轮盘赌选择在遗传算法

Question

任何人都可以提供一些伪代码一轮选择的功能？我将如何实现这一点：

我真的不知道如何解读这个数学符号。我从来没有采取任何概率或统计数据。

Answer 1

自从我自己完成这项工作已有几年了，但是谷歌上发现了以下伪代码。

for all members of population
    sum += fitness of this individual
end for

for all members of population
    probability = sum of probabilities + (fitness / sum)
    sum of probabilities += probability
end for

loop until new population is full
    do this twice
        number = Random between 0 and 1
        for all members of population
            if number > probability but less than next probability 
                then you have been selected
        end for
    end
    create offspring
end loop

如果您需要更多信息，请访问此处的网站。的信息。

Answer 2

已经有很多正确的解决方案，但我认为这段代码更清晰。

def select(fs):
    p = random.uniform(0, sum(fs))
    for i, f in enumerate(fs):
        if p <= 0:
            break
        p -= f
    return i

此外，如果累积fs，您可以生成更有效的解决方案。

cfs = [sum(fs[:i+1]) for i in xrange(len(fs))]

def select(cfs):
    return bisect.bisect_left(cfs, random.uniform(0, cfs[-1]))

这既快又简单，代码非常简洁。如果这是您正在使用的语言，则C ++中的STL具有类似的二分算法。

Answer 3

发布的伪代码包含一些不清楚的元素，它增加了生成后代的复杂性，而不是执行纯粹的选择。这是一个伪代码的简单python实现：

def roulette_select(population, fitnesses, num):
    """ Roulette selection, implemented according to:
        <http://stackoverflow.com/questions/177271/roulette
        -selection-in-genetic-algorithms/177278#177278>
    """
    total_fitness = float(sum(fitnesses))
    rel_fitness = [f/total_fitness for f in fitnesses]
    # Generate probability intervals for each individual
    probs = [sum(rel_fitness[:i+1]) for i in range(len(rel_fitness))]
    # Draw new population
    new_population = []
    for n in xrange(num):
        r = rand()
        for (i, individual) in enumerate(population):
            if r <= probs[i]:
                new_population.append(individual)
                break
    return new_population

Answer 4

这就是所谓赌轮选择通过随机的接受：

/// \param[in] f_max maximum fitness of the population
///
/// \return index of the selected individual
///
/// \note Assuming positive fitness. Greater is better.

unsigned rw_selection(double f_max)
{
  for (;;)
  {
    // Select randomly one of the individuals
    unsigned i(random_individual());

    // The selection is accepted with probability fitness(i) / f_max
    if (uniform_random_01() < fitness(i) / f_max)
      return i;
  }   
}

平均数量的尝试，需要一个单一的选择是：

τ=f_max /avg(f)

• f_max 是最大的健身的人口
• avg(f)的平均健身

τ并不取决于明确的数量上的个人在人口(N)，但该比率可以改变与N.

然而，在许多应用程序(在健身仍然是界定和平均健身并不减少到0增加N)τ不增加unboundedly N并因此 一个典型的复杂的算法是O(1) (赌轮选择使用的搜索算法已O(N)或O(日志N)的复杂性).

概率分布的这一程序确实是一样的，在典型的轮盘赌轮选择。

为进一步的详细信息见：

• 轮盘赌轮选择通过随机的接受 (Adam Liposki,Dorota Lipowska-2011年)

Answer 5

以下是C中的一些代码：

// Find the sum of fitnesses. The function fitness(i) should 
//return the fitness value   for member i**

float sumFitness = 0.0f;
for (int i=0; i < nmembers; i++)
    sumFitness += fitness(i);

// Get a floating point number in the interval 0.0 ... sumFitness**
float randomNumber = (float(rand() % 10000) / 9999.0f) * sumFitness;

// Translate this number to the corresponding member**
int memberID=0;
float partialSum=0.0f;

while (randomNumber > partialSum)
{
   partialSum += fitness(memberID);
   memberID++;
} 

**// We have just found the member of the population using the roulette algorithm**
**// It is stored in the "memberID" variable**
**// Repeat this procedure as many times to find random members of the population**

Answer 6

从上面的回答中，我得到了以下内容，这对我来说比答案本身更清晰。

举个例子：

随机（和）::随机（12） 迭代人口，我们检查以下内容：random <！> lt;总和

让我们选择7作为随机数。

Index   |   Fitness |   Sum |   7 < Sum
0       |   2   |   2       |   false
1       |   3   |   5       |   false
2       |   1   |   6       |   false
3       |   4   |   10      |   true
4       |   2   |   12      |   ...

通过这个例子，最合适（指数3）的选择百分比最高（33％）;因为随机数只需要在6 - <！> gt; 10之内着陆，它就会被选中。

    for (unsigned int i=0;i<sets.size();i++) {
        sum += sets[i].eval();
    }       
    double rand = (((double)rand() / (double)RAND_MAX) * sum);
    sum = 0;
    for (unsigned int i=0;i<sets.size();i++) {
        sum += sets[i].eval();
        if (rand < sum) {
            //breed i
            break;
        }
    }

Answer 7

教授。斯坦福人工智能实验室的Thrun还在他的CS373 of Udacity期间在python中提出了一个快速（呃？）重采样代码。谷歌搜索结果导致以下链接：

http://www.udacity-forums.com/ cs373 /问题/ 20194 /快重新采样算法

希望这有帮助

Answer 8

这是我最近为轮盘选择写的一个紧凑的java实现，希望可以使用。

public static gene rouletteSelection()
{
    float totalScore = 0;
    float runningScore = 0;
    for (gene g : genes)
    {
        totalScore += g.score;
    }

    float rnd = (float) (Math.random() * totalScore);

    for (gene g : genes)
    {   
        if (    rnd>=runningScore &&
                rnd<=runningScore+g.score)
        {
            return g;
        }
        runningScore+=g.score;
    }

    return null;
}

Answer 9

MatLab中的轮盘赌轮选择：

TotalFitness=sum(Fitness);
    ProbSelection=zeros(PopLength,1);
    CumProb=zeros(PopLength,1);

    for i=1:PopLength
        ProbSelection(i)=Fitness(i)/TotalFitness;
        if i==1
            CumProb(i)=ProbSelection(i);
        else
            CumProb(i)=CumProb(i-1)+ProbSelection(i);
        end
    end

    SelectInd=rand(PopLength,1);

    for i=1:PopLength
        flag=0;
        for j=1:PopLength
            if(CumProb(j)<SelectInd(i) && CumProb(j+1)>=SelectInd(i))
                SelectedPop(i,1:IndLength)=CurrentPop(j+1,1:IndLength);
                flag=1;
                break;
            end
        end
        if(flag==0)
            SelectedPop(i,1:IndLength)=CurrentPop(1,1:IndLength);
        end
    end

Answer 10

Based on my research ,Here is another implementation in C# if there is a need for it:


//those with higher fitness get selected wit a large probability 
//return-->individuals with highest fitness
        private int RouletteSelection()
        {
            double randomFitness = m_random.NextDouble() * m_totalFitness;
            int idx = -1;
            int mid;
            int first = 0;
            int last = m_populationSize -1;
            mid = (last - first)/2;

            //  ArrayList's BinarySearch is for exact values only
            //  so do this by hand.
            while (idx == -1 && first <= last)
            {
                if (randomFitness < (double)m_fitnessTable[mid])
                {
                    last = mid;
                }
                else if (randomFitness > (double)m_fitnessTable[mid])
                {
                    first = mid;
                }
                mid = (first + last)/2;
                //  lies between i and i+1
                if ((last - first) == 1)
                    idx = last;
            }
            return idx;
        }

Answer 11

好的，轮盘选择实施有两种方法：常用和随机接受。

常用算法：

# there will be some amount of repeating organisms here.
mating_pool = []

all_organisms_in_population.each do |organism|
  organism.fitness.times { mating_pool.push(organism) }
end

# [very_fit_organism, very_fit_organism, very_fit_organism, not_so_fit_organism]
return mating_pool.sample #=> random, likely fit, parent!

随机接受算法：

max_fitness_in_population = all_organisms_in_population.sort_by(:fitness)[0]
loop do
  random_parent = all_organisms_in_population.sample
  probability = random_parent.fitness/max_fitness_in_population * 100
  # if random_parent's fitness is 90%,
  # it's very likely that rand(100) is smaller than it.
  if rand(100) < probability
    return random_parent #=> random, likely fit, parent!
  else
    next #=> or let's keep on searching for one.
  end
end

您可以选择其中之一，他们将返回相同的结果。

---

有用的资源：

http://natureofcode.com/book/chapter-9 - 代码演变 - 关于遗传算法的初学者友好且清晰的章节。解释轮盘赌轮选择作为一桶木制字母（你输入的越多 - 选择A，常用算法的机会就越大）。

https://en.wikipedia.org/wiki/Fitness_proportionate_selection - 介绍随机接受算法。

Answer 12

这个 Swift 4 数组扩展实现了加权随机选择，a.k.a从其元素中选择轮盘：

public extension Array where Element == Double {

    /// Consider the elements as weight values and return a weighted random selection by index.
    /// a.k.a Roulette wheel selection.
    func weightedRandomIndex() -> Int {
        var selected: Int = 0
        var total: Double = self[0]

        for i in 1..<self.count { // start at 1
            total += self[i]
            if( Double.random(in: 0...1) <= (self[i] / total)) { selected = i }
        }

        return selected
    }
}

例如给定两个元素数组：

[0.9, 0.1]

weightedRandomIndex()将在90％的时间和10％的时间内返回零。

这是一个更完整的测试：

let weights = [0.1, 0.7, 0.1, 0.1]
var results = [Int:Int]()
let n = 100000
for _ in 0..<n {
    let index = weights.weightedRandomIndex()
    results[index] = results[index, default:0] + 1
}
for (key,val) in results.sorted(by: { a,b in weights[a.key] < weights[b.key] }) {
    print(weights[key], Double(val)/Double(n))
}

输出：

0.1 0.09906
0.1 0.10126
0.1 0.09876
0.7 0.70092

这个答案与Andrew Mao在这里的答案基本相同： https://stackoverflow.com/a/15582983/74975

Answer 13

我在C＃中写了一个版本，我真的在寻找确认它确实是正确的：

（roulette_selector是一个随机数，范围在0.0到1.0之间）

private Individual Select_Roulette(double sum_fitness)
    {
        Individual ret = new Individual();
        bool loop = true;

        while (loop)
        {
            //this will give us a double within the range 0.0 to total fitness
            double slice = roulette_selector.NextDouble() * sum_fitness;

            double curFitness = 0.0;

            foreach (Individual ind in _generation)
            {
                curFitness += ind.Fitness;
                if (curFitness >= slice)
                {
                    loop = false;
                    ret = ind;
                    break;
                }
            }
        }
        return ret;

    }