为什么列表理解在 os.walk 函数结果的列表的情况下不起作用？

Question

在 os.walk 函数的结果中，我得到了这个列表，其中包含文件夹中的文件名：

['tr-02-10-18.pdf', 'tr-02-11-18.pdf', 'tr-02-12-18.pdf', 'tr-03-11-18.pdf', 'tr-03-12-18.pdf', 'tr-04-10-18.pdf', 'tr-04-11-18.pdf', 'tr-04-12-18.pdf']

我的目标是将列表变成那种形式：

['tr/02/10/18.pdf', 'tr/02/11/18.pdf','tr/02/12/18.pdf', 'tr/03/11/18.pdf', 'tr/03/12/18.pdf', 'tr/04/10/18.pdf', 'tr/04/11/18.pdf', 'tr/04/12/18.pdf']

我尝试使用如下所示的代码：

import os
for file_name in os.walk(input()):
    egg1=(str(file_name[-1]))
    egg2= []
    for mark in egg1:
        mark=[i.replace("-","/") for i in mark]
        egg2.append(mark)
print(egg2)

我得到的是这个列表：

[['['], ["'"], ['t'], ['r'], ['/'], ['0'], ['2'], ['/'], ['1'], ['0'], ['/'], ['1'], ['8'], ['.'], ['p'], ['d'], ['f'], ["'"], [','], [' '], ["'"], ['t'], ['r'], ['/'], ['0'], ['2'], ['/'], ['1'], ['1'], ['/'], ['1'], ['8'], ['.'], ['p'], ['d'], ['f'], ["'"], [','], [' '], ["'"], ['t'], ['r'], ['/'], ['0'], ['2'], ['/'], ['1'], ['2'], ['/'], ['1'], ['8'], ['.'], ['p'], ['d'], ['f'], ["'"], [','], [' '], ["'"], ['t'], ['r'], ['/'], ['0'], ['3'], ['/'], ['1'], ['1'], ['/'], ['1'], ['8'], ['.'], ['p'], ['d'], ['f'], ["'"], [','], [' '], ["'"], ['t'], ['r'], ['/'], ['0'], ['3'], ['/'], ['1'], ['2'], ['/'], ['1'], ['8'], ['.'], ['p'], ['d'], ['f'], ["'"], [','], [' '], ["'"], ['t'], ['r'], ['/'], ['0'], ['4'], ['/'], ['1'], ['0'], ['/'], ['1'], ['8'], ['.'], ['p'], ['d'], ['f'], ["'"], [','], [' '], ["'"], ['t'], ['r'], ['/'], ['0'], ['4'], ['/'], ['1'], ['1'], ['/'], ['1'], ['8'], ['.'], ['p'], ['d'], ['f'], ["'"], [','], [' '], ["'"], ['t'], ['r'], ['/'], ['0'], ['4'], ['/'], ['1'], ['2'], ['/'], ['1'], ['8'], ['.'], ['p'], ['d'], ['f'], ["'"], [']']]

P.S。这是我的第一个问题，对于任何不一致之处，我们深表歉意

Answer 1

你犯了几个错误：

os.walk() 生成一个包含 3 个元素的元组，即当前目录的路径、文件名列表和目录名列表。您将元组命名为 file_name:
```
for file_name in os.walk(input()):
```
最好将其拆分成单独的组件，所以 for dirpath, dirnames, filenames in os.walk(...):。由于您忽略了前两个参数，因此您还可以使用 for *_, filenames in os.walk(...):
您将文件名列表变成了一个字符串：
```
egg1=(str(file_name[-1]))
```
file_name[-1] 是文件名列表。你把 list 变成了一个字符串，所以现在你有字符串表示和 [、] 和 , 逗号字符：
```
>>> egg1
"['tr-02-10-18.pdf', 'tr-02-11-18.pdf', 'tr-02-12-18.pdf', 'tr-03-11-18.pdf', 'tr-03-12-18.pdf', 'tr-04-10-18.pdf', 'tr-04-11-18.pdf', 'tr-04-12-18.pdf']"
```
然后您从上面的字符串中取出每个单独的字符，并将破折号替换为斜杠。不要把列表变成字符串，直接遍历列表。
您 将 egg2 列表替换为每个目录的新空列表。在 os.walk() 循环之前创建列表。

您将每个 str.replace() 调用的结果放入一个包含单个元素的新列表中，然后将该列表附加到 egg2 列表中。不要添加更多列表。

此代码有效：

result = [] for *_, filenames in os.walk(input()): for filename in filenames: replaced = filename.replace('-', '/') result.append(replaced)

这可以组合成一个列表理解：

result = [f.replace('-', '/') for *_, fs in os.walk(input()) for f in fs]

为什么列表理解在 os.walk 函数结果的列表的情况下不起作用？

Why list comprehension doesn't work in case of list whih is result of os.walk function?

python-3.x

python-os