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"""遗传算法实现求函数极大值—Zjh"""

import numpy as np

import random

import matplotlib.pyplot as plt

class Ga():

"""求出二进制编码的长度"""

def __init__(self):

self.boundsbegin = -2

self.boundsend = 3

precision = 0.0001 # 运算精确度

self.Bitlength = int(np.log2((self.boundsend - self.boundsbegin)/precision))+1#%染色体长度

self.popsize = 50# 初始种群大小

self.Generationmax = 12# 最大进化代数

self.pcrossover = 0.90# 交叉概率

self.pmutation = 0.2# 变异概率

self.population=np.random.randint(0,2,size=(self.popsize,self.Bitlength))

"""计算出适应度"""

def fitness(self,population):

Fitvalue=[]

cumsump = []

for i in population:

x=self.transform2to10(i)#二进制对应的十进制

xx=self.boundsbegin + x * (self.boundsend - self.boundsbegin) / (pow(2,self.Bitlength)-1)

s=self.targetfun(xx)

Fitvalue.append(s)

fsum=sum(Fitvalue)

everypopulation=[x/fsum for x in Fitvalue]

cumsump.append(everypopulation[0])

everypopulation.remove(everypopulation[0])

for j in everypopulation:

p=cumsump[-1]+j

cumsump.append(p)

return Fitvalue,cumsump

"""选择两个基因，准备交叉"""

def select(self,cumsump):

seln=[]

for i in range(2):

j = 1

r=np.random.uniform(0,1)

prand =[x-r for x in cumsump]

while prand[j] < 0:

j = j + 1

seln.append(j)

return seln

"""交叉"""

def crossover(self, seln, pc):

d=self.population[seln[1]].copy()

f=self.population[seln[0]].copy()

r=np.random.uniform()

if r

print('yes')

c=np.random.randint(1,self.Bitlength-1)

print(c)

a=self.population[seln[1]][c:]

b=self.population[seln[0]][c:]

d[c:]=b

f[c:]=a

print(d)

print(f)

g=d

h=f

else:

g=self.population[seln[1]]

h=self.population[seln[0]]

return g,h

"""变异操作"""

def mutation(self,scnew,pmutation):

r=np.random.uniform(0, 1)

if r < pmutation:

v=np.random.randint(0,self.Bitlength)

scnew[v]=abs(scnew[v]-1)

else:

scnew=scnew

return scnew

"""二进制转换为十进制"""

def transform2to10(self,population):

#x=population[-1] #最后一位的值

x=0

#n=len(population)

n=self.Bitlength

p=population.copy()

p=p.tolist()

p.reverse()

for j in range(n):

x=x+p[j]*pow(2,j)

return x #返回十进制的数

"""目标函数"""

def targetfun(self,x):

#y = x∗(np.sin(10∗(np.pi)∗x))+ 2

y=x*(np.sin(10*np.pi*x))+2

return y

if __name__ == '__main__':

Generationmax=12

gg=Ga()

scnew=[]

ymax=[]

#print(gg.population)

Fitvalue, cumsump=gg.fitness(gg.population)

Generation = 1

while Generation < Generationmax +1:

Fitvalue, cumsump = gg.fitness(gg.population)

for j in range(0,gg.popsize,2):

seln = gg.select( cumsump) #返回选中的2个个体的序号

scro = gg.crossover(seln, gg.pcrossover) #返回两条染色体

s1=gg.mutation(scro[0],gg.pmutation)

s2=gg.mutation(scro[1],gg.pmutation)

scnew.append(s1)

scnew.append(s2)

gg.population = scnew

Fitvalue, cumsump = gg.fitness(gg.population)

fmax=max(Fitvalue)

d=Fitvalue.index(fmax)

ymax.append(fmax)

x = gg.transform2to10(gg.population[d])

xx = gg.boundsbegin + x * (gg.boundsend - gg.boundsbegin) / (pow(2, gg.Bitlength) - 1)

Generation = Generation + 1

Bestpopulation = xx

Targetmax = gg.targetfun(xx)

print(xx)

print(Targetmax)

x=np.linspace(-2,3,30)

y=x*(np.sin(10*np.pi*x))+2

plt.scatter(2.65,4.65,c='red')

plt.xlim(0,5)

plt.ylim(0,6)

plt.plot(x,y)

plt.annotate('local max', xy=(2.7,4.8), xytext=(3.6, 5.2),arrowprops=dict(facecolor='black', shrink=0.05))

plt.show()

一个函数求极值的仿真的作业，参考了别人的matlab代码，用python复现了一遍，加深印象！

以上这篇python 遗传算法求函数极值的实现代码就是小编分享给大家的全部内容了，希望能给大家一个参考，也希望大家多多支持脚本之家。