C# 分布式概率随机数发生器
我想根据分布概率生成一个数字。例如,只需假设每个数字都出现以下情况:C# 分布式概率随机数发生器,c#,random,probability,probability-theory,C#,Random,Probability,Probability Theory,我想根据分布概率生成一个数字。例如,只需假设每个数字都出现以下情况: Number| Count 1 | 150 2 | 40 3 | 15 4 | 3 with a total of (150+40+15+3) = 208 then the probability of a 1 is 150/208= 0.72 and the probabili
Number| Count
1 | 150
2 | 40
3 | 15
4 | 3
with a total of (150+40+15+3) = 208
then the probability of a 1 is 150/208= 0.72
and the probability of a 2 is 40/208 = 0.192
我如何制作一个随机数生成器,它返回的是基于这个概率分布的数字
我很高兴现在能够基于一个静态的、硬编码的集合,但我最终希望它能够从数据库查询中得到概率分布
我见过类似的例子,如,但它们不是很通用。有什么建议吗?通常的方法是将均匀分布的随机数从0..1的间隔输入到所需的分布中 因此,在您的例子中,只需从0..1(例如,使用)中根据其返回值绘制一个随机数x
- 1如果0仅执行一次:
- 编写一个函数,计算给定pdf数组的cdf数组。在您的示例中,pdf数组为[150,40,15,3],cdf数组为[150190205208]李>
- 在[0,1]中得到一个随机数,乘以208,向上截断(或向下截断:我让你考虑一下角的情况)在1..208中得到一个整数。将其命名为r
- 对cdf数组中的r执行二进制搜索。返回包含r的单元格的索引
运行时间将与给定pdf数组大小的对数成正比。这很好。但是,如果数组大小总是很小(在您的示例中为4),则执行线性搜索会更容易,也会执行得更好。有许多方法可以生成具有自定义分布的随机整数(也称为离散分布)。选择取决于许多因素,包括可供选择的整数数量、分布的形状以及分布是否会随时间而变化 使用自定义权重函数选择整数的最简单方法之一
是拒绝采样方法。以下假设f(x)
的最大可能值为f
。拒绝采样的时间复杂度平均为常数,但在很大程度上取决于分布的形状,并且存在永远运行的最坏情况。要在[1,max
中选择整数使用拒绝抽样:k
- 在[1,
]中选择一个统一的随机整数k
i
- 使用概率
其他算法的平均采样时间不太依赖于分布(通常为常数或对数),但通常要求您在设置步骤中预先计算权重并将其存储在数据结构中。其中一些算法在平均使用的随机位数方面也很经济。这些算法包括alias方法、Knuth–Yao算法、MVN数据结构等。有关调查,请参阅我的“”部分
,返回f(i)/max
。否则,转到步骤1i
下面的C#代码实现了Michael Vose版本的alias方法,如中所述;另请参见。为了方便起见,我编写了此代码,并在此处提供public class LoadedDie { // Initializes a new loaded die. Probs // is an array of numbers indicating the relative // probability of each choice relative to all the // others. For example, if probs is [3,4,2], then // the chances are 3/9, 4/9, and 2/9, since the probabilities // add up to 9. public LoadedDie(int probs){ this.prob=new List<long>(); this.alias=new List<int>(); this.total=0; this.n=probs; this.even=true; } Random random=new Random(); List<long> prob; List<int> alias; long total; int n; bool even; public LoadedDie(IEnumerable<int> probs){ // Raise an error if nil if(probs==null)throw new ArgumentNullException("probs"); this.prob=new List<long>(); this.alias=new List<int>(); this.total=0; this.even=false; var small=new List<int>(); var large=new List<int>(); var tmpprobs=new List<long>(); foreach(var p in probs){ tmpprobs.Add(p); } this.n=tmpprobs.Count; // Get the max and min choice and calculate total long mx=-1, mn=-1; foreach(var p in tmpprobs){ if(p<0)throw new ArgumentException("probs contains a negative probability."); mx=(mx<0 || p>mx) ? P : mx; mn=(mn<0 || p<mn) ? P : mn; this.total+=p; } // We use a shortcut if all probabilities are equal if(mx==mn){ this.even=true; return; } // Clone the probabilities and scale them by // the number of probabilities for(var i=0;i<tmpprobs.Count;i++){ tmpprobs[i]*=this.n; this.alias.Add(0); this.prob.Add(0); } // Use Michael Vose's alias method for(var i=0;i<tmpprobs.Count;i++){ if(tmpprobs[i]<this.total) small.Add(i); // Smaller than probability sum else large.Add(i); // Probability sum or greater } // Calculate probabilities and aliases while(small.Count>0 && large.Count>0){ var l=small[small.Count-1];small.RemoveAt(small.Count-1); var g=large[large.Count-1];large.RemoveAt(large.Count-1); this.prob[l]=tmpprobs[l]; this.alias[l]=g; var newprob=(tmpprobs[g]+tmpprobs[l])-this.total; tmpprobs[g]=newprob; if(newprob<this.total) small.Add(g); else large.Add(g); } foreach(var g in large) this.prob[g]=this.total; foreach(var l in small) this.prob[l]=this.total; } // Returns the number of choices. public int Count { get { return this.n; } } // Chooses a choice at random, ranging from 0 to the number of choices // minus 1. public int NextValue(){ var i=random.Next(this.n); return (this.even || random.Next((int)this.total)<this.prob[i]) ? I : this.alias[i]; } }
我将此代码放在公共域中。感谢各位的解决方案!非常感谢 @Menjaraz我尝试实现您的解决方案,因为它看起来非常资源友好,但是在语法方面有一些困难 现在,我只是使用LINQ SelectMany()和Enumerable.Repeat()将我的摘要转换成一个简单的值列表
我知道这是一篇老文章,但我也搜索了这样一个生成器,对我找到的解决方案不满意。所以我写了自己的,并想与全世界分享 在调用“NextItem(…)”之前,只需多次调用“Add(…)”公共类InventoryItemQuantityDomainGenerator { 私有只读随机\u随机; 私人只读可计量单位数量; 公共InventoryItemQuantityDomainGenerator(IRepository数据库,int max) { _数量=数据库。AsQueryable() .其中(x=>x.数量x.数量) .选择(x=>new { 数量=x.键, Count=x.Count() }) .SelectMany(x=>Enumerable.Repeat(x.Quantity,x.Count)); _随机=新随机(); } 公共int Next() { 返回_数量.ElementAt(_random.Next(0,_数量.Count()-1)); } }
///将返回具有指定可能性的给定项之一的类。 ///要返回的类型。 ///如果生成器只有一个项目,它将始终返回该项目。 ///如果有两个项目的可能性为0.4和0.6(也可以使用4和6或2和3) ///它将十次返回第一项4次,十次返回第二项6次。 公共类随机数生成器 { 私有列表_items=新列表(); 私有随机_Random=新随机(); /// ///所有项目的可能性总和。 /// 私有双精度_totalprobability=0; /// ///添加要返回的新项目。 /// ///返回此项目的可能性。与传入的其他可能性相关。 ///要返回的项目。 公共无效添加(双重可能性,T项) { _添加(新元组(可能性,项)); _总可能性+=可能性; } /// ///从列表中返回具有指定相对可能性的随机项。 /// ///如果没有要返回的项目。 公共T NextItem() { var rand=\u random.NextDouble()*\u totalprobability; 双值=0; foreach(var项目在_项目中) { 值+=项目1;
如果(rand使用我的方法,它简单易懂。 我不计算0…1范围内的部分,我只使用“概率池”(听起来很酷,var loadedDie=new LoadedDie(new int[]{150,40,15,3}); // list of probabilities for each number: // 0 is 150, 1 is 40, and so on int number=loadedDie.nextValue(); // return a number from 0-3 according to given probabilities; // the number can be an index to another array, if needed
public class InventoryItemQuantityRandomGenerator { private readonly Random _random; private readonly IQueryable<int> _quantities; public InventoryItemQuantityRandomGenerator(IRepository database, int max) { _quantities = database.AsQueryable<ReceiptItem>() .Where(x => x.Quantity <= max) .GroupBy(x => x.Quantity) .Select(x => new { Quantity = x.Key, Count = x.Count() }) .SelectMany(x => Enumerable.Repeat(x.Quantity, x.Count)); _random = new Random(); } public int Next() { return _quantities.ElementAt(_random.Next(0, _quantities.Count() - 1)); } }
/// <summary> A class that will return one of the given items with a specified possibility. </summary> /// <typeparam name="T"> The type to return. </typeparam> /// <example> If the generator has only one item, it will always return that item. /// If there are two items with possibilities of 0.4 and 0.6 (you could also use 4 and 6 or 2 and 3) /// it will return the first item 4 times out of ten, the second item 6 times out of ten. </example> public class RandomNumberGenerator<T> { private List<Tuple<double, T>> _items = new List<Tuple<double, T>>(); private Random _random = new Random(); /// <summary> /// All items possibilities sum. /// </summary> private double _totalPossibility = 0; /// <summary> /// Adds a new item to return. /// </summary> /// <param name="possibility"> The possibility to return this item. Is relative to the other possibilites passed in. </param> /// <param name="item"> The item to return. </param> public void Add(double possibility, T item) { _items.Add(new Tuple<double, T>(possibility, item)); _totalPossibility += possibility; } /// <summary> /// Returns a random item from the list with the specified relative possibility. /// </summary> /// <exception cref="InvalidOperationException"> If there are no items to return from. </exception> public T NextItem() { var rand = _random.NextDouble() * _totalPossibility; double value = 0; foreach (var item in _items) { value += item.Item1; if (rand <= value) return item.Item2; } return _items.Last().Item2; // Should never happen } }
` // Some c`lass or struct for represent items you want to roulette public class Item { public string name; // not only string, any type of data public int chance; // chance of getting this Item } public class ProportionalWheelSelection { public static Random rnd = new Random(); // Static method for using from anywhere. You can make its overload for accepting not only List, but arrays also: // public static Item SelectItem (Item[] items)... public static Item SelectItem(List<Item> items) { // Calculate the summa of all portions. int poolSize = 0; for (int i = 0; i < items.Count; i++) { poolSize += items[i].chance; } // Get a random integer from 0 to PoolSize. int randomNumber = rnd.Next(0, poolSize) + 1; // Detect the item, which corresponds to current random number. int accumulatedProbability = 0; for (int i = 0; i < items.Count; i++) { accumulatedProbability += items[i].chance; if (randomNumber <= accumulatedProbability) return items[i]; } return null; // this code will never come while you use this programm right :) } } // Example of using somewhere in your program: static void Main(string[] args) { List<Item> items = new List<Item>(); items.Add(new Item() { name = "Anna", chance = 100}); items.Add(new Item() { name = "Alex", chance = 125}); items.Add(new Item() { name = "Dog", chance = 50}); items.Add(new Item() { name = "Cat", chance = 35}); Item newItem = ProportionalWheelSelection.SelectItem(items); }
using System; using System.Linq; // ... private static readonly Random RandomGenerator = new Random(); private int GetDistributedRandomNumber() { double totalCount = 208; var number1Prob = 150 / totalCount; var number2Prob = (150 + 40) / totalCount; var number3Prob = (150 + 40 + 15) / totalCount; var randomNumber = RandomGenerator.NextDouble(); int selectedNumber; if (randomNumber < number1Prob) { selectedNumber = 1; } else if (randomNumber >= number1Prob && randomNumber < number2Prob) { selectedNumber = 2; } else if (randomNumber >= number2Prob && randomNumber < number3Prob) { selectedNumber = 3; } else { selectedNumber = 4; } return selectedNumber; }
int totalNumber1Count = 0; int totalNumber2Count = 0; int totalNumber3Count = 0; int totalNumber4Count = 0; int testTotalCount = 100; foreach (var unused in Enumerable.Range(1, testTotalCount)) { int selectedNumber = GetDistributedRandomNumber(); Console.WriteLine($"selected number is {selectedNumber}"); if (selectedNumber == 1) { totalNumber1Count += 1; } if (selectedNumber == 2) { totalNumber2Count += 1; } if (selectedNumber == 3) { totalNumber3Count += 1; } if (selectedNumber == 4) { totalNumber4Count += 1; } } Console.WriteLine(""); Console.WriteLine($"number 1 -> total selected count is {totalNumber1Count} ({100 * (totalNumber1Count / (double) testTotalCount):0.0} %) "); Console.WriteLine($"number 2 -> total selected count is {totalNumber2Count} ({100 * (totalNumber2Count / (double) testTotalCount):0.0} %) "); Console.WriteLine($"number 3 -> total selected count is {totalNumber3Count} ({100 * (totalNumber3Count / (double) testTotalCount):0.0} %) "); Console.WriteLine($"number 4 -> total selected count is {totalNumber4Count} ({100 * (totalNumber4Count / (double) testTotalCount):0.0} %) ");
selected number is 1 selected number is 1 selected number is 1 selected number is 1 selected number is 2 selected number is 1 ... selected number is 2 selected number is 3 selected number is 1 selected number is 1 selected number is 1 selected number is 1 selected number is 1 number 1 -> total selected count is 71 (71.0 %) number 2 -> total selected count is 20 (20.0 %) number 3 -> total selected count is 8 (8.0 %) number 4 -> total selected count is 1 (1.0 %)