blocks|key|5260122|text|实际上，有一个function可以精确地返回您想要的内容|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|5260123|import+os
print(os.path.basename(your_path))|code-block|syntax|javascript|5260124|警告:当在POSIX系统上使用os.path.basename()从Windows样式的路径(例如"C:\\my\\file.txt"__)中获取基本名称时，将返回整个路径。|style|CODE|5260125|下面是运行在Linux主机上的交互式python+shell示例：|5260126|Python+3.8.2+(default,+Mar+13+2020,+10:14:16)
[GCC+9.3.0]+on+linux
Type+"help",+"copyright",+"credits"+or+"license"+for+more+information.
>>>+import+os
>>>+filepath+=+"C:\\my\\path\\to\\file.txt"+#+A+Windows+style+file+path.
>>>+os.path.basename(filepath)
'C:\\my\\path\\to\\file.txt'|5260127|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/library/os.path.html#os.path.basename^0|7|8|0|0|0|F|I|1D|I|0|0|0^^$0|@$1|2|3|4|5|6|7|Y|8|@]|9|@$A|Z|B|10|1|11]]|C|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|13|8|@$A|14|B|15|K|L]|$A|16|B|17|K|L]]|9|@]|C|$]]|$1|M|3|N|5|6|7|18|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|19|8|@]|9|@]|C|$G|H]]|$1|Q|3|-4|5|6|7|1A|8|@]|9|@]|C|$]]]|R|$S|$5|T|U|V|C|$W|X]]]]

Actually, there's a <a href="https://docs.python.org/2/library/os.path.html#os.path.basename" rel="noreferrer">function</a> that returns exactly what you want
<pre><code>import os
print(os.path.basename(your_path))
</code></pre>
WARNING: When <code>os.path.basename()</code> is used on a POSIX system to get the base name from a Windows styled path (e.g. <code>&quot;C:\\my\\file.txt&quot;</code>), the entire path will be returned.
Example below from interactive python shell running on a Linux host:
<pre><code>Python 3.8.2 (default, Mar 13 2020, 10:14:16)
[GCC 9.3.0] on linux
Type &quot;help&quot;, &quot;copyright&quot;, &quot;credits&quot; or &quot;license&quot; for more information.
&gt;&gt;&gt; import os
&gt;&gt;&gt; filepath = &quot;C:\\my\\path\\to\\file.txt&quot; # A Windows style file path.
&gt;&gt;&gt; os.path.basename(filepath)
'C:\\my\\path\\to\\file.txt'
</code></pre>

blocks|key|34552|text|os.path.split是您要查找的函数|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|34553|head,+tail+=+os.path.split("/tmp/d/a.dat")

>>>+print(tail)
a.dat
>>>+print(head)
/tmp/d|code-block|syntax|javascript|34554|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/os.path.html#os.path.split^0|0|D|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

<a href="http://docs.python.org/library/os.path.html#os.path.split" rel="noreferrer">os.path.split</a>
is the function you are looking for

<pre><code>head, tail = os.path.split("/tmp/d/a.dat")

&gt;&gt;&gt; print(tail)
a.dat
&gt;&gt;&gt; print(head)
/tmp/d
</code></pre>

blocks|key|34971|text|在python+3中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|34972|>>>+from+pathlib+import+Path++++
>>>+Path("/tmp/d/a.dat").name
'a.dat'|code-block|syntax|javascript|34973|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

In python 3

<pre><code>&gt;&gt;&gt; from pathlib import Path 
&gt;&gt;&gt; Path("/tmp/d/a.dat").name
'a.dat'
</code></pre>

blocks|key|34763|text|import+os
head,+tail+=+os.path.split('path/to/file.exe')|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|34764|tail是您想要的文件名。|unstyled|34765|有关详情，请参阅python+os+module+docs|offset|length|34766|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/os.path.html#os.path.split^0|0|0|8|L|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|T|8|@]|9|@]|A|$]]|$1|G|3|H|5|F|7|U|8|@]|9|@$I|V|J|W|1|X]]|A|$]]|$1|K|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|L|$M|$5|N|O|P|A|$Q|R]]]]

<pre><code>import os
head, tail = os.path.split('path/to/file.exe')
</code></pre>

tail is what you want, the filename.

See <a href="http://docs.python.org/library/os.path.html#os.path.split" rel="noreferrer">python os module docs</a> for detail

blocks|key|3646578|text|import+os
file_location+=+'/srv/volume1/data/eds/eds_report.csv'
file_name+=+os.path.basename(file_location+)++#eds_report.csv
location+=+os.path.dirname(file_location+)++++#/srv/volume1/data/eds|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|3646579|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre class="lang-py prettyprint-override"><code>import os
file_location = '/srv/volume1/data/eds/eds_report.csv'
file_name = os.path.basename(file_location ) #eds_report.csv
location = os.path.dirname(file_location ) #/srv/volume1/data/eds
</code></pre>

blocks|key|1052349|text|如果你想自动获取文件名，你可以这样做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1052350|import+glob

for+f+in+glob.glob('/your/path/*'):
++++print(os.path.split(f)[-1])|code-block|syntax|javascript|1052351|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you want to get the filename automatically you can do

<pre><code>import glob

for f in glob.glob('/your/path/*'):
 print(os.path.split(f)[-1])
</code></pre>

blocks|key|5260235|text|fname+=+str("C:\Windows\paint.exe").split('\\')[-1:][0]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|5260236|这将返回：paint.exe|unstyled|offset|length|style|BOLD|5260237|5260238|根据您的路径或操作系统更改拆分函数的SEP值。|blockquote|5260239|5260240|entityMap^0|0|5|9|0|0|0|0^^$0|@$1|2|3|4|5|6|7|R|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|S|8|@$G|T|H|U|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|V|8|@]|9|@]|A|$]]|$1|L|3|M|5|N|7|W|8|@]|9|@]|A|$]]|$1|O|3|-4|5|F|7|X|8|@]|9|@]|A|$]]|$1|P|3|-4|5|F|7|Y|8|@]|9|@]|A|$]]]|Q|$]]

<pre><code>fname = str("C:\Windows\paint.exe").split('\\')[-1:][0]
</code></pre>

this will return : paint.exe

<blockquote>
 change the sep value of the split function regarding your path or OS.
</blockquote>

blocks|key|1054825|text|我个人最喜欢的是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1054826|filename+=+fullname.split(os.sep)[-1]|code-block|syntax|javascript|1054827|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My personal favourite is:
<pre><code>filename = fullname.split(os.sep)[-1]
</code></pre>

blocks|key|34821|text|在您的示例中，您还需要从右侧去掉斜杠以返回c|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|34822|>>>+import+os
>>>+path+=+'a/b/c/'
>>>+path+=+path.rstrip(os.sep)+#+strip+the+slash+from+the+right+side
>>>+os.path.basename(path)
'c'|code-block|syntax|javascript|34823|第二级：|34824|>>>+os.path.filename(os.path.dirname(path))
'b'|34825|更新:我认为lazyr提供了正确的答案。我的代码不能在unix系统上使用类windows路径，反之亦然。|34826|entityMap^0|L|1|0|0|0|0|6|5|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|W|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|R|$]]

In your example you will also need to strip slash from right the right side to return <code>c</code>:

<pre><code>&gt;&gt;&gt; import os
&gt;&gt;&gt; path = 'a/b/c/'
&gt;&gt;&gt; path = path.rstrip(os.sep) # strip the slash from the right side
&gt;&gt;&gt; os.path.basename(path)
'c'
</code></pre>

Second level:

<pre><code>&gt;&gt;&gt; os.path.filename(os.path.dirname(path))
'b'
</code></pre>

update: I think <code>lazyr</code> has provided the right answer. My code will not work with windows-like paths on unix systems and vice versus with unix-like paths on windows system.

blocks|key|34701|text|这适用于linux和windows以及标准库。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|34702|paths+=+['a/b/c/',+'a/b/c',+'\\a\\b\\c',+'\\a\\b\\c\\',+'a\\b\\c',
+++++++++'a/b/../../a/b/c/',+'a/b/../../a/b/c']

def+path_leaf(path):
++++return+path.strip('/').strip('\\').split('/')[-1].split('\\')[-1]

[path_leaf(path)+for+path+in+paths]|code-block|syntax|javascript|34703|结果：|34704|['c',+'c',+'c',+'c',+'c',+'c',+'c']|34705|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

This is working for linux and windows as well with standard library

<pre><code>paths = ['a/b/c/', 'a/b/c', '\\a\\b\\c', '\\a\\b\\c\\', 'a\\b\\c',
 'a/b/../../a/b/c/', 'a/b/../../a/b/c']

def path_leaf(path):
 return path.strip('/').strip('\\').split('/')[-1].split('\\')[-1]

[path_leaf(path) for path in paths]
</code></pre>

Results:

<pre><code>['c', 'c', 'c', 'c', 'c', 'c', 'c']
</code></pre>

blocks|key|1054804|text|如果您的文件路径不是以"/“结尾，目录以"/”分隔，则使用以下代码。正如我们所知，路径通常不会以"/“结尾。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1054805|import+os
path_str+=+"/var/www/index.html"
print(os.path.basename(path_str))|code-block|syntax|javascript|1054806|但在某些情况下，如URL以"/“结尾，则使用以下代码|1054807|import+os
path_str+=+"/home/some_str/last_str/"
split_path+=+path_str.rsplit("/",1)
print(os.path.basename(split_path[0]))|1054808|但是，当您的路径由"\“表示时，您可以使用以下代码|1054809|import+os
path_str+=+"c:\\var\www\index.html"
print(os.path.basename(path_str))

import+os
path_str+=+"c:\\home\some_str\last_str\\"
split_path+=+path_str.rsplit("\\",1)
print(os.path.basename(split_path[0]))|1054810|通过检查OS类型并返回结果，您可以将这两个函数合并为一个函数。|1054811|entityMap^0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|Y|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

If your file path not ended with "/" and directories separated by "/" then use the following code. As we know generally path doesn't end with "/".

<pre><code>import os
path_str = "/var/www/index.html"
print(os.path.basename(path_str))
</code></pre>

But in some cases like URLs end with "/" then use the following code

<pre><code>import os
path_str = "/home/some_str/last_str/"
split_path = path_str.rsplit("/",1)
print(os.path.basename(split_path[0]))
</code></pre>

but when your path sperated by "\" which you generally find in windows paths then you can use the following codes

<pre><code>import os
path_str = "c:\\var\www\index.html"
print(os.path.basename(path_str))

import os
path_str = "c:\\home\some_str\last_str\\"
split_path = path_str.rsplit("\\",1)
print(os.path.basename(split_path[0]))
</code></pre>

You can combine both into one function by check OS type and return the result.

blocks|key|1052254|text|这是一个仅支持regex的解决方案，它似乎适用于任何操作系统上的任何OS路径。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1052255|不需要其他模块，也不需要预处理：|1052256|import+re

def+extract_basename(path):
++"""Extracts+basename+of+a+given+path.+Should+Work+with+any+OS+Path+on+any+OS"""
++basename+=+re.search(r'[%5E\\/]%2B(?=[\\/]?$)',+path)
++if+basename:
++++return+basename.group(0)


paths+=+['a/b/c/',+'a/b/c',+'\\a\\b\\c',+'\\a\\b\\c\\',+'a\\b\\c',
+++++++++'a/b/../../a/b/c/',+'a/b/../../a/b/c']

print([extract_basename(path)+for+path+in+paths])
#+['c',+'c',+'c',+'c',+'c',+'c',+'c']


extra_paths+=+['C:\\',+'alone',+'/a/space+in+filename',+'C:\\multi\nline']

print([extract_basename(path)+for+path+in+extra_paths])
#+['C:',+'alone',+'space+in+filename',+'multi\nline']|code-block|syntax|javascript|1052257|更新：|1052258|如果您只想要一个潜在的文件名(如果存在)(例如，/a/b/是一个目录，c:\windows\也是一个目录)，请将正则表达式更改为：r'[%5E\\/]%2B(?![\\/])$'。对于“正则表达式挑战”，这会将某种斜杠的正向前视更改为负向前视，从而导致以该斜杠结尾的路径名不返回任何内容，而不是返回路径名中的最后一个子目录。当然，不能保证潜在的文件名确实指的是文件，因此需要使用os.path.is_dir()或os.path.is_file()。|offset|length|style|CODE|1052259|这将匹配如下：|1052260|/a/b/c/+++++++++++++#+nothing,+pathname+ends+with+the+dir+'c'
c:\windows\+++++++++#+nothing,+pathname+ends+with+the+dir+'windows'
c:hello.txt+++++++++#+matches+potential+filename+'hello.txt'
~it_s_me/.bashrc++++#+matches+potential+filename+'.bashrc'
c:\windows\system32+#+matches+potential+filename+'system32',+except
++++++++++++++++++++#+that+is+obviously+a+dir.+os.path.is_dir()
++++++++++++++++++++#+should+be+used+to+tell+us+for+sure|1052261|可以对正则表达式进行here测试。|1052262|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/r/PEALFz/2^0|0|0|0|0|O|5|Z|B|1T|K|55|G|5M|H|0|0|0|A|4|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|17|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|18|8|@$M|19|N|1A|O|P]|$M|1B|N|1C|O|P]|$M|1D|N|1E|O|P]|$M|1F|N|1G|O|P]|$M|1H|N|1I|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1J|8|@]|9|@]|A|$]]|$1|S|3|T|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1L|8|@]|9|@$M|1M|N|1N|1|1O]]|A|$]]|$1|W|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]]]

Here's a regex-only solution, which seems to work with any OS path on any OS.

No other module is needed, and no preprocessing is needed either :

<pre><code>import re

def extract_basename(path):
 """Extracts basename of a given path. Should Work with any OS Path on any OS"""
 basename = re.search(r'[^\\/]+(?=[\\/]?$)', path)
 if basename:
 return basename.group(0)


paths = ['a/b/c/', 'a/b/c', '\\a\\b\\c', '\\a\\b\\c\\', 'a\\b\\c',
 'a/b/../../a/b/c/', 'a/b/../../a/b/c']

print([extract_basename(path) for path in paths])
# ['c', 'c', 'c', 'c', 'c', 'c', 'c']


extra_paths = ['C:\\', 'alone', '/a/space in filename', 'C:\\multi\nline']

print([extract_basename(path) for path in extra_paths])
# ['C:', 'alone', 'space in filename', 'multi\nline']
</code></pre>

Update: 

If you only want a potential filename, if present (i.e., <code>/a/b/</code> is a dir and so is <code>c:\windows\</code>), change the regex to: <code>r'[^\\/]+(?![\\/])$'</code> . For the "regex challenged," this changes the positive forward lookahead for some sort of slash to a negative forward lookahead, causing pathnames that end with said slash to return nothing instead of the last sub-directory in the pathname. Of course there is no guarantee that the potential filename actually refers to a file and for that <code>os.path.is_dir()</code> or <code>os.path.is_file()</code> would need to be employed.

This will match as follows:

<pre><code>/a/b/c/ # nothing, pathname ends with the dir 'c'
c:\windows\ # nothing, pathname ends with the dir 'windows'
c:hello.txt # matches potential filename 'hello.txt'
~it_s_me/.bashrc # matches potential filename '.bashrc'
c:\windows\system32 # matches potential filename 'system32', except
 # that is obviously a dir. os.path.is_dir()
 # should be used to tell us for sure
</code></pre>

The regex can be tested <a href="https://regex101.com/r/PEALFz/2" rel="noreferrer">here</a>.

blocks|key|1052380|text|这是工作！|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1052381|os.path.basename(name)|code-block|syntax|javascript|1052382|但是你不能在Linux下使用Windows文件路径来获取文件名。Windows也是。os.path在不同的操作系统上加载不同的模块：|1052383|npath+Linux-+posixpath|1052384|Windows+-|unordered-list-item|1052385|1052386|1052387|1052388|因此你可以使用os.path来获得正确结果|1052389|entityMap^0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|Y|8|@]|9|@]|A|$]]|$1|N|3|-4|5|M|7|Z|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|10|8|@]|9|@]|A|$]]|$1|P|3|-4|5|6|7|11|8|@]|9|@]|A|$]]|$1|Q|3|R|5|6|7|12|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|T|$]]

It’s work!
<pre class="lang-py prettyprint-override"><code>os.path.basename(name)
</code></pre>
But you can’t get file name in Linux with Windows file path. Windows too.
os.path load different module on different operator system :
<ul>
<li>Linux - posixpath</li>
<li>Windows - npath</li>
</ul>
So you can use os.path get correct result always

blocks|key|5260196|text|也许只是我的一体化解决方案，而不是重要的一些新的(关于创建临时文件的临时文件:D+)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5260197|import+tempfile
abc+=+tempfile.NamedTemporaryFile(dir='/tmp/')
abc.name
abc.name.replace("/",+"+").split()[-1]+|code-block|syntax|javascript|5260198|获取abc.name的值将是如下所示的字符串：'/tmp/tmpks5oksk7'，这样我就可以用一个空格.replace("/",+"+")替换/，然后调用split()。这将返回一个列表，并使用[-1]获取该列表的最后一个元素|offset|length|style|CODE|5260199|不需要导入任何模块。|5260200|entityMap^0|0|0|2|8|N|I|1H|I|21|1|27|7|2R|4|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@$I|T|J|U|K|L]|$I|V|J|W|K|L]|$I|X|J|Y|K|L]|$I|Z|J|10|K|L]|$I|11|J|12|K|L]|$I|13|J|14|K|L]]|9|@]|A|$]]|$1|M|3|N|5|6|7|15|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|16|8|@]|9|@]|A|$]]]|P|$]]

Maybe just my all in one solution without important some new(regard the tempfile for creating temporary files :D )

<pre><code>import tempfile
abc = tempfile.NamedTemporaryFile(dir='/tmp/')
abc.name
abc.name.replace("/", " ").split()[-1] 
</code></pre>

Getting the values of <code>abc.name</code> will be a string like this: <code>'/tmp/tmpks5oksk7'</code>
So I can replace the <code>/</code> with a space <code>.replace("/", " ")</code> and then call <code>split()</code>. That will return a list and I get the
last element of the list with <code>[-1]</code>

No need to get any module imported.

blocks|key|3646640|text|如果您在一个目录中有多个文件，并希望将这些文件名存储到一个列表中。使用下面的代码。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3646641|import+os+as+os
import+glob+as+glob
path+=+'mypath'
file_list=+[]
for+file+in+glob.glob(path):
++++data_file_list+=+os.path.basename(file)
++++file_list.append(data_file_list)|code-block|syntax|javascript|3646642|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

If you have a number of files in a directory and want to store those file names into a list. Use the below code.
<pre><code>import os as os
import glob as glob
path = 'mypath'
file_list= []
for file in glob.glob(path):
 data_file_list = os.path.basename(file)
 file_list.append(data_file_list)
</code></pre>

blocks|key|1054668|text|我从来没有见过双反斜杠的路径，它们存在吗？python模块os的内置特性无法满足这些要求。所有其他方法都可以工作，还有您对os.path.normpath()给出的警告|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1054669|paths+=+['a/b/c/',+'a/b/c',+'\\a\\b\\c',+'\\a\\b\\c\\',+'a\\b\\c',+
...+++++'a/b/../../a/b/c/',+'a/b/../../a/b/c',+'a/./b/c',+'a\b/c']
for+path+in+paths:
++++os.path.basename(os.path.normpath(path))|code-block|syntax|javascript|1054670|entityMap^0|T|2|1P|I|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

I have never seen double-backslashed paths, are they existing? The built-in feature of python module <code>os</code> fails for those. All others work, also the caveat given by you with <code>os.path.normpath()</code>:

<pre><code>paths = ['a/b/c/', 'a/b/c', '\\a\\b\\c', '\\a\\b\\c\\', 'a\\b\\c', 
... 'a/b/../../a/b/c/', 'a/b/../../a/b/c', 'a/./b/c', 'a\b/c']
for path in paths:
 os.path.basename(os.path.normpath(path))
</code></pre>

blocks|key|5260328|text|Windows分隔符可以是Unix文件名或Windows路径。Unix分隔符只能存在于Unix路径中。Unix分隔符的存在表示非Windows路径。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5260329|下面的代码将通过操作系统特定的分隔符剥离(剪切尾随分隔符)，然后拆分并返回最右边的值。它很难看，但基于上面的假设很简单。如果假设不正确，请更新，我将更新此响应以匹配更准确的条件。|5260330|a.rstrip("\\\\"+if+a.count("/")+==+0+else+'/').split("\\\\"+if+a.count("/")+==+0+else+'/')[-1]|code-block|syntax|javascript|5260331|示例代码：|5260332|b+=+['a/b/c/','a/b/c','\\a\\b\\c','\\a\\b\\c\\','a\\b\\c','a/b/../../a/b/c/','a/b/../../a/b/c']

for+a+in+b:

++++print+(a,+a.rstrip("\\"+if+a.count("/")+==+0+else+'/').split("\\"+if+a.count("/")+==+0+else+'/')[-1])|5260333|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

The Windows separator can be in a Unix filename or Windows Path. The Unix separator can only exist in the Unix path. The presence of a Unix separator indicates a non-Windows path.

The following will strip (cut trailing separator) by the OS specific separator, then split and return the rightmost value. It's ugly, but simple based on the assumption above. If the assumption is incorrect, please update and I will update this response to match the more accurate conditions.

<pre><code>a.rstrip("\\\\" if a.count("/") == 0 else '/').split("\\\\" if a.count("/") == 0 else '/')[-1]
</code></pre>

sample code:

<pre><code>b = ['a/b/c/','a/b/c','\\a\\b\\c','\\a\\b\\c\\','a\\b\\c','a/b/../../a/b/c/','a/b/../../a/b/c']

for a in b:

 print (a, a.rstrip("\\" if a.count("/") == 0 else '/').split("\\" if a.count("/") == 0 else '/')[-1])
</code></pre>

blocks|key|3646470|text|为了完整起见，下面是针对python+3.2%2B的pathlib解决方案：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3646471|>>>+from+pathlib+import+PureWindowsPath

>>>+paths+=+['a/b/c/',+'a/b/c',+'\\a\\b\\c',+'\\a\\b\\c\\',+'a\\b\\c',+
...++++++++++'a/b/../../a/b/c/',+'a/b/../../a/b/c']

>>>+[PureWindowsPath(path).name+for+path+in+paths]
['c',+'c',+'c',+'c',+'c',+'c',+'c']|code-block|syntax|javascript|3646472|这在Windows和Linux上都有效。|3646473|entityMap^0|O|7|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

For completeness sake, here is the <code>pathlib</code> solution for python 3.2+:

<pre><code>&gt;&gt;&gt; from pathlib import PureWindowsPath

&gt;&gt;&gt; paths = ['a/b/c/', 'a/b/c', '\\a\\b\\c', '\\a\\b\\c\\', 'a\\b\\c', 
... 'a/b/../../a/b/c/', 'a/b/../../a/b/c']

&gt;&gt;&gt; [PureWindowsPath(path).name for path in paths]
['c', 'c', 'c', 'c', 'c', 'c', 'c']
</code></pre>

This works on both Windows and Linux.

blocks|key|1054779|text|在Python2和3中，使用模块pathlib2|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1054780|import+posixpath++#+to+generate+unix+paths
from+pathlib2+import+PurePath,+PureWindowsPath,+PurePosixPath

def+path2unix(path,+nojoin=True,+fromwinpath=False):
++++"""From+a+path+given+in+any+format,+converts+to+posix+path+format
++++fromwinpath=True+forces+the+input+path+to+be+recognized+as+a+Windows+path+(useful+on+Unix+machines+to+unit+test+Windows+paths)"""
++++if+not+path:
++++++++return+path
++++if+fromwinpath:
++++++++pathparts+=+list(PureWindowsPath(path).parts)
++++else:
++++++++pathparts+=+list(PurePath(path).parts)
++++if+nojoin:
++++++++return+pathparts
++++else:
++++++++return+posixpath.join(*pathparts)|code-block|syntax|javascript|1054781|用法：|1054782|In+[9]:+path2unix('lala/lolo/haha.dat')
Out[9]:+['lala',+'lolo',+'haha.dat']

In+[10]:+path2unix(r'C:\lala/lolo/haha.dat')
Out[10]:+['C:\\',+'lala',+'lolo',+'haha.dat']

In+[11]:+path2unix(r'C:\lala/lolo/haha.dat')+#+works+even+with+malformatted+cases+mixing+both+Windows+and+Linux+path+separators
Out[11]:+['C:\\',+'lala',+'lolo',+'haha.dat']|1054783|使用你的测试用例：|1054784|In+[12]:+testcase+=+paths+=+['a/b/c/',+'a/b/c',+'\\a\\b\\c',+'\\a\\b\\c\\',+'a\\b\\c',
++++...:+...+++++'a/b/../../a/b/c/',+'a/b/../../a/b/c']

In+[14]:+for+t+in+testcase:
++++...:+++++print(path2unix(t)[-1])
++++...:
++++...:
c
c
c
c
c
c
c|1054785|这里的想法是将所有路径转换为pathlib2的统一内部表示，并根据平台使用不同的解码器。幸运的是，pathlib2包含一个名为PurePath的通用解码器，它可以在任何路径上工作。如果这不起作用，您可以使用fromwinpath=True强制识别windows路径。这将把输入字符串分成几个部分，最后一个部分是您要查找的叶，因此是path2unix(t)[-1]。|style|CODE|1054786|如果参数为nojoin=False，则路径将被连接回去，因此输出只是转换为Unix格式的输入字符串，这对于比较跨平台的子路径非常有用。|1054787|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/mcmtroffaes/pathlib2^0|G|8|0|0|0|0|0|0|0|E|8|1D|8|1R|8|2V|G|4L|G|0|5|C|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@$A|15|B|16|1|17]]|C|$]]|$1|D|3|E|5|F|7|18|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|19|8|@]|9|@]|C|$]]|$1|K|3|L|5|F|7|1A|8|@]|9|@]|C|$G|H]]|$1|M|3|N|5|6|7|1B|8|@]|9|@]|C|$]]|$1|O|3|P|5|F|7|1C|8|@]|9|@]|C|$G|H]]|$1|Q|3|R|5|6|7|1D|8|@$A|1E|B|1F|S|T]|$A|1G|B|1H|S|T]|$A|1I|B|1J|S|T]|$A|1K|B|1L|S|T]|$A|1M|B|1N|S|T]]|9|@]|C|$]]|$1|U|3|V|5|6|7|1O|8|@$A|1P|B|1Q|S|T]]|9|@]|C|$]]|$1|W|3|-4|5|6|7|1R|8|@]|9|@]|C|$]]]|X|$Y|$5|Z|10|11|C|$12|13]]]]

In both Python 2 and 3, using the module <a href="https://github.com/mcmtroffaes/pathlib2" rel="nofollow noreferrer">pathlib2</a>:

<pre><code>import posixpath # to generate unix paths
from pathlib2 import PurePath, PureWindowsPath, PurePosixPath

def path2unix(path, nojoin=True, fromwinpath=False):
 """From a path given in any format, converts to posix path format
 fromwinpath=True forces the input path to be recognized as a Windows path (useful on Unix machines to unit test Windows paths)"""
 if not path:
 return path
 if fromwinpath:
 pathparts = list(PureWindowsPath(path).parts)
 else:
 pathparts = list(PurePath(path).parts)
 if nojoin:
 return pathparts
 else:
 return posixpath.join(*pathparts)
</code></pre>

Usage:

<pre><code>In [9]: path2unix('lala/lolo/haha.dat')
Out[9]: ['lala', 'lolo', 'haha.dat']

In [10]: path2unix(r'C:\lala/lolo/haha.dat')
Out[10]: ['C:\\', 'lala', 'lolo', 'haha.dat']

In [11]: path2unix(r'C:\lala/lolo/haha.dat') # works even with malformatted cases mixing both Windows and Linux path separators
Out[11]: ['C:\\', 'lala', 'lolo', 'haha.dat']
</code></pre>

With your testcase:

<pre><code>In [12]: testcase = paths = ['a/b/c/', 'a/b/c', '\\a\\b\\c', '\\a\\b\\c\\', 'a\\b\\c',
 ...: ... 'a/b/../../a/b/c/', 'a/b/../../a/b/c']

In [14]: for t in testcase:
 ...: print(path2unix(t)[-1])
 ...:
 ...:
c
c
c
c
c
c
c
</code></pre>

The idea here is to convert all paths into the unified internal representation of <code>pathlib2</code>, with different decoders depending on the platform. Fortunately, <code>pathlib2</code> includes a generic decoder called <code>PurePath</code> that should work on any path. In case this does not work, you can force the recognition of windows path using <code>fromwinpath=True</code>. This will split the input string into parts, the last one is the leaf you are looking for, hence the <code>path2unix(t)[-1]</code>.

If the argument <code>nojoin=False</code>, the path will be joined back, so that the output is simply the input string converted to a Unix format, which can be useful to compare subpaths across platforms.

blocks|key|1052472|text|扩展名为的文件名|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1052473|filepath+=+'./dir/subdir/filename.ext'
basename+=+os.path.basename(filepath)
print(basename)
#+filename.ext

print(type(basename))
#+<class+'str'>|code-block|syntax|javascript|1052474|不带扩展名的文件名|1052475|basename_without_ext+=+os.path.splitext(os.path.basename(filepath))[0]
print(basename_without_ext)
#+filename|1052476|entityMap^0|5|3|0|0|0|5|6|3|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|10|8|@]|D|@]|E|$]]]|P|$]]

File name with extension
<pre><code>filepath = './dir/subdir/filename.ext'
basename = os.path.basename(filepath)
print(basename)
# filename.ext

print(type(basename))
# &lt;class 'str'&gt;
</code></pre>
File name without extension
<pre><code>basename_without_ext = os.path.splitext(os.path.basename(filepath))[0]
print(basename_without_ext)
# filename
</code></pre>

blocks|key|5260744|text|我在Windows和Ubuntu+(WSL)上使用了这个方法，并且它只使用'import+os‘就能像预期的那样工作:所以基本上，根据你当前的操作系统平台，替换()放置正确的路径分隔符。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5260745|如果路径以斜杠'/‘结尾，那么它不是一个文件，而是一个目录，所以它返回一个空字符串。|5260746|import+os

my_fullpath+=+r"D:\MY_FOLDER\TEST\20201108\20201108_073751.DNG"
os.path.basename(my_fullpath.replace('\\',os.sep))

my_fullpath+=+r"/MY_FOLDER/TEST/20201108/20201108_073751.DNG"
os.path.basename(my_fullpath.replace('\\',os.sep))

my_fullpath+=+r"/MY_FOLDER/TEST/20201108/"
os.path.basename(my_fullpath.replace('\\',os.sep))

my_fullpath+=+r"/MY_FOLDER/TEST/20201108"
os.path.basename(my_fullpath.replace('\\',os.sep))|code-block|syntax|javascript|5260747|在Windows+(左)和Ubuntu+(通过WSL，右)上：|5260748|📷|atomic|offset|length|5260749|​|5260750|entityMap|0|IMAGE|mutability|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/9f2d2f9c49c7afdc545849d87ec94143.png|imageAlt^0|0|0|0|0|0|1|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|M|7|14|8|@]|9|@$N|15|O|16|1|17]]|A|$]]|$1|P|3|Q|5|6|7|18|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|19|8|@]|9|@]|A|$]]]|S|$T|$5|U|V|W|A|$X|Y|Z|-4]]]]

I use this method on Windows and Ubuntu (WSL) and it works as (I) expected only using 'import os':
So basically, replace() put the right path seperator based on your current os platform.
If the path finished by a slash '/', then it's not a file but a directory, so it returns an empty string.
<pre class="lang-py prettyprint-override"><code>import os

my_fullpath = r&quot;D:\MY_FOLDER\TEST\20201108\20201108_073751.DNG&quot;
os.path.basename(my_fullpath.replace('\\',os.sep))

my_fullpath = r&quot;/MY_FOLDER/TEST/20201108/20201108_073751.DNG&quot;
os.path.basename(my_fullpath.replace('\\',os.sep))

my_fullpath = r&quot;/MY_FOLDER/TEST/20201108/&quot;
os.path.basename(my_fullpath.replace('\\',os.sep))

my_fullpath = r&quot;/MY_FOLDER/TEST/20201108&quot;
os.path.basename(my_fullpath.replace('\\',os.sep))
</code></pre>
On Windows (Left) and Ubuntu (via WSL, Right):
<a href="https://i.stack.imgur.com/oGJMQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/oGJMQ.png" alt="enter image description here" /></a>

Which Python library can I use to extract filenames from paths, no matter what the operating system or path format could be?

For example, I'd like all of these paths to return me <code>c</code>:

<pre><code>a/b/c/
a/b/c
\a\b\c
\a\b\c\
a\b\c
a/b/../../a/b/c/
a/b/../../a/b/c
</code></pre>

Extract file name from path, no matter what the os/path format

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我可以使用哪个Python库从路径中提取文件名，而不管操作系统或路径格式是什么？例如，我希望所有这些路径都返回ca/b/c/a/b/c\a\b\c\a\b\c\a\b\ca/b/../../a/b/c/a/b/../../a/b/c

问从path中提取文件名，不管os/path格式是什么
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问从path中提取文件名，不管os/path格式是什么EN

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问从path中提取文件名，不管os/path格式是什么
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