[Numpy-discussion] Faster array version of ndindex
Stefan van der Walt
Fri Dec 14 07:32:53 CST 2007
N.fromiter only works on 1D arrays. I thought the following may work,
but it doesn't:
This kind of loop is probably best implemented in C, although I think
Jonathan's version is rather clever.
On Fri, Dec 14, 2007 at 10:31:56AM +0100, Sebastian Haase wrote:
> Do you know about
> -Sebastian Haase
> On Dec 14, 2007 12:33 AM, Jonathan Taylor <firstname.lastname@example.org> wrote:
> > I was needing an array representation of ndindex since ndindex only
> > gives an iterator but array(list(ndindex)) takes too long. There is
> > prob some obvious way to do this I am missing but if not feel free to
> > include this code which is much faster.
> > In : time a=np.array(list(np.ndindex(10,10,10,10,10,10)))
> > CPU times: user 11.61 s, sys: 0.09 s, total: 11.70 s
> > Wall time: 11.82
> > In : time a=ndtuples(10,10,10,10,10,10)
> > CPU times: user 0.32 s, sys: 0.21 s, total: 0.53 s
> > Wall time: 0.60
> > def ndtuples(*dims):
> > """Fast implementation of array(list(ndindex(*dims)))."""
> > # Need a list because we will go through it in reverse popping
> > # off the size of the last dimension.
> > dims = list(dims)
> > # N will keep track of the current length of the indices.
> > N = dims.pop()
> > # At the beginning the current list of indices just ranges over the
> > # last dimension.
> > cur = np.arange(N)
> > cur = cur[:,np.newaxis]
> > while dims != :
> > d = dims.pop()
> > # This repeats the current set of indices d times.
> > # e.g. [0,1,2] -> [0,1,2,0,1,2,...,0,1,2]
> > cur = np.kron(np.ones((d,1)),cur)
> > # This ranges over the new dimension and 'stretches' it by N.
> > # e.g. [0,1,2] -> [0,0,...,0,1,1,...,1,2,2,...,2]
> > front = np.arange(d).repeat(N)[:,np.newaxis]
> > # This puts these two together.
> > cur = np.column_stack((front,cur))
> > N *= d
> > return cur
More information about the Numpy-discussion