Calculate the average of a series
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[edit] C
This particular function averages all integers until a INT_MAX is passed as an argument. INT_MAX is used because it is a very large number that is most likely to not be naturally part of an average. Since INT_MAX is defined by nearly all compilers, you should be able to call this function relatively easily, e.g. average(4,5,6,INT_MAX);
//stdarg.h must be included for this type of function to be declared
#include <stdarg.h>
double average( int first, ... ) {
int count = 0, sum = 0, i = first;
va_list marker;
va_start( marker, first ); /* Initialize variable arguments. */
while( i != INT_MAX )
{
sum += i; //increases the sum
count++; //increases the count
i = va_arg( marker, int); //Gets the next argument
}
va_end( marker ); // Resets the list
return( sum ? ((double)sum / count) : 0.0 );
}
[edit] C++
#include <numeric>
template <typename Iterator>
double average(Iterator begin, Iterator end) {
return std::accumulate(begin, end, 0.0) / (end - begin);
}
[edit] Common Lisp
> (defun average (list) (/ (reduce #'+ list) (length list))) AVERAGE > (average '(1 2 3 4)) 5/2 > (average '(1 2 3 4.0)) 2.5
[edit] Assembly
[edit] Integers
Following the description for the ANSI C version, except that INT_MAX here is always 2147483647 (0x7FFFFFFF). Formatted in MASM, only returns an integer value.
avg proc mov edx,esp add edx,4 avg_loop1: add eax,DWORD PTR [edx] add edx,4 cmp DWORD PTR [edx],7fffffffh jne avg_loop1 mov ecx,DWORD PTR [esp] xchg esp,edx sub edx,esp add esp,4 push ecx neg edx shr edx,2 mov ecx,edx xor edx,edx sub ecx,1 div ecx ret avg endp
[edit] Using the FPU
This leaves the result in st(0) - caller has to free it.
avg proc mov ecx,DWORD PTR [esp] add esp,4 mov edx,esp fldz avg_loop1: fild DWORD PTR [esp] fadd add esp,4 cmp DWORD PTR [esp],7fffffffh jne avg_loop1 sub edx,esp neg edx shr edx,2 mov DWORD PTR [esp],edx fild DWORD PTR [esp] fdiv mov DWORD PTR [esp],ecx ret avg endp
[edit] Haskell
Sum a list of numbers and divide by the length of the list:
avg l = sum l / List.genericLength l
[edit] Java
This implementation can only be used with java version 1.5.0 (a.k.a java 5.0) since this new version implements varargs.
import java.util.*;
public class Averages {
public static int average(int... values) {
double sum = 0;
for (int i : values) {
if (i == Integer.MAX_VALUE) break;
sum += i;
}
return (int) Math.round(sum/(values.length-1)); //Math.round is used for rounding
}
public static void main(String[] args) {
System.out.printf("Average: %d: ", average(1,2,4,40,Integer.MAX_VALUE));
}
}
[edit] OCaml
Sum a list of integers and divide by the length of the list:
# let avg l = float(List.fold_left ( + ) 0 l) /. float(List.length l);; val avg : int list -> float = <fun>
For example:
# avg [1;2;4;8;16];; - : float = 6.2
[edit] Perl
use List::Util 'sum';
# Pass the list directly as the argument list
print average(1 .. 3);
sub average {
my @array = @_;
return sum(@array) / @array;
}
[edit] Python
This python function returns the average of an arbitrary sequence, a wikipedia:Generator (computer science) too:
def average(seq, total=0.0):
num = 0
for item in seq:
total += item
num += 1
return total / num
If seq is sequence type such as a list or a tuple, you can calculate its average just like this:
def average(seq): return float(sum(seq)) / len(seq)
The float() call is required to coerce the division into being a floating point, and not integer division. In python 3 all numeric divisions will be floating point divisions, and the // operator will be used to mean purely integer division.
[edit] Ruby
def average(sequence) sequence.inject(:+).to_f / sequence.length end
(Although it Ruby also knows @for ... in@, you should not use it unless you know why you're doing it.)
[edit] Scheme
(define (average list) (/ (apply + list) (length list)))
[edit] REXX
/* Calculating average in REXX */ sum = 0.0 n = 0 do loop until figure = 0 pull figure sum = sum + figure n = n + 1 end do n = n - 1 average = sum / n say "Average =" average
[edit] Zsh
average() {
local -a array
local -F1 sum
array=({1..9})
(( sum = (${(j:+:)array}.0) / ${(w)#array} ))
print "$sum"
}
[edit] Excel
=AVERAGE([cells])
