If I have a filename like one of these:
1.1.1.1.1.jpg 1.1.jpg 1.jpg How could I get only the filename, without the extension? Would a regex be appropriate?
- Does this answer your question? How to get the filename without the extension from a path in Python?Seanny123– Seanny1232020-01-10 21:02:17 +00:00Commented Jan 10, 2020 at 21:02
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6 Answers
In most cases, you shouldn't use a regex for that.
os.path.splitext(filename)[0] This will also handle a filename like .bashrc correctly by keeping the whole name.
2 Comments
blueyed
Does not work properly with "git-1.7.8.tar.gz", where it only removes the ".gz". I use
basename[:-len(".tar.gz")] for this.
Marcelo Cantos
@blueyed: "Does not work properly" is a matter of perspective. The file is a gzip file, who's base name is
git-1.7.8.tar. There is no way to correctly guess how many dots the caller wants to strip off, so splitext() only strips the last one. If you want to handle edge-cases like .tar.gz, you'll have to do it by hand. Obviously, you can't strip all the dots, since you'll end up with git-1.>>> import os >>> os.path.splitext("1.1.1.1.1.jpg") ('1.1.1.1.1', '.jpg') 
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You can use stem method to get file name.
Here is an example:
from pathlib import Path p = Path(r"\\some_directory\subdirectory\my_file.txt") print(p.stem) # my_file 
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If I had to do this with a regex, I'd do it like this:
s = re.sub(r'\.jpg$', '', s) 
Comments
No need for regex. os.path.splitext is your friend:
os.path.splitext('1.1.1.jpg') >>> ('1.1.1', '.jpg') 
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One can also use the string slicing.
>>> "1.1.1.1.1.jpg"[:-len(".jpg")] '1.1.1.1.1' 