在Python中提取一部分文件路径(目录) [英] Extract a part of the filepath (a directory) in Python

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问题描述

我需要提取特定路径的父目录的名称。看起来像这样:

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

我正在使用 directory_i_need 名称(而不是路径)。我创建了一个函数,该函数将给我所有文件的列表,然后...

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.

我该怎么办?

推荐答案

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
...



目录的目录

您可以根据需要继续执行多次...

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

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

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.

这篇关于在Python中提取一部分文件路径(目录)的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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