Question

Say, I have two absolute paths. I need to check if the location referring to by one of the paths is a descendant of the other. If true, I need to find out the relative path of the descendant from the ancestor. What’s a good way to implement this in Python? Any library that I can benefit from?

Answer 1

os.path.commonprefix() and os.path.relpath() are your friends:

>>> print os.path.commonprefix(['/usr/var/log', '/usr/var/security'])
'/usr/var'
>>> print os.path.commonprefix(['/tmp', '/usr/var'])  # No common prefix: the root is the common prefix
'/'

You can thus test whether the common prefix is one of the paths, i.e. if one of the paths is a common ancestor:

paths = […, …, …]
common_prefix = os.path.commonprefix(list_of_paths)
if common_prefix in paths:
    …

You can then find the relative paths:

relative_paths = [os.path.relpath(path, common_prefix) for path in paths]

You can even handle more than two paths, with this method, and test whether all the paths are all below one of them.

PS: depending on how your paths look like, you might want to perform some normalization first (this is useful in situations where one does not know whether they always end with ‘/’ or not, or if some of the paths are relative). Relevant functions include os.path.abspath() and os.path.normpath().

PPS: as Peter Briggs mentioned in the comments, the simple approach described above can fail:

>>> os.path.commonprefix(['/usr/var', '/usr/var2/log'])
'/usr/var'

even though /usr/var is not a common prefix of the paths. Forcing all paths to end with ‘/’ before calling commonprefix() solves this (specific) problem.

PPPS: as bluenote10 mentioned, adding a slash does not solve the general problem. Here is his followup question: How to circumvent the fallacy of Python's os.path.commonprefix?

PPPPS: starting with Python 3.4, we have pathlib, a module that provides a saner path manipulation environment. I guess that the common prefix of a set of paths can be obtained by getting all the prefixes of each path (with PurePath.parents()), taking the intersection of all these parent sets, and selecting the longest common prefix.

PPPPPS: Python 3.5 introduced a proper solution to this question: os.path.commonpath(), which returns a valid path.