Extract a part of the filepath (a directory) in Python

import os
## first file in current dir (with full path)
file = os.path.join(os.getcwd(), os.listdir(os.getcwd())[0])
file
os.path.dirname(file) ## directory of file
os.path.dirname(os.path.dirname(file)) ## directory of directory of file
...

And you can continue doing this as many times as necessary...

Edit: from os.path, you can use either os.path.split or os.path.basename:

dir = os.path.dirname(os.path.dirname(file)) ## dir of dir of file
## once you're at the directory level you want, with the desired directory as the final path node:
dirname1 = os.path.basename(dir) 
dirname2 = os.path.split(dir)[1] ## if you look at the documentation, this is exactly what os.path.basename does.

For Python 3.4+, try the pathlib module:

>>> from pathlib import Path

>>> p = Path('C:\\Program Files\\Internet Explorer\\iexplore.exe')

>>> str(p.parent)
'C:\\Program Files\\Internet Explorer'

>>> p.name
'iexplore.exe'

>>> p.suffix
'.exe'

>>> p.parts
('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')

>>> p.relative_to('C:\\Program Files')
WindowsPath('Internet Explorer/iexplore.exe')

>>> p.exists()
True

All you need is parent part if you use pathlib.

from pathlib import Path
p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parent) 

Will output:

C:\Program Files\Internet Explorer    

Case you need all parts (already covered in other answers) use parts:

p = Path(r'C:\Program Files\Internet Explorer\iexplore.exe')
print(p.parts) 

Then you will get a list:

('C:\\', 'Program Files', 'Internet Explorer', 'iexplore.exe')

Saves tone of time.