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?

Asked By: JFB

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Source

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:\route44baz` 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)))])

Answered By: Carel

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)

Answered By: Brett Ryland

Answer 3

A recursive version of @Brett_Ryland’s relpath for pathlib. I find this to be a tad more readable and it is going to succeed on first try in most cases so it should have similar performance as the original relative_to function:

def relative(target: Path, origin: Path):
    """ return path of target relative to origin """
    try:
        return Path(target).resolve().relative_to(Path(origin).resolve())
    except ValueError as e: # target does not start with origin
        # recursion with origin (eventually origin is root so try will succeed)
        return Path('..').joinpath(relative(target, Path(origin).parent))

Answered By: Felix B.

Using pathlib's relative_to for directories on the same level

Question:

Answers:

Relative Paths

Common Path

Common Prefix