二进制数的Python遗传算法

我被要求做一个遗传算法，目标是确定一个8位的字符串，最多有1个和0个。eval函数应返回更改数加1。例如00000000返回10011100返回3，01100101返回6。这就是我所拥有的：

def random_population():
    from random import choice

    pop = ''.join(choice(('0','1')) for _ in range(8))
    return pop   

def mutate(dna):   
    """   For each gene in the DNA, there is a 1/mutation_chance chance 
    that it will be   switched out with a random character. This ensures 
    diversity in the   population, and ensures that is difficult to get stuck in 
    local minima.   """   
    dna_out = ""   
    mutation_chance = 100   
    for c in xrange(DNA_SIZE):
        if int(random.random()*mutation_chance) == 1:
            dna_out += random_char()
        else:
            dna_out += dna[c]   return dna_out

def crossover(dna1, dna2):   
    """   Slices both dna1 and dna2 into two parts at a random index within their   
    length and merges them. Both keep their initial sublist up to the crossover   
    index, but their ends are swapped.   """   
    pos = int(random.random()*DNA_SIZE)
    return (dna1[:pos]+dna2[pos:], dna2[:pos]+dna1[pos:])

def eval(dna):
    changes = 0
    for index, bit in enumerate(dna):
        if(index == 0):
            prev = bit
        else:
            if(bit != prev):
                changes += 1
        prev = bit
    return changes+1


#============== End Functions =======================#


#============== Main ================# changes = 0

prev = 0
dna = random_population()
print "dna: "
print dna
print eval(dna)

---

我很难真正理解遗传算法的部分（交叉/变异）。我应该随机配对数字，然后随机选择一对，让一对保持不变，然后在一个随机点交叉。然后，它将通过在整个种群中随机变异一位来结束。当前的交叉和变异代码只是从我发现的一个遗传算法示例中提取出来的，我正试图理解它。欢迎任何帮助。

我建议的一部分：

代码不起作用，但它可能传输信息

# a population consists of many individuals
def random_population(population_size = 10):
    from random import choice

    pop = [''.join(choice(('0','1')) for _ in range(8)) for i in range(population_size)]
    return pop   

# you need a fitness function
def fitness(individual):
    return # a value from 0 up

def algorithm():
    # a simple algorithm somehow alike
    # create population
    population = random_population()
    # this loop must stop after some rounds because the best result may never be reached
    while goal_not_reached(population) and not time_is_up():
        # create the roulette wheel
        roulette_wheel = map(fitness, population)
        # highest value of roulette wheel
        max_value = sum(roulette_wheel)
        # the new generation
        new_population = []
        for i in range(len(population) - len(new_population)):
             # create children from the population
                 # choose 2 random values from 0 to max_value and find the individuals
                 # for it in the roulette wheel, combine them to new individuals 
             new_population.append(new_individual)
        # mutate the population
        population = map(mutate, new_population)             # a new generation is created

---

我发现我喜欢做的一件事是：

从上一批中选出最优秀的候选人，比如说5名

让1只与2,3,4,5配对

让2个与3、4、5配对

让3人与4人和5人交配

4人与5人交配。一般来说，如果你让原来的5只进入下一代，并且每次交配产生2只后代，那么在你达到这一点之前，你的种群已经满了。一对交配的双胞胎

至于实际的杂交，我喜欢在长度的40%到60%之间随机切割染色体，然后在下一次交配时，我选择范围内的另一个随机点

在我和它们交配后，我检查了我染色体上的每一个片段，并给它一个大致的数字！5%的翻转或突变几率

有时，我也会让最差的两个人中的一些人交配，以降低我获得局部最大值或最小值的机会

我希望这对你有点帮助

-杰夫

---

编辑：哦，天哪，这是四月份问的。很抱歉挖坟墓。

一个群体由许多个体组成——我只看到一个“dna”。交叉有助于基因“子程序”的收敛，变异有助于产生达到目标所需的错误。此外，你需要一个适应力函数，确定个体交叉和重组的可能性。你可以使用轮盘赌来决定哪些人可以交叉创造下一代的孩子。