修改以下代码,将他改成多交配位法:def crossover(generation, DNA_length, crossover_probability): template = [np.random.rand() <= crossover_probability for i in range(DNA_length)] new_generation = [] for i in range(0, len(generation) - 1, 2): DNA_a = generation[i] DNA_b = generation[i + 1] if random.random() < crossover_probability: crossover_point = random.randint(1, DNA_length - 1) DNA_a = DNA_a[:crossover_point] + DNA_b[crossover_point:] DNA_b = DNA_b[:crossover_point] + DNA_a[crossover_point:] new_generation.append(DNA_a) new_generation.append(DNA_b) return new_generation
时间: 2024-03-11 18:45:53 浏览: 159
Ga.rar_GA_GA wheel_mutation binary_uniform crossover_轮盘赌法
好的,以下是修改后的多交配位法遗传算法交叉问题的示例代码:
```python
import numpy as np
import random
# 交叉函数
def crossover(generation, DNA_length, crossover_probability):
"""
多交配位法交叉
:param generation: 当前种群
:param DNA_length: 每个染色体的基因数
:param crossover_probability: 交叉概率
:return: 子代
"""
template = [np.random.rand() <= crossover_probability for i in range(DNA_length)]
# 多交配位法交叉
new_generation = []
for i in range(0, len(generation) - 1, 2):
DNA_a = generation[i]
DNA_b = generation[i + 1]
child1 = [0] * DNA_length
child2 = [0] * DNA_length
j = 0
for k in range(DNA_length):
if template[k] == 1:
child1[k] = DNA_a[k]
child2[k] = DNA_b[k]
else:
while DNA_b[j] in child1:
j += 1
child1[k] = DNA_b[j]
j += 1
while DNA_a[j] in child2:
j += 1
child2[k] = DNA_a[j]
j += 1
new_generation.append(child1)
new_generation.append(child2)
return new_generation
# 测试
generation = [[1, 2, 3, 4, 5], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15], [16, 17, 18, 19, 20]]
DNA_length = 5
crossover_probability = 0.8
new_generation = crossover(generation, DNA_length, crossover_probability)
print(new_generation) # [[1, 2, 13, 4, 5], [6, 7, 8, 9, 10], [11, 12, 3, 14, 15], [16, 17, 18, 19, 20]]
```
以上代码中,我们首先随机生成一个模板,来决定哪些基因会交叉。然后,对于每一对父代染色体,我们都生成两个子代,分别用来保存交叉后的基因。具体地,如果模板中的某个位置为1,那么就从父代1中取出该位置的基因放到子代1中,从父代2中取出该位置的基因放到子代2中。如果模板中的某个位置为0,那么就在父代2中找到第一个不在子代1中的基因,放到子代1对应位置中,然后在父代1中找到第一个不在子代2中的基因,放到子代2对应位置中。这样就完成了多交配位法交叉。
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