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最近在做遗传算法的作业,题目是一个分段函数求最大值的问题(图片附下方),但是在函数分段这块碰到问题。本人比较新手,代码不是自己写的,也是用的别人的,但是改了函数以后,并不是想要的函数图像,想请大家帮我看一下。
附代码:
import numpy as np
import matplotlib.pyplot as plt
from matplotlib import cm
from mpl_toolkits.mplot3d import Axes3D
DNA_SIZE = 24
POP_SIZE = 200
CROSSOVER_RATE = 0.8
MUTATION_RATE = 0.005
N_GENERATIONS = 50
X_BOUND = [-3, 3]
Y_BOUND = [-3, 3]
def F(x, y):
if (-3 < x.all() < -2.5 and -3 < y.all() < -2.5) or (2.5 < x.all() < 3 and 2.5 < y.all() < 3):
return np.sin((2 * np.pi * x) / 3) ** 2 * np.sin((2 * np.pi * y) / 3) ** 2 * np.exp((x + y) / (5 * np.pi))
else:
return -(np.sin((2 * np.pi * x) / 3) ** 2 * np.sin((2 * np.pi * y) / 3) ** 2 * np.exp((x + y) / (5 * np.pi))) #自己写的部分
def plot_3d(ax):
X = np.linspace(*X_BOUND, 100)
Y = np.linspace(*Y_BOUND, 100)
X, Y = np.meshgrid(X, Y)
Z = F(X, Y)
ax.plot_surface(X, Y, Z, rstride=1, cstride=1, cmap=cm.coolwarm)
ax.set_zlim(-1, 1)
ax.set_xlabel('x')
ax.set_ylabel('y')
ax.set_zlabel('z')
plt.pause(3)
plt.show()
def get_fitness(pop):
x, y = translateDNA(pop)
pred = F(x, y)
return (pred - np.min(
pred)) + 1e-3 # 减去最小的适应度是为了防止适应度出现负数,通过这一步fitness的范围为[0, np.max(pred)-np.min(pred)],最后在加上一个很小的数防止出现为0的适应度
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[:, 1::2] # 奇数列表示X
y_pop = pop[:, ::2] # 偶数列表示y
# pop:(POP_SIZE,DNA_SIZE)*(DNA_SIZE,1) --> (POP_SIZE,1)
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]
return x, y
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 * 2) # 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_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 * 2) # 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 select(pop, fitness): # nature selection wrt pop's fitness
idx = np.random.choice(np.arange(POP_SIZE), size=POP_SIZE, replace=True,
p=(fitness) / (fitness.sum()))
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 = translateDNA(pop)
print("最优的基因型:", pop[max_fitness_index])
print("(x, y):", (x[max_fitness_index], y[max_fitness_index]))
if __name__ == "__main__":
fig = plt.figure()
ax = Axes3D(fig)
plt.ion() # 将画图模式改为交互模式,程序遇到plt.show不会暂停,而是继续执行
plot_3d(ax)
pop = np.random.randint(2, size=(POP_SIZE, DNA_SIZE * 2)) # matrix (POP_SIZE, DNA_SIZE)
for _ in range(N_GENERATIONS): # 迭代N代
x, y = translateDNA(pop)
if 'sca' in locals():
sca.remove()
sca = ax.scatter(x, y, F(x, y), c='black', marker='o');
plt.show();
plt.pause(0.1)
pop = np.array(crossover_and_mutation(pop, CROSSOVER_RATE))
# F_values = F(translateDNA(pop)[0], translateDNA(pop)[1])#x, y --> Z matrix
fitness = get_fitness(pop)
pop = select(pop, fitness) # 选择生成新的种群
print_info(pop)
plt.ioff()
plot_3d(ax)
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