我想根据分布式概率生成一个数字。例如，只要说每个数字都有以下几种情况：
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 probability of a 2 is 40/208 = 0.192
如何根据此概率分布制作一个返回数字的随机数生成器？
我很高兴现在基于静态的硬编码集，但我最终希望它从数据库查询中获得概率分布。
我见过像this one这样的类似例子，但它们并不是非常通用的。有什么建议吗？
答案 0 :(得分：33)
一般方法是将均匀分布的随机数从0..1区间提供到所需分布的the inverse of the cumulative distribution function。
因此，在您的情况下，只需从0..1中绘制一个随机数x（例如使用Random.NextDouble()）并根据其值返回
答案 1 :(得分：4)
This question解释了生成具有不同概率的随机数的各种方法。根据问题所示的this article，最好的方法（就时间复杂度而言）是Michael Vose所谓的“别名方法”。
为方便起见，我已经编写并提供了一个实现别名方法的随机数生成器的C＃实现：
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];
}
}
示例：
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
我将此代码放在公共领域。
答案 2 :(得分：3)
只做一次：
每次都这样做：
运行时间将与给定pdf数组大小的对数成比例。这很好。但是，如果您的数组大小总是如此之小（在您的示例中为4），那么执行线性搜索会更容易，也会表现得更好。
答案 3 :(得分：1)
我知道这是一个老帖子，但我也搜索了这样一个发生器，并且对我找到的解决方案不满意。所以我写了自己的想法并希望与全世界分享。
在调用“NextItem（...）”之前，只需调用“添加（...）”一段时间
/// <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
}
}
答案 4 :(得分：0)
感谢所有解决方案的人！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，但是语法有些困难。
所以现在，我只是使用LINQ SelectMany（）和Enumerable.Repeat（）将我的摘要转换为一个平面的值列表。
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));
}
}
答案 5 :(得分：0)
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“Probabilityp Pool”（听起来很酷，是吗？）
At circle diagram you can see weight of every element in pool
Here you can see an implementing of accumulative probability for roulette
`
// 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);
}
答案 6 :(得分：0)
这是使用Inverse distribution function的实现：
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;
}
验证随机分布的示例：
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} %) ");
示例输出：
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 %)