用C＃从id集合中获取4个（或任意数量）随机唯一ID的最简单方法是什么？

Question

Assume that you have an idCollection IList<long> and you have a method to get 4 unique ids.Every time you call it, it gives you random 4 unique ids ?

var idCollec = new[] {1,2,3,4,5,6,7,8,9,10,11,12}.ToList();

For example {2,6,11,12}
            {3,4,7,8}
            {5,8,10,12}
            ...
            ..

What is the smartest way to do it ?

Thanks

Answer 1

Seems like easiest way would be to have something like:

if(idCollection.Count <4)
{
    throw new ArgumentException("Source array not long enough");
}
List<long> FourUniqueIds = new List<long>(4);
while(FourUniqueIds.Count <4)
{
    long temp = idCollection[random.Next(idCollection.Count)];
    if(!FourUniqueIds.Contains(temp))
    {
        FourUniqueIds.add(temp);
    }
}

Answer 2

It can be done with a nice LINQ query. The key to doing it without the risk of getting duplicates, is to create a never ending IEnumerable of random integers. Then you can take n distinct values from it, and use them as indexes into the list.

Sample program:

using System;
using System.Collections.Generic;
using System.Linq;

namespace TestRandom
{
    class Program
{
    static void Main(string[] args)
    {
        // Just to prepopulate a list.
        var ids = (from n in Enumerable.Range(0, 100)
                   select (long)rand.Next(0, 1000)).ToList();

        // Example usage of the GetRandomSet method.
        foreach(long id in GetRandomSet(ids, 4))
            Console.WriteLine(id);
    }

    // Get count random entries from the list.
    public static IEnumerable<long> GetRandomSet(IList<long> ids, int count)
    {
        // Can't get more than there is in the list.
        if ( count > ids.Count)
            count = ids.Count;

        return RandomIntegers(0, ids.Count)
            .Distinct()
            .Take(count)
            .Select(index => ids[index]);
    }

    private static IEnumerable<int> RandomIntegers(int min, int max)
    {
        while (true)
            yield return rand.Next(min, max);
    }

    private static readonly Random rand = new Random();
}

}

If you use this approach, make sure you do not try to take more distinct values than there are available in the range passed to RandomIntegers.

Answer 3

What about shuffling the set then just taking the first four each time?

public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> e)
{
    var r = new Random();
    return e.OrderBy(x => r.Next());
}

Then something like this? It would probably be faster to use a for loop instead of Take and Except.

 var ordered = new List<int> {1, 2, 3, 4, 5, 6, 7, 8, 10};

 var random = ordered.Shuffle();
 while(random.Count() > 0)
 {
     var ourSet = random.Take(4).ToList();            
     random = random.Except(ourSet);
 }

Answer 4

If the IList <long> is populated with non unique values, you could use LINQ's Distinct() in combination with Take(), if it already has unique values, just use Take().

List<long> myUniqueIds = //prepoulation
var first4UniqueUnused = myUniqueIds.Take(4);

var next4UniqueUnused = myUniqueIds.Where(l=>!first4UniqueUnused.Contains(l)).Take(4);

another way that is too easy, i think we've been making it too hard:

List<long> myIDs = //prepopulation;
List<long> my4Random = new List<long>();
Random r = new Random();

for(int i=0; i< 4; i++)
{
     int j = r.Next();
     while(j>myIDs.Count || my4Random.Contains(myIDs[j]))
          j = r.Next();

     my4Random.Add(myIDs[j]);
}

Answer 5

Random random = new Random(); 

long firstOne = idCollection[random.Next(idCollection.Count)];
long secondOne = idCollection[random.NExt(idCollection.Count)];

...and so on