Say there's a directory hierarchy like this: .../A/B/main.py.
main.py is the file being executed, and I need to get the full path of folder A, how can I do that?
- stackoverflow.com/questions/3430372/…apomene– apomene2014-05-09 15:44:25 +00:00Commented May 9, 2014 at 15:44
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3 Answers
Use the os module:
import os a_path = os.path.dirname(os.path.dirname(os.path.abspath(__file__))) 
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In Python 3:
from pathlib import Path mypath = Path().absolute().parent.parent # each '.parent' goes one level up - vary as required print(mypath) 
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To get the path of the current directory, you can use:
import os print os.path.realpath('.') since . represents the current directory in the file system. So to get the path of the higher level directory, just replace . with .., which represents parent directory
import os print os.path.realpath('..') You can also use the os.getcwd() method to get the current working directory and then get its parent directory with the .. representation.
3 Comments
apomene
Won't this give the dir one level higher? What if Parent is 3 or 4 levels deep?
anirudh
in that case, we can still use the normal unix or windows path traversing. Just use `../..'to the the path of a directory, two levels higher.
Tony Suffolk 66
The OP doesn't ask for the current directory - he wants the parent of the parent of the current executing file - using "." or cwd wont do that.