On Tue, Sep 20, 2005 at 10:20:44AM -0700, Josiah Carlson wrote:
Try using the code I offered. It allows the cross of an aribitrary number of restartable iterables, in the same order as an equivalent list comprehension or generator expression.
list(cross([1,2], [3,4], [5,6])) [(1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6)]
There were a few hoops I had to jump through in cross in order to be able to hande single iterables as well as tuples embedded in the passed iterables, but they work as they should.
list(cross([(1,1),(2,2)], [(3,3),(4,4)], [(5,5),(6,6)])) [((1, 1), (3, 3), (5, 5)), ((1, 1), (3, 3), (6, 6)), ((1, 1), (4, 4), (5, 5)), ((1, 1), (4, 4), (6, 6)), ((2, 2), (3, 3), (5, 5)), ((2, 2), (3, 3), (6, 6)), ((2, 2), (4, 4), (5, 5)), ((2, 2), (4, 4), (6, 6))]
This comes up on c.l.p every month or two. Folks offer their own solutions optimized for speed, memory, or golfed for char length. I'll throw in my same two bits as always, sprat:~# python Python 2.4.1 (#2, Mar 30 2005, 21:51:10) [GCC 3.3.5 (Debian 1:3.3.5-8ubuntu2)] on linux2 Type "help", "copyright", "credits" or "license" for more information.
import probstat c = probstat.Cartesian([[1,2], [3,4], [5,6]]) list(c) [[1, 3, 5], [2, 3, 5], [1, 4, 5], [2, 4, 5], [1, 3, 6], [2, 3, 6], [1, 4, 6], [2, 4, 6]] c = probstat.Cartesian([[(1,1),(2,2)], [(3,3),(4,4)], [(5,5),(6,6)]]) list(c) [[(1, 1), (3, 3), (5, 5)], [(2, 2), (3, 3), (5, 5)], [(1, 1), (4, 4), (5, 5)], [(2, 2), (4, 4), (5, 5)], [(1, 1), (3, 3), (6, 6)], [(2, 2), (3, 3), (6, 6)], [(1, 1), (4, 4), (6, 6)], [(2, 2), (4, 4), (6, 6)]]
The signature is slightly different (list of lists) but otherwise does what you want. Unchanged since 2002! http://probstat.sourceforge.net/ -jackdied