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()