How to get folder name, in which given file resides, from pathlib.path?

Question

Is there something similar to os.path.dirname(path), but in pathlib?

Answer 1

It looks like there is a parents element that contains all the parent directories of a given path. E.g., if you start with:

>>> import pathlib
>>> p = pathlib.Path('/path/to/my/file')

Then p.parents[0] is the directory containing file:

>>> p.parents[0]
PosixPath('/path/to/my')

...and p.parents[1] will be the next directory up:

>>> p.parents[1]
PosixPath('/path/to')

Etc.

p.parent is another way to ask for p.parents[0]. You can convert a Path into a string and get pretty much what you would expect:

>>> str(p.parent)
'/path/to/my'

And also on any Path you can use the .absolute() method to get an absolute path:

>>> os.chdir('/etc')
>>> p = pathlib.Path('../relative/path')
>>> str(p.parent)
'../relative'
>>> str(p.parent.absolute())
'/etc/../relative'

Note that os.path.dirname and pathlib treat paths with a trailing slash differently. The pathlib parent of some/path/ is some:

>>> p = pathlib.Path('some/path/')
>>> p.parent
PosixPath('some')

While os.path.dirname on some/path/ returns some/path:

>>> os.path.dirname('some/path/')
'some/path'

Answer 2

summary

from pathlib import Path
file_path = Path("/Users/yuanz/PycharmProjects/workenv/little_code/code09/sample.csv")

1. get dir path

file_path.parent
# >>> /Users/yuanz/PycharmProjects/workenv/little_code/code09

2. get filename

file_path.name
# >>> sample.csv

3. get file type

file_path.suffix

# >>> .csv

4.add new file in this dir path

file_path.parent.joinpath("dd.png")

Answer 3

I came here looking for something very similar. My solution, based on the above by @larsks, and assuming you want to preserve the entire path except the filename, is to do:

>>> import pathlib
>>> p = pathlib.Path('/path/to/my/file')
>>> pathlib.Path('/'.join(list(p.parts)[1:-1])+'/')

Essentially, list(p.parts)[1:-1] creates a list of Path elements, starting from the second to n-1th, and you join them with a '/' and make a path of the resulting string. Edit The final +'/' adds in the trailing slash - adjust as required.