On Apr 13, 1:17 pm, Mensanator <mensana...@aol.com> wrote: > On Apr 13, 6:36 am, "skorpi...@gmail.com" <skorpi...@gmail.com> wrote: > > > > > On Apr 13, 7:13 am, Chris Rebert <c...@rebertia.com> wrote: > > > > On Mon, Apr 13, 2009 at 4:05 AM, skorpi...@gmail.com > > > > <skorpi...@gmail.com> wrote: > > > > I am trying to generate all possible permutations of length three from > > > > elements of [0,1]. i.e in this scenario there are a total of 8 > > > > distinct permutations: > > > > > [0,0,0] > > > > [0,0,1] > > > > [0,1,0] > > > > . > > > > . > > > > . > > > > [1,1,1] > > > > > Does numpy define a function to achieve this ? > > > > No idea, but the Python standard library already has this covered with > > > itertools.permutations() > > > [http://docs.python.org/library/itertools.html]. > > > > Cheers, > > > Chris > > > > -- > > > I have a blog:http://blog.rebertia.com > > > Thanks Chris, > > > That looks promising, however I am still stuck at python 2.5 (I need > > numpy). And the 2.5 version does not look as nice as the 2.6 itertool. > > So, if there is a numpy method ... please let me know .. > > This isn't dependent on numpy or Python version: > > >>> e = [0,1] > >>> for i in e: > > for j in e: > for k in e: > print [i,j,k] > > [0, 0, 0] > [0, 0, 1] > [0, 1, 0] > [0, 1, 1] > [1, 0, 0] > [1, 0, 1] > [1, 1, 0] > [1, 1, 1]
Much appreciated. This is exactly what was needed... -- http://mail.python.org/mailman/listinfo/python-list
