Python语言、操作系统、相对路径，使文件名成为一个列表[重复]

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Python Glob without the whole path - only the filename 3 answers
Python通配符，没有完整的路径-只有文件名3

I am trying to make a list of all files in a directory with filenames in a that end in .root.

我正在尝试列出一个目录中的所有文件的列表，其中的文件名以.root结尾。

After reading some writings in the forum I tried to basic strategies using glob and os.listdir but I got into trouble for both of them

在论坛上阅读了一些文章后，我尝试了使用glob和os的基本策略。但是他们两个我都惹上麻烦了

First, when I use

首先,当我使用

import glob
filelist = glob.glob('/home/usr/dir/*.root')

It does make a list of string with all filenames that end in .root but I still face a problem.

它确实创建了一个字符串列表，所有文件名以.root结尾，但是我仍然面临一个问题。

I would like to be the list of string to have filenames as '/dir/.root' but the string has full path '/home/usr/dir/.root'

我想要成为有文件名的字符串列表。但字符串有完整路径'/home/usr/dir/.root'

Second, if I use os.listdir, I get into the trouble that

其次，如果我使用os。李斯特，我有麻烦了

  path = '/home/usr/'
  filelist = os.listdir(path + 'dir/*.root')
  syntax error

which tells me that I can not only get the list of files for .root.

这告诉我，我不仅可以得到.root的文件列表。

In summary, I would like to make a list of filenames, that end in .root and are in my /home/usr/dir, while cutting off the '/home/usr' part. If I use globe, I get into the trouble of having /home/usr/. If I use os.listdir, I can't specify ".root" endling.

总之，我想列出一个文件名列表，它以.root结尾，在my /home/usr/dir中，同时删除“/home/usr”部分。如果我用地球仪，我就会有/home/usr/的麻烦。如果我使用操作系统。listdir，我不能指定"。“无尽的根源。

2 个解决方案

#1

glob will return paths in a format matching your query, so that

glob将以与查询匹配的格式返回路径

glob.glob("/home/usr/dir/*.root")
# ['home/usr/dir/foo.root', 'home/usr/dir/bar.root', ...]

glob.glob("*.root")
# ['foo.root', 'bar.root', ...]

glob.glob("./*.root")
# ['./foo.root', './bar.root', ...]

...and so forth.

…等等。

To get only the filename, you can use path.basename of the os module, something like this:

要获取文件名，可以使用path。操作系统模块的basename:

from glob import glob
from os import path

pattern = "/home/usr/dir/*.root"
files = [path.basename(x) for x in glob(pattern)]
# ['foo.root', 'bar.root', ...]

...or, if you want to prepend the dir part:

…或者，如果您想在dir部分之前:

pattern = "/home/usr/dir/*.root"
files = [path.join('dir', path.basename(x)) for x in glob(pattern)]
# ['dir/foo.root', 'dir/bar.root', ...]

...or, if you really want the path separator at the start:

…或者，如果您真的想在开始时使用路径分隔符:

from glob import glob
import os

pattern = "/home/usr/dir/*.root"
files = [os.sep + os.path.join('dir', os.path.basename(x)) for x in glob(pattern)]
# ['/dir/foo.root', '/dir/bar.root', ...]

Using path.join and path.sep will make sure that the correct path syntax is used, depending on your OS (i.e. / or \ as a separator).

使用路径。加入和路径。sep将确保使用正确的路径语法，这取决于您的操作系统(即/或作为分隔符)。

Depending on what you are really trying to do here, you might want to look at os.path.relpath, for the relative path. The title of your question indicates that relative paths might be what you are actually after:

根据您在这里真正要做的事情，您可能需要查看os.path。关系路径。你的问题的标题表明，相对路径可能是你真正想要的:

pattern = "/home/usr/dir/*.root"
files = [os.path.relpath(x) for x in glob(pattern)]
# files will now contain the relative path to each file, from the current working directory

#2

just use glob for getting the list you want
and then use os.path.relpath on each file

只需使用glob获取所需的列表，然后使用os.path。relpath在每个文件

import glob
files_names = []
for file in glob.glob('/home/usr/dir/*.root'):
    files_names.append(os.path.relpath(file, "/home/usr"))

You can also use regex

您还可以使用regex

import re
files_names.append(re.sub(r'//home//usr//','', file, flags=re.I))

#1