Computing the Relative Path from One Directory to Another(Python cookbook 2-22)

Problem

You need to know the relative path from one directory to anotherfor example, to create a symbolic link or a relative reference in a URL.

Solution

Precondition

itertools.izip(*iterables)

Make an iterator that aggregates elements from each of the iterables. Like zip()except that it returns an iterator instead of a list. Used for lock-step iteration over several iterables at a time.

 1>>> import itertools
 2>>> s1='/a/b/c/d'
 3>>> s2='/a/b/c1/d1'
 4>>> t1=s1.split('/')
 5>>> t2=s2.split('/')
 6>>> t1
 7['', 'a', 'b', 'c', 'd']
 8>>> t2
 9['', 'a', 'b', 'c1', 'd1']
>>> test = itertools.izip(t1, t2)
>>> for s in test:
    print s
('', '')
('a', 'a')
('b', 'b')
('c', 'c1')
('d', 'd1')

 

Code

all_equal: return True if all the elements are equal, otherwise False.

 1>>> def all_equal(elements):
 2    first = elements[0]
 3    for e in elements[1:]:
 4        if first != e:
 5            return False
 6    return True
 7>>> all_equal(('a','a','a'))
 8True
 9>>> all_equal(['a','c'])
False

 

common_prefix: return a list of common elements at the start of all sequences, then a list of lists that are the unique tails of each sequence.

 1>>> def common_prefix(*sequences):
 2    if not sequences:
 3        return [],[]
 4    common = []
 5    for elements in itertools.izip(*sequences):
 6        if not all_equal(elements):
 7            break
 8        common.append(elements[0])
 9    return common, [sequence[len(common):] for sequence in sequences]
>>> s1='/a/b/c/d'
>>> s2='/a/b/c1/d1'
>>> t1=s1.split('/')
>>> t2=s2.split('/')
>>> common_prefix(t1, t2)
(['', 'a', 'b'], [['c', 'd'], ['c1', 'd1']])

 

relpath: return a relative path from p1 equivalent to path p2. In particular: the empty string, if p1 == p2; p2, if p1 and p2 have no common prefix.

>>> import os
>>> def relpath(p1, p2, sep=os.path.sep, pardir=os.path.pardir):
    common, (u1, u2) = common_prefix(p1.split(sep), p2.split(sep))
    if not common:
        return p2
    return sep.join([pardir]*len(u1) + u2)
>>> print 'from', p1, 'to', p2, ' -> ', relpath(p1,p2, '/')
from /a/b/c/d to /a/b/c1/d1  ->  ../../c1/d1