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遗传算法实现(Matlab实现/PYTHON 实现)

作者:互联网

1.整体流程(有不懂的地方可以在评论区留言)

  初始化种群,计算适应性,选择,交叉,变异,更新并评估,判断是否达到最大迭代次数,输出最佳目标函数

2.求9*sin(5*x)+8*cos(4*x)最优值(matlab)

clear all;
popsize=20;      %群体大小
chromlength=10;  %字符串长度(个体长度)
pc=0.7;  %交叉概率
pm=0.005;  %变异概率
pop=initpop(popsize,chromlength);  %随机产生初始群体
for i=1:20   %20为迭代次数
    value=calobjvalue(pop);  %计算目标函数
    fitvalue=calfitvalue(value);  %计算群体中每个个体的适应度
    newpop=selection(pop,fitvalue);  %复制
    newpop=crossover(pop,pc); %交叉
    newpop=mutation(pop,pc);  %变异
  [bestindividual,bestfit]=best(pop,fitvalue);%求出群体中适应值最大的个体及其适应值
    y(i)=max(bestfit);
    n(i)=i;
    pop3=bestindividual;
    x(i)=decodechrom(pop3,1,chromlength)*10/1023;
    pop=newpop;
end
fplot('9*sin(5*x)+8*cos(4*x)');
hold on;
plot(x,y,'r+');
hold off;

3.求y=sin(10*x)*x + cos(2*x)*x 在【0, 5】区间最大值(Python)

import numpy as np

DNA_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 = 200
X_BOUND = [0, 5]  # x upper and lower bounds

def F(x): 
    return np.sin(10*x)*x + np.cos(2*x)*x     # to find the maximum of this function
    
def get_fitness(pred):
    return pred - np.min(pred) + 1e-3  # 防止出现非正值,因为 select 时概率非负。
    
def translateDNA(pop):
    # np.arange(DNA_SIZE)[::-1] --> array([9, 8, 7, 6, 5, 4, 3, 2, 1, 0])
    # 2 ** np.arange(DNA_SIZE)[::-1] --> array([512, 256, 128,  64,  32,  16,   8,   4,   2,   1], dtype=int32)
    # pop.dot(2 ** np.arange(DNA_SIZE)[::-1]) --> 将二进制转换为十进制
    # (2 ** DNA_SIZE - 1) --> 1023
    # pop.dot(2 ** np.arange(DNA_SIZE)[::-1]) / (2 ** DNA_SIZE - 1) --> 计算染色体值占最大值(1023)的比例
    # pop.dot(2 ** np.arange(DNA_SIZE)[::-1]) / (2 ** DNA_SIZE - 1) * X_BOUND[1] --> 将范围从[0, 1023]缩放到[0, 5]
    
    return pop.dot(2 ** np.arange(DNA_SIZE)[::-1]) / (2 ** DNA_SIZE - 1) * X_BOUND[1]
    
def select(pop, fitness):
    # 在 np.arange(POP_SIZE) 中依概率 p 选出 size 个染色体代号
    # replace=True 表示选出的染色体可以有重复的
    idx = np.random.choice(np.arange(POP_SIZE), size=POP_SIZE, replace=True, p=fitness / fitness.sum())
    return pop[idx]
    
def crossover(parent, pop):
    if np.random.rand() < CROSS_RATE:
        i_ = np.random.randint(0, POP_SIZE, size=1)  # 选定 pop 中需要交叉的染色体,如 i_ = 1,表示第二条染色体
        
        # 选择需要改变的基因,如 cross_points = [False  True  True False False  True False  True  True False]
        # 表示第2,3,6,8,9个位置上的基因会改变
        cross_points = np.random.randint(0, 2, size=DNA_SIZE).astype(np.bool)
        
        # pop[i_, cross_points] 会得到 pop 种群中 i_ 这条染色体上的 cross_points 中 True 位置上的基因
        # parent[cross_points] = pop[i_, cross_points] 将刚刚得到的基因放到 parent 这条染色体的 cross_points 中 True 的位置上
        parent[cross_points] = pop[i_, cross_points]
    return parent
    
def mutate(child):
    for point in range(DNA_SIZE):
        if np.random.rand() < MUTATION_RATE:
            child[point] = 1 if child[point] == 0 else 0
    return child
    
pop = np.random.randint(2, size=(POP_SIZE, DNA_SIZE))   # initialize the pop DNA
print(pop.shape)

for _ in range(N_GENERATIONS):
    F_values = F(translateDNA(pop))    # compute function value by extracting DNA

    # GA part (evolution)
    fitness = get_fitness(F_values)
    print("Most fitted DNA: ", pop[np.argmax(fitness), :], "适应度为:", round(fitness[np.argmax(fitness)], 3))
    pop = select(pop, fitness)
    pop_copy = pop.copy()
    for parent in pop:
        child = crossover(parent, pop_copy)
        child = mutate(child)
        parent = child       # 用经过交叉变异的下一代(新的染色体)来代替上一代(pop 种群中选出的)
print('最大值为:', F(translateDNA(pop[np.argmax(fitness), :])))

标签:DNA,实现,cross,pop,PYTHON,fitness,np,遗传算法,SIZE
来源: https://blog.csdn.net/qq_40840797/article/details/118685117