Averages/Median
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.
You are encouraged to solve this task according to the task description, using any language you may know.
There are several approaches to this. One is to sort the elements, and then pick the one in the middle. Sorting would take at least O(n log n). Another would be to build a priority queue from the elements, and then extract half of the elements to get to the middle one(s). This would also take O(n log n). The best solution is to use the selection algorithm to find the median in O(n) time.
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>
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 (*(float *)a > *(float *)b) return 1; else if (*(float *)a < *(float *)b) return -1; else return 0;
}
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>
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>
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>
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>
Java
Sorting: <lang java5>// Note: this function modifies the input list public static double median(List<Double> list){
Collections.sort(list); if(list.size() % 2 == 0){ return (list.get(list.size() / 2) + list.get(list.size() / 2 - 1)) / 2; } return list.get((list.size()-1)/2);
}</lang>
Using priority queue: <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>
Mathematica
Built-in function: <lang Mathematica>
Median[{1, 5, 3, 2, 4}] Median[{1, 5, 3, 6, 4, 2}]
</lang> gives back: <lang Mathematica>
3 7/2
</lang> 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> gives back: <lang Mathematica>
3 7/2
</lang>
OCaml
<lang ocaml>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>
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>
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>
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>