang Fortran> K = N/2
MEDIAN = FINDELEMENT(K + 1,A,N)
IF (MOD(N,2).EQ.0) MEDIAN = (FINDELEMENT(K,A,N) + MEDIAN)/2 </lang>
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
<lang 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</lang>
- Output:
Median for all 10 elements : 4.45 Median for first 9 elements : 4.4
GAP
<lang 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</lang>
Go
<lang go>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
}</lang>
Groovy
Solution (brute force sorting, with arithmetic averaging of dual midpoints (even sizes)): <lang groovy>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
}</lang>
Test: <lang groovy>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]})""" }</lang>
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([])
Haskell
This uses a quick select algorithm and runs in expected O(n) time. <lang haskell>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</lang>
- Output:
(not defined) 7.0 4.0 3.5 2.1
Or
<lang haskell>> Math.Statistics.median [1,9,2,4] 3.0</lang>
HicEst
If the input has an even number of elements, median is the mean of the middle two values: <lang HicEst>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</lang>
Icon and Unicon
A quick and dirty solution: <lang>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</lang>
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:
<lang j> require 'stats/base'
median 1 9 2 4
3</lang> The definition given in the addon script is: <lang j>midpt=: -:@<:@# median=: -:@(+/)@((<. , >.)@midpt { /:~)</lang>
If, for an even number of elements, both values were desired when those two values are distinct, then the following implementation would suffice: <lang j> median=: ~.@(<. , >.)@midpt { /:~
median 1 9 2 4
2 4</lang>
Java
Sorting: <lang java5>// 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;
}</lang>
Using priority queue (which sorts under the hood): <lang java5>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++)
pq.poll(); // discard first half
if (n % 2 != 0) // odd length
return pq.poll();
else
return (pq.poll() + pq.poll()) / 2.0;
}</lang>
JavaScript
ES5
<lang javascript>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</lang>
ES6
Using a quick select algorithm
<lang JavaScript>(() => {
'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);
})();</lang>
- Output:
<lang JavaScript>[
null, 4, 3.5, 2.1
]</lang>
jq
<lang 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 ;</lang>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 <lang sh>[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]</lang>Then invoking the jq program yields a stream of values:<lang sh>$ jq -f median.jq in.dat 4.4 4.25</lang>
Julia
Julia has a built-in median() function <lang julia> function median2(n) s = sort(n) len = length(n) len%2 == 0 && return (s[ifloor(len/2)+1] + s[ifloor(len/2)])/2 return s[ifloor(len/2)+1] 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] @assert median(a) == median2(a) @assert median(b) == median2(b)</lang>
K
<lang k>
med:{a:x@<x; i:(#a)%2; :[(#a)!2; a@i; {(+/x)%#x} a@i,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 </lang>
Kotlin
<lang scala>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
}</lang>
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 )" <lang Lasso>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</lang>
Liberty BASIC
<lang lb>
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
</lang>
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
LiveCode
LC has median as a built-in function <lang LiveCode>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</lang>
To make our own, we need own own floor function first
<lang LiveCode>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</lang>
LSL
<lang 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));
}
}</lang> 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
<lang 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}))</lang>
Maple
Builtin
This works for numeric lists or arrays, and is designed for large data sets. <lang Maple> > Statistics:-Median( [ 1, 5, 3, 2, 4 ] );
3.
> Statistics:-Median( [ 1, 5, 3, 6, 2, 4 ] );
3.50000000000000
</lang>
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. <lang Maple> median1 := proc()
local L := sort( [ args ] );
( L[ iquo( 1 + nargs, 2 ) ] + L[ 1 + iquo( nargs, 2 ) ] ) / 2
end proc: </lang> For example: <lang Maple> > median1( 1, 5, 3, 2, 4 ); # 3
3
> median1( 1, 5, 3, 6, 4, 2 ); # 7/2
7/2
</lang>
Mathematica / Wolfram Language
Built-in function: <lang Mathematica>Median[{1, 5, 3, 2, 4}] Median[{1, 5, 3, 6, 4, 2}]</lang>
- Output:
3 7/2
Custom function: <lang Mathematica>mymedian[x_List]:=Module[{t=Sort[x],L=Length[x]},
If[Mod[L,2]==0, (tL/2+tL/2+1)/2 , t(L+1)/2 ]
]</lang> Example of custom function: <lang Mathematica>mymedian[{1, 5, 3, 2, 4}] mymedian[{1, 5, 3, 6, 4, 2}]</lang>
- Output:
3 7/2
MATLAB
If the input has an even number of elements, function returns the mean of the middle two values: <lang Matlab>function medianValue = findmedian(setOfValues)
medianValue = median(setOfValues);
end</lang>
Maxima
<lang maxima>/* built-in */ median([41, 56, 72, 17, 93, 44, 32]); /* 44 */ median([41, 72, 17, 93, 44, 32]); /* 85/2 */</lang>
MUMPS
<lang 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)
</lang>
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
<lang NetRexx>/* 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
</lang> Output:
Meadian: 10.000000 , Vector: [10.0] Meadian: 5.500000 , Vector: [10.0,9.0,8.0,7.0,6.0,5.0,4.0,3.0,2.0,1.0] Meadian: 5.000000 , Vector: [9,8,7,6,5,4,3,2,1] Meadian: 3.000000 , Vector: [1.0,9,2.0,4.0] Meadian: 4.500000 , Vector: [3.0,1,4,1.0,5.0,9,7.0,6.0] Meadian: 2.100000 , Vector: [3,4,1,-8.4,7.2,4,1,1.2] Meadian: 1.150000E+700, Vector: [-1.2345678E+99,2.3e+700] Meadian: 4.400000 , Vector: [4.1,5.6,7.2,1.7,9.3,4.4,3.2] Meadian: 4.250000 , Vector: [4.1,7.2,1.7,9.3,4.4,3.2] Meadian: 51.695000 , Vector: [28.207,74.916,51.695,72.486,51.118,3.241,73.807] Meadian: 50.924000 , Vector: [27.984,89.172,0.250,66.316,41.805,60.043] Meadian: 4.850000 , Vector: [5.1,2.6,6.2,8.8,4.6,4.1] Meadian: 4.600000 , Vector: [5.1,2.6,8.8,4.6,4.1] Meadian: 4.450000 , Vector: [4.4,2.3,-1.7,7.5,6.6,0.0,1.9,8.2,9.3,4.5] Meadian: 3.000000 , Vector: [10,9,8,7,6,5,4,3,2,1,0,0,0,0,0.11] Meadian: 15.000000 , Vector: [10,20,30,40,50,-100,4.7,-1100] Meadian: 4.050000 , Vector: [9.3,-2.0,4.0,7.3,8.1,4.1,-6.3,4.2,-1.0,-8.4] Meadian: 5.700000 , Vector: [8.3,-3.6,5.7,2.3,9.3,5.4,-2.3,6.3,9.9]
NewLISP
<lang 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)</lang>
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
<lang nim>import algorithm, strutils
proc median(xs): 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)</lang>
Oberon-2
Oxford Oberon-2 <lang oberon2> 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.
</lang>
Output:
0.00 4.00 3.50 2.10
Objeck
<lang 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);
}
}
} </lang>
OCaml
<lang 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;;</lang>
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. <lang octave>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))</lang>
ooRexx
<lang 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
</lang>
Oz
<lang 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]}}</lang>
PARI/GP
Sorting solution. <lang parigp>median(v)={
vecsort(v)[#v\2]
};</lang>
Linear-time solution, mostly proof-of-concept but perhaps suitable for large lists. <lang parigp>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) ) };</lang>
Pascal
<lang 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.</lang> 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
<lang perl>sub median {
my @a = sort {$a <=> $b} @_;
return ($a[$#a/2] + $a[@a/2]) / 2;
}</lang>
Perl 6
<lang perl6>sub median {
my @a = sort @_; return (@a[(*-1) div 2] + @a[* div 2]) / 2;
}</lang>
Notes:
- The div operator does integer division. The / operator (rational number division) would work too, since the array subscript automatically coerces to Int, but using div is more explicit (i.e. clearer to readers) as well as faster, and thus recommended in cases like this.
- The * inside the subscript stands for the array's length (see documentation).
In a slightly more compact way:
<lang perl6>sub median { @_.sort[(*-1)/2, */2].sum / 2 }</lang>
Phix
The obvious simple way: <lang Phix>function median(sequence s) atom res=0 integer l = length(s), k = floor((l+1)/2)
if l then
s = sort(s)
res = s[k]
if remainder(l,2)=0 then
res = (res+s[k+1])/2
end if
end if
return res
end function</lang> It is also possible to use the quick_select routine for a small (20%) performance improvement, which as suggested below may with luck be magnified by retaining any partially sorted results. <lang Phix>function medianq(sequence s) atom res=0, tmp integer l = length(s), k = floor((l+1)/2)
if l then
{s,res} = quick_select(s,k)
if remainder(l,2)=0 then
{s,tmp} = quick_select(s,k+1)
res = (res+tmp)/2
end if
end if
return res -- (or perhaps return {s,res})
end function</lang>
PHP
This solution uses the sorting method of finding the median. <lang php> function median($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 $median;
}
echo median(array(4.1, 5.6, 7.2, 1.7, 9.3, 4.4, 3.2)) . "\n"; // 4.4 echo median(array(4.1, 7.2, 1.7, 9.3, 4.4, 3.2)) . "\n"; // 4.25 </lang>
PicoLisp
<lang PicoLisp>(de median (Lst)
(let N (length Lst)
(if (bit? 1 N)
(get (sort Lst) (/ (inc N) 2))
(setq Lst (nth (sort Lst) (/ N 2)))
(/ (+ (car Lst) (cadr Lst)) 2) ) ) )
(scl 2) (prinl (round (median (1.0 2.0 3.0)))) (prinl (round (median (1.0 2.0 3.0 4.0)))) (prinl (round (median (5.1 2.6 6.2 8.8 4.6 4.1)))) (prinl (round (median (5.1 2.6 8.8 4.6 4.1))))</lang> Output:
2.00 2.50 4.85 4.60
PL/I
<lang pli>call sort(A); n = dimension(A,1); if iand(n,1) = 1 then /* an odd number of elements */
median = A(n/2);
else /* an even number of elements */
median = (a(n/2) + a(trunc(n/2)+1) )/2;</lang>
PowerShell
This function returns an object containing the minimal amount of statistical data, including Median, and could be modified to take input directly from the pipeline.
All statistical properties could easily be added to the output object.
<lang PowerShell> function Measure-Data {
[CmdletBinding()]
[OutputType([PSCustomObject])]
Param
(
[Parameter(Mandatory=$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
}
} </lang> <lang PowerShell> $statistics = Measure-Data 4, 5, 6, 7, 7, 7, 8, 1, 1, 1, 2, 3 $statistics </lang>
- 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: <lang PowerShell> $statistics.Median </lang>
- Output:
4.5
Prolog
<lang 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.</lang>
Pure
Inspired by the Haskell version. <lang 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;</lang>
Output:
> median [1, 3, 5]; 3.0 > median [1, 2, 3, 4]; 2.5
PureBasic
<lang 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
Procedure.i readArray(Array values.d(1))
Protected length.i, i.i Read.i length ReDim values(length - 1) For i = 0 To length - 1 Read.d values(i) Next ProcedureReturn i
EndProcedure
Dim floats.d(0) Restore array1 length.i = readArray(floats()) Debug median(floats(), length) Restore array2 length.i = readArray(floats()) 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</lang>
Python
<lang 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)</lang>
R
R has its built-in median function.
<lang rsplus>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))</lang>
Racket
<lang 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</lang>
REBOL
<lang 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 [
pick block add 1 mid
][
(block/:mid) + (pick block add 1 mid) / 2
]
] </lang>
REXX
<lang 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>l*/ /*end of exchange sort*/
return
/*──────────────────────────────────────────────────────────────────────────────────────*/ median: procedure; call eSORT arg(1) /*obtain the elements of the vector.*/
m=# % 2 /* % is REXX's integer division.*/
n=m+1 /*N: the next element after M. */
if #//2 then return @.n /*(odd?) // is REXX's ÷ remainder.*/
return (@.m + @.n) / 2 /*process an even─element vector. */</lang>
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
<lang 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
</lang>
Ruby
<lang 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</lang>
Alternately: <lang ruby>def median(aray)
srtd = aray.sort alen = srtd.length (srtd[(alen-1)/2] + srtd[alen/2]) / 2.0
end</lang>
Run BASIC
<lang Runbasic>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</lang>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:
<lang rust>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))
}</lang>
- Output:
9
Scala
(See the Scala discussion on Mean for more information.)
<lang scala>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
}</lang>
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
Using Rosetta Code's bubble-sort function <lang Scheme>(define (median l)
(* (+ (list-ref (bubble-sort l >) (round (/ (- (length l) 1) 2)))
(list-ref (bubble-sort l >) (round (/ (length l) 2)))) 0.5))</lang>
Using SRFI-95: <lang Scheme>(define (median l)
(* (+ (list-ref (sort l less?) (round (/ (- (length l) 1) 2)))
(list-ref (sort l less?) (round (/ (length l) 2)))) 0.5))</lang>
Seed7
<lang 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;</lang>
Sidef
<lang ruby>func median(arry) {
var srtd = arry.sort; var alen = srtd.length; srtd[(alen-1)/2]+srtd[alen/2] / 2;
}</lang>
Slate
<lang slate>s@(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]]
].</lang>
<lang slate>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.</lang>
Smalltalk
<lang 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 ]
]
].</lang>
<lang smalltalk>{ 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.</lang>
Tcl
<lang 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 $idx2])/2.0
}]
}
puts [median 3.0 4.0 1.0 -8.4 7.2 4.0 1.0 1.2]; # --> 2.1</lang>
TI-83 BASIC
Using the built-in function: <lang ti83b>median({1.1, 2.5, 0.3241})</lang>
TI-89 BASIC
<lang ti89b>median({3, 4, 1, -8.4, 7.2, 4, 1, 1})</lang>
Ursala
the simple way (sort first and then look in the middle) <lang Ursala>#import std
- import flo
median = fleq-<; @K30K31X eql?\~&rh div\2.+ plus@lzPrhPX</lang> test program, once with an odd length and once with an even length vector <lang Ursala>#cast %eW
examples =
median~~ (
<9.3,-2.0,4.0,7.3,8.1,4.1,-6.3,4.2,-1.0,-8.4>, <8.3,-3.6,5.7,2.3,9.3,5.4,-2.3,6.3,9.9>)</lang>
output:
(4.050000e+00,5.700000e+00)
Vala
Requires --pkg posix -X -lm compilation flags in order to use POSIX qsort, and to have access to math library.
<lang vala>int compare_numbers(void* a_ref, void* b_ref) {
double a = *(double*) a_ref; double b = *(double*) b_ref; return a > b ? 1 : a < b ? -1 : 0;
}
double median(double[] elements) {
double[] clone = elements; Posix.qsort(clone, clone.length, sizeof(double), compare_numbers); double middle = clone.length / 2.0; int first = (int) Math.floor(middle); int second = (int) Math.ceil(middle); return (clone[first] + clone[second]) / 2;
} void main() {
double[] array1 = {2, 4, 6, 1, 7, 3, 5};
double[] array2 = {2, 4, 6, 1, 7, 3, 5, 8};
print(@"$(median(array1)) $(median(array2))\n");
}</lang>
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.
<lang vedit>Sort(0, File_Size, NOCOLLATE+NORESTORE) EOF Goto_Line(Cur_Line/2) Reg_Copy(10, 1)</lang>
Wortel
<lang 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] ]]
}</lang>
Returns:
[2 3 2 3]
zkl
Using the Quickselect algorithm#zkl for O(n) time: <lang zkl>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;
}</lang> <lang zkl>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</lang>
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