[Numpy-discussion] Faster array version of ndindex

Sebastian Haase haase@msg.ucsf....
Fri Dec 14 03:31:56 CST 2007


Do you know about
N.fromiter()
?

-Sebastian Haase


On Dec 14, 2007 12:33 AM, Jonathan Taylor <jonathan.taylor@utoronto.ca> 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 [252]: 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 [253]: 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
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