How to get mask from numpy array according to all values on third axis

Question

I have numpy array like:

A = np.zeros((X,Y,Z))

Later I fill the array with values and I need to find x,y coordinates according to values in arrays on z axis to replace it with different array.

Example withZ=2 I have array

A = [ [ [1,2], [2,1], [0, 0] ],
      [ [1,1], [1,1], [0, 0] ], ]

I need something like

A[ where A[x,y,:] == [1,2] ] = [2,1]

What will produce array

A = [ [ [2,1], [2,1], [0, 0] ],
      [ [1,1], [1,1], [0, 0] ], ]

Is it possible to achieve that somehow simple? I would like to avoid iteration over the x,y coordinates.

Could you add a sample case and the expected output?

Divakar
– Divakar

2016-10-08 10:43:08 +00:00
Commented Oct 8, 2016 at 10:43 — Divakar
– Divakar, Commented Oct 8, 2016 at 10:43
@Divakar Done. If it is not enough, then let me know.

matousc
– matousc

2016-10-08 10:49:37 +00:00
Commented Oct 8, 2016 at 10:49 — matousc
– matousc, Commented Oct 8, 2016 at 10:49

Divakar · Accepted Answer · 2016-10-08 10:50:55Z

3

One vectorized approach -

A[(A == [1,2]).all(-1)] = [2,1]

Sample run -

In [15]: A
Out[15]: 
array([[[1, 2],
        [2, 1],
        [0, 0]],

       [[1, 1],
        [1, 2],
        [0, 0]]])

In [16]: A[(A == [1,2]).all(-1)] = [2,1]

In [17]: A
Out[17]: 
array([[[2, 1],
        [2, 1],
        [0, 0]],

       [[1, 1],
        [2, 1],
        [0, 0]]])

answered Oct 8, 2016 at 10:50

Divakar

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