蚁群算法Python3可运行代码

原理就不再赘述了，直接上代码：

import numpy as np
import matplotlib.pyplot as plt
import 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]
    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 = 250
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
        unvisited = set(range(numcity))  # 未访问的城市,以集合的形式存储{}
        unvisited.remove(visiting)  # 删除元素；利用集合的remove方法删除存储的数据内容
        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()
            k = listunvisited[(np.where(cumsumprobtrans > 0)[0])[0]]  # python3中原代码运行bug，类型问题；鉴于此特找到其他方法
            # 通过where（）方法寻找矩阵大于0的元素的索引并返回ndarray类型，然后接着载使用[0]提取其中的元素，用作listunvisited列表中
            # 元素的提取（也就是下一轮选的城市）
            pathtable[i, j] = k  # 添加到路径表中（也就是蚂蚁走过的路径)
            unvisited.remove(k)  # 然后在为访问城市set中remove（）删除掉该城市
            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
        # 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]]  # 计算信息素增量
        changepheromonetable[pathtable[i, j + 1]][pathtable[i, 0]] += Q / distmat[pathtable[i, j + 1]][pathtable[i, 0]]
    pheromonetable = (1 - rho) * pheromonetable + changepheromonetable  # 计算信息素公式
    iter += 1  # 迭代次数指示器+1
    print("iter:", iter)

# 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.show()

# 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])

for i in range(numcity - 1):
    m = int(bestpath[i])
    n = int(bestpath[i + 1])
    plt.plot([coordinates[m][0], coordinates[n][0]], [coordinates[m][1], coordinates[n][1]], 'k')
plt.plot([coordinates[int(bestpath[0])][0],coordinates[int(n)][0]],[coordinates[int(bestpath[0])][1],coordinates[int(n)][1]],'b')
ax = plt.gca()
ax.set_title("Best Path")
ax.set_xlabel('X axis')
ax.set_ylabel('Y_axis')

plt.savefig('best path.png', dpi=500, bbox_inches='tight')
plt.show()

结果：