Average function without overflow exception

Asked 2010-05-24T07:58:25.973

.NET Framework 3.5.
I'm trying to calculate the average of some pretty large numbers.
For instance:

using System;
using System.Linq;

class Program
{
    static void Main(string[] args)
    {
        var items = new long[]
                        {
                            long.MaxValue - 100, 
                            long.MaxValue - 200, 
                            long.MaxValue - 300
                        };
        try
        {
            var avg = items.Average();
            Console.WriteLine(avg);
        }
        catch (OverflowException ex)
        {
            Console.WriteLine("can't calculate that!");
        }
        Console.ReadLine();
    }
}

Obviously, the mathematical result is 9223372036854775607 (long.MaxValue - 200), but I get an exception there. This is because the implementation (on my machine) to the Average extension method, as inspected by .NET Reflector is:

public static double Average(this IEnumerable<long> source)
{
    if (source == null)
    {
        throw Error.ArgumentNull("source");
    }
    long num = 0L;
    long num2 = 0L;
    foreach (long num3 in source)
    {
        num += num3;
        num2 += 1L;
    }
    if (num2 <= 0L)
    {
        throw Error.NoElements();
    }
    return (((double) num) / ((double) num2));
}

I know I can use a BigInt library (yes, I know that it is included in .NET Framework 4.0, but I'm tied to 3.5).

But I still wonder if there's a pretty straight forward implementation of calculating the average of integers without an external library. Do you happen to know about such implementation?

Thanks!!

UPDATE:

The previous example, of three large integers, was just an example to illustrate the overflow issue. The question is about

Edit Report

2 Answers

You may try the following approach:

let number of elements is N, and numbers are arr[0], .., arr[N-1].

You need to define 2 variables:

mean and remainder.

initially mean = 0, remainder = 0.

at step i you need to change mean and remainder in the following way:

mean += arr[i] / N;
remainder += arr[i] % N;
mean += remainder / N;
remainder %= N;

after N steps you will get correct answer in mean variable and remainder / N will be fractional part of the answer (I am not sure you need it, but anyway)

answered 2010-05-24T11:05:59.807

Here is my version of an extension method that can help with this.

    public static long Average(this IEnumerable<long> longs)
    {
        long mean = 0;
        long count = longs.Count();
        foreach (var val in longs)
        {
            mean += val / count;
        }
        return mean;
    }

answered 2013-04-03T15:42:04.317

Average function without overflow exception

2 Answers

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