Python——粒子群算法求解背包问题（详细代码）_粒子群算法 python 背包问题

一、粒子群算法简介

二、详细代码

import random
import numpy as np
import matplotlib.pyplot as plt
 
'''个人认为参数设置非常的合理，但是效果却是最不好的'''
def init_population(n):
    '''生成一个种群'''
    population = []
    for i in range(100):
        cs = [i for i in range(1,n+1)]
        random.shuffle(cs)
        population.append(cs)
    return population
 
def init_v(n):
    '''生成一个初始速度的列表，对应一个种群'''
    v = []
    for i in range(100):
        in1 = []
        for j in range(n):  #n维
            x = random.random()
            in1.append(x)
        v.append(in1)
    return v
 
def Map(lis):
    '''这是一个映射函数，将一个列表变成全排列'''
    lis_dup = lis[:]
    lis.sort()  
    #使用两个列表，对其合理的进行排序
    location = []
    for i in lis_dup:
        index = lis.index(i) + 1
        location.append(index)
    return location
 
def ff(population,n,v1,v2):
    '''传入一个种群，返回不同个体对应函数值的列表'''
    y_s = []
    for i in population:
        location = Map(i)
        cost_sum = 0
        for j in range(n):
            for k in range(n):
                loca1 = location.index(j+1)
                loca2 = location.index(k+1)
                cost = v2[j][k]*v1[loca1][loca2] 
                cost_sum = cost_sum + cost
        y_s.append(cost_sum)
        index = y_s.index(min(y_s))
        best = population[index]  #best为种群中表现最好的个体
    return y_s,best
 
 
 
def ff_solo(i,n,v1,v2):
    '''传入一个个体，得到这个个体的函数值'''
    location = Map(i)
    cost_sum = 0
    for j in range(n):
        for k in range(n):
            loca1 = location.index(j+1)
            loca2 = location.index(k+1)
            cost = v2[j][k]*v1[loca1][loca2] 
            cost_sum = cost_sum + cost
    return cost_sum
 
def trans_v(v,population,p,g):
    '''速度改变函数'''
    vs = []
    for i in range(len(v)):
        r1 = random.random()
        r2 = random.random()
        j = 10*np.array(v[i]) + 5*r1*(np.array(p[i]) - np.array(population[i])) + 5*r2*(np.array(g) - np.array(population[i]))
        j = list(j)
        vs.append(j)
    return vs
 
def trans_popu(population,v,n,v1,v2):
    '''种群改变函数'''
    population_new = np.array(population) + np.array(v)
    
    for i in range(len(population)):
        if ff_solo(list(population_new[i]), n, v1, v2) < ff_solo(population[i], n, v1, v2):
            population[i] =  list(population_new[i])
    return population
                                  
 
def trans_p(population,p,n,v1,v2):
    '''p也为一个种群，记录粒子到访的最好位置'''
    for i in range(len(population)):
        if ff_solo(population[i], n, v1, v2)  <  ff_solo(p[i], n, v1, v2):
            p[i] = population[i]
    return p
def read():
    with open('D:/学习文件/大三上/科研课堂/qap-problems/QAP12.dat','r',encoding='utf-8') as f:   
        comments = f.read().splitlines()
    n = eval(comments[0])
    v11 = comments[2:2+n]
    v22 = comments[3+n:3+n+n]
    v1 = []
    v2 = []
    for i in v11:
        int_list = list(map(int, i.split()))
        v1.append(int_list)
    for i in v22:
        int_list = list(map(int, i.split()))
        v2.append(int_list)
    return v1,v2,n
 
def main():
    v1,v2,n = read()
    population = init_population(n)
    v = init_v(n)
    
    y_s_before,g = ff(population,n,v1,v2) #一开始就定义一个全局最优位置
   
    p = population
    ans = []
    for i in range(1200):
        y_s,best = ff(population,n,v1,v2)
        #下面这个判断是对全局最优进行判断
        if ff_solo(best, n, v1, v2) < ff_solo(g, n, v1, v2):
            g = best
        
        #下面更新个体最优
        trans_p(population,p,n,v1,v2)
        
        
        print(min(y_s))
        ans.append(min(y_s))
        
        
        #print(len(v),len(population),len(p))
        v = trans_v(v, population, p, best)
        population = trans_popu(population,v,n,v1,v2)
        
        plt.plot(ans)
 
main()