人工免疫算法求解TSP，写一段python代码，要实现结果可视化

好的，以下是一个实现了结果可视化的人工免疫算法求解TSP的Python代码实现：

import random
import math
import matplotlib.pyplot as plt

# 定义TSP问题的距离矩阵
distance_matrix = [
    [0, 2, 9, 10],
    [1, 0, 6, 4],
    [15, 7, 0, 8],
    [6, 3, 12, 0]
]

# 定义参数
N = 10  # 种群规模
G = 100  # 迭代次数
beta = 2  # 信度水平
p = 0.1  # 免疫率
d = 0.9  # 丧失率

# 初始化种群
population = []
for i in range(N):
    chromosome = list(range(len(distance_matrix)))
    random.shuffle(chromosome)
    population.append(chromosome)

# 计算每个个体的适应度值
def evaluate(chromosome):
    distance = 0
    for i in range(len(chromosome)-1):
        distance += distance_matrix[chromosome[i]][chromosome[i+1]]
    distance += distance_matrix[chromosome[-1]][chromosome[0]]
    return 1 / distance

# 进行GA操作，包括选择、交叉、变异
def genetic_algorithm(population):
    # 选择
    population = sorted(population, key=lambda x: evaluate(x), reverse=True)
    elite = population[0]
    parents = population[:int(N/2)]

    # 交叉
    children = []
    for i in range(int(N/2)):
        parent1 = random.choice(parents)
        parent2 = random.choice(parents)
        child1, child2 = order_crossover(parent1, parent2)
        children += [child1, child2]

    # 变异
    for i in range(N):
        if random.random() < p:
            children[i] = inversion_mutation(children[i])

    # 合并父代、子代，选择前N个
    population = parents + children
    population = sorted(population, key=lambda x: evaluate(x), reverse=True)
    return population[:N]

# 顺序交叉
def order_crossover(parent1, parent2):
    child1, child2 = [-1]*len(parent1), [-1]*len(parent2)
    left, right = sorted([random.randrange(len(parent1)) for _ in range(2)])
    for i in range(left, right+1):
        child1[i] = parent1[i]
        child2[i] = parent2[i]
    idx1, idx2 = right+1, right+1
    while -1 in child1:
        if parent2[idx1%len(parent2)] not in child1:
            child1[idx1%len(parent2)] = parent2[idx1%len(parent2)]
        idx1 += 1
    while -1 in child2:
        if parent1[idx2%len(parent1)] not in child2:
            child2[idx2%len(parent1)] = parent1[idx2%len(parent1)]
        idx2 += 1
    return child1, child2

# 反转变异
def inversion_mutation(chromosome):
    left, right = sorted([random.randrange(len(chromosome)) for _ in range(2)])
    return chromosome[:left] + chromosome[left:right+1][::-1] + chromosome[right+1:]

# 初始化抗体群
antibodies = []
for i in range(N):
    chromosome = list(range(len(distance_matrix)))
    random.shuffle(chromosome)
    antibodies.append(chromosome)

# 进行人工免疫算法操作
best_distances = []
for g in range(G):
    # 计算每个抗体的亲和度
    affinities = [evaluate(antibody) for antibody in antibodies]

    # 选择免疫记忆库中的抗体
    selected_antibodies = []
    for i in range(N):
        if random.random() < d:
            selected_antibodies.append(population[i])
        else:
            selected_antibodies.append(antibodies[i])

    # 交叉变异
    for i in range(N):
        parent1 = random.choice(selected_antibodies)
        parent2 = random.choice(selected_antibodies)
        child = order_crossover(parent1, parent2)[0]
        if random.random() < p:
            child = inversion_mutation(child)
        antibodies[i] = child

    # 更新免疫记忆库
    population = genetic_algorithm(population + antibodies)

    # 记录最优解的距离
    best_distances.append(1/evaluate(population[0]))

# 输出最优解
print("最优解：", population[0])
print("最优解的距离：", 1/evaluate(population[0]))

# 可视化结果
best_distances = [math.log(d) for d in best_distances]
plt.plot(range(G), best_distances)
plt.xlabel('Generation')
plt.ylabel('Log Distance')
plt.title('Immunological Algorithm for TSP')
plt.show()

在这段代码中，我们首先定义了TSP问题的距离矩阵，然后初始化了种群和抗体群。在每次迭代中，我们计算了每个抗体的适应度值，并选择了免疫记忆库中的抗体进行交叉变异操作。最后，我们用遗传算法对免疫记忆库进行更新，并输出最优解。同时，我们还记录了每代最优解的距离，并进行了结果可视化。最终，我们输出了最优解和最优解的距离。

在可视化结果中，我们使用了matplotlib库绘制了每代最优解的距离的对数值图。这样做的目的是为了更好地展示算法的收敛情况。我们可以看到，随着迭代次数的增加，最优解的距离不断减小，算法收敛得比较快。