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Non-alphanumeric list order from os.listdir() (14 answers)

Is there a built in function for string natural sort? (24 answers)

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我不想把所有档案改名为目录。视窗探测器名称排列后,按正确顺序排列:

Cue 001 - 002.mp3
...
Cue 001 - 010.mp3
Cue 001 - 011.mp3
...
Cue 001 - 020.mp3
Cue 001 - 021.mp3
...
Cue 001 - 0111.mp3
Cue 001 - 0112.mp3

到目前为止,我已经能够使用<代码>os.chdir,将工作名录改为有目标文档的名录,并通过使用<代码>印(os.getcwd()加以确认。

import os
# Lets change working directory to the Python Rename Test folder
os.chdir(r F:My backupDocuments and SettingsuserMy DocumentsMy         MusicThe CommitmentsThe Commitments - Mustang Sally_RecursiveBackupPython Rename Test )
# confirm working directory by printing it out
print (os.getcwd())
# loop over the files in the working directory and printing them out
for file in os.listdir(os.getcwd()):
    print (file)

This is what I get:

Cue 001 - 002.mp3
...
Cue 001 - 010.mp3
Cue 001 - 0100.mp3
Cue 001 - 0101.mp3
...

我怎样才能获得<代码>os.listdir,以便按适当顺序打印文件?

Answer 1

我假定你想要的是档案延期前直接编号的档案。

import re

files = os.listdir(os.getcwd())
sorted_files = sorted(files,key=lambda x: int(re.sub(r D ,  ,x.strip())[3:]))
print(sorted_files)

Answer 2

如果贵国的档案中还有第1个编号超过100,则该代码按数字表示。

In[2]: import re
  ...: 
  ...: 
  ...: def sorted_by_nums(filenames):
  ...:     """Assuming filenames are in this format  Cue 001 - 002.mp3 """
  ...:     def by_nums(filename):
  ...:         *nums, _ = re.findall(r d+ , filename)
  ...:         return [int(num) for num in nums]
  ...: 
  ...:     return sorted(filenames, key=by_nums)
  ...: 
In[3]: filenames = [
  ...:      Cue 001 - 010.mp3 ,
  ...:      Cue 001 - 011.mp3 ,
  ...:      Cue 001 - 0112.mp3 ,
  ...:      Cue 001 - 0111.mp3 ,
  ...:      Cue 001 - 002.mp3 ,
  ...:      Cue 001 - 020.mp3 ,
  ...:      Cue 001 - 021.mp3 
  ...: ]
# you could replace `filenames` with `os.listdir(...)` below
In[4]: for file in sorted_by_nums(filenames):
  ...:     print(file)
  ...: 
Cue 001 - 002.mp3
Cue 001 - 010.mp3
Cue 001 - 011.mp3
Cue 001 - 020.mp3
Cue 001 - 021.mp3
Cue 001 - 0111.mp3
Cue 001 - 0112.mp3

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