遗传算法中的轮盘赌选择

我自己做这件事已经有几年了,但是下面的伪代码在google上很容易找到。

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

here 如果你需要进一步的细节。

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,则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]))

这不仅速度更快,而且是非常简洁的代码。C++中的STL有一个类似的二分算法,如果你使用的是这种语言。

发布的伪代码包含一些不清楚的元素,这增加了生成伪代码的复杂性 孩子 而不是进行纯粹的选择。下面是该伪代码的简单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

/// \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;
  }   
}

单个选择所需的平均尝试次数为:

τ=f _最大值 /平均值(f)

• F _最大值 是人口的最大适应度
• 平均值(f)是平均适合度

τ并不明确地依赖于群体中个体的数量(N),但比率可以随N而变化。

该算法的典型复杂性为O(1) (使用搜索算法的轮盘赌轮选择具有O(N)或O(logn)复杂性)。

这个过程的概率分布确实与经典轮盘赌选择中的概率分布相同。

有关更多详细信息,请参阅:

• 基于随机接受的轮盘赌轮选择 (亚当·利波斯基,多罗塔·利波夫斯卡——2011)

以下是C语言中的一些代码:

// 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**

从上面的答案中,我得到了下面的答案,这比答案本身更清楚。

随机(和):随机(12) 通过遍历总体,我们检查以下内容:随机<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      |   ...

    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;
        }
    }

斯坦福人工智能实验室(Stanford AI lab)的Thrun教授在其Udacity的CS373中还介绍了python中的快速(呃?)重新采样代码。谷歌搜索结果导致以下链接:

http://www.udacity-forums.com/cs373/questions/20194/fast-resampling-algorithm

这是我最近为轮盘赌选择编写的一个紧凑的java实现,希望能有所帮助。

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;
}

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

轮盘赌选择 实施: 通常的 随机接受 一

通常的

# 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

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,它们将返回相同的结果。

有用资源:

http://natureofcode.com/book/chapter-9-the-evolution-of-code -对初学者友好且清晰的遗传算法章节。解释 轮盘赌选择 作为一桶木头字母(你放的越多,选择a的机会就越大, 通常的

https://en.wikipedia.org/wiki/Fitness_proportionate_selection -描述 随机接受 算法。

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;
        }

这 斯威夫特4

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,给定两元素数组:

[0.9, 0.1]

weightedRandomIndex()

下面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

这个答案与毛主席在这里的答案基本相同: https://stackoverflow.com/a/15582983/74975

下面是python中的代码。此代码还可以处理适应度的负值。

from numpy import min, sum, ptp, array 
from numpy.random import uniform 

list_fitness1 = array([-12, -45, 0, 72.1, -32.3])
list_fitness2 = array([0.5, 6.32, 988.2, 1.23])

def get_index_roulette_wheel_selection(list_fitness=None):
    """ It can handle negative also. Make sure your list fitness is 1D-numpy array"""
    scaled_fitness = (list_fitness - min(list_fitness)) / ptp(list_fitness)
    minimized_fitness = 1.0 - scaled_fitness
    total_sum = sum(minimized_fitness)
    r = uniform(low=0, high=total_sum)
    for idx, f in enumerate(minimized_fitness):
        r = r + f
        if r > total_sum:
            return idx

get_index_roulette_wheel_selection(list_fitness1)
get_index_roulette_wheel_selection(list_fitness2)

1. 确保你的健身清单是1D numpy阵列
2. 将适应度列表缩放到范围[0,1]
3. 通过1.0-比例适应度列表将最大问题转换为最小问题
5. 继续在最小化适应度列表中添加元素,直到我们得到的值大于总和
6. 您可以查看适应度是否很小-->它在最小化适应度方面有更大的价值-->它有更大的机会增加,使价值大于总和。

我用C#编写了一个版本,我真的希望确认它确实是正确的:

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;

    }