c# - 随机加权选择

Question

考虑下面代表代理的类：

public class Broker
{
    public string Name = string.Empty;
    public int Weight = 0;

    public Broker(string n, int w)
    {
        this.Name = n;
        this.Weight = w;
    }
}

我想从数组中随机选择一个经纪人，考虑到他们的权重。

你觉得下面的代码怎么样？

class Program
    {
        private static Random _rnd = new Random();

        public static Broker GetBroker(List<Broker> brokers, int totalWeight)
        {
            // totalWeight is the sum of all brokers' weight

            int randomNumber = _rnd.Next(0, totalWeight);

            Broker selectedBroker = null;
            foreach (Broker broker in brokers)
            {
                if (randomNumber <= broker.Weight)
                {
                    selectedBroker = broker;
                    break;
                }

                randomNumber = randomNumber - broker.Weight;
            }

            return selectedBroker;
        }


        static void Main(string[] args)
        {
            List<Broker> brokers = new List<Broker>();
            brokers.Add(new Broker("A", 10));
            brokers.Add(new Broker("B", 20));
            brokers.Add(new Broker("C", 20));
            brokers.Add(new Broker("D", 10));

            // total the weigth
            int totalWeight = 0;
            foreach (Broker broker in brokers)
            {
                totalWeight += broker.Weight;
            }

            while (true)
            {
                Dictionary<string, int> result = new Dictionary<string, int>();

                Broker selectedBroker = null;

                for (int i = 0; i < 1000; i++)
                {
                    selectedBroker = GetBroker(brokers, totalWeight);
                    if (selectedBroker != null)
                    {
                        if (result.ContainsKey(selectedBroker.Name))
                        {
                            result[selectedBroker.Name] = result[selectedBroker.Name] + 1;
                        }
                        else
                        {
                            result.Add(selectedBroker.Name, 1);
                        }
                    }
                }


                Console.WriteLine("A\t\t" + result["A"]);
                Console.WriteLine("B\t\t" + result["B"]);
                Console.WriteLine("C\t\t" + result["C"]);
                Console.WriteLine("D\t\t" + result["D"]);

                result.Clear();
                Console.WriteLine();
                Console.ReadLine();
            }
        }
    }

我没那么自信。当我运行这个时，经纪人 A 总是比经纪人 D 获得更多的命中，并且它们具有相同的权重。

有没有更准确的算法？

谢谢！

score 42 · Accepted Answer

你的算法几乎是正确的。但是，测试应该<代替<=：

if (randomNumber < broker.Weight)

这是因为 0 包含在随机数中，而 0totalWeight是排除的。换句话说，权重为 0 的经纪人仍然有很小的机会被选中——根本不是你想要的。这说明经纪人 A 的点击次数比经纪人 D 多。

除此之外，您的算法很好，实际上是解决此问题的规范方法。

score 19 · Accepted Answer

可以用于任何数据类型的更通用的东西怎么样？

using System;
using System.Linq;
using System.Collections;
using System.Collections.Generic;

public static class IEnumerableExtensions {
    
    public static T RandomElementByWeight<T>(this IEnumerable<T> sequence, Func<T, float> weightSelector) {
        float totalWeight = sequence.Sum(weightSelector);
        // The weight we are after...
        float itemWeightIndex =  (float)new Random().NextDouble() * totalWeight;
        float currentWeightIndex = 0;

        foreach(var item in from weightedItem in sequence select new { Value = weightedItem, Weight = weightSelector(weightedItem) }) {
            currentWeightIndex += item.Weight;
            
            // If we've hit or passed the weight we are after for this item then it's the one we want....
            if(currentWeightIndex >= itemWeightIndex)
                return item.Value;
            
        }
        
        return default(T);
        
    }
    
}

只需致电

    Dictionary<string, float> foo = new Dictionary<string, float>();
    foo.Add("Item 25% 1", 0.5f);
    foo.Add("Item 25% 2", 0.5f);
    foo.Add("Item 50%", 1f);
    
    for(int i = 0; i < 10; i++)
        Console.WriteLine(this, "Item Chosen {0}", foo.RandomElementByWeight(e => e.Value));

score 13 · Accepted Answer

class Program
{
    static void Main(string[] args)
    {
        var books = new List<Book> {
        new Book{Isbn=1,Name="A",Weight=1},
        new Book{Isbn=2,Name="B",Weight=100},
        new Book{Isbn=3,Name="C",Weight=1000},
        new Book{Isbn=4,Name="D",Weight=10000},
        new Book{Isbn=5,Name="E",Weight=100000}};

        Book randomlySelectedBook = WeightedRandomization.Choose(books);
    }
}

public static class WeightedRandomization
{
    public static T Choose<T>(List<T> list) where T : IWeighted
    {
        if (list.Count == 0)
        {
            return default(T);
        }

        int totalweight = list.Sum(c => c.Weight);
        Random rand = new Random();
        int choice = rand.Next(totalweight);
        int sum = 0;

        foreach (var obj in list)
        {
            for (int i = sum; i < obj.Weight + sum; i++)
            {
                if (i >= choice)
                {
                    return obj;
                }
            }
            sum += obj.Weight;
        }

        return list.First();
    }
}

public interface IWeighted
{
    int Weight { get; set; }
}

public class Book : IWeighted
{
    public int Isbn { get; set; }
    public string Name { get; set; }
    public int Weight { get; set; }
}

score 5 · Accepted Answer

由于这是 Google 上的最高结果：

我为随机选择的加权项目创建了一个 C# 库。

它实现了树选择和 walker 别名方法算法，为所有用例提供最佳性能。
它经过单元测试和优化。
它具有 LINQ 支持。
它是免费和开源的，在 MIT 许可下获得许可。

一些示例代码：

IWeightedRandomizer<string> randomizer = new DynamicWeightedRandomizer<string>();
randomizer["Joe"] = 1;
randomizer["Ryan"] = 2;
randomizer["Jason"] = 2;

string name1 = randomizer.RandomWithReplacement();
//name1 has a 20% chance of being "Joe", 40% of "Ryan", 40% of "Jason"

string name2 = randomizer.RandomWithRemoval();
//Same as above, except whichever one was chosen has been removed from the list.

score 3 · Accepted Answer

在通过内存使用情况下选择代理时，替代方法有利于速度。基本上，我们创建的列表包含与指定权重相同数量的代理实例引用。

List<Broker> brokers = new List<Broker>();
for (int i=0; i<10; i++)
    brokers.Add(new Broker("A", 10));
for (int i=0; i<20; i++)
    brokers.Add(new Broker("B", 20));
for (int i=0; i<20; i++)
    brokers.Add(new Broker("C", 20));
for (int i=0; i<10; i++)
    brokers.Add(new Broker("D", 10));

然后，选择一个随机加权的实例是一个 O(1) 操作：

int randomNumber = _rnd.Next(0, brokers.length);
selectedBroker = brokers[randomNumber];

score 2 · Accepted Answer

有点太晚了，但这里是 C#7 示例。它非常小并且可以提供正确的分布。

public static class RandomTools
{
    public static T PickRandomItemWeighted<T>(IList<(T Item, int Weight)> items)
    {
        if ((items?.Count ?? 0) == 0)
        {
            return default;
        }

        int offset = 0;
        (T Item, int RangeTo)[] rangedItems = items
            .OrderBy(item => item.Weight)
            .Select(entry => (entry.Item, RangeTo: offset += entry.Weight))
            .ToArray();

        int randomNumber = new Random().Next(items.Sum(item => item.Weight)) + 1;
        return rangedItems.First(item => randomNumber <= item.RangeTo).Item;
    }
}

score 1 · Accepted Answer

如果您想要更快的速度，您可以考虑加权水库采样，您不必提前找到总重量（但您可以更频繁地从随机数生成器中采样）。代码可能看起来像

Broker selected = null;
int s = 0;
foreach(Broker broker in brokers) {
    s += broker.Weight;
    if (broker.Weight <= _rnd.Next(0,s)) {
        selected = broker;
    }
}

这需要通过列表经纪人一次。但是，如果经纪人列表是固定的或不经常更改，您可以保留一个累积和数组，即 A[i] 是所有经纪人的权重总和 0,..,i-1。那么 A[n] 是总权重，如果你选择一个介于 1 和 A[n-1] 之间的数字，比如说 x，你会找到经纪人 j st A[j-1] <= x < A[j]。为方便起见，您让 A[0] = 0。您可以使用二进制搜索在 log(n) 步中找到这个经纪人编号 j，我将把代码留作一个简单的练习。如果您的数据经常更改，这可能不是一个好方法，因为每次一些权重更改时，您可能需要更新数组的大部分。

score 0 · Accepted Answer

我想出了这个解决方案的通用版本：

public static class WeightedEx
{
    /// <summary>
    /// Select an item from the given sequence according to their respective weights.
    /// </summary>
    /// <typeparam name="TItem">Type of item item in the given sequence.</typeparam>
    /// <param name="a_source">Given sequence of weighted items.</param>
    /// <returns>Randomly picked item.</returns>
    public static TItem PickWeighted<TItem>(this IEnumerable<TItem> a_source)
        where TItem : IWeighted
    {
        if (!a_source.Any())
            return default(TItem);

        var source= a_source.OrderBy(i => i.Weight);

        double dTotalWeight = source.Sum(i => i.Weight);

        Random rand = new Random();

        while (true)
        {
            double dRandom = rand.NextDouble() * dTotalWeight;

            foreach (var item in source)
            {
                if (dRandom < item.Weight)
                    return item;

                dRandom -= item.Weight;
            }
        }
    }
}

/// <summary>
/// IWeighted: Implementation of an item that is weighted.
/// </summary>
public interface IWeighted
{
    double Weight { get; }
}

score 0 · Accepted Answer

只是为了分享我自己的实现。希望你会发现它有用。

    // Author: Giovanni Costagliola <giovanni.costagliola@gmail.com>

    using System;
    using System.Collections.Generic;
    using System.Linq;

    namespace Utils
    {
    /// <summary>
    /// Represent a Weighted Item.
    /// </summary>
    public interface IWeighted
    {
        /// <summary>
        /// A positive weight. It's up to the implementer ensure this requirement
        /// </summary>
        int Weight { get; }
    }

    /// <summary>
    /// Pick up an element reflecting its weight.
    /// </summary>
    /// <typeparam name="T"></typeparam>
    public class RandomWeightedPicker<T> where T:IWeighted
    {
        private readonly IEnumerable<T> items;
        private readonly int totalWeight;
        private Random random = new Random();

        /// <summary>
        /// Initiliaze the structure. O(1) or O(n) depending by the options, default O(n).
        /// </summary>
        /// <param name="items">The items</param>
        /// <param name="checkWeights">If <c>true</c> will check that the weights are positive. O(N)</param>
        /// <param name="shallowCopy">If <c>true</c> will copy the original collection structure (not the items). Keep in mind that items lifecycle is impacted.</param>
        public RandomWeightedPicker(IEnumerable<T> items, bool checkWeights = true, bool shallowCopy = true)
        {
            if (items == null) throw new ArgumentNullException("items");
            if (!items.Any()) throw new ArgumentException("items cannot be empty");
            if (shallowCopy)
                this.items = new List<T>(items);
            else
                this.items = items;
            if (checkWeights && this.items.Any(i => i.Weight <= 0))
            {
                throw new ArgumentException("There exists some items with a non positive weight");
            }
            totalWeight = this.items.Sum(i => i.Weight);
        }
        /// <summary>
        /// Pick a random item based on its chance. O(n)
        /// </summary>
        /// <param name="defaultValue">The value returned in case the element has not been found</param>
        /// <returns></returns>
        public T PickAnItem()
        {
            int rnd = random.Next(totalWeight);
            return items.First(i => (rnd -= i.Weight) < 0);
        }

        /// <summary>
        /// Resets the internal random generator. O(1)
        /// </summary>
        /// <param name="seed"></param>
        public void ResetRandomGenerator(int? seed)
        {
            random = seed.HasValue ? new Random(seed.Value) : new Random();
        }
    }
}

要点：https ://gist.github.com/MrBogomips/ae6f6c9af8032392e4b93aaa393df447

score 0 · Accepted Answer

原始问题中的实现对我来说似乎有点奇怪；

列表的总权重为 60，因此随机数为 0-59。它总是根据权重检查随机数，然后递减它。在我看来，它会根据它们的顺序来支持列表中的东西。

这是我正在使用的通用实现 - 症结在 Random 属性中：

using System;
using System.Collections.Generic;
using System.Linq;

public class WeightedList<T>
{
    private readonly Dictionary<T,int> _items = new Dictionary<T,int>();

    // Doesn't allow items with zero weight; to remove an item, set its weight to zero
    public void SetWeight(T item, int weight)
    {
        if (_items.ContainsKey(item))
        {
            if (weight != _items[item])
            {
                if (weight > 0)
                {
                    _items[item] = weight;
                }
                else
                {
                    _items.Remove(item);
                }

                _totalWeight = null; // Will recalculate the total weight later
            }
        }
        else if (weight > 0)
        {
            _items.Add(item, weight);

            _totalWeight = null; // Will recalculate the total weight later
        }
    }

    public int GetWeight(T item)
    {
        return _items.ContainsKey(item) ? _items[item] : 0;
    }

    private int? _totalWeight;
    public int totalWeight
    {
        get
        {
            if (!_totalWeight.HasValue) _totalWeight = _items.Sum(x => x.Value);

            return _totalWeight.Value;
        }
    }

    public T Random
    {
        get
        {
            var temp = 0;
            var random = new Random().Next(totalWeight);

            foreach (var item in _items)
            {
                temp += item.Value;

                if (random < temp) return item.Key;
            }

            throw new Exception($"unable to determine random {typeof(T)} at {random} in {totalWeight}");
        }
    }
}

score 0 · Accepted Answer

另一种选择是这个

private static Random _Rng = new Random();
public static Broker GetBroker(List<Broker> brokers){
    List<Broker> weightedBrokerList = new List<Broker>();
    foreach(Broker broker in brokers) {
        for(int i=0;i<broker.Weight;i++) {
            weightedBrokerList.Add(broker);
        }
    }
    return weightedBrokerList[_Rng.Next(weightedBrokerList.Count)];
}

c# - 随机加权选择

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