Hi,
this should work:
import numpy as np
ndim = 20
cube = np.random.rand(32,ndim, ndim)
result = np.zeros([ndim, ndim], np.float32)
def combine(cube, result):
for ii in range(ndim):
for jj in range(ndim):
result[ii, jj] = np.sqrt((cube[:,ii, jj])).sum()
return result
def combine2(cube, result):
r2 = np.sqrt(cube)
r3 = r2.sum(axis=0)
return r3
r3 = combine2(cube,result)
r1 = combine(cube, result)
print np.allclose(r3, r1)
When I time it in ipython with ndim=20, I get:
In [8]: timeit combine2(cube, result)
1000 loops, best of 3: 246 us per loop
In [9]: timeit combine(cube, result)
100 loops, best of 3: 3.86 ms per loop
Best regards,
Hanno
Am 06.03.2012 um 13:00 schrieb Jose Miguel Ibáñez:
> Hello everyone,
>
> does anyone know of an efficient implementation (maybe using
> numpy.where statement) of the next code for data cube (3d array)
> combining ?
>
> import numpy as np
>
> def combine( )
>
> cube = np.random.rand(32,2048,2048)
> result = np.zeros([2048,2048], np.float32)
>
> for ii in range(2048):
> for jj in range(2048):
> result[, ii, jj] = np.sqrt((cube[:,ii, jj])).sum()
>
>
> It takes long time to run, however,
>
>
>>> result = np.median(cube,0)
>
>
> only around one second ! where is the point ? any suggestions ?
>
>
>
> Thanks !
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