multidimensional take
Johannes Nix
jnix at aster.homelinux.net
Thu Mar 20 00:59:37 CET 2003
Johannes Nix <jnix at medi.physik.uni-oldenburg.de> writes:
> Now I want to have a function defined as following:
>
> >>> a
> array([[140, 455, 325, 360, 498],
> [372, 647, 636, 365, 462],
> [893, 141, 776, 238, 259]])
>
> >>> mtake(a, [[1, 0], [ 1, 1], [2, 4]])
> array([372, 647, 259])
>
>
Below is the solution I got after some hours - i does not has any
interpreted loop. It differs a bit from the list comprehension
definition in that if a e.g. has shape (1, 2, 3, 5, 7) and b has shape
(9, 11, 13), mtake(a, b) will have shape (1, 2, 3, 9) and not (9, 1,
2, 3) , ergo
>>> a2 = array((a, 10 * a))
>>> mtake(a2, [[1, 0], [ 1, 1], [2, 4]])
array([[ 372, 647, 259],
[3720, 6470, 2590]])
>>> a2.shape
(2, 3, 5)
I've omitted the axis parameter because it seems confusing to
me, if necessary the axes can be changed by Numeric.transpose . In my
application, shapes of a and b will be (10000, 3, 3) and
(2, 43, 24, 50) so this will save a large amount of computing time.
Perhaps this would be a useful addition to Numeric ?
Johannes
-------------------------------------------------------------------------
def mtake(ar, indices):
"""similar to take but uses last dimension of indices for
coordinate lookup along last axes of ar.
"""
indices = array(indices)
ar = array(ar)
if len(indices.shape) == 1:
return a[indices]
elif (len(indices.shape) == 2) and (indices.shape[-1] == 1):
return take(a, indices)
numrdims = len(ar.shape) - indices.shape[-1]
strides = multiply.accumulate(array(ar.shape[numrdims+1:] +
(1,))[::-1])[::-1]
newshape = ar.shape[:numrdims] + (strides[0] * ar.shape[numrdims],)
ar = reshape(ar, newshape)
return take(ar, add.reduce(strides * indices, -1), -1)
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