C＃中：从一个IEnumerable得到一个随机值优雅的代码 [英] C#: Elegant code for getting a random value from an IEnumerable

问题描述

在Python中，我可以这样做：

In Python, I can do this:

>>> import random
>>> ints = [1,2,3]
>>> random.choice(ints)
3

在C＃我做的第一件事是：

In C# the first thing I did was:

var randgen = new Random();
var ints = new int[] { 1, 2, 3 };
ints[randgen.Next(ints.Length)];

不过，这需要建立索引， int的也重复困扰我。所以，我想出了这一点：

But this requires indexing, also the duplication of ints bothers me. So, I came up with this:

var randgen = new Random();
var ints = new int[] { 1, 2, 3 };
ints.OrderBy(x=> randgen.Next()).First();

还是不太好，效率高。 ？有没有得到来自IEnumberable一个随机值更优雅的方式。

Still not very nice and efficient. Is there a more elegant way of getting a random value from an IEnumberable?

推荐答案

下面是给你一对夫妇的扩展方法：

Here's a couple extension methods for you:

public static T RandomElement<T>(this IEnumerable<T> enumerable)
{
    return enumerable.RandomElementUsing<T>(new Random());
}

public static T RandomElementUsing<T>(this IEnumerable<T> enumerable, Random rand)
{
    int index = rand.Next(0, enumerable.Count());
    return enumerable.ElementAt(index);
}

// Usage:
var ints = new int[] { 1, 2, 3 };
int randomInt = ints.RandomElement();

// If you have a preexisting `Random` instance, rand, use it:
// this is important e.g. if you are in a loop, because otherwise you will create new
// `Random` instances every time around, with nearly the same seed every time.
int anotherRandomInt = ints.RandomElementUsing(rand);

对于一般的IEnumerable< T> ，这将是O（ N ），因为这是的复杂性.Count之间的（）和随机 .ElementAt（ ）通话;然而，这两种特殊情况下为数组和列表，所以在这些情况下，这将是O（1）。

For a general IEnumerable<T>, this will be O(n), since that is the complexity of .Count() and a random .ElementAt() call; however, both special-case for arrays and lists, so in those cases it will be O(1).

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