numpy permutations with replacement

[skorpio11 at gmail.com]

On Apr 13, 1:17 pm, Mensanator <mensana... at aol.com> wrote:
> On Apr 13, 6:36 am, "skorpi... at gmail.com" <skorpi... at gmail.com> wrote:
>
>
>
> > On Apr 13, 7:13 am, Chris Rebert <c... at rebertia.com> wrote:
>
> > > On Mon, Apr 13, 2009 at 4:05 AM, skorpi... at gmail.com
>
> > > <skorpi... at 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...