遗传算法实战1

命题： 来自不同地方的人们去同一个地方，需要安排他们的飞机行程，希望同一天在纽约会面，同时在同一天返回。要求总成本最小。

航班样例数据如下：

起点，终点，起飞时间，到达时间，价格
LGA,OMA,6:19,8:13,239
OMA,LGA,6:11,8:31,249
LGA,OMA,8:04,10:59,136
OMA,LGA,7:39,10:24,219
LGA,OMA,9:31,11:43,210
OMA,LGA,9:15,12:03,99
LGA,OMA,11:07,13:24,171
OMA,LGA,11:08,13:07,175
LGA,OMA,12:31,14:02,234
OMA,LGA,12:18,14:56,172
LGA,OMA,14:05,15:47,226
OMA,LGA,13:37,15:08,250
LGA,OMA,15:07,17:21,129
OMA,LGA,15:03,16:42,135
LGA,OMA,16:35,18:56,144
OMA,LGA,16:51,19:09,147
LGA,OMA,18:25,20:34,205
OMA,LGA,18:12,20:17,242
LGA,OMA,20:05,21:44,172
OMA,LGA,20:05,22:06,261
……..

import time
import random
import numpy as np

#创建每个人对应的出发城市
people = [('Seymour', 'BOS'),
          ('Franny', 'DAL'),
          ('Zooey', 'CAK'),
          ('Walt', 'MIA'),
          ('Buddy', 'ORD'),
          ('Les', 'OMA')]
# 定义目的地
destination = 'LGA'

#定义一个函数解析原始数据txt文件
flights = {}
def extract_file(filename):
    flights = {}
    for line in open(filename):
        origin, dest, depart, arrive, price = line.strip().split(',')
        flights.setdefault((origin, dest), [])
        flights[(origin, dest)].append((depart, arrive, int(price)))
    return flights
flights = extract_file('schedule.txt')
list(flights.keys())[0]  

Out[277]: (‘LGA’, ‘DAL’)

[python3不支持字典索引，先转成list]
(http://blog.csdn.net/a070220106/article/details/9089379)

#定义一个方便获取分钟数的函数
def getminutes(t):
    x = time.strptime(t, '%H:%M')
    return x[3] * 60 + x[4]

#定义打印出题解对应的人员航班信息的函数
def printschedule(r):
    for d in range(int(len(r) / 2)):
        name = people[d][0]
        origin = people[d][1]
        out = flights[(origin, destination)][(r[d])]
        ret = flights[(destination, origin)][(r[d + 1])]
        print('%10s%10s %5s-%5s $%3s %5s-%5s $%3s' % (name, origin,
                                                      out[0], out[1], out[2],                                                      ret[0], ret[1], ret[2]))

s = [1, 4, 3, 2, 7, 3, 6, 3, 2, 4, 5, 3]
printschedule(s)

Seymour BOS 8:04-10:11 9512:08−14:05$9512:08-14:05$ 142
Franny DAL 12:19-15:25 34210:51−14:16$34210:51-14:16$ 256
Zooey CAK 10:53-13:36 1899:58−12:56$1899:58-12:56$ 249
Walt MIA 9:15-12:29 22516:50−19:26$22516:50-19:26$ 304
Buddy ORD 16:43-19:00 24610:33−13:11$24610:33-13:11$ 132
Les OMA 11:08-13:07 17515:07−17:21$17515:07-17:21$ 129

# 定义成本函数（是优化问题的核心）
def costf(sol):
    totalprice = 0
    latestarrival = 0
    earliestdep = 24 * 60

    for d in range(int(len(sol) / 2)):
        # Get the inbound and outbound flights
        origin = people[d][1]
        #print('origin', origin)
        outbound = flights[(origin, destination)][(sol[d])]
        #print('outbound', outbound)
        returnf = flights[(destination, origin)][(sol[d + 1])]
        #print('returnf', returnf)

        # Total price is the price of all outbound and return flights
        totalprice += outbound[2]
        totalprice += returnf[2]

        # Track the latest arrival and earliest departure
        if latestarrival < getminutes(outbound[1]): latestarrival = getminutes(outbound[1])
        if earliestdep > getminutes(returnf[0]): earliestdep = getminutes(returnf[0])

    # Every person must wait at the airport until the latest person arrives.
    # They also must arrive at the same time and wait for their flights.
    totalwait = 0
    for d in range(int(len(sol) / 2)):
        origin = people[d][1]
        outbound = flights[(origin, destination)][(sol[d])]
        returnf = flights[(destination, origin)][(sol[d + 1])]
        totalwait += latestarrival - getminutes(outbound[1])
        totalwait += getminutes(returnf[0]) - earliestdep

        # Does this solution require an extra day of car rental? That'll be $50!
    if latestarrival > earliestdep: totalprice += 50

    return totalprice + totalwait

# 定义遗传算法
def geneticoptimize(domain, costf, popsize=50, step=1 ,
                    mutprob=0.2, elite=0.2, maxiter=100):
    #popsize:种群大小
    #step: 变异步长（一个种群中的单个DNA变异的步长）
    #mutprob:种群的新成员由变异而非交叉得来的概率
    #elite:种群中被认为是最优解且被允许传入下一代的部分
    #maxiter: 需要运行多少代就停止

    # 定义变异参数
    def mutate(vec):
        i = random.randint(0, len(domain) - 1)
        if random.random() < 0.5 and vec[i] > domain[i][0]:
            return vec[0:i] + [vec[i] - step] + vec[i + 1:]
        elif vec[i] < domain[i][1]:
            return vec[0:i] + [vec[i] + step] + vec[i + 1:]

    # 定义交叉（或者叫配对）函数
    def crossover(r1, r2):
        i = random.randint(1, len(domain) - 2)
        return r1[0:i] + r2[i:]

    # 创建随机初始种群(50个不同的种群)
    pop = []
    for i in range(popsize):
        vec = [random.randint(domain[i][0], domain[i][1])
               for i in range(len(domain))]
        pop.append(vec)

    #每一代胜出的个数（这里定义了20%的上一代优胜者会进入下一代）
    topelite = int(elite * popsize)

    # 主循环
    for i in range(100): #range(maxiter)
        scores = [(costf(v), v) for v in pop] #计算50个初始种群每个的成本及排列方式
        scores.sort() #按照成本从小到大排序
        ranked = [v for (s, v) in scores]

        #从纯粹的胜出者开始
        pop = ranked[0:topelite] #只保留20%的初始种群

        # 添加变异和配对后的胜出者
        while len(pop) < popsize:
            if random.random() < mutprob: 
                #random.random()产生0-1之间的一个随机数
                #当这个随机数小于变异概率系数0.2的时候，开始变异
                #也就意味着有20%的机率是突变

                #变异（基因在未配对之前突变） 
                c = random.randint(0, topelite)#指定那个位置的基因发生突变
                pop.append(mutate(ranked[c])) #添加突变的种群到被选中的纯粹胜出的10个种群中
            else:

                # 配对（2个初始纯粹胜出的种群配对）
                #这里有机会c1与c2是相同的
                c1 = random.randint(0, topelite)
                c2 = random.randint(0, topelite)
                pop.append(crossover(ranked[c1], ranked[c2]))

        # 打印当前最佳的最小成本函数
        print(scores[0][0])

    #返回成本函数对应的解列表
    return scores[0][1] 

算法解释

标准的题解数据格式应该是一个长度为12的一元列表
e.g: sol = [8, 0, 3, 5, 4, 8, 5, 4, 8, 9, 4, 1]
定义一个与原始题解相同长度的二元列表，这样子定义的目的是每天都有往返10趟航班，每个人往返都可以从这10趟航班中任意选择，有利创造出任意的初始种题解
domain = [(0, 9)] * len(people) * 2

[(0, 9),
(0, 9),
(0, 9),
(0, 9),
(0, 9),
(0, 9),
(0, 9),
(0, 9),
(0, 9),
(0, 9),
(0, 9),
(0, 9)]

• 1，创建初始种群

pop ：创建50个初始种群的题解

for i in range(popsize):
    vec = [random.randint(domain[i][0], domain[i][1])
           for i in range(len(domain))]
    pop.append(vec)

pop
Out[301]:
[[0, 5, 6, 6, 6, 4, 7, 2, 3, 6, 7, 6],
[8, 9, 3, 1, 1, 8, 2, 7, 4, 2, 5, 8],
[0, 2, 0, 5, 4, 4, 8, 4, 1, 4, 0, 3],
[4, 6, 3, 5, 3, 6, 7, 5, 9, 4, 5, 9],
[4, 5, 6, 6, 5, 2, 8, 6, 5, 9, 9, 8],
[0, 9, 9, 8, 9, 1, 8, 8, 5, 1, 7, 7],
[2, 4, 8, 2, 5, 4, 8, 0, 0, 4, 8, 1],
[7, 8, 8, 7, 9, 5, 9, 5, 5, 5, 0, 9],
[0, 6, 1, 0, 5, 5, 5, 9, 0, 1, 6, 0],
[3, 8, 1, 7, 7, 5, 3, 0, 7, 7, 4, 5],
[4, 4, 9, 3, 2, 9, 7, 8, 6, 2, 0, 2],
[4, 9, 4, 8, 5, 9, 0, 6, 9, 8, 8, 5],
[9, 6, 6, 1, 2, 1, 4, 8, 8, 1, 5, 1],
[9, 9, 9, 9, 0, 8, 0, 1, 1, 8, 5, 7],
[7, 2, 7, 6, 5, 0, 8, 8, 1, 8, 1, 1],
[0, 6, 5, 9, 2, 7, 0, 8, 7, 0, 9, 7],
[1, 3, 4, 9, 8, 2, 9, 1, 0, 8, 2, 3],
[4, 3, 4, 9, 7, 2, 4, 7, 0, 7, 2, 2],
[8, 6, 6, 1, 8, 7, 6, 5, 8, 5, 2, 9],
[1, 8, 0, 1, 8, 1, 4, 5, 0, 4, 9, 1],
[1, 0, 2, 3, 0, 1, 4, 0, 1, 6, 2, 2],
[8, 1, 9, 0, 7, 9, 1, 2, 4, 2, 9, 3],
[4, 7, 8, 4, 6, 8, 8, 4, 5, 4, 5, 7],
[7, 1, 1, 0, 7, 2, 0, 0, 9, 6, 5, 5],
[0, 5, 4, 8, 9, 9, 9, 3, 4, 9, 7, 1],
[0, 9, 9, 4, 6, 8, 3, 8, 7, 1, 2, 2],
[6, 2, 3, 7, 3, 7, 5, 7, 4, 9, 5, 4],
[3, 4, 8, 7, 8, 6, 9, 4, 8, 2, 2, 6],
[4, 6, 8, 8, 7, 7, 1, 6, 0, 7, 5, 5],
[7, 3, 6, 0, 4, 9, 3, 5, 8, 7, 7, 7],
[5, 9, 9, 8, 1, 1, 1, 0, 0, 3, 9, 4],
[5, 8, 8, 9, 8, 6, 8, 8, 8, 1, 6, 5],
[6, 0, 7, 5, 9, 9, 1, 6, 7, 6, 4, 8],
[8, 7, 4, 8, 3, 9, 2, 9, 8, 7, 1, 7],
[9, 9, 2, 8, 6, 6, 3, 8, 8, 9, 8, 2],
[1, 4, 6, 5, 9, 5, 6, 5, 3, 6, 4, 7],
[1, 8, 3, 7, 3, 3, 3, 6, 7, 9, 6, 3],
[8, 3, 2, 9, 2, 5, 0, 7, 4, 0, 9, 8],
[8, 3, 0, 0, 5, 2, 3, 4, 9, 4, 6, 5],
[5, 0, 1, 3, 8, 8, 6, 9, 5, 0, 8, 9],
[1, 4, 3, 3, 3, 0, 5, 2, 9, 5, 7, 4],
[2, 4, 0, 2, 4, 8, 5, 5, 7, 0, 3, 9],
[0, 7, 5, 5, 1, 0, 2, 4, 7, 8, 0, 7],
[5, 4, 4, 4, 5, 7, 7, 7, 2, 8, 4, 9],
[3, 5, 4, 1, 7, 2, 5, 1, 4, 8, 5, 7],
[4, 9, 8, 8, 0, 0, 1, 5, 5, 9, 1, 2],
[3, 1, 0, 5, 7, 9, 4, 4, 0, 2, 9, 1],
[8, 8, 3, 5, 1, 7, 7, 9, 1, 8, 8, 1],
[5, 9, 6, 7, 6, 9, 9, 0, 4, 7, 7, 9],
[8, 0, 3, 5, 4, 8, 5, 4, 8, 9, 4, 1]]

• 2，topelite， 对初始种群排序，选择前20%的初始种群作为下一代
  通过选拔后，第二代只剩下了10个种群

for i in range(100): #range(maxiter)
        scores = [(costf(v), v) for v in pop] #计算50个初始种群每个的成本及排列方式
        scores.sort() #按照成本从小到大排序
        ranked = [v for (s, v) in scores]

        #从纯粹的胜出者开始
        pop = ranked[0:topelite] #只保留20%的初始种群`

pop #10个种群
Out[453]:
Out[453]:
[[6, 2, 2, 1, 5, 2, 6, 5, 9, 8, 3, 5],
[6, 8, 5, 6, 5, 7, 4, 3, 5, 7, 1, 2],
[3, 3, 5, 6, 3, 5, 7, 4, 4, 3, 7, 0],
[0, 6, 3, 4, 4, 5, 4, 2, 9, 4, 6, 7],
[9, 9, 8, 7, 8, 5, 6, 3, 9, 1, 5, 5],
[8, 4, 4, 3, 3, 3, 9, 6, 5, 5, 6, 6],
[2, 0, 5, 4, 5, 6, 2, 0, 8, 8, 7, 3],
[8, 3, 7, 6, 5, 1, 2, 9, 4, 2, 3, 8],
[2, 6, 2, 5, 7, 3, 8, 9, 1, 9, 9, 6],
[5, 8, 5, 7, 7, 4, 8, 5, 4, 3, 9, 2]]

• 3，在第二代种群的基础上进行变异（这里定义了20%的机会突变）或者配对，迭代100次，每次迭代要么发生变异，要么配对，变异或者配对。每次迭代总会保留初始种群（这里是50）的20%的个体进入下一代，然后继续有20%的机率突变，或者80%的机率配对。达到设定的迭代次数后，选择成本函数最小的一个题解，算法停止。

第一次迭代的时候，会从50个随机初始种群中产生10个种群，然后突变一个或者配对产生一个，第二代总共11个种群时候的pop变量：
（这里出来的结果显示是突变还是配对？）
观察最后的一个结果就是突变或者配对产生新的一个后代，它跟倒数第三个前面一段类似，后面一段跟顺数第四个一样，所以是配对了的）
Out[459]:
[[6, 2, 2, 1, 5, 2, 6, 5, 9, 8, 3, 5],
[6, 8, 5, 6, 5, 7, 4, 3, 5, 7, 1, 2],
[3, 3, 5, 6, 3, 5, 7, 4, 4, 3, 7, 0],
[0, 6, 3, 4, 4, 5, 4, 2, 9, 4, 6, 7],
[9, 9, 8, 7, 8, 5, 6, 3, 9, 1, 5, 5],
[8, 4, 4, 3, 3, 3, 9, 6, 5, 5, 6, 6],
[2, 0, 5, 4, 5, 6, 2, 0, 8, 8, 7, 3],
[8, 3, 7, 6, 5, 1, 2, 9, 4, 2, 3, 8],
[2, 6, 2, 5, 7, 3, 8, 9, 1, 9, 9, 6],
[5, 8, 5, 7, 7, 4, 8, 5, 4, 3, 9, 2],
[2, 6, 2, 5, 7, 3, 4, 2, 9, 4, 6, 7]]

11个种群对应的成本函数排序

scores
Out[462]:
[(4802, [6, 2, 2, 1, 5, 2, 6, 5, 9, 8, 3, 5]),
(4931, [6, 8, 5, 6, 5, 7, 4, 3, 5, 7, 1, 2]),
(4952, [3, 3, 5, 6, 3, 5, 7, 4, 4, 3, 7, 0]),
(5019, [0, 6, 3, 4, 4, 5, 4, 2, 9, 4, 6, 7]),
(5098, [9, 9, 8, 7, 8, 5, 6, 3, 9, 1, 5, 5]),
(5151, [8, 4, 4, 3, 3, 3, 9, 6, 5, 5, 6, 6]),
(5194, [2, 6, 2, 5, 7, 3, 4, 2, 9, 4, 6, 7]),
(5223, [2, 0, 5, 4, 5, 6, 2, 0, 8, 8, 7, 3]),
(5423, [8, 3, 7, 6, 5, 1, 2, 9, 4, 2, 3, 8]),
(5519, [2, 6, 2, 5, 7, 3, 8, 9, 1, 9, 9, 6]),
(5709, [5, 8, 5, 7, 7, 4, 8, 5, 4, 3, 9, 2])]

经过对此迭代后，每次总保留10个pop进行变异或者配对，迭代完成后，返回成本最小的一个结果作为最终输出。
最终的题解以及最小的成本在迭代12次就得到了。。

s = geneticoptimize(domain, costf)
[5, 3, 5, 3, 4, 3, 3, 0, 0, 0, 0, 0] #题解
4288
3901
3646
3646
3646
3490
3337
3303
2994
2994
2994
2912
2912
2912
2912
2912
2912
2912
2912
2912
2912
2912
2912