On 06.02.2012 22:27, Sturla Molden wrote:
>
>
>>
>> # Make a 4D view of this data, such that b[i,j]
>> # is a 2D block with shape (4,4) (e.g. b[0,0] is
>> # the same as a[:4, :4]).
>> b = as_strided(a, shape=(a.shape[0]/4, a.shape[1]/4, 4, 4),
>> strides=(4*a.strides[0], 4*a.strides[1], a.strides[0],
>> a.strides[1]))
>>
>
> Yes :-) Being used to Fortran (and also MATLAB) this is the kind of mapping
> It never occurs for me to think about. What else but NumPy is flexible enough
> to do this? :-)
Actually, using as_strided is not needed. We can just reshape like this:
(m,n) ---> (m//4, 4, n//4, 4)
and then use np.any along the two length-4 dimensions.
m,n = data.shape
cond = lamda x : (x <= t1) & (x >= t2)
x = cond(data).reshape((m//4, 4, n//4, 4))
found = np.any(np.any(x, axis=1), axis=2)
Sturla
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