How do I get the child folder name of the path besides these methods?

Question

Of the given path like "level1/level2/level3/", I'd like pass it through some operation and get the result like "level3/". So I made two trials like these:

TRIAL 1: After finding parent property within the Path object, I looked for something close to a child property, but could not.

>>> from pathlib import Path
>>> path = Path("level1/level2/level3/")
>>> path.parent
WindowsPath('level1/level2')
>>> str(path.parent)
'level1\\level2'

TRIAL 2: I used the os module like this:

>>> import os
>>> os.path.basename("level1/level2/level3/".strip("/")) + "/"
'level3/'

Is there an alternative to TRIAL 2, or can I make something work within TRIAL 1 from the pathlib package or Path class?

score 3 · Accepted Answer · edited Jan 25 '22 at 13:28

3

Try using pathlib.parts

>>> from pathlib import Path
>>> path = Path("level1/level2/level3/")
>>> path.parts[-1]
'level3'

You can then append the "/" character if needed.

edited Jan 25 '22 at 13:28

Neuron

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answered Nov 18 '19 at 10:48

CDJB

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Thank you, it looks like ultimately I'll have to add this `+ "/"` at the end – Kiran Racherla Nov 18 '19 at 10:51
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I have been looking for this answer for so long :))) – JVGD Oct 14 '21 at 13:16

score 0 · Answer 2 · answered Aug 15 '22 at 15:31

0

I think you're looking for pathlib.name

>>> from pathlib import Path
>>> path = Path("level1/level2/level3/")
>>> path.name + "/"
'level3/'

answered Aug 15 '22 at 15:31

Torran Green

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How do I get the child folder name of the path besides these methods?

2 Answers2