On Thu, Mar 17, 2011 at 03:29:08PM +0100, Jeroen Demeyer wrote:
> On 2011-03-17 15:04, Jason Grout wrote:
> > def hash(self):
> > h=0
> > for i,j,entry in m.nonzero_entries(): # nonzero entries for sparse
> > matrices
> > h^=hash(entry)^i^j
> > return h
>
> Since you're only xorring, this will give a lot of collisions. I think
> something like
>
> def hash(self):
> n = m.ncols()
> h = (m.nrows() + m.ncols())**2 + n
> for i,j,entry in m.nonzero_entries():
> h ^= hash(entry)*(i*n + j + 1)
> return h
>
> has less chance of collisions. Of course everything should be computed
> as a C long.
Btw: any reason not to use Python's builtin hash function for tuples?
I assume it as been designed with precisely this kind of issues in
mind? Or do we want that a matrix gets the same hash in sparse and a
dense representation?
Cheers,
Nicolas
--
Nicolas M. Thiéry "Isil" <nthi...@users.sf.net>
http://Nicolas.Thiery.name/
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