Comments on: Generating all combinations from an arbitrary number of sequences

By: Brandon Mintern

Brandon Mintern — Sat, 24 May 2008 05:01:44 +0000

FYI if you're coming from Google or whatever, this is in Python 2.6 and 3K as itertools.product (implemented in C).

And I tested the various implementations along with a couple others, and ecombinations was (surprisingly) the clear winner. It makes me want to eschew Python for Lisp... :-)

By: Peter Hosey

Peter Hosey — Thu, 25 Oct 2007 18:27:28 +0000

Colin: I wonder whether the list comprehension could be turned into a generator comprehension somehow, possibly relieving the need to call reversed…

By: Colin Barrett

Colin Barrett — Thu, 25 Oct 2007 18:24:24 +0000

This actually makes it work the way I want it to:

def combinations(*list):
    r=[[]]
    for x in reversed(list):
        r = [[y]+i for y in x for i in r]
    return r

Adding print r inside the loop reveals how it works:

>>> combinations([1,2],[3,4],[5,6]) and 'Done'
[[5], [6]]
[[3, 5], [3, 6], [4, 5], [4, 6]]
[[1, 3, 5], [1, 3, 6], [1, 4, 5], [1, 4, 6], [2, 3, 5], [2, 3, 6], [2, 4, 5], [2, 4, 6]]
'Done'

By: Wensheng Wang

Wensheng Wang — Thu, 25 Oct 2007 17:37:18 +0000

from: http://aspn.activestate.com/ASPN/Cookbook/Python/Recipe/496807

def combinations(*list):
    r=[[]]
    for x in list:
        r = [i+[y] for y in x for i in r]
    return r

print combinations([1,2,3],[4,5,6])
print
print combinations([1,2],[3,4],[5,6])
print
print combinations([1,2,3],[4,5],[6],[7,8,9,10])

By: Arnar Birgisson

Arnar Birgisson — Thu, 25 Oct 2007 00:06:03 +0000

On a slightly related note, here's how to generate the powerset (all possible subsets) of some set:

In [103]:
def powerset(S):
	for b in range(1 << len(S)):
		yield set(e for (i,e) in enumerate(S) if (1<<i)&b)

In [104]: myset = set(range(1,5))

In [105]: list(powerset(myset))
Out[105]: 
[set([]),
 set([1]),
 set([2]),
 set([1, 2]),
 set([3]),
 set([1, 3]),
 set([2, 3]),
 set([1, 2, 3]),
 set([4]),
 set([1, 4]),
 set([2, 4]),
 set([1, 2, 4]),
 set([3, 4]),
 set([1, 3, 4]),
 set([2, 3, 4]),
 set([1, 2, 3, 4])]

(Problem suggested by Árni Már)

By: Arnar Birgisson

Arnar Birgisson — Wed, 24 Oct 2007 09:45:10 +0000

It's interesting to see some timings. We basically have two dimensions that can vary, the number of sequences and the lengths of the sequences. Don't have time right now to do it myself, but I propose the following test cases:

(range(2) for i in range(10))
(range(2) for i in range(100))
(range(2) for i in range(1000))
(range(2) for i in range(10000)) # perhaps to much?
(range(10) for i in range(10))
(range(100) for i in range(10))
(range(10000) for i in range(2))
(range(i) for i in range(10))
(range(i) for i in range(100))

etc.

Arnar

By: Carl

Carl — Wed, 24 Oct 2007 08:39:26 +0000

Non-scientific tests:

>>> c=1
>>> t = time(); list(gcombs(*(range(2) for i in range(18)))) and time() - t
0.87823700904846191
>>> c=1
>>> t = time(); list(gcombs(*(range(2) for i in range(18)))) and time() - t
0.67177414894104004
>>> t = time(); list(icombinations(*(range(2) for i in range(18)))) and time() 
- t
0.68789410591125488
>>> t = time(); list(icombinations(*(range(2) for i in range(18)))) and time() 
- t
0.72254085540771484
>>> t = time(); list(ecombinations(*(range(2) for i in range(18)))) and time() 
- t
0.58095097541809082
>>> t = time(); list(ecombinations(*(range(2) for i in range(18)))) and time() 
- t
0.59889411926269531

I wouldn't take these numbers seriously though. In some of my tests, icombinations beat ecombinations.

By: Gary Godfrey

Gary Godfrey — Wed, 24 Oct 2007 05:38:34 +0000

I suspect I could add a reverse in there to make it match the for loops. To be honest, I didn't use this for work - it's far too difficult to read. It just occurred to me when I was thinking about the problem and I thought it was cute.

I finally ran some profiling. It seems to be about 16x slower than the recursive solution. I don't think it's worth updating the variable names :-).

Gary Godfrey
Austin, TX, USA

By: Peter Hosey

Peter Hosey — Wed, 24 Oct 2007 04:20:05 +0000

Gary: Looks cool, but Colin complains that the order in which combinations are yielded does not match the original nested for loops.

Any chance we could get a version with clearer variable names? ☺

By: Gary Godfrey

Gary Godfrey — Wed, 24 Oct 2007 02:14:50 +0000

I needed something like this for work today. Here's my version - no recursion. I've not profiled it.

from itertools import *

c = 1
def cumul(n):
    global c
    r = c
    c *= len(n)
    return r

def gcombs(*s):
    tary = [ cycle(chain(*[ repeat(nn, cumval) for nn in news])) for (news,cumval) in [ (ns, cumul(ns)) for ns in s]]
    tary.append(xrange(0,c))
    t = izip( *tary)
    for v in t:
        yield v[:-1]

print list(gcombs([1,2],[4,5,6],[7,8,9]))

By: Carl

Carl — Wed, 24 Oct 2007 00:54:24 +0000

(This is fun to mess around with.) The Lisp way to do it:

def ecombinations(*args):
    n = len(args)
    defun = "def inner(args):"
    for_loops = "\n".join("%sfor v%s in args[%s]:" % ("    "*(i+1), i, i) 
        for i in range(n))
    yield_line = "v%s, " * n % tuple(range(n))
    yield_line = "%syield (%s)" % ("    "*(n+1), yield_line)
    exec '\n'.join([defun, for_loops, yield_line])
    return inner(args)

By: Carl

Carl — Tue, 23 Oct 2007 23:50:01 +0000

(hey, my previous comment got double-escaped since I ignored the CAPTCHA the first time. feel free to delete it, and this parenthetical remark.)

Actually, the previous version goes through items in the wrong order:

>>> print '\n'.join(repr(i) for i in icombinations("12", "ab", "yz")) ('1', 'a', 'y') ('2', 'a', 'y') ('1', 'b', 'y') ('2', 'b', 'y') ('1', 'a', 'z') ('2', 'a', 'z') ('1', 'b', 'z') ('2', 'b', 'z')

What we actually want is this:

def icombinations(*seqs): if len(seqs) == 1: for i in seqs[0]: yield (i, ) else: for rest in icombinations(*seqs[:-1]): for i in seqs[-1]: yield rest + (i, )

In order to get this output:

>>> print '\n'.join(repr(i) for i in icombinations("12", "ab", "yz")) ('1', 'a', 'y') ('1', 'a', 'z') ('1', 'b', 'y') ('1', 'b', 'z') ('2', 'a', 'y') ('2', 'a', 'z') ('2', 'b', 'y') ('2', 'b', 'z')

By: Carl

Carl — Tue, 23 Oct 2007 23:48:37 +0000

Actually, the previous version goes through items in the wrong order:

>>> print \'\\n\'.join(repr(i) for i in icombinations(\"12\", \"ab\", \"yz\")) (\'1\', \'a\', \'y\') (\'2\', \'a\', \'y\') (\'1\', \'b\', \'y\') (\'2\', \'b\', \'y\') (\'1\', \'a\', \'z\') (\'2\', \'a\', \'z\') (\'1\', \'b\', \'z\') (\'2\', \'b\', \'z\')

What we actually want is this:

def icombinations(*seqs): if len(seqs) == 1: for i in seqs[0]: yield (i, ) else: for rest in icombinations(*seqs[:-1]): for i in seqs[-1]: yield rest + (i, )

In order to get this output:

>>> print \'\\n\'.join(repr(i) for i in icombinations(\"12\", \"ab\", \"yz\")) (\'1\', \'a\', \'y\') (\'1\', \'a\', \'z\') (\'1\', \'b\', \'y\') (\'1\', \'b\', \'z\') (\'2\', \'a\', \'y\') (\'2\', \'a\', \'z\') (\'2\', \'b\', \'y\') (\'2\', \'b\', \'z\')

By: Carl

Carl — Tue, 23 Oct 2007 19:12:02 +0000

Canonical version?

def icombinations(*seqs):
    if len(seqs) == 1:
        for i in seqs[0]: yield (i, )
    else:
        for rest in icombinations(*seqs[1:]):
            for i in seqs[0]:
                yield (i, ) + rest

By: Arnar Birgisson

Arnar Birgisson — Tue, 23 Oct 2007 15:45:23 +0000

Brian, that's nice. I would have used tuples if I were to use it in my code - but left it as lists here to match the interface in the OP. Didn't realize the speed difference though, but that certainly makes sense.