Let me be the first of many (i like this game :-) to give you
(hopefully) the final solution:
def partition(v,n,pad=0):
return [(v+[pad]*(n-len(v)%n))[i:i+n] for i in range(0,len(v),n)]
-vgermrk-
On 24 Jan., 01:34, Jason Grout <[EMAIL PROTECTED]> wrote:
> [EMAIL PROTECTED] wrote:
>
> > On Wed, 23 Jan 2008, William Stein wrote:
>
> >> On Jan 23, 2008 4:12 PM, Jason Grout <[EMAIL PROTECTED]> wrote:
> >>> Does anyone know the best way to partition a list into sublists of a
> >>> specific length, similar to the Partition command in Mathematica? I'm
> >>> thinking of something like:
>
> >>> sage: partition([1,2,3,4],2)
> >>> [[1,2],[3,4]]
> >>> sage: partition([1,2,3,4,5],2,pad=0)
> >>> [[1,2],[3,4],[5,0]]
>
> >>> It seems like this is a problem that python would have solved millions
> >>> of years ago, but I can't find anything when searching online. I can
> >>> whip it up quickly, but I'm sure it's already been invented, which is
> >>> why I'm asking.
> >> Let me be the first of many to post a one liner:
>
> >> sage: def partition(v, n):
> >> ... return [v[n*i:n*i+n] for i in range(len(v)//n)]
>
> >> sage: partition([1,2,3,4],2)
> >> [[1, 2], [3, 4]]
>
> > Let me be the first of many to post a counterexample:
> > sage: partition([1,2,3,4,5],2)
> > [[1, 2], [3, 4]]
>
> > and a fix:
>
> > sage: def partition(v,n):
> > ... return [v[i:i+n] for i in range(0,len(v),n)]
> > sage: partition([1,2,3,4,5],2)
> > [[1, 2], [3, 4], [5]]
>
> And let me be the first of many to say that that was the solution that I
> finally found as I continued to search. Is there a nice, fast function
> that provides some of the pretty extensive functionality of the
> Partition function in Mathematica? Or how about even just a default
> padding, as in my example?
>
> Seehttp://reference.wolfram.com/mathematica/ref/Partition.html
>
> Or is it time to write such a function? :)
>
> As it is, the one-liner above is all I need for now.
>
> Thanks,
>
> Jason
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