C# 如何访问列表中的随机项目？
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How to access random item in list?
提问by jay_t55
I have an ArrayList, and I need to be able to click a button and then randomly pick out a string from that list and display it in a messagebox.
我有一个 ArrayList，我需要能够单击一个按钮，然后从该列表中随机挑选一个字符串并将其显示在消息框中。
How would I go about doing this?
我该怎么做呢？
采纳答案by Mehrdad Afshari
Create an instance of
Randomclass somewhere. Note that it's pretty important not to create a new instance each time you need a random number. You should reuse the old instance to achieve uniformity in the generated numbers. You can have astaticfield somewhere (be careful about thread safety issues):static Random rnd = new Random();Ask the
Randominstance to give you a random number with the maximum of the number of items in theArrayList:int r = rnd.Next(list.Count);Display the string:
MessageBox.Show((string)list[r]);
在
Random某处创建类的实例。请注意，每次需要随机数时不要创建新实例非常重要。您应该重用旧实例以实现生成数字的一致性。你可以在static某处有一个字段（注意线程安全问题）：static Random rnd = new Random();要求
Random实例为您提供一个随机数，其中包含 中项目数的最大值ArrayList：int r = rnd.Next(list.Count);显示字符串：
MessageBox.Show((string)list[r]);
回答by Joey
Create a Randominstance:
创建Random实例：
Random rnd = new Random();
Fetch a random string:
获取随机字符串：
string s = arraylist[rnd.Next(arraylist.Count)];
Remember though, that if you do this frequently you should re-use the Randomobject. Put it as a static field in the class so it's initialized only once and then access it.
但请记住，如果您经常这样做，您应该重新使用该Random对象。将它作为类中的静态字段，这样它只初始化一次，然后访问它。
回答by Mark Seemann
I usually use this little collection of extension methods:
我通常使用这个小小的扩展方法集合：
public static class EnumerableExtension
{
public static T PickRandom<T>(this IEnumerable<T> source)
{
return source.PickRandom(1).Single();
}
public static IEnumerable<T> PickRandom<T>(this IEnumerable<T> source, int count)
{
return source.Shuffle().Take(count);
}
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> source)
{
return source.OrderBy(x => Guid.NewGuid());
}
}
For a strongly typed list, this would allow you to write:
对于强类型列表，这将允许您编写：
var strings = new List<string>();
var randomString = strings.PickRandom();
If all you have is an ArrayList, you can cast it:
如果你只有一个 ArrayList，你可以投射它：
var strings = myArrayList.Cast<string>();
回答by Felipe Fujiy Pessoto
You can do:
你可以做：
list.OrderBy(x => Guid.NewGuid()).FirstOrDefault()
回答by Rajesh Varma
ArrayList ar = new ArrayList();
ar.Add(1);
ar.Add(5);
ar.Add(25);
ar.Add(37);
ar.Add(6);
ar.Add(11);
ar.Add(35);
Random r = new Random();
int index = r.Next(0,ar.Count-1);
MessageBox.Show(ar[index].ToString());
回答by Dave_cz
Or simple extension class like this:
或者像这样的简单扩展类：
public static class CollectionExtension
{
private static Random rng = new Random();
public static T RandomElement<T>(this IList<T> list)
{
return list[rng.Next(list.Count)];
}
public static T RandomElement<T>(this T[] array)
{
return array[rng.Next(array.Length)];
}
}
Then just call:
然后只需调用：
myList.RandomElement();
Works for arrays as well.
也适用于数组。
I would avoid calling OrderBy()as it can be expensive for larger collections. Use indexed collections like List<T>or arrays for this purpose.
我会避免打电话，OrderBy()因为对于较大的集合来说它可能很昂贵。List<T>为此目的使用索引集合，如或 数组。
回答by Carlos Toledo
I have been using this ExtensionMethod for a while:
我一直在使用这个 ExtensionMethod 一段时间：
public static IEnumerable<T> GetRandom<T>(this IEnumerable<T> list, int count)
{
if (count <= 0)
yield break;
var r = new Random();
int limit = (count * 10);
foreach (var item in list.OrderBy(x => r.Next(0, limit)).Take(count))
yield return item;
}
回答by bafsar
I needed to more item instead of just one. So, I wrote this:
我需要更多的项目而不是一个。所以，我写了这个：
public static TList GetSelectedRandom<TList>(this TList list, int count)
where TList : IList, new()
{
var r = new Random();
var rList = new TList();
while (count > 0 && list.Count > 0)
{
var n = r.Next(0, list.Count);
var e = list[n];
rList.Add(e);
list.RemoveAt(n);
count--;
}
return rList;
}
With this, you can get elements how many you want as randomly like this:
有了这个，你可以像这样随机获得你想要的元素数量：
var _allItems = new List<TModel>()
{
// ...
// ...
// ...
}
var randomItemList = _allItems.GetSelectedRandom(10);
回答by Lucas
Why not:
为什么不：
public static T GetRandom<T>(this IEnumerable<T> list)
{
return list.ElementAt(new Random(DateTime.Now.Millisecond).Next(list.Count()));
}
回答by Хидеки Матосува
Why not[2]:
为什么不[2]：
public static T GetRandom<T>(this List<T> list)
{
return list[(int)(DateTime.Now.Ticks%list.Count)];
}
