How to get the filename without the extension from a path in Python?

Question

How to get the filename without the extension from a path in Python?

For instance, if I had "/path/to/some/file.txt", I would want "

Answer 1

https://docs.python.org/3/library/os.path.html

In python 3 pathlib "The pathlib module offers high-level path objects." so,

>>> from pathlib import Path
>>> p = Path("/a/b/c.txt")
>>> print(p.with_suffix(''))
\a\b\c
>>> print(p.stem)
c

Answer 2

If you want to keep the path to the file and just remove the extension

>>> file = '/root/dir/sub.exten/file.data.1.2.dat'
>>> print ('.').join(file.split('.')[:-1])
/root/dir/sub.exten/file.data.1.2

Answer 3

import os
list = []
def getFileName( path ):
for file in os.listdir(path):
    #print file
    try:
        base=os.path.basename(file)
        splitbase=os.path.splitext(base)
        ext = os.path.splitext(base)[1]
        if(ext):
            list.append(base)
        else:
            newpath = path+"/"+file
            #print path
            getFileName(newpath)
    except:
        pass
return list

getFileName("/home/weexcel-java3/Desktop/backup")
print list

Answer 4

We could do some simple split / pop magic as seen here (https://stackoverflow.com/a/424006/1250044), to extract the filename (respecting the windows and POSIX differences).

def getFileNameWithoutExtension(path):
  return path.split('\\').pop().split('/').pop().rsplit('.', 1)[0]

getFileNameWithoutExtension('/path/to/file-0.0.1.ext')
# => file-0.0.1

getFileNameWithoutExtension('\\path\\to\\file-0.0.1.ext')
# => file-0.0.1

Answer 5

I didn't look very hard but I didn't see anyone who used regex for this problem.

I interpreted the question as "given a path, return the basename without the extension."

e.g.

"path/to/file.json" => "file"

"path/to/my.file.json" => "my.file"

In Python 2.7, where we still live without pathlib...

def get_file_name_prefix(file_path):
    basename = os.path.basename(file_path)

    file_name_prefix_match = re.compile(r"^(?P<file_name_pre fix>.*)\..*$").match(basename)

    if file_name_prefix_match is None:
        return file_name
    else:
        return file_name_prefix_match.group("file_name_prefix")

get_file_name_prefix("path/to/file.json")
>> file

get_file_name_prefix("path/to/my.file.json")
>> my.file

get_file_name_prefix("path/to/no_extension")
>> no_extension

Answer 6

import os filename, file_extension = os.path.splitext('/d1/d2/example.cs') filename is '/d1/d2/example' file_extension is '.cs'