问题:在Python中提取一部分文件路径(目录)

我需要提取某个路径的父目录的名称。看起来是这样的:

c:\stuff\directory_i_need\subdir\file

我正在使用使用文件directory_i_need名(而不是路径)的东西来修改“文件”的内容。我创建了一个函数,该函数会给我所有文件的列表,然后…

for path in file_list:
   #directory_name = os.path.dirname(path)   # this is not what I need, that's why it is commented
   directories, files = path.split('\\')

   line_replace_add_directory = line_replace + directories  
   # this is what I want to add in the text, with the directory name at the end 
   # of the line.

我怎样才能做到这一点?

I need to extract the name of the parent directory of a certain path. This is what it looks like:

c:\stuff\directory_i_need\subdir\file

I am modifying the content of the “file” with something that uses the directory_i_need name in it (not the path). I have created a function that will give me a list of all the files, and then…

for path in file_list:
   #directory_name = os.path.dirname(path)   # this is not what I need, that's why it is commented
   directories, files = path.split('\\')

   line_replace_add_directory = line_replace + directories  
   # this is what I want to add in the text, with the directory name at the end 
   # of the line.

How can I do that?


回答 0

import os
## first file in current dir (with full path)
file = os.path.join(os.getcwd(), os.listdir(os.getcwd())[0])
file
os.path.dirname(file) ## directory of file
os.path.dirname(os.path.dirname(file)) ## directory of directory of file
...

而且您可以根据需要继续执行多次…

编辑:os.path,您可以使用os.path.split或os.path.basename:

dir = os.path.dirname(os.path.dirname(file)) ## dir of dir of file
## once you're at the directory level you want, with the desired directory as the final path node:
dirname1 = os.path.basename(dir) 
dirname2 = os.path.split(dir)[1] ## if you look at the documentation, this is exactly what os.path.basename does.
import os
## first file in current dir (with full path)
file = os.path.join(os.getcwd(), os.listdir(os.getcwd())[0])
file
os.path.dirname(file) ## directory of file
os.path.dirname(os.path.dirname(file)) ## directory of directory of file
...

And you can continue doing this as many times as necessary…

Edit: from os.path, you can use either os.path.split or os.path.basename:

dir = os.path.dirname(os.path.dirname(file)) ## dir of dir of file
## once you're at the directory level you want, with the desired directory as the final path node:
dirname1 = os.path.basename(dir) 
dirname2 = os.path.split(dir)[1] ## if you look at the documentation, this is exactly what os.path.basename does.

回答 1

在Python 3.4中,您可以使用pathlib模块

>>> from pathlib import Path
>>> p = Path('C:\Program Files\Internet Explorer\iexplore.exe')
>>> p.name
'iexplore.exe'
>>> p.suffix
'.exe'
>>> p.root
'\\'
>>> p.parts
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
>>> p.relative_to('C:\Program Files')
WindowsPath('Internet Explorer/iexplore.exe')
>>> p.exists()
True

In Python 3.4 you can use the pathlib module:

>>> from pathlib import Path
>>> p = Path('C:\Program Files\Internet Explorer\iexplore.exe')
>>> p.name
'iexplore.exe'
>>> p.suffix
'.exe'
>>> p.root
'\\'
>>> p.parts
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')
>>> p.relative_to('C:\Program Files')
WindowsPath('Internet Explorer/iexplore.exe')
>>> p.exists()
True

回答 2

parent如果您使用,您所需要的只是一部分pathlib

from pathlib import Path
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parent) 

将输出:

C:\Program Files\Internet Explorer    

如果您需要所有部分(已经包含在其他答案中),请使用parts

p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parts) 

然后,您将获得一个列表:

('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')

节省时间。

All you need is parent part if you use pathlib.

from pathlib import Path
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parent) 

Will output:

C:\Program Files\Internet Explorer    

Case you need all parts (already covered in other answers) use parts:

p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parts) 

Then you will get a list:

('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')

Saves tone of time.


回答 3

首先,查看中是否有splitunc()可用功能os.path。返回的第一项应该是您想要的…但是我在Linux上,并且在导入os并尝试使用它时没有此功能。

否则,完成工作的一种半丑陋的方法是使用:

>>> pathname = "\\C:\\mystuff\\project\\file.py"
>>> pathname
'\\C:\\mystuff\\project\\file.py'
>>> print pathname
\C:\mystuff\project\file.py
>>> "\\".join(pathname.split('\\')[:-2])
'\\C:\\mystuff'
>>> "\\".join(pathname.split('\\')[:-1])
'\\C:\\mystuff\\project'

该图显示了检索文件正上方的目录以及该目录正上方的目录。

First, see if you have splitunc() as an available function within os.path. The first item returned should be what you want… but I am on Linux and I do not have this function when I import os and try to use it.

Otherwise, one semi-ugly way that gets the job done is to use:

>>> pathname = "\\C:\\mystuff\\project\\file.py"
>>> pathname
'\\C:\\mystuff\\project\\file.py'
>>> print pathname
\C:\mystuff\project\file.py
>>> "\\".join(pathname.split('\\')[:-2])
'\\C:\\mystuff'
>>> "\\".join(pathname.split('\\')[:-1])
'\\C:\\mystuff\\project'

which shows retrieving the directory just above the file, and the directory just above that.


回答 4

这是我提取目录的一部分的工作:

for path in file_list:
  directories = path.rsplit('\\')
  directories.reverse()
  line_replace_add_directory = line_replace+directories[2]

谢谢您的帮助。

This is what I did to extract the piece of the directory:

for path in file_list:
  directories = path.rsplit('\\')
  directories.reverse()
  line_replace_add_directory = line_replace+directories[2]

Thank you for your help.


回答 5

import os

directory = os.path.abspath('\\') # root directory
print(directory) # e.g. 'C:\'

directory = os.path.abspath('.') # current directory
print(directory) # e.g. 'C:\Users\User\Desktop'

parent_directory, directory_name = os.path.split(directory)
print(directory_name) # e.g. 'Desktop'
parent_parent_directory, parent_directory_name = os.path.split(parent_directory)
print(parent_directory_name) # e.g. 'User'

这也应该可以解决问题。

import os

directory = os.path.abspath('\\') # root directory
print(directory) # e.g. 'C:\'

directory = os.path.abspath('.') # current directory
print(directory) # e.g. 'C:\Users\User\Desktop'

parent_directory, directory_name = os.path.split(directory)
print(directory_name) # e.g. 'Desktop'
parent_parent_directory, parent_directory_name = os.path.split(parent_directory)
print(parent_directory_name) # e.g. 'User'

This should also do the trick.


回答 6

您必须将整个路径作为os.path.split的参数。请参阅文档。它不像字符串拆分那样工作。

You have to put the entire path as a parameter to os.path.split. See The docs. It doesn’t work like string split.


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