在c#中随机化泛型列表顺序的最佳方法是什么?我在一个列表中有一个有限的75个数字集,我想随机分配一个顺序,以便为彩票类型的应用程序绘制它们。
当前回答
这是我最喜欢的shuffle方法,当不需要修改原始的时候。它是Fisher-Yates“由内到外”算法的变体,适用于任何可枚举序列(源的长度不需要从一开始就知道)。
public static IList<T> NextList<T>(this Random r, IEnumerable<T> source)
{
var list = new List<T>();
foreach (var item in source)
{
var i = r.Next(list.Count + 1);
if (i == list.Count)
{
list.Add(item);
}
else
{
var temp = list[i];
list[i] = item;
list.Add(temp);
}
}
return list;
}
该算法还可以通过分配一个从0到length - 1的范围来实现,并通过将随机选择的索引与最后一个索引交换来随机耗尽索引,直到所有索引都被选中一次。上面的代码完成了完全相同的事情,但没有额外的分配。非常简洁。
With regards to the Random class it's a general purpose number generator (and If I was running a lottery I'd consider using something different). It also relies on a time based seed value by default. A small alleviation of the problem is to seed the Random class with the RNGCryptoServiceProvider or you could use the RNGCryptoServiceProvider in a method similar to this (see below) to generate uniformly chosen random double floating point values but running a lottery pretty much requires understanding randomness and the nature of the randomness source.
var bytes = new byte[8];
_secureRng.GetBytes(bytes);
var v = BitConverter.ToUInt64(bytes, 0);
return (double)v / ((double)ulong.MaxValue + 1);
生成随机双精度(仅在0和1之间)的目的是用于扩展到整数解。如果你需要从一个基于随机双x的列表中选择一个东西,它总是0 <= x && x < 1是很简单的。
return list[(int)(x * list.Count)];
享受吧!
其他回答
public Deck(IEnumerable<Card> initialCards)
{
cards = new List<Card>(initialCards);
public void Shuffle()
}
{
List<Card> NewCards = new List<Card>();
while (cards.Count > 0)
{
int CardToMove = random.Next(cards.Count);
NewCards.Add(cards[CardToMove]);
cards.RemoveAt(CardToMove);
}
cards = NewCards;
}
public IEnumerable<string> GetCardNames()
{
string[] CardNames = new string[cards.Count];
for (int i = 0; i < cards.Count; i++)
CardNames[i] = cards[i].Name;
return CardNames;
}
Deck deck1;
Deck deck2;
Random random = new Random();
public Form1()
{
InitializeComponent();
ResetDeck(1);
ResetDeck(2);
RedrawDeck(1);
RedrawDeck(2);
}
private void ResetDeck(int deckNumber)
{
if (deckNumber == 1)
{
int numberOfCards = random.Next(1, 11);
deck1 = new Deck(new Card[] { });
for (int i = 0; i < numberOfCards; i++)
deck1.Add(new Card((Suits)random.Next(4),(Values)random.Next(1, 14)));
deck1.Sort();
}
else
deck2 = new Deck();
}
private void reset1_Click(object sender, EventArgs e) {
ResetDeck(1);
RedrawDeck(1);
}
private void shuffle1_Click(object sender, EventArgs e)
{
deck1.Shuffle();
RedrawDeck(1);
}
private void moveToDeck1_Click(object sender, EventArgs e)
{
if (listBox2.SelectedIndex >= 0)
if (deck2.Count > 0) {
deck1.Add(deck2.Deal(listBox2.SelectedIndex));
}
RedrawDeck(1);
RedrawDeck(2);
}
List<T> OriginalList = new List<T>();
List<T> TempList = new List<T>();
Random random = new Random();
int length = OriginalList.Count;
int TempIndex = 0;
while (length > 0) {
TempIndex = random.Next(0, length); // get random value between 0 and original length
TempList.Add(OriginalList[TempIndex]); // add to temp list
OriginalList.RemoveAt(TempIndex); // remove from original list
length = OriginalList.Count; // get new list <T> length.
}
OriginalList = new List<T>();
OriginalList = TempList; // copy all items from temp list to original list.
对已接受答案的简单修改,返回一个新的列表,而不是原地工作,并像许多其他Linq方法一样接受更通用的IEnumerable<T>。
private static Random rng = new Random();
/// <summary>
/// Returns a new list where the elements are randomly shuffled.
/// Based on the Fisher-Yates shuffle, which has O(n) complexity.
/// </summary>
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> list) {
var source = list.ToList();
int n = source.Count;
var shuffled = new List<T>(n);
shuffled.AddRange(source);
while (n > 1) {
n--;
int k = rng.Next(n + 1);
T value = shuffled[k];
shuffled[k] = shuffled[n];
shuffled[n] = value;
}
return shuffled;
}
肯定是旧帖子,但我只是使用GUID。
Items = Items.OrderBy(o => Guid.NewGuid().ToString()).ToList();
GUID总是唯一的,因为它每次都会重新生成,所以每次结果都会改变。
这里有一个线程安全的方法来做到这一点:
public static class EnumerableExtension
{
private static Random globalRng = new Random();
[ThreadStatic]
private static Random _rng;
private static Random rng
{
get
{
if (_rng == null)
{
int seed;
lock (globalRng)
{
seed = globalRng.Next();
}
_rng = new Random(seed);
}
return _rng;
}
}
public static IEnumerable<T> Shuffle<T>(this IEnumerable<T> items)
{
return items.OrderBy (i => rng.Next());
}
}
推荐文章
- 实体框架核心:在上一个操作完成之前,在此上下文中开始的第二个操作
- 如何为构造函数定制Visual Studio的私有字段生成快捷方式?
- 如何使用JSON确保字符串是有效的JSON。网
- AppSettings从.config文件中获取值
- 通过HttpClient向REST API发布一个空体
- 如何检查IEnumerable是否为空或空?
- 自动化invokerrequired代码模式
- 在c#代码中设置WPF文本框的背景颜色
- 在c#中,什么是单子?
- c#和Java中的泛型有什么不同?和模板在c++ ?
- c#线程安全快速(est)计数器
- 如何将此foreach代码转换为Parallel.ForEach?
- 如何分裂()一个分隔字符串到一个列表<字符串>
- 如何转换列表<字符串>列表<int>?
- c#对象列表,我如何得到一个属性的和