Hi Sebastian N.fromiter only works on 1D arrays. I thought the following may work, but it doesn't: np.fromiter(np.ndindex(10,10,10),N.dtype((int,3))) This kind of loop is probably best implemented in C, although I think Jonathan's version is rather clever. Regards Stéfan On Fri, Dec 14, 2007 at 10:31:56AM +0100, Sebastian Haase wrote:

Do you know about N.fromiter() ?

-Sebastian Haase

On Dec 14, 2007 12:33 AM, Jonathan Taylor 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