这次代码的修改点在交叉变异 函数中，做了其他修改：
def crossover_and_mutation(pop, CROSSOVER_RATE=0.015): new_pop = [] for i in range(POP_SIZE//2-20): fatherpoint = np.random.randint(low=0, high=POP_SIZE) child=pop[fatherpoint] motherpoint = np.random.randint(low=0, high=POP_SIZE) cross_points = np.random.randint(low=0, high=DNA_SIZE * 3) # 随机产生交叉的点 child[cross_points] = pop[motherpoint][cross_points] new_pop.append(child) for i in range(20): fatherpoint = np.random.randint(low=0, high=POP_SIZE) child = pop[fatherpoint] mutation(child, MUTATION_RATE=1) new_pop.append(child) # for father in pop: # 遍历种群中的每一个个体，将该个体作为父亲 # child = father # 孩子先得到父亲的全部基因 # if np.random.rand() < CROSSOVER_RATE: # 产生子代时不是必然发生交叉，而是以一定的概率发生交叉 # # mother = pop[np.random.randint(POP_SIZE)] # 再种群中选择另一个个体，并将该个体作为母亲 # # mother = pop[np.random.randint(POP_SIZE)] # # # cross_points = np.random.randint(low=0, high=DNA_SIZE * 3) # 随机产生交叉的点 # # child[cross_points:] = mother[cross_points:] # 孩子得到位于交叉点后的母亲的基因 # child[cross_points] = mother[cross_points] # else: # mutation(child, MUTATION_RATE) # mutation(child,MUTATION_RATE)每个后代有一定的机率发生变异 # new_pop.append(child) return new_pop完整代码：
# x1*x1-x1*x2+x3 import numpy as np import random DNA_SIZE = 1 POP_SIZE = 100 CROSSOVER_RATE = 0.8 MUTATION_RATE = 0.015 N_GENERATIONS = 200 X_BOUND = [3.0, 5.0] # x1 Y_BOUND = [2.1, 6.7] # x2 Z_BOUND = [1.2, 9.6] # x3 sum=0#测试变量 def F(x, y, z): val = x * x - x * y + z # print(val.shape) #(100,) 100 <class 'numpy.ndarray'> return val def get_fitness(pop): x, y, z = translateDNA(pop) pred = F(x, y, z) return pred def translateDNA(pop): # pop表示种群矩阵，一行表示一个二进制编码表示的DNA，矩阵的行数为种群数目 # x_pop = pop[:,0:DNA_SIZE]#这样的写法shape 是(3, 1) 3行1列 ndim维度是2(行，列 矩阵 ) # 也可以认为是二维数组，有3行，每行有1个元素 size为3 [[3.18796615]\n [3.32110516]\n [4.34665405]] '''因为这样写x_pop = pop[:, 0:DNA_SIZE] shape是(3,1)是二维数组，所以会报"对象太深，无法容纳所需的数组"的错误， 第一种解决方法是进行reshape，比如reshape(3,)即变成了一维数组，元素个数是3个，即语法是x_pop=pop[:,0:DNA_SIZE].reshape(POP_SIZE,) 这时x_pop就变为[4.96893731 3.24515899 3.51500566] 一维数组 第二种方法是在矩阵（二维数组）pop中直接选择某一列元素，比如 pop[:, 0],表示选择pop第0列所有的元素 ''' x_pop = pop[:,0] # 取前DNA_SIZE个列表示x 这样的写法shape是(3,) ndim维度是1 一维数组 ，数组元素有3个 size为3 [4.28040552 3.25412449 4.61336022] # print(x_pop.shape) y_pop = pop[:, 1] # 取中间DNA_SIZE个列表示y z_pop = pop[:, 2] # 取后DNA_SIZE个列表示z # print(x_pop) return x_pop, y_pop, z_pop def mutation(child, MUTATION_RATE): if np.random.rand() < MUTATION_RATE: # 以MUTATION_RATE的概率进行变异 mutate_point = np.random.randint(0, DNA_SIZE * 3) # 随机产生一个实数，代表要变异基因的位置 if mutate_point == 0: child[mutate_point] = np.random.uniform(3.0, 5.0) elif mutate_point == 1: child[mutate_point] = np.random.uniform(2.1, 6.7) else: child[mutate_point] = np.random.uniform(1.2, 9.6) def crossover_and_mutation(pop, CROSSOVER_RATE=0.015): new_pop = [] for i in range(POP_SIZE//2-20): fatherpoint = np.random.randint(low=0, high=POP_SIZE) child=pop[fatherpoint] motherpoint = np.random.randint(low=0, high=POP_SIZE) cross_points = np.random.randint(low=0, high=DNA_SIZE * 3) # 随机产生交叉的点 child[cross_points] = pop[motherpoint][cross_points] new_pop.append(child) for i in range(20): fatherpoint = np.random.randint(low=0, high=POP_SIZE) child = pop[fatherpoint] mutation(child, MUTATION_RATE=1) new_pop.append(child) # for father in pop: # 遍历种群中的每一个个体，将该个体作为父亲 # child = father # 孩子先得到父亲的全部基因 # if np.random.rand() < CROSSOVER_RATE: # 产生子代时不是必然发生交叉，而是以一定的概率发生交叉 # # mother = pop[np.random.randint(POP_SIZE)] # 再种群中选择另一个个体，并将该个体作为母亲 # # mother = pop[np.random.randint(POP_SIZE)] # # # cross_points = np.random.randint(low=0, high=DNA_SIZE * 3) # 随机产生交叉的点 # # child[cross_points:] = mother[cross_points:] # 孩子得到位于交叉点后的母亲的基因 # child[cross_points] = mother[cross_points] # else: # mutation(child, MUTATION_RATE) # mutation(child,MUTATION_RATE)每个后代有一定的机率发生变异 # new_pop.append(child) return new_pop def getbest(pop, fitness): best_indiv = [] fitness = get_fitness(pop) max_fitness_index = np.argmax(fitness) pop_copy_max = pop[max_fitness_index] best_indiv.append(pop_copy_max) # print(best_indiv) return max_fitness_index def getworst(pop,fitness): worst_indiv = [] fitness = get_fitness(pop) min_fitness_index = np.argmin(fitness) pop_copy_min = pop[min_fitness_index] worst_indiv.append(pop_copy_min) return min_fitness_index def choicebyyang(arr, size, replace, p): for i in range(size): aa = np.random.rand() sum = 0 for j in range(size): sum = sum + p[j] # 累加概率 if sum >= aa: break arr[i] = j return arr def select(pop, fitness): # nature selection wrt pop's fitness # fitnew=fitness.copy() #深拷贝 fitnew = fitness # print('fitness',id(fitness))# 赋值操作在Python里是浅拷贝，两个变量地址一样 # print('fitnew',id(fitnew)) # print(id(fitnew)==id(fitness)) #True fitnew = fitnew + 1e-3 - np.min(fitnew) p = (fitnew) / (fitnew.sum()) # print(np.arange(POP_SIZE)) #产生的是一维数组[0 1 2 ...100] # idx = np.random.choice(np.arange(POP_SIZE), size=POP_SIZE, replace=True, p=p) # 从0~POP_SIZE这个序列里随机取样 如果[pop_size]是一维数组，就表示从这个一维数组中随机采样,采size个，上面这行是全采用 idx = choicebyyang(np.arange(POP_SIZE), size=POP_SIZE, replace=True, p=p) #一维数组 bestindex = getbest(pop, fitness) #最优个体下标 worstindex=getworst(pop,fitness) #最差个体下标 # new_idx=[bestindex,worstindex] #<class 'list'>: [0, 1] # new_idx=[idx]+[bestindex,worstindex] #<class 'list'>: [array([0, 0, 0]), 0, 2] # new_idx = list(idx) + [bestindex, worstindex] #<class 'list'>: [0, 1, 1, 1, 2] new_idx=list(idx) #<class 'list'>: [2, 1, 2] half_pop_idx=new_idx[:POP_SIZE//2-2] half_pop_idx.append(bestindex) half_pop_idx.append(worstindex) half_pop_idx2=new_idx[POP_SIZE//2:] return pop[half_pop_idx],pop[half_pop_idx2] # 尽量选择适应值高的函数值的个体 ''' 如果POP_SIZE=3,即种群个数是3，则从交叉，变异后的种群中，选择3个适应值高 pop[idx]=[2 0 0]的新个体去 更新pop种群，之后再进行不断的迭代，直到达到迭代次数终止。 ''' def print_info(pop): fitness = get_fitness(pop) max_fitness_index = np.argmax(fitness) print("max_fitness:", fitness[max_fitness_index]) x, y, z = translateDNA(pop) print("最优的基因型：", pop[max_fitness_index]) print("(x, y, z):", (x[max_fitness_index], y[max_fitness_index], z[max_fitness_index])) if __name__ == "__main__": # pop1 = np.random.uniform(3.0, 5.0, size=(POP_SIZE, DNA_SIZE)) # matrix (POP_SIZE, DNA_SIZE) # pop2 = np.random.uniform(2.1, 6.7, size=(POP_SIZE, DNA_SIZE)) # pop3 = np.random.uniform(1.2, 9.6, size=(POP_SIZE, DNA_SIZE)) # # # print(type(pop1))# (100,1) 维度是2(行列 矩阵) <class 'numpy.ndarray'> # # pop={pop1,pop2,pop3} # pop = np.hstack((pop1, pop2, pop3)) #水平拼接 # print(pop) ''' [[3.44603448 4.51707625 7.90178727] [4.57616299 5.11309286 4.86911781] [3.24273815 2.9253602 4.45149325] ... [4.39321276 3.1657492 5.16654786]] ''' # print(type(pop)) #<class 'numpy.ndarray'> n维数组 # print(pop.shape) #(100,3) 矩阵有 100行，3列 # print(pop.ndim) # 2 因为矩阵有行和列两个维度 # print(pop.size) #300 矩阵共有300个元素 # print(pop.dtype) #float64 矩阵元素类型是float64 # for i in range(100):#测试代码 pop1 = np.random.uniform(3.0, 5.0, size=(POP_SIZE, DNA_SIZE)) # matrix (POP_SIZE, DNA_SIZE) pop2 = np.random.uniform(2.1, 6.7, size=(POP_SIZE, DNA_SIZE)) pop3 = np.random.uniform(1.2, 9.6, size=(POP_SIZE, DNA_SIZE)) # print(type(pop1))# (100,1) 维度是2(行列 矩阵) <class 'numpy.ndarray'> # pop={pop1,pop2,pop3} pop = np.hstack((pop1, pop2, pop3)) #水平拼接 for _ in range(N_GENERATIONS): # 迭代N代 x, y, z = translateDNA(pop) # 这句代码，我觉得没啥作用 # print(x) #(100,) [4.82264692 4.04610252 4.92107325 4.49556859 3.1322498 3.60757363...] 一维数组100个数据 # pop = np.array(crossover_and_mutation(pop, CROSSOVER_RATE)) # print(pop.dtype)#<class 'numpy.ndarray'> (100, 3) 2 300 float64 fitness = get_fitness(pop) # print(fitness) #<class 'numpy.ndarray'> (100,) 一维数组 100 pop_half,pop_half2 = select(pop, fitness) # 选择生成新的种群 50行3列 pop_half3=np.array(crossover_and_mutation(pop, CROSSOVER_RATE)) #50行3列 pop=np.vstack((pop_half,pop_half3)) #纵向（上下拼接）拼接 100行 3列 print_info(pop) #测试 运行100次 的最大值平均值代码 # i=i+1 # print(i) # fitness = get_fitness(pop) # max_fitness_index = np.argmax(fitness) # sum=sum+fitness[max_fitness_index] # print(sum/100.0) '''#随机数-随机值 for i in range(10): i=np.random.rand() print(i) i=i+1 '''