模拟退火之函数最值

利用 Python 实现模拟退火算法解决函数最值问题, 并画出最优函数点

代码

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# -*- coding: cp936 -*-
import math
import random
import time,os
from matplotlib.pyplot import *
import numpy as np

class Metal():
def __init__(self, lower_bound, upper_bound, temperature, descending, formula):

self.lower_bound = lower_bound
self.upper_bound = upper_bound
self.temperature = temperature
self.descending = descending
self.x = self.RandX()
self.y = 0.0
self.history_x = []
self.history_y = []
self.formula = formula
self.best = [0,0]
#self.Cm = cm.get_cmap('Greys')

def RandX(self):
"""
获取随机 x
"""
return random.uniform(self.lower_bound, self.upper_bound)


def Smelt(self):
"""
冶炼开始
"""

x = self.RandX()
NewY = eval(self.formula)

dE = NewY - self.y

if dE >= 0 or math.e ** (dE / self.temperature) > random.uniform(0, 1):
self.y = NewY
self.x = x

if self.best[1] < self.y:
self.best[0] = self.x
self.best[1] = self.y
os.system('cls')
print '当金属温度为', self.temperature, '℃时,x 到达历史最佳, 为', self.best[0], 'y 为', self.best[1], '\n', '-'*100

#self.PlotAndSave() #每次退火都画图的话速度较慢

self.history_x += [x]
self.history_y += [NewY]

self.temperature -= self.descending

def PlotAndSave(self):
'''
冶炼结束时画出每温度下的 x 与 y
'''
x = np.arange(self.lower_bound, self.upper_bound, 0.01)
y = eval(self.formula.replace('math','np'))
plot(x,y)
plot(self.x, self.y, 'o')
#sc = scatter(self.history_x, self.history_y, c = range(len(self.history_x)) , vmin = 0, vmax = len(self.history_x), s=10, cmap = self.Cm)
#colorbar(sc)
savefig('Best.jpg')
close('all')

time.clock()
#------------------------------------------------------------------参数设置------------------------------------------------------------------

lower_bound = 0 #函数定义域的下限
upper_bound = 9 #函数定义域的上限
temperature = 100 #金属初始温度
descending = 0.01 #降温幅度
formula = 'x + 10 * math.sin(5 * x) + 7 * math.cos(4 * x)' #函数式

#------------------------------------------------------------------参数设置------------------------------------------------------------------

metal = Metal(lower_bound, upper_bound, temperature, descending, formula)
while metal.temperature > 1:

t = time.clock()
minute, second = divmod(t, 60)
hour, minute = divmod(minute, 60)

metal.Smelt()

print '当前金属温度为', metal.temperature, '℃', '此温度下最优 x 为', metal.x, 'y 为', metal.y, ' 花费时间 %02d:%02d:%02ds' %(hour, minute, second), ' '*5,'\r',

print
metal.PlotAndSave()
os.system('pause')

注意

  • 双击运行
  • 降温幅度较小, 会有更大概率得到较优值, 但是相应花费的时间也就越长.
  • 结果截图
    运行结果
    最值


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