我正在用python编写一个简单的遗传算法,当我试图找到一组单倍体的平均适应度时,它给我的结果是去掉剩余部分,我不明白它为什么要这样做。以下是完整的来源
import random
def generate_random_haploid(haploid_length):
haploid = []
for x in range(haploid_length):
haploid.append(random.randint(0,1))
return haploid
def crossover_haploid(haploid_1, haploid_2):
locus = random.randint(1, len(haploid_1))
for x in range(locus - 1):
haploid_1[x] = haploid_2[x]
for x in range(len(haploid_2) - locus):
haploid_2[x + locus] = haploid_1[x + locus]
return [haploid_1, haploid_2]
def crossover_diploid(diploid_1, diploid_2):
children_1, children_2 = crossover_haploid(diploid_1[0], diploid_1[1]), crossover_haploid(diploid_2[0], diploid_2[1])
return crossover_haploid(children_1[0], children_2[1])
def flipbit(bit):
if bit == 1:
bit = 0
elif bit == 0:
bit = 1
return bit
def mutate_haploid(haploid, mutate_prob):
for x in haploid:
if random.randint(0, mutate_prob) <= mutate_prob:
haploid[x] = flipbit(haploid[x])
return haploid
def average_fitness(haploid_list):
return sum(haploid_list[0]) / len(haploid_list)
def fitness(haploid):
fitness = 0
for x in range(len(haploid)):
if haploid[x] == 1:
fitness += 1
return fitness
def print_haploid(haploid):
print(haploid, "Fitness: ", fitness(haploid))
x = generate_random_haploid(4)
y = generate_random_haploid(4)
print_haploid(x)
print_haploid(y)
print("-------------------------------")
children = crossover_haploid(x, y)
print_haploid(children[0])
print_haploid(children[1])
print("-------------------------------")
print("Parent Fitness: ", average_fitness([x, y]) )
print("-------------------------------")
print("Children Fitness: ", average_fitness([children[0], children[1]]) )
这是因为python integer要截短到最小值。你有很多选择来解决这个问题:
选项1
将其中一个值强制转换为float,python将自动向上转换所有其他int类型
^{pr2}$对代码的更正:
方案2
将此添加到脚本的顶部
现在,
5 / 2
将生成2.5
,并且您不需要如选项1所示更改您的average_fitness
方法。__future__
是指python3,其中/
操作符默认执行浮点除法。通过导入该特性,现在您将在代码中到处使用float
除法运算符/
,而不是python2的int
除法运算符选项3
您可以将
/
替换为//
运算符对代码的更正:
相关问题 更多 >
编程相关推荐