遗传算法求解旅行商（TSP）问题 -- python
参考资料：
from itertools import permutations
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
from itertools import combinations, permutations
#%matplotlib inline
def fitnessFunction(pop,num,city_num,x_position_add_end,y_position_add_end):
'''适应度函数，计算每个排列的适应度，并保存到pop矩阵第二维的最后一项'''
for x1 in range(num):
square_sum = 0
for x2 in range(city_num):
square_sum += (x_position_add_end[int(pop[x1][x2])] - x_position_add_end[int(pop[x1][x2+1])])**2 + (y_position_add_end[int(pop[x1][x2])] - y_position_add_end[int(pop[x1][x2+1])])**2
# print(round(1/np.sqrt(square_sum),7))
pop[x1][-1] = round(1/np.sqrt(square_sum),7)
def choiceFuction(pop):
'''
这里的做法：比如A当前种群中的最优解，B为经过交叉、变异后的最差解，把A作为最当前代中的最优解保存下来作为这一代的最优解，同时A也参与交叉
和变异。经过交叉、变异后的最差解为B，那么我再用A替代B。
:argument pop矩阵
:return 本代适应度最低的个体的索引值和本代适应度最高的个体
'''
yield np.argmin(pop[:, -1])
yield pop[np.argmax(pop[:, -1])]
def choice(pop,num,city_num,x_position_add_end,y_position_add_end,b):
fitnessFunction(pop,num,city_num,x_position_add_end,y_position_add_end)
c,d =choiceFuction(pop)
# 上一代的最优值替代本代中的最差值
pop[c] = b
return pop
def drawPic(maxFitness,x_position,y_position,i):
index = np.array(maxFitness[:-1],dtype=np.int32)
x_position_add_end = np.append(x_position[index],x_position[[index[0]]])
y_position_add_end = np.append(y_position[index],y_position[[index[0]]])
fig = plt.figure()
plt.plot(x_position_add_end,y_position_add_end,'-o')
plt.xlabel('x',fontsize = 16)
plt.ylabel('y',fontsize = 16)
plt.title('{iter}'.format(iter=i))
def matuingFuction(pop,pc,city_num,pm,num):
mating_matrix =np.array(1-(np.random.rand(num)>pc),dtype=np.bool) # 交配矩阵，如果为true则进行交配
a = list(pop[mating_matrix][:,:-1])# 进行交配的个体
b = list(pop[np.array(1-mating_matrix,dtype=bool)][:,:-1]) # 未进行交配的个体,直接放到下一代
b = [list(i) for i in b] # 对b进行类型转换，避免下面numpy.array 没有index属性
# print(a)
if len(a)%2 !=0:
b.append(a.pop())
# print('ab的长度：',len(a),len(b))
for i in range(int(len(a)/2)):
# 随机初始化两个交配点,这里写得不好，这边的两个点初始化都是一个在中间位置偏左，一个在中间位置偏右
p1 = np.random.randint(1,int(city_num/2)+1)
p2 = np.random.randint(int(city_num/2)+1,city_num)
x1 = list(a.pop())
x2 = list(a.pop())
matuting(x1,x2,p1,p2)
# 交配之后产生的个体进行一定概率上的变异
variationFunction(x1,pm,city_num)
variationFunction(x2,pm,city_num)
b.append(x1)
b.append(x2)
zero = np.zeros((num,1))
# print('b的形状：',len(b))
temp = np.column_stack((b, zero))
return temp
def matuting(x1,x2,p1,p2):
# 以下进行交配
# 左边交换位置
temp = x1[:p1]
x1[:p1] = x2[:p1]
x2[:p1] = temp
# 右边交换位置
temp = x1[p2:]
x1[p2:] = x2[p2:]
x2[p2:] = temp
# 寻找重复的元素
center1 = x1[p1:p2]
center2 = x2[p1:p2]
while True: # x1左边
for i in x1[:p1]:
if i in center1:
# print(center1.index(i)) # 根据值找到索引
x1[x1[:p1].index(i)] = center2[center1.index(i)]
break
if np.intersect1d(x1[:p1],center1).size == 0: # 如果不存在交集，则循环结束
break
while True: # x1右边
for i in x1[p2:]:
if i in center1:
# print(center1.index(i)) # 根据值找到索引
x1[x1[p2:].index(i) + p2] = center2[center1.index(i)]
# print(x1)
break
if np.intersect1d(x1[p2:],center1).size == 0: # 如果不存在交集，则循环结束
break
while True: # x2左边
for i in x2[:p1]:
if i in center2:
# print(center2.index(i)) # 根据值找到索引
x2[x2[:p1].index(i)] = center1[center2.index(i)]
break
if np.intersect1d(x2[:p1],center2).size == 0: # 如果不存在交集，则循环结束
break
while True: # x2右边
for i in x2[p2:]:
if i in center2:
# print(center2.index(i)) # 根据值找到索引
x2[x2[p2:].index(i) + p2] = center1[center2.index(i)]
# print(x2)
break
if np.intersect1d(x2[p2:],center2).size == 0: # 如果不存在交集，则循环结束
break
def variationFunction(list_a,pm,city_num):
'''变异函数'''
if np.random.rand() < pm:
p1 = np.random.randint(1,int(city_num/2)+1)
p2 = np.random.randint(int(city_num/2)+1,city_num)
# print(p1,p2)
temp = list_a[p1:p2]
temp.reverse()
list_a[p1:p2] = temp
# print(list_a)
def main():
# 初始化
pop = [] # 存放访问顺序和每个个体适应度
num = 250 # 初始化群体的数目
city_num = 10 # 城市数目
pc = 0.9 # 每个个体的交配概率
pm = 0.2 # 每个个体的变异概率
x_position = np.random.randint(0,100,size=city_num)
y_position = np.random.randint(0,100,size=city_num)
x_position_add_end = np.append(x_position,x_position[0])
y_position_add_end = np.append(y_position,y_position[0])
for i in range(num):
pop.append(np.random.permutation(np.arange(0,city_num))) # 假设有5个城市，初始群体的数目为60个
# 初始化化一个60*1的拼接矩阵，值为0
zero = np.zeros((num,1))
pop = np.column_stack((pop, zero)) # 矩阵的拼接
fitnessFunction(pop,num,city_num,x_position_add_end,y_position_add_end)
for i in range(180):
a,b = choiceFuction(pop) # a 为当代适应度最小的个体的索引，b为当代适应度最大的个体,这边要保留的是b
# print('索引值和适应度最大的个体：',a,b)
# pop[a]=b
if (i+1)%10==0:
drawPic(b,x_position,y_position,i+1) # 根据本代中的适应度最大的个体画图
pop_temp = matuingFuction(pop,pc,city_num,pm,num) #交配变异
pop = choice(pop_temp,num,city_num,x_position_add_end,y_position_add_end,b)
main()
