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Suppose your application needs to know the relative path from one path to another (say because you want to create a symbolic link, a relative reference in a URL, etc). These functions may be of help.

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#!/usr/bin/env python
#
# Author: Cimarron Taylor
# Date: July 6, 2003
# File Name: relpath.py
# Program Description: Print relative path from /a/b/c/d to /a/b/c1/d1

#
# helper functions for relative paths
#
import os

def pathsplit(p, rest=[]):
    (h,t) = os.path.split(p)
    if len(h) < 1: return [t]+rest
    if len(t) < 1: return [h]+rest
    return pathsplit(h,[t]+rest)

def commonpath(l1, l2, common=[]):
    if len(l1) < 1: return (common, l1, l2)
    if len(l2) < 1: return (common, l1, l2)
    if l1[0] != l2[0]: return (common, l1, l2)
    return commonpath(l1[1:], l2[1:], common+[l1[0]])

def relpath(p1, p2):
    (common,l1,l2) = commonpath(pathsplit(p1), pathsplit(p2))
    p = []
    if len(l1) > 0:
        p = [ '../' * len(l1) ]
    p = p + l2
    return os.path.join( *p )

def test(p1,p2):
    print "from", p1, "to", p2, " -> ", relpath(p1, p2)

if __name__ == '__main__':
    test('/a/b/c/d', '/a/b/c1/d1')

2 comments

Alan Ezust 10 years, 2 months ago  # | flag

abs2rel and rel2abs. ignore previous comment - characters didn't get escaped properly.

&quot;&quot;&quot; Authors: Cimarron Taylor and Alan Ezust
 Version: 2.0
 Date: July 01, 2004
 (relpath.py v1 originally from Oreilly/Activestate Python cookbook 2003)

 helper functions for relative paths.
 This package includes rel2abs() and abs2rel(),
 based on the perl functions from cpan File::Spec
&quot;&quot;&quot;

import os
import os.path
import re

# matches http:// and ftp:// and mailto://
protocolPattern = re.compile(r&apos^\w+://&apos)

def isabs(string):
    &quot;&quot;&quot;

    @return true if string is an absolute path or protocoladdress
    for addresses beginning in http:// or ftp:// or ldap:// -
    they are considered &quot;absolute&quot; paths.
    &quot;&quot;&quot;
    if protocolPattern.match(string): return 1
    return os.path.isabs(string)

def rel2abs(path, base = os.curdir):
    &quot;&quot;&quot; converts a relative path to an absolute path.

    @param path the path to convert - if already absolute, is returned
    without conversion.
    @param base - optional. Defaults to the current directory.
    The base is intelligently concatenated to the given relative path.
    @return the relative path of path from base
    &quot;&quot;&quot;
    if isabs(path): return path
    retval = os.path.join(base,path)
    return os.path.abspath(retval)


def pathsplit(p, rest=[]):
    (h,t) = os.path.split(p)
    if len(h) &lt; 1: return [t]+rest
    if len(t) &lt; 1: return [h]+rest
    return pathsplit(h,[t]+rest)

def commonpath(l1, l2, common=[]):
    if len(l1) &lt; 1: return (common, l1, l2)
    if len(l2) &lt; 1: return (common, l1, l2)
    if l1[0] != l2[0]: return (common, l1, l2)
    return commonpath(l1[1:], l2[1:], common+[l1[0]])


def relpath(p1, p2):
    (common,l1,l2) = commonpath(pathsplit(p1), pathsplit(p2))
    p = []
    if len(l1) &gt; 0:
        p = [ &apos../&apos * len(l1) ]
    p = p + l2
    if len(p) is 0:
        return &quot;.&quot;
    return os.path.join( *p )


def abs2rel(path, base = os.curdir):
    &quot;&quot;&quot; @return a relative path from base to path.

    base can be absolute, or relative to curdir, or defaults
    to curdir.
    &quot;&quot;&quot;
    if protocolPattern.match(path): return path
    base = rel2abs(base)
    path = rel2abs(path) # redundant - should already be absolute
    return relpath(base, path)
Alan Ezust 10 years, 1 month ago  # | flag

Better pathsplit.

def pathsplit(path):
    """ This version, in contrast to the original version, permits trailing
    slashes in the pathname (in the event that it is a directory).
    It also uses no recursion """
    return path.split(os.path.sep)

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