python求函数极值_python 遗传算法求函数极值的实现代码
废话不多说，大家直接看代码吧！
"""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。