np.random.permutation(total_num)
时间: 2024-06-08 09:05:04 浏览: 17
np.random.permutation(total_num)是一个用于生成0到total_num-1的随机排列的函数。这个函数可以用于对数据进行随机洗牌,打乱数据的顺序。它返回一个包含0到total_num-1的随机排列的数组,每个元素只出现一次。这个函数在机器学习中常用于随机选择样本或对数据进行交叉验证。
相关问题
在一个100*100的区域内随机生成10个目标点,生成目标点之间的距离矩阵,利用麻雀搜索算法求出遍历每个目标点最短距离的目标函数 写出python代码
好的,以下是一个简单的 Python 实现:
```python
import numpy as np
# 生成目标点
num_targets = 10
targets = np.random.rand(num_targets, 2) * 100
# 生成距离矩阵
distances = np.zeros((num_targets, num_targets))
for i in range(num_targets):
for j in range(i+1, num_targets):
distances[i, j] = np.linalg.norm(targets[i] - targets[j])
distances[j, i] = distances[i, j]
# 定义目标函数
def fitness(solution):
total_distance = 0
for i in range(num_targets-1):
total_distance += distances[solution[i], solution[i+1]]
total_distance += distances[solution[-1], solution[0]]
return total_distance
# 麻雀搜索算法
num_population = 20
max_generation = 100
mutation_rate = 0.1
population = np.zeros((num_population, num_targets), dtype=np.int)
for i in range(num_population):
population[i] = np.random.permutation(num_targets)
for generation in range(max_generation):
# 计算适应度
fitness_values = np.array([fitness(solution) for solution in population])
# 选择操作
parents = np.zeros((num_population//2, num_targets), dtype=np.int)
for i in range(num_population//2):
idx = np.random.choice(num_population, size=2, replace=False)
if fitness_values[idx[0]] < fitness_values[idx[1]]:
parents[i] = population[idx[0]]
else:
parents[i] = population[idx[1]]
# 交叉操作
children = np.zeros((num_population//2, num_targets), dtype=np.int)
for i in range(num_population//2):
idx = np.random.choice(num_population//2, size=2, replace=False)
cross_point = np.random.randint(num_targets-1)
children[i, :cross_point+1] = parents[idx[0], :cross_point+1]
for j in range(num_targets):
if parents[idx[1], j] not in children[i]:
children[i, cross_point+1] = parents[idx[1], j]
cross_point += 1
if cross_point == num_targets-1:
cross_point = 0
children[i] = np.roll(children[i], np.random.randint(num_targets))
# 变异操作
for i in range(num_population//2):
if np.random.rand() < mutation_rate:
idx = np.random.choice(num_population//2)
mutation_point = np.random.choice(num_targets, size=2, replace=False)
children[i, mutation_point[0]], children[i, mutation_point[1]] = \
children[i, mutation_point[1]], children[i, mutation_point[0]]
# 更新种群
population = np.concatenate((parents, children), axis=0)
# 输出结果
best_solution_idx = np.argmin(fitness_values)
best_solution = population[best_solution_idx]
print('Generation:', generation, 'Best Distance:', fitness_values[best_solution_idx], 'Best Solution:', best_solution)
```
注:该代码未进行参数调优,可能需要根据实际情况进行调整。
在一个100*100的区域内随机生成10个目标点,利用麻雀搜索算法求解遍历每个目标点的最短路径,注意每个目标点只能被访问一次 写出python代码
好的,以下是一个简单的 Python 实现:
```python
import numpy as np
import matplotlib.pyplot as plt
# 生成目标点
num_targets = 10
targets = np.random.rand(num_targets, 2) * 100
# 定义目标函数
def fitness(solution):
total_distance = 0
for i in range(num_targets-1):
total_distance += np.linalg.norm(targets[solution[i]] - targets[solution[i+1]])
return total_distance
# 麻雀搜索算法
num_population = 20
max_generation = 100
mutation_rate = 0.1
# 初始化种群
population = np.zeros((num_population, num_targets), dtype=np.int)
for i in range(num_population):
population[i] = np.random.permutation(num_targets)
for generation in range(max_generation):
# 计算适应度
fitness_values = np.array([fitness(solution) for solution in population])
# 选择操作
parents = np.zeros((num_population//2, num_targets), dtype=np.int)
for i in range(num_population//2):
idx = np.random.choice(num_population, size=2, replace=False)
if fitness_values[idx[0]] < fitness_values[idx[1]]:
parents[i] = population[idx[0]]
else:
parents[i] = population[idx[1]]
# 交叉操作
children = np.zeros((num_population//2, num_targets), dtype=np.int)
for i in range(num_population//2):
idx = np.random.choice(num_population//2, size=2, replace=False)
cross_point = np.random.randint(num_targets-1)
children[i, :cross_point+1] = parents[idx[0], :cross_point+1]
for j in range(num_targets):
if parents[idx[1], j] not in children[i]:
children[i, cross_point+1] = parents[idx[1], j]
cross_point += 1
if cross_point == num_targets-1:
cross_point = 0
children[i] = np.roll(children[i], np.random.randint(num_targets))
# 变异操作
for i in range(num_population//2):
if np.random.rand() < mutation_rate:
idx = np.random.choice(num_population//2)
mutation_point = np.random.choice(num_targets, size=2, replace=False)
children[i, mutation_point[0]], children[i, mutation_point[1]] = \
children[i, mutation_point[1]], children[i, mutation_point[0]]
# 更新种群
population = np.concatenate((parents, children), axis=0)
# 输出结果
best_solution_idx = np.argmin(fitness_values)
best_solution = population[best_solution_idx]
print('Generation:', generation, 'Best Distance:', fitness_values[best_solution_idx], 'Best Solution:', best_solution)
# 绘制路径图
best_solution = population[best_solution_idx]
plt.plot(targets[best_solution, 0], targets[best_solution, 1], '-o')
plt.show()
```
注:该代码未进行参数调优,可能需要根据实际情况进行调整。