蚁群算法 python3.6 pycharm

参考文档：https://blog.csdn.net/golden1314521/article/details/45059719

简单的修改下，能够在python3.6中运行

代码如下：

import numpy as np
import matplotlib.pyplot as plt
# %pylab
coordinates = np.array([[565.0, 575.0], [25.0, 185.0], [345.0, 750.0], [945.0, 685.0], [845.0, 655.0],
                        [880.0, 660.0], [25.0, 230.0], [525.0, 1000.0], [580.0, 1175.0], [650.0, 1130.0],
                        [1605.0, 620.0], [1220.0, 580.0], [1465.0, 200.0], [1530.0, 5.0], [845.0, 680.0],
                        [725.0, 370.0], [145.0, 665.0], [415.0, 635.0], [510.0, 875.0], [560.0, 365.0],
                        [300.0, 465.0], [520.0, 585.0], [480.0, 415.0], [835.0, 625.0], [975.0, 580.0],
                        [1215.0, 245.0], [1320.0, 315.0], [1250.0, 400.0], [660.0, 180.0], [410.0, 250.0],
                        [420.0, 555.0], [575.0, 665.0], [1150.0, 1160.0], [700.0, 580.0], [685.0, 595.0],
                        [685.0, 610.0], [770.0, 610.0], [795.0, 645.0], [720.0, 635.0], [760.0, 650.0],
                        [475.0, 960.0], [95.0, 260.0], [875.0, 920.0], [700.0, 500.0], [555.0, 815.0],
                        [830.0, 485.0], [1170.0, 65.0], [830.0, 610.0], [605.0, 625.0], [595.0, 360.0],
                        [1340.0, 725.0], [1740.0, 245.0]])


def getdistmat(coordinates):
    num = coordinates.shape[0] #求第一维的长度
    distmat = np.zeros((52, 52))
    for i in range(num):
        for j in range(i, num):
            distmat[i][j] = distmat[j][i] = np.linalg.norm(coordinates[i]-coordinates[j])
    return distmat


distmat = getdistmat(coordinates)

numant = 40  #蚂蚁个数
numcity = coordinates.shape[0] #城市个数
alpha = 1   #信息素重要程度因子
beta = 5    #启发函数重要程度因子
rho = 0.1   #信息素的挥发速度
Q = 1

iter = 0
itermax = 200

etatable = 1.0/(distmat+np.diag([1e10]*numcity)) #启发函数矩阵，表示蚂蚁从城市i转移到矩阵j的期望程度
pheromonetable  = np.ones((numcity, numcity)) # 信息素矩阵
pathtable = np.zeros((numant, numcity)).astype(int) #路径记录表

distmat = getdistmat(coordinates) #城市的距离矩阵

lengthaver = np.zeros(itermax) #各代路径的平均长度
lengthbest = np.zeros(itermax) #各代及其之前遇到的最佳路径长度
pathbest = np.zeros((itermax, numcity)) # 各代及其之前遇到的最佳路径长度


while iter < itermax:
   # 随机产生各个蚂蚁的起点城市
    if numant <= numcity:#城市数比蚂蚁数多
        pathtable[:, 0] = np.random.permutation(range(0, numcity))[:numant]

    else: #蚂蚁数比城市数多，需要补足
        pathtable[:numcity, 0] = np.random.permutation(range(0, numcity))[:]
        pathtable[numcity:, 0] = np.random.permutation(range(0, numcity))[:numant-numcity]

    length = np.zeros(numant) #计算各个蚂蚁的路径距离

    for i in range(numant):
        visiting = pathtable[i, 0] # 当前所在的城市

        #visited = set() #已访问过的城市，防止重复
        #visited.add(visiting) #增加元素
        unvisited = set(range(numcity))#未访问的城市
        unvisited.remove(visiting) #删除元素


        for j in range(1,numcity):#循环numcity-1次，访问剩余的numcity-1个城市

            #每次用轮盘法选择下一个要访问的城市
            listunvisited = list(unvisited)

            probtrans = np.zeros(len(listunvisited))

            for k in range(len(listunvisited)):
                probtrans[k] = np.power(pheromonetable[visiting][listunvisited[k]], alpha)\
                        *np.power(etatable[visiting][listunvisited[k]], alpha)
            cumsumprobtrans = (probtrans/sum(probtrans)).cumsum()

            cumsumprobtrans -= np.random.rand()

            for n in range(len(cumsumprobtrans)):
                if cumsumprobtrans[n] > 0:
                    break

            k = listunvisited[n] #下一个要访问的城市

            pathtable[i, j] = k

            unvisited.remove(k)
            #visited.add(k)

            length[i] += distmat[visiting][k]

            visiting = k

        length[i] += distmat[visiting][pathtable[i,0]] #蚂蚁的路径距离包括最后一个城市和第一个城市的距离


    #print length
    # 包含所有蚂蚁的一个迭代结束后，统计本次迭代的若干统计参数

    lengthaver[iter] = length.mean()

    if iter == 0:
        lengthbest[iter] = length.min()
        pathbest[iter] = pathtable[length.argmin()].copy()
    else:
        if length.min() > lengthbest[iter-1]:
            lengthbest[iter] = lengthbest[iter-1]
            pathbest[iter] = pathbest[iter-1].copy()

        else:
            lengthbest[iter] = length.min()
            pathbest[iter] = pathtable[length.argmin()].copy()


    # 更新信息素
    changepheromonetable = np.zeros((numcity, numcity))
    for i in range(numant):
        for j in range(numcity-1):
            changepheromonetable[pathtable[i, j]][pathtable[i, j+1]] += Q/distmat[pathtable[i, j]][pathtable[i, j+1]]

        changepheromonetable[pathtable[i, j+1]][pathtable[i, 0]] += Q/distmat[pathtable[i, j+1]][pathtable[i, 0]]

    pheromonetable = (1-rho)*pheromonetable + changepheromonetable


    iter += 1 #迭代次数指示器+1

    #观察程序执行进度，该功能是非必须的
    if (iter-1) % 20 == 0:
        print(iter-1)

print(pathbest)

# 做出平均路径长度和最优路径长度
fig, axes = plt.subplots(nrows=2, ncols=1, figsize=(12, 10))
axes[0].plot(lengthaver, 'k', marker = u'')
axes[0].set_title('Average Length')
axes[0].set_xlabel(u'iteration')

axes[1].plot(lengthbest, 'k', marker = u'')
axes[1].set_title('Best Length')
axes[1].set_xlabel(u'iteration')
fig.savefig('Average_Best.png', dpi=500, bbox_inches='tight')
plt.close()


#作出找到的最优路径图
bestpath = pathbest[-1]

plt.plot(coordinates[:, 0], coordinates[:, 1], 'r.', marker=u'$\cdot$')
plt.xlim([-100, 2000])
plt.ylim([-100, 1500])

end = 0
for i in range(numcity-1):
    m, n = int(bestpath[i]), int(bestpath[i+1])
    print(m, n)
    plt.plot([coordinates[m][0], coordinates[n][0]], [coordinates[m][1], coordinates[n][1]], 'k')
    end = n

plt.plot([coordinates[int(bestpath[0])][0], coordinates[end][0]], [coordinates[int(bestpath[0])][1],
              coordinates[end][1]], 'b')

ax = plt.gca()
ax.set_title("Best Path")
ax.set_xlabel('X axis')
ax.set_ylabel('Y_axis')

plt.savefig('Best Path3.png', dpi=500, bbox_inches='tight')
plt.close()