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二进制编码求三元函数极值
所求函数为

要想使用遗传算法，首要任务是进行编码

传统的 GA 中, DNA 我们能用一串二进制来表示, 比如:
DNA1 = [1, 1, 0, 1, 0, 0, 1]
DNA2 = [1, 0, 1, 1, 0, 1, 1]
这里，我们仍然使用二进制编码，但是如何与我们的问题对应起来呢？

我们知道二进制很容易转十进制，再区间压缩以下，这样一个DNA和一个解一一映射。

def translateDNA(pop):return pop.dot(2 ** np.arange(DNA_SIZE)[::-1]) / float(2**DNA_SIZE-1) * X_BOUND[1]

例如，1 0 1 0 1 0 0 1 0 0 => (4+32+128+256)/(1024-1)*(5-0)=3.3
上面针对的单变量的编码，求解单变量的函数极值
其完整代码如下，这里照搬了别人的代码，后面我会给出引用地址，
用它主要是参考，并修改为求解三元函数极值的代码
单变量求函数极值：

"""
Visualize Genetic Algorithm to find a maximum point in a function.
Visit my tutorial website for more: https://morvanzhou.github.io/tutorials/
"""
import numpy as np
import matplotlib.pyplot as pltDNA_SIZE = 10            # DNA length
POP_SIZE = 100           # population size
CROSS_RATE = 0.8         # mating probability (DNA crossover)
MUTATION_RATE = 0.003    # mutation probability
N_GENERATIONS = 100
X_BOUND = [0, 5]         # x upper and lower boundsdef F(x): return np.sin(10*x)*x + np.cos(2*x)*x     # to find the maximum of this function
#def F(x):  return -x*(x-2)# find non-zero fitness for selection
def get_fitness(pred): return pred + 1e-3 - np.min(pred)# convert binary DNA to decimal and normalize it to a range(0, 5)
# 1 0 1 0 1 0 0 1 0 0 => (4+32+128+256)/(1024-1)*(5-0)=3.3
def translateDNA(pop): return pop.dot(2 ** np.arange(DNA_SIZE)[::-1]) / float(2**DNA_SIZE-1) * (X_BOUND[1]-X_BOUND[0])def select(pop, fitness):    # nature selection wrt pop's fitnessidx = np.random.choice(np.arange(POP_SIZE), size=POP_SIZE, replace=True,p=fitness/fitness.sum())return pop[idx]def crossover(parent, pop):     # mating process (genes crossover)if np.random.rand() < CROSS_RATE:i_ = np.random.randint(0, POP_SIZE, size=1)                             # select another individual from popcross_points = np.random.randint(0, 2, size=DNA_SIZE).astype(np.bool)   # choose crossover pointsparent[cross_points] = pop[i_, cross_points]                            # mating and produce one childreturn parentdef mutate(child):for point in range(DNA_SIZE):if np.random.rand() < MUTATION_RATE:child[point] = 1 if child[point] == 0 else 0return child# pop是100*10的随机01矩阵
pop = np.random.randint(2, size=(POP_SIZE, DNA_SIZE))   # initialize the pop DNA# 绘制原始函数图像
plt.ion()       # something about plotting
x = np.linspace(*X_BOUND, 200)
plt.plot(x, F(x))for _ in range(N_GENERATIONS):F_values = F(translateDNA(pop))    # compute function value by extracting DNA# something about plottingif 'sca' in globals(): sca.remove()sca = plt.scatter(translateDNA(pop), F_values, s=200, lw=0, c='red', alpha=0.5); plt.pause(0.05)# GA part (evolution)fitness = get_fitness(F_values)print("Generation: ", _)print("Most fitted DNA: ", pop[np.argmax(fitness), :])print("Most fitted Value: ", translateDNA(pop[np.argmax(fitness), :]))pop = select(pop, fitness)pop_copy = pop.copy()for parent in pop:child = crossover(parent, pop_copy)child = mutate(child)parent[:] = child       # parent is replaced by its childplt.ioff(); plt.show()

然后下面是我的求解三元函数极值的完整代码：

# x1*x2-x1*x2+x3
import numpy as np
import randomDNA_SIZE = 24
POP_SIZE = 100
CROSSOVER_RATE = 0.8
MUTATION_RATE = 0.015
N_GENERATIONS = 100
X_BOUND = [3.0,5.0]#x1
Y_BOUND = [2.1,6.7]#x2
Z_BOUND = [1.2,9.6]#x3def F(x, y,z):val=x*x-x*y+z#print(val)'''for index in range(len(val)):if val[index] <0.2:val[index]=0.2'''return valdef get_fitness(pop): x,y ,z= translateDNA(pop)pred = F(x, y,z)return preddef 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]#取前DNA_SIZE个列表示xy_pop = pop[:,DNA_SIZE:DNA_SIZE*2] ##取中间DNA_SIZE个列表示yz_pop = pop[:,DNA_SIZE*2:]取后DNA_SIZE个列表示z#print(x_pop)#print(y_pop)#print(z_pop)#print("\n")'''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,y,zdef 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child[mutate_point] = child[mutate_point]^1 	#将变异点的二进制为反转def crossover_and_mutation(pop, CROSSOVER_RATE = 0.8):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（这里我把一串二进制串的那些0，1称为基因）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)	#每个后代有一定的机率发生变异new_pop.append(child)return new_popdef select(pop, fitness):    # nature selection wrt pop's fitness#idx = np.random.choice(np.arange(POP_SIZE), size=POP_SIZE, replace=True,fitnew = fitnessfor index in range(len(fitnew)):if fitnew[index] <0:fitnew[index]=0p=(fitnew)/(fitnew.sum())idx = np.random.choice(np.arange(POP_SIZE), size=POP_SIZE, replace=True,p=p)#print(idx)return pop[idx]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__":pop = np.random.randint(2, size=(POP_SIZE, DNA_SIZE*3)) #matrix (POP_SIZE, DNA_SIZE)#print(pop)for _ in range(N_GENERATIONS):#迭代N代x,y,z = translateDNA(pop)pop = np.array(crossover_and_mutation(pop, CROSSOVER_RATE))fitness = get_fitness(pop)pop = select(pop, fitness) #选择生成新的种群print_info(pop)

参考文献：
https://www.cnblogs.com/lfri/p/12240355.html
https://blog.csdn.net/zxxxxxxy/article/details/105287743?utm_medium=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-3.nonecase&depth_1-utm_source=distribute.pc_relevant.none-task-blog-BlogCommendFromMachineLearnPai2-3.nonecase