Could you provide more details about this to the ticket I created
based on your email:
http://projects.scipy.org/scipy/numpy/ticket/636
Thanks,
On Thu, Dec 13, 2007 at 3:33 PM, Jonathan Taylor
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 _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
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