Question

我正在尝试编写一个Python函数，它存储网格上两点之间的所有路径。 Here是最初的问题，我受到答案的启发，并试图在Python中创建一个类似的版本。

我能够成功打印出通过应用递归函数找到的所有路径，但问题是当我需要将它们存储在列表中时，我得到了一个空列表列表。我已经尝试使用全局变量和局部变量作为存储列表，并且它们给出了相同的结果。有谁可以请评论下面我的代码，并建议我做错了什么？

非常感谢！

import numpy as np

#initialize a matrix
grid = np.arange(0, 1225).reshape((35,35))
m = 10
n = 5
#starting point
s = (0, 0)
#target point
d = (n, m)
results = []

#a function to find the next step in the path
def next_step(x, y):
    step = []
    dx = [0, 1]
    dy = [1, 0]

    for i in range(2):
        next_x = x + dx[i]
        next_y = y + dy[i]
        if next_x >= 0 and next_x <= n and next_y >= 0 and next_y <= m:
            step.append((next_x, next_y))
    return step

def print_all_path(s, d, visited, path):
    global results
    visited[s[0]][s[1]] = True
    path.append(grid[s[0]][s[1]])

    if s == d:
        results.append(path)
        print(path)
    else:
        for step in next_step(s[0], s[1]):
            if visited[step[0],step[1]] == False:
                print_all_path(step, d, visited, path)
    path.pop()
    visited[s[0]][s[1]] = False

def print_path(s, d):
    visited = np.zeros((35,35), dtype = bool)
    path = []
    print_all_path(s, d, visited, path)

print_path(s, d)

Answer 1

问题可能是当你追加时，你只追加path我怀疑你做了这样的事情：

# global 
all_lists = []

# your functions
...
def print_all_path(s, d, visited, path):
    global results
    visited[s[0]][s[1]] = True
    path.append(grid[s[0]][s[1]])

    if s == d:
        results.append(path)
        print(path)
        all_lists.append(path)
...

但是，路径仍然与原始path变量相关联。

您可以使用以下方法解决此问题：

all_lists.append(path + [])

这会复制列表并删除链接

所以整个程序现在是

import numpy as np

#initialize a matrix
grid = np.arange(0, 1225).reshape((35,35))
m = 10
n = 5
#starting point
s = (0, 0)
#target point
d = (n, m)
results = []
all_paths = []

#a function to find the next step in the path
def next_step(x, y):
    step = []
    dx = [0, 1]
    dy = [1, 0]

    for i in range(2):
        next_x = x + dx[i]
        next_y = y + dy[i]
        if next_x >= 0 and next_x <= n and next_y >= 0 and next_y <= m:
            step.append((next_x, next_y))
    return step

def print_all_path(s, d, visited, path):
    global results
    visited[s[0]][s[1]] = True
    path.append(grid[s[0]][s[1]])

    if s == d:
        results.append(path)
        all_paths.append(path + [])
        #print(path)
    else:
        for step in next_step(s[0], s[1]):
            if visited[step[0],step[1]] == False:
                print_all_path(step, d, visited, path)
    path.pop()
    visited[s[0]][s[1]] = False

def print_path(s, d):
    visited = np.zeros((35,35), dtype = bool)
    path = []
    print_all_path(s, d, visited, path)


print_path(s, d)
print all_paths

将值存储到Python中的递归函数中的列表时出现问题

1 个答案: