math stats with Linq

Question

I have a collection of person objects (IEnumerable) and each person has an age property.

I want to generate stats on the collection such as Max, Min, Average, Median, etc on this age property.

What is the most elegant way of doing this using LINQ?

Be careful if the data is comming from a database, as Linq may read the data more then once, however in your case Linq should be a good tool as you seem to have your collection in ram. — Ian Ringrose, Nov 09 '10 at 14:14

score 45 · Answer 1 · answered May 24 '12 at 13:25

Here is a complete, generic implementation of Median that properly handles empty collections and nullable types. It is LINQ-friendly in the style of Enumerable.Average, for example:

    double? medianAge = people.Median(p => p.Age);

This implementation returns null when there are no non-null values in the collection, but if you don't like the nullable return type, you could easily change it to throw an exception instead.

public static double? Median<TColl, TValue>(
    this IEnumerable<TColl> source,
    Func<TColl, TValue>     selector)
{
    return source.Select<TColl, TValue>(selector).Median();
}

public static double? Median<T>(
    this IEnumerable<T> source)
{
    if(Nullable.GetUnderlyingType(typeof(T)) != null)
        source = source.Where(x => x != null);

    int count = source.Count();
    if(count == 0)
        return null;

    source = source.OrderBy(n => n);

    int midpoint = count / 2;
    if(count % 2 == 0)
        return (Convert.ToDouble(source.ElementAt(midpoint - 1)) + Convert.ToDouble(source.ElementAt(midpoint))) / 2.0;
    else
        return Convert.ToDouble(source.ElementAt(midpoint));
}

This, AFAIK, enumerates the source 2 or 3 times: first when `Count()` is called, second (and possibly third time) - when ElementAt is called. — DarkWanderer, May 24 '14 at 09:51
You're absolutely right. And -- as always with LINQ -- the impact of this will range from trivial to prohibitive, depending on the nature of your collection. Remember the Rules of Optimization: Rule 1: Don't do it. Rule 2 (for experts only): Don't do it yet. (See http://blog.codinghorror.com/why-arent-my-optimizations-optimizing/) — Rand Scullard, May 24 '14 at 14:22

score 32 · Accepted Answer · edited Apr 06 '19 at 07:06

32

var max = persons.Max(p => p.age);
var min = persons.Min(p => p.age);
var average = persons.Average(p => p.age);

Fix for median in case of even number of elements

int count = persons.Count();
var orderedPersons = persons.OrderBy(p => p.age);
float median = orderedPersons.ElementAt(count/2).age + orderedPersons.ElementAt((count-1)/2).age;
median /= 2;

edited Apr 06 '19 at 07:06

Sai Manoj Kumar Yadlapati

330
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answered Nov 09 '10 at 13:55

Itay Karo

17,924
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1

Little note here, that this algorithm is not time-optimal. Median can be calculated in O(n) time. – Max May 13 '17 at 11:09
6

@Max Where is your time-optimal solution? – crush Aug 28 '19 at 19:23
Note that this method may give wrong results if the list contains null values. One needs to filter out those null values before using it! – draz Nov 03 '22 at 21:42

Aliostad · Answer 3 · 2010-11-09T14:02:08.507

14

Max, Min, Average are part of Linq:

int[] ints = new int[]{3,4,5};
Console.WriteLine(ints.Max());
Console.WriteLine(ints.Min());
Console.WriteLine(ints.Average());

Median is easy:

UPDATE

I have added order:

ints.OrderBy(x=>x).Skip(ints.Count()/2).First();

BEWARE

All these operations are done in a loop. For example, ints.Count() is a loop so if you already get ints.Length and stored to a variable or simply use it as it is, would be better.

edited Nov 09 '10 at 14:02

answered Nov 09 '10 at 13:55

Aliostad

80,612
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`ints.ElementAt(ints.Count()/2)` – Cheng Chen Nov 09 '10 at 13:58
for the Median to be right you need to order the array first. – Itay Karo Nov 09 '10 at 13:59
1

Your implementation of `Median` also assumes that the input has an odd number of elements. It will fail for `{0, 1}` (it will give 0 instead of 0.5). – Mark Byers Nov 09 '10 at 14:00
1

@Mark - AFAIK Median will always return one of the source array elements – Itay Karo Nov 09 '10 at 14:04
4

@Itay: No, that's usually called the medoid. *A related concept, in which the outcome is forced to correspond to a member of the sample is the medoid.* http://en.wikipedia.org/wiki/Median The way you are calculating the medoid is biased as it will often return a value that is lower than the median, and never a value that is higher. – Mark Byers Nov 09 '10 at 14:04

Josh Withee · Answer 4 · 2020-06-12T10:58:57.083

7

Get median using Linq (works for even or odd number of elements)

int count = persons.Count();

if (count % 2 == 0)
    var median = persons.Select(x => x.Age).OrderBy(x => x).Skip((count / 2) - 1).Take(2).Average();
else
    var median = persons.Select(x => x.Age).OrderBy(x => x).ElementAt(count / 2);

edited Jun 12 '20 at 10:58

answered Jun 08 '20 at 13:11

Josh Withee

9,922
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score 1 · Answer 5 · answered May 15 '23 at 11:07

My Median LINQ extension, using Generic Math (.NET 7+)

public static TSource Median<TSource>(this IEnumerable<TSource> source)
    where TSource : struct, INumber<TSource>
    => Median<TSource, TSource>(source);

public static TResult Median<TSource, TResult>(this IEnumerable<TSource> source)
    where TSource : struct, INumber<TSource>
    where TResult : struct, INumber<TResult>
{
    var array = source.ToArray();
    var length = array.Length;
    if (length == 0)
    {
        throw new InvalidOperationException("Sequence contains no elements.");
    }
    Array.Sort(array);
    var index = length / 2;
    var value = TResult.CreateChecked(array[index]);
    if (length % 2 == 1)
    {
        return value;
    }
    var sum = value + TResult.CreateChecked(array[index - 1]);
    return sum / TResult.CreateChecked(2);
}

math stats with Linq

5 Answers5

Get median using Linq (works for even or odd number of elements)

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