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我正在尝试运行我用 python 编写的遗传算法。不幸的是，当发生突变时，尽管使用精英主义将最适合的解决方案从上一代传递给新的解决方案，但最适合的解决方案可能比上一代的最适合解决方案更差。像这样：

There are 86825 generations left
Invalid count is:  0
The fittest individual has a fitness of 16.9094. 
The least fit individual has a fitness  of 36.6535
*******************************************************************************
mutation on 107
There are 86824 generations left
Invalid count is:  3
The fittest individual has a fitness of 19.8637. 
The least fit individual has a fitness of 1.1618e+09

我试图实施精英主义，我认为这会避免这种情况发生，但它仍然会发生。我的算法按以下顺序执行：

NUM_GEN = 100000
print "Running Genetic Algorithm for %d generations." % NUM_GEN
gen_count = 0
while gen_count < NUM_GEN:
    #genetic.add_fitness_key()
    genetic.add_fitness_fast()
    fittest_list = np.append(fittest_list, genetic.fittest_fitness)
    least_fit_list = np.append(least_fit_list, genetic.least_fitness)
    genetic.sort_pop()
    genetic.make_new_pop()
    genetic.elitism()
    genetic.population = genetic.new_pop
    print "There are %g generations left" %(NUM_GEN-gen_count) 
    gen_count+=1

被调用的函数是：

def select_parent_from_tournament(self):
    x = random.randint(0, 19)
    player1 = self.population[x]
    y = random.randint(0, 19)
    player2 = self.population[y]
    if player1['fitness'] <= player2['fitness']:
        parent = player1['chrom_list']
    else:
        parent = player2['chrom_list']
    return parent

def crossover(self):
    crossover_point = random.randint(0, self.chromosome_size)*(self.string_length)
    parent1 = self.select_parent_from_tournament()
    parent2 = self.select_parent_from_tournament()
    parent1 = self.mutate(parent1)
    parent2 = self.mutate(parent2)
    child1 = parent1[:crossover_point] + parent2[crossover_point:]
    child2 = parent1[crossover_point:] + parent2[:crossover_point]
    return child1, child2

def mutate(self, chromosome):
    for i in range(len(chromosome)):
        if random.random() < self.mutation_rate:
            print 'mutation on %i' % i
            if chromosome[i] =='0':
                chromosome[i] = '1'
            else:
                chromosome[i] = '0'
    return chromosome

def make_new_pop(self):
    self.new_pop = []
    for i in range(10):
        dictionary1= {}
        dictionary2 = {}
        dictionary1['chrom_list'], dictionary2['chrom_list'] = \
        self.crossover()
        self.new_pop = np.append(self.new_pop, [dictionary1, dictionary2])

def elitism(self):
    r = random.randint(0, 19)
    self.new_pop[r] = self.population[0]

所以我不明白为什么如果发生突变，旧人口中最适合的解决方案不会传递给新人口？

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1 回答 1

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在您的交叉方法中，您执行parent1 = self.select_parent_from_tournament()返回对原始种群中染色体列表的引用。在那之后，你做一个修改列表的突变（在原始人口中）。只有在突变后，您才能通过child1 = parent1[:crossover_point] + parent2[crossover_point:]. 正如@Xavier 建议的那样，您需要从mutation().

实际上，您的mutation()方法改变了原始人口。

作为旁注，通常交叉和变异是两种不同的操作。此外，文献中的参考资料表明突变概率应保持在非常低的水平。否则它会让你无法收敛。

于 2013-01-14T16:57:06.310 回答