Averages/Arithmetic mean

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

Write a program to find the mean (arithmetic average) of a numeric vector. The program should work on a zero-length vector (with an answer of 0).

Ada

This example shows how to pass a zero length vector as well as a larger vector.

with Ada.Float_Text_Io; use Ada.Float_Text_Io;
with Ada.Text_IO; use Ada.Text_IO;

procedure Mean_Main is
   type Vector is array(Positive range <>) of Float;
   function Mean(Item : Vector) return Float is
      Sum : Float := 0.0;
      Result : Float := 0.0;
   begin
      for I in Item'range loop
         Sum := Sum + Item(I);
      end loop;
      if Item'Length > 0 then
         Result := Sum / Float(Item'Length);
      end if;
      return Result;
   end Mean;
   A : Vector := (3.0, 1.0, 4.0, 1.0, 5.0, 9.0);
begin
   Put(Item => Mean(A), Fore => 1, Exp => 0);
   New_Line;
   -- test for zero length vector
   Put(Item => Mean(A(1..0)), Fore => 1, Exp => 0);
   New_Line;
end Mean_Main;

Output:

3.83333
0.00000

BASIC

Works with: QuickBasic version 4.5

Assume the numbers are in a DIM named nums.

mean = 0
sum = 0;
FOR i = LBOUND(nums) TO UBOUND(nums)
   sum = sum + nums(i);
NEXT i
size = UBOUND(nums) - LBOUND(nums) + 1
PRINT "The mean is: ";
IF size <> 0 THEN
   PRINT (sum / size)
ELSE
   PRINT 0
END IF

C++

Library: STL

double mean(std::vector<double> const& vNumbers)
{
     double sum = 0;
     for( std::vector<double>::iterator i = vNumbers.begin(); vNumbers.end() != i; ++i )
          sum += *i;

     if( 0 == vNumbers.size() )
          return 0;
     else
          return sum / vNumbers.size();
}

Shorter (and more idiomatic) version:

#include <vector>
#include <algorithm>

double mean(std::vector<double> const& numbers)
{
  if (numbers.empty())
    return 0;
  return std::accumulate(numbers.begin(), numbers.end(), 0.0) / numbers.size();
}

Common Lisp

(defun mean (sequence)
  (let ((length (length sequence)))
    (if (zerop length)
      0
      (/ (reduce #'+ sequence) length))))

D

Using template to make the mean function work for higher-rank array.

module mean ;
import std.stdio ;

real mean(T)(T[] a) {
  static if(is(T U : U[])) {
    // recursively unfold the multi-array
    T u ;
    foreach(e ; a)
      u ~= e ;
    return u.mean() ;   

  } else {
    // do the math
    if(a.length == 0) return 0.0 ;
    real sum = 0.0 ;
    foreach(e ; a)
      sum += e ;
    return sum / a.length ;    
  }
}
void main() {
  int[] array = [3,1,4,1,5,9];    
  real[][][] 
    multi = [[[1,2,2],[2,3,4],[4,5,7]],
            [[4,1,3],[0,3,1],[4,4,6]],
            [[1,3,3],[2,7,8],[9,1,5]]] ;
  writefln("array : ", array.mean()) ;
  writefln("multi : ", multi.mean()) ;
}

Forth

: fmean ( addr n -- f )
  0e
  dup 0= if 2drop exit then
  tuck floats bounds do
    i f@ f+
  1 floats +loop
  0 d>f f/ ;

create test 3e f, 1e f, 4e f, 1e f, 5e f, 9e f,
test 6 fmean f.     \ 3.83333333333333

Fortran

In ISO Fortran 90 or later, use the SUM intrinsic, the SIZE intrinsic and the MAX intrinsic (to avoid divide by zero):

 real, target, dimension(100) :: a = (/ (i, i=1, 100) /)
 real, dimension(5,20) :: b = reshape( a, (/ 5,20 /) )
 real, pointer, dimension(:) :: p => a(2:1)       ! pointer to zero-length array
 real :: mean, zmean, bmean
 real, dimension(20) :: colmeans
 real, dimension(5) :: rowmeans
 
 mean = sum(a)/size(a)                ! SUM of A's elements divided by SIZE of A
 mean = sum(a)/max(size(a),1)         ! Same result, but safer code
                                      ! MAX of SIZE and 1 prevents divide by zero if SIZE == 0 (zero-length array)
 
 zmean = sum(p)/max(size(p),1)        ! Here the safety check pays off. Since P is a zero-length array,
                                      ! expression becomes "0 / MAX( 0, 1 ) -> 0 / 1 -> 0", rather than "0 / 0 -> NaN"
 
 bmean = sum(b)/max(size(b),1)        ! multidimensional SUM over multidimensional SIZE
 
 rowmeans = sum(b,1)/max(size(b,2),1) ! SUM elements in each row (dimension 1)
                                      ! dividing by the length of the row, which is the number of columns (SIZE of dimension 2)
 colmeans = sum(b,2)/max(size(b,1),1) ! SUM elements in each column (dimension 2)
                                      ! dividing by the length of the column, which is the number of rows (SIZE of dimension 1)

Haskell

mean xs = sum xs / Data.List.genericLength xs

IDL

If truly only the mean is wanted, one could use

 x = [3,1,4,1,5,9]
 print,mean(x)

But mean() is just a thin wrapper returning the zeroth element of moment() :

print,moment(x)
; ==>
  3.83333      8.96667     0.580037     -1.25081

which are mean, variance, skewness and kurtosis.

There are no zero-length vectors in IDL. Every variable has at least one value or otherwise it is <Undefined>.

J

mean=: +/ % #

That is, sum divided by the number of items. The verb also works on higher-ranked arrays. For example:

   mean 3 1 4 1 5 9
3.83333
   mean $0         NB. $0 is a zero-length vector
0
   x=: 20 4 ?@$ 0  NB. a 20-by-4 table of random (0,1) numbers
   mean x
0.58243 0.402948 0.477066 0.511155

The computation can also be written as a loop. It is shown here for comparison only and is highly non-preferred compared to the version above.

mean1=: 3 : 0
 z=. 0
 for_i. i.#y do. z=. z+i{y end.
 z % #y
)
   mean1 3 1 4 1 5 9
3.83333
   mean1 $0
0
   mean1 x
0.58243 0.402948 0.477066 0.511155

Java

Assume the numbers are in a double array called "nums".

...
double mean = 0;
double sum = 0;
for(double i : nums){
  sum += i;
}
System.out.println("The mean is: " + ((nums.length != 0) ? (sum / nums.length) : 0));
...

JavaScript

function mean(array) {
  var sum = 0;
  for(var i in array)
    sum += array[i];
  return array.length ? sum / array.length : 0;
}

alert( mean( [1,2,3,4,5] ) );   // 3

Library: Functional

function mean(a) {
  return a.length ? Functional.reduce('+', 0, a) / a.length : 0;
}

Logo

to average :l
  if empty? :l [output 0]
  output quotient apply "sum :l count :l
end
print average [1 2 3 4]    ; 2.5

Lucid

[1]

avg(x)
 where 
    sum = first(x) fby sum + next(x);
    n = 1 fby n + 1;
    avg = sum / n;
 end

MAXScript

fn mean data =
(
    total = 0
    for i in data do
    (
        total += i
    )
    if data.count == 0 then 0 else total as float/data.count
)

print (mean #(3, 1, 4, 1, 5, 9))

OCaml

These functions return a float:

let mean_floats xs =
  if xs = [] then
    0.
  else
    List.fold_left (+.) 0. xs /. float_of_int (List.length xs)

let mean_ints xs = mean_floats (List.map float_of_int xs)

Perl

sub avg(@_) {
  $count = 0;
  $sum = 0;
  foreach (@_) {
    $sum += $_;
    $count++;
  }
  return $count > 0 ? $sum / $count : 0;
}

print avg(qw(3 1 4 1 5 9))."\n";

Output:

3.83333333333333

Library: Data::AverageAverage

With module Data::Average. (For zero-length array returns ().)

use Data::Average;

my $d = Data::Average->new;
$d->add($_) foreach (qw(3 1 4 1 5 9));
print $d->avg."\n"

Output:

3.83333333333333

Python

def avg(data):
    return sum(data)/float(len(data)) if len(data)!=0 else 0
print avg([3,1,4,1,5,9])

Output:

3.83333333333333

Works with: Python version 2.4

def avg(data):
    if len(data)==0:
        return 0
    else:
        return sum(data)/float(len(data))
print avg([3,1,4,1,5,9])

Output:

3.83333333333333

Ruby

nums = [3, 1, 4, 1,  5, 9]
nums.empty? ? 0 : nums.inject(:+) / Float(nums.size)

Scheme

(define (mean l)
  (if (null? l)
      0
      (/ (apply + l) (length l))))

> (mean (list 3 1 4 1 5 9))
3 5/6

UnixPipes

term() {
   b=$1;res=$2
   echo "scale=5;$res+$b" | bc
}

sum() {
  (read B; res=$1;
  test -n "$B" && (term $B $res) || (term 0 $res))
}

fold() {
  func=$1
  (while read a ; do
      fold $func | $func $a
  done)
}

mean() {
   tee >(wc -l > count) | fold sum | xargs echo "scale=5;(1/" $(cat count) ") * " | bc
}

(echo 3; echo 1; echo 4) | mean

V

[mean
   [sum 0 [+] fold].
   dup sum
   swap size [[1 <] [1]] when /
].