[fixed] distinguish file name from the path

[paul18fr]

Hi,

I'm digging into the os.path lib to find a very simple way to get the path without the file name., but with no sucess (I'm missing something)
Basic example: i've the '/X/Y/Z/file.txt' and i need to get '/X/Y/Z'
(opposite of os.path.joint)

Thanks

Paul

[Gribouillis]

Use os.path.split()

>>> import os
>>> os.path.split('/X/Y/Z/file.txt')
('/X/Y/Z', 'file.txt')
>>> 

[noisefloor]

Hi,

os.path is kind of legacy, the pathlib is the current state for dealing with pathes:

>>> from pathlib import Path
>>> p = Path('/X/Y/Z/file.txt')
>>> p
WindowsPath('/X/Y/Z/file.txt')
>>> p.parents[0]
WindowsPath('/X/Y/Z')
>>> p.name
'file.txt'
>>> p.parts
('\\', 'X', 'Y', 'Z', 'file.txt')
>>>

Regards, noisefloor

[Pedroski55]

I agree with noisefloor, pathlib is superior for paths.

from pathlib import Path

# https://docs.python.org/3/library/pathlib.html

path2files = Path('/home/peterr/temp') 
filelist = [filename for filename in path2files.iterdir() if filename.is_file()]
for filename in filelist:
    print(f"\nfilename: {filename.name}")
    print(f"file suffix: {filename.suffix}")
    print(f"full path: {filename.resolve()}")
    print(f"filepath parts: {filename.parts}")

Long old output, here just 1 part of it:

Output:
filename: vivelafrance.txt
file suffix: .txt
full path: /home/peterr/temp/vivelafrance.txt
filepath parts: ('/', 'home', 'peterr', 'temp', 'vivelafrance.txt')

[paul18fr]

thanks to all contributors

[DeaD_EyE]

Old way to get the dirname:

import os


dirname = os.path.dirname(r"/home/user/.local/directory/some_file.txt")
print(dirname)

Output:
/home/user/.local/directory

But this should not be used directly. Use Path instead. The pathlib uses the os.path module.