python如何求二元函数的极值 python求多元函数极值

遗传算法求三元函数极值(python)-采用实数编码

想看二进制编码编码的博客地址在这

遗传算法求三元函数极值(python)-采用二进制编码 本文的遗传算法采用实数编码求三元函数极值

所求函数为

`

其完整代码如下：

x1x1-x1x2+x3
import numpy as np
 import randomDNA_SIZE =1
 POP_SIZE =100
 CROSSOVER_RATE = 0.8
 MUTATION_RATE = 0.015
 N_GENERATIONS = 500
 X_BOUND = [3.0,5.0]#x1Y_BOUND = [2.1,6.7]#x2
 Z_BOUND = [1.2,9.6]#x3def F(x, y,z):
 val=xx-xy+z
 ‘’’
 for index in range(len(val)):if val[index] <0.2:
     val[index]=0.2
     '''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)

'''pop:(POP_SIZE,DNA_SIZE)*(DNA_SIZE,1) --> (POP_SIZE,1)'''#二进制--->十进制
# x = x_pop.dot(2**np.arange(DNA_SIZE)[::-1])/float(2**DNA_SIZE-1)*(X_BOUND[1]-X_BOUND[0])+X_BOUND[0]
# y = y_pop.dot(2**np.arange(DNA_SIZE)[::-1])/float(2**DNA_SIZE-1)*(Y_BOUND[1]-Y_BOUND[0])+Y_BOUND[0]
# z = z_pop.dot(2**np.arange(DNA_SIZE)[::-1])/float(2**DNA_SIZE-1)*(Z_BOUND[1]-Z_BOUND[0])+Z_BOUND[0]
# print(x,z)
return x_pop,y_pop,z_pop

def mutation(child, MUTATION_RATE=0.003):
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_point0:
child[mutate_point] =np.random.uniform(3.0,5.0)
elif mutate_point1:
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 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
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
mutation(child) #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 select(pop, fitness): # nature selection wrt pop’s fitness
# fitnew=fitness.copy() #浅拷贝
fitnew = fitness
# for index in range(len(fitnew)):
# if fitnew[index] <0:
# fitnew[index]=0
fitnew=fitnew + 1e-3 - np.min(fitnew)
p=(fitnew)/(fitnew.sum())
# print(fitnew)
# print(fitnew.sum())

# print(np.arange(POP_SIZE))  #0,1,2,3,...,99
idx = np.random.choice(np.arange(POP_SIZE), size=POP_SIZE,replace=True,p=p)
# print(pop[idx].shape) #shape (3, 3)
'''
[[3.43993966 5.19547684 5.17821964]
 [4.0803077  5.19547684 5.17821964]
 [3.43993966 5.19547684 5.17821964]]
'''
return pop[idx]  #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))
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 _ 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 = select(pop, fitness) #选择生成新的种群
 # print(pop.shape) (100,3)print_info(pop)
`