Reverse a string in Python two characters at a time (Network byte order)

A concise way to do this is:

"".join(reversed([a[i:i+2] for i in range(0, len(a), 2)]))

This works by first breaking the string into pairs:

>>> [a[i:i+2] for i in range(0, len(a), 2)]['AB', 'CD', 'EF', 'GH']

then reversing that, and finally concatenating the result back together.

Lots of fun ways to do this

>>> s="ABCDEFGH">>> "".join(map(str.__add__, s[-2::-2] ,s[-1::-2]))'GHEFCDAB'

If anybody is interested, this is the timing for all* the answers.

EDIT (had got it wrong the first time):

import timeitimport structstring = "ABCDEFGH"# Expected resutlt => GHEFCDABdef rev(a):    new = ""    for x in range(-1, -len(a), -2):        new += a[x-1] + a[x]    return newdef rev2(a):    return "".join(reversed([a[i:i+2] for i in range(0, len(a), 2)]))def rev3(a):    return "".join(map(str.__add__, a[-2::-2] ,a[-1::-2]))def rev4(a):    return "".join(map("".join, reversed(zip(*[iter(a)]*2))))def rev5(a):    n = len(a) / 2    fmt = '%dh' % n    return struct.pack(fmt, *reversed(struct.unpack(fmt, a)))def rev6(a):    return "".join([a[x:x+2] for x in range(0,len(a),2)][::-1])print "Greg Hewgill %f" %timeit.Timer("rev2(string)", "from __main__ import rev2, string").timeit(100000)print "gnibbler %f" %timeit.Timer("rev3(string)", "from __main__ import rev3, string").timeit(100000)print "gnibbler second %f" %timeit.Timer("rev4(string)", "from __main__ import rev4, string").timeit(100000)print "Alok %f" %timeit.Timer("rev5(string)", "from __main__ import rev5, struct, string").timeit(100000)print "elliot42 %f" %timeit.Timer("rev6(string)", "from __main__ import rev6, struct, string").timeit(100000)print "me %f" %timeit.Timer("rev(string)", "from __main__ import rev, string").timeit(100000)

results for string = "ABCDEFGH":

Greg Hewgill 0.853000gnibbler 0.428000gnibbler second 0.707000Alok 0.763000elliot42 0.237000me 0.200000

results for string = "ABCDEFGH"*5:

Greg Hewgill 2.246000gnibbler 0.811000gnibbler second 1.205000Alok 0.972000elliot42 0.594000me 0.584000

results for string = "ABCDEFGH"*10:

Greg Hewgill 2.058000gnibbler 1.178000gnibbler second 1.926000Alok 1.210000elliot42 0.935000me 1.082000

results for string = "ABCDEFGH"*100:

Greg Hewgill 9.762000gnibbler 9.134000gnibbler second 14.782000Alok 5.775000elliot42 7.351000me 18.140000

*Sorry @Lacrymology could not make your's work!