Using pathlib's relative_to for directories on the same level

Question

The python library pathlib provides Path.relative_to. This function works fine if one path is a subpath of the other one, like this:

from pathlib import Path foo = Path("C:\\foo") bar = Path("C:\\foo\\bar") bar.relative_to(foo)  > WindowsPath('bar') 

However, if two paths are on the same level, relative_to does not work.

baz = Path("C:\\baz") foo.relative_to(baz)  > ValueError: 'C:\\foo' does not start with 'C:\\baz' 

I would expect the result to be

WindowsPath("..\\baz") 

The function os.path.relpath does this correctly:

import os foo = "C:\\foo" bar = "C:\\bar" os.path.relpath(foo, bar)  > '..\\foo' 

Is there a way to achieve the functionality of os.path.relpath using pathlib.Path?

Answer 1

The first section solves the OP's problem, though if like me, he really wanted the solution relative to a common root then the second section solves it for him. The third section describes how I originally approached it and is kept for interest sake.

Relative Paths

Recently, as in Python 3.4-6, the os.path module has been extended to accept pathlib.Path objects. In the following case however it does not return a Path object and one is forced to wrap the result.

foo = Path("C:\\foo") baz = Path("C:\\baz") Path(os.path.relpath(foo, baz))  > Path("..\\foo") 

Common Path

My suspicion is that you're really looking a path relative to a common root. If that is the case the following, from EOL, is more useful

Path(os.path.commonpath([foo, baz]))  > Path('c:/root') 

Common Prefix

Before I'd struck upon os.path.commonpath I'd used os.path.comonprefix.

foo = Path("C:\\foo") baz = Path("C:\\baz") baz.relative_to(os.path.commonprefix([baz,foo]))  > Path('baz') 

But be forewarned you are not supposed to use it in this context (See commonprefix : Yes, that old chestnut)

foo = Path("C:\\route66\\foo") baz = Path("C:\\route44\\baz") baz.relative_to(os.path.commonprefix([baz,foo]))  > ... > ValueError : `c:\\route44\baz` does not start with `C:\\route` 

but rather the following one from J. F. Sebastian.

Path(*os.path.commonprefix([foo.parts, baz.parts]))  > Path('c:/root') 

... or if you're feeling verbose ...

from itertools import takewhile Path(*[set(i).pop() for i in (takewhile(lambda x : x[0]==x[1], zip(foo.parts, baz.parts)))]) 

Answer 2

This was bugging me, so here's a pathlib-only version that I think does what os.path.relpath does.

def relpath(path_to, path_from):     path_to = Path(path_to).resolve()     path_from = Path(path_from).resolve()     try:         for p in (*reversed(path_from.parents), path_from):             head, tail = p, path_to.relative_to(p)     except ValueError:  # Stop when the paths diverge.         pass     return Path('../' * (len(path_from.parents) - len(head.parents))).joinpath(tail)