2

我被要求制作一个遗传算法，目标是确定一个 1 和 0 最多的 8 位字符串。eval 函数应该返回更改的数量加 1。例如，00000000 返回 1，00011100 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)

我实际上无法弄清楚遗传算法部分（交叉/突变）。我应该随机配对数字，然后随机选择一对，让一对保持不变，然后在随机点交叉。然后它将通过随机突变整个种群中的一位而结束。当前的交叉和变异代码取自我发现并试图理解的遗传算法示例。欢迎任何帮助。

4

2 回答 2

1

我建议的部分内容：

该代码不起作用，但它可能会传输信息。

# 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
于 2013-04-13T20:07:44.140 回答
0

我发现我喜欢做的一件事是：

  1. 选择最后一批的最佳候选人，比如说 5。
  2. 与 2、3、4、5 配对 1 个。
  3. 与 3, 4, 5 配对 2
  4. 有 3 与 4 和 5 交配
  5. 有 4 与 5 交配。通常，如果您将原始的 5 放入下一代并且每次交配产生 2 个后代，那么您的种群在达到这一点之前就已经满了。一个交配和它的相反双胞胎。
  6. 至于实际的交配，我喜欢在长度的 40% 到 60% 之间的一点上随机切割染色体，然后在下一次交配时，我选择该范围内的另一个随机点。
  7. 与它们交配后，我会检查染色体中的每一位，并给它大约 5% 的翻转或变异机会
  8. 有时我也会让一些最差的两个配对，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，我的这个是在四月份被问到的。对不起挖坟。

于 2013-07-13T23:13:21.533 回答