Getting a specific parent folder with Python

Question

I already have this piece of functioning code, but after writing it I did feel the urge to scream "It's alive, it's alive!".

What I want to do is get the folder which has the folder "modules" as its parent folder, e.g. from /home/user/puppet/modules/impuls-test/templates/apache22/ I want /home/user/puppet/modules/impuls-test/

What I came up with is the following:

user@server:~/puppet/modules/impuls-test/templates/apache22$ python
Python 2.4.2 (#1, Apr 13 2007, 15:38:32)
[GCC 4.1.0 (SUSE Linux)] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> import os
>>> cwd = os.getcwd()
>>> path = cwd
>>> print "cwd: %s" % cwd
cwd: /home/user/puppet/modules/impuls-test/templates/apache22
>>> for i in xrange(len(cwd.split('/'))):
...     (head, tail) = os.path.split(path)
...     print "head: %s" % head
...     print "tail: %s" % tail
...     if tail == 'modules':
...         moduleDir = head + '/modules/' + cwd.split('/')[i+2] + '/'
...         print "moduleDir: %s" % moduleDir
...         break
...     else:
...         path = head
...
head: /home/user/puppet/modules/impuls-test/templates
tail: apache22
head: /home/user/puppet/modules/impuls-test
tail: templates
head: /home/user/puppet/modules
tail: impuls-test
head: /home/user/puppet
tail: modules
moduleDir: /home/user/puppet/modules/impuls-test/

I get the current working directory and use the os.path.split so long until it reaches the modules folder. Using the normal string.split function to iterate over the cwd, I can then append the moduleDir from the original cwd.split('/') array to the current head.

Can someone tell me a better/pythonic way to do this? Sure I can check if the current head ends with modules and then append the current tail, but that would only make the loop break faster and would still be ugly.

Answer 1

path = "/home/user/puppet/modules/impuls-test/templates"
components = path.split(os.sep)
print str.join(os.sep, components[:components.index("modules")+2])

prints

/home/user/puppet/modules/impuls-test

Answer 2

Since os.path.normpath handles the ".." operator, you could just tack on a ".." and let normpath do the work:

>>> path = "/home/user/puppet/modules/impuls-test/templates"
>>> os.path.normpath(os.path.join(path, ".."))
'/home/user/puppet/modules/impuls-test'

Answer 3

You could use a regular expression

>>> import re
>>> print re.findall(r'(/.*?/modules/(.*?)/).*','/hello/modules/foo/bye')[0]
('/hello/modules/foo/', 'foo')

Answer 4

Similar to Sven's (original) answer but more portable and with error handling...

>>> import os
>>> cwd = '/home/user/puppet/modules/impuls-test/templates/apache22'
>>> directories = cwd.split(os.sep)
>>> try:
...     modules_depth = directories.index('modules')
...     print(os.sep.join(directories[:modules_depth + 2]))
... except ValueError:
...     print('Could not find "modules"')
... 
/home/user/puppet/modules/impuls-test