如何用RandomOrDefault替换FirstOrDefault来“平衡”调用？

Question

I have such code

    senders.FirstOrDefault(sender => !sender.IsBusy);

This line is called pretty often.

The problem is that it always returns the first non-busy sender ; the same first sender is returned pretty often, but the last one is returned very rarely. How to easily balance this?

Ideally, on every call I should return the most rarely used sender. Ie between all non-busy senders select the one that was selected the least number of times during the last second.

Answer 1

Maybe something like:

public static T RandomOrDefault<T>(this IEnumerable<T> dataSet)
{ 
    return dataSet.RandomOrDefault(y => true);
}

public static T RandomOrDefault<T>(this IEnumerable<T> dataSet, Func<T, bool> filter)
{
    var elems = dataSet.Where(filter).ToList();
    var count = elems.Count;
    if (count == 0)
    {
        return default(T);
    }
    var random = new Random();
    var index = random.Next(count - 1);
    return elems[index];
}

then you can call it with:

senders.RandomOrDefault(sender => !sender.IsBusy);

Answer 2

If you want to get the least used one efficiently you will be probably good with the following non-Linq 'list rotation' solution, which is O(n) effiency and O(1) space unlike most of others:

// keep track of these
List<Sender> senders;
int nSelected = 0;          // number of already selected senders

// ...

// solution
int total = senders.Count;  // total number of senders
// looking for next non-busy sender
Sender s = null;
for (int i = 0; i < total; i++)
{
    int ind = (i + nSelected) % total; // getting the one 'after' previous
    if (!senders[ind].IsBusy)
    {
        s = senders[ind];
        ++nSelected;
        break;
    }
}

Of course this adds the must-be-indexable constraint on senders container.

Answer 3

您可以在FirstOrDefault之前使用此帖子中的“Shuffle”扩展方法

Answer 4

Keep a look-up of senders that you have used and the time when they were used.

var recentlyUsed = new Dictionary<Sender, DateTime>();
var sender = senders.FirstOrDefault(sender => !sender.IsBusy && (!recentlyUsed.ContainsKey(sender) || recentlyUsed[sender] < DateTime.Now.AddSeconds(-1)));
if (sender != null)
    recentlyUsed[sender] = DateTime.Now;

Answer 5

You could use Skip with a random number less than the total number of non-busy senders.

senders.Where(sender => !sender.IsBusy).Skip(randomNumber).FirstOrDefault();

Identifying a sensible limit for the random number might be a bit tricky though.

Answer 6

You could easily reorder by a new Guid, like this:

senders.Where(sender => !sender.IsBusy).OrderBy(x => Guid.NewGuid()).FirstOrDefault();

You don't mess with random numbers, you don't have to identify a "range" for these numbers. It just plain works and it's pretty elegant, I think.

Answer 7

Based on algorithm from the "Real world functional programming" book here's the O(n) implementation of extension method for taking random or default value from IEnumearble.

public static class SampleExtensions
{
    // The Random class is instantiated as singleton 
    // because it would give bad random values 
    // if instantiated on every call to RandomOrDefault method
    private static readonly Random RandomGenerator = new Random(unchecked((int)DateTime.Now.Ticks));

    public static T RandomOrDefault<T>(this IEnumerable<T> source, Func<T, bool> predicate)
    {
        IEnumerable<T> filtered = source.Where(predicate);

        int count = 0;
        T selected = default(T);
        foreach (T current in filtered)
        {
            if (RandomGenerator.Next(0, ++count) == 0)
            {
                selected = current;
            }
        }

        return selected;
    }

    public static T RandomOrDefault<T>(this IEnumerable<T> source)
    {
        return RandomOrDefault(source, element => true);
    }
}

Here's the code to ensure that this algorithm really gives the Uniform distribution:

[Test]
public void TestRandom()
{
    IEnumerable<int> source = Enumerable.Range(1, 10);
    Dictionary<int, int> result = source.ToDictionary(element => element, element => 0);
    result[0] = 0;

    const int Limit = 1000000;
    for (int i = 0; i < Limit; i++)
    {
        result[source.RandomOrDefault()]++;
    }

    foreach (var pair in result)
    {
        Console.WriteLine("{0}: {1:F2}%", pair.Key, pair.Value * 100f / Limit);
    }

    Console.WriteLine(Enumerable.Empty<int>().RandomOrDefault());
}

The output of TestRandom method is:

1: 9,92%
2: 10,03%
3: 10,04%
4: 9,99%
5: 10,00%
6: 10,01%
7: 9,98%
8: 10,03%
9: 9,97%
10: 10,02%
0: 0,00%
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