廢話不多說，大家直接看程式碼吧！

"""遺傳演算法實現求函式極大值—Zjh"""
import numpy as np
import random
import matplotlib.pyplot as plt
class Ga():
 """求出二進位制編碼的長度"""
 def __init__(self):
  self.boundsbegin = -2
  self.boundsend = 3
  precision = 0.0001 # 運算精確度
  self.Bitlength = int(np.log2((self.boundsend - self.boundsbegin)/precision))+1#%染色體長度
  self.popsize = 50# 初始種群大小
  self.Generationmax = 12# 最大進化代數
  self.pcrossover = 0.90# 交叉概率
  self.pmutation = 0.2# 變異概率
  self.population=np.random.randint(0,2,size=(self.popsize,self.Bitlength))
 
 """計算出適應度"""
 def fitness(self,population):
  Fitvalue=[]
  cumsump = []
  for i in population:
   x=self.transform2to10(i)#二進位制對應的十進位制
   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
 """選擇兩個基因，準備交叉"""
 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<pc:
   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
 """變異操作"""
 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
 
 """二進位制轉換為十進位制"""
 def transform2to10(self,population):
  #x=population[-1] #最後一位的值
  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 #返回十進位制的數
 """目標函式"""
 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) #返回選中的2個個體的序號
   scro = gg.crossover(seln,gg.pcrossover) #返回兩條染色體
   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()

一個函式求極值的模擬的作業，參考了別人的matlab程式碼，用python復現了一遍，加深印象！

以上這篇python 遺傳演算法求函式極值的實現程式碼就是小編分享給大家的全部內容了，希望能給大家一個參考，也希望大家多多支援我們。