numpy where operation on 2D array

Question

I have a numpy array 'A' of size 571x24 and I am trying to find the index of zeros in it so I do:

>>>A.shape
(571L, 24L)

import numpy as np
z1 = np.where(A==0)

z1 is a tuple with following size:

>>> len(z1)
2
>>> len(z1[0])
29
>>> len(z1[1])
29

I was hoping to create a z1 of same size as A. How do I achieve that?

Edit: I want to create array z1 of booleans for presence of zero in A such that:

>>>z1.shape
    (571L, 24L)

"I was hoping to create a z1 of same size as A." - what would you expect that z1 to mean? You seem to be looking for a data representation other than what np.where provides. What data representation are you looking for? — user2357112
– user2357112, Commented Jun 17, 2016 at 16:28
If I'm understanding this correctly, you want to find the indices of where zeros occur in the array? — Scratch'N'Purr
– Scratch'N'Purr, Commented Jun 17, 2016 at 16:29

Martin Hallén · Accepted Answer · 2016-06-17 16:50:45Z

3

You can just check this with the equality operator in python with numpy. Example:

>>> A = np.array([[0,2,2,1],[2,0,0,3]])
>>> A == 0
array([[ True, False, False, False],
       [False,  True,  True, False]], dtype=bool)

np.where() does something else, see documentation. Although, it is possible to achieve this with np.where() using broadcasting. See documentation.

>>> np.where(A == 0, True, False)
array([[ True, False, False, False],
       [False,  True,  True, False]], dtype=bool)

edited Jun 17, 2016 at 16:50

answered Jun 17, 2016 at 16:37

Martin Hallén

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Comments

Scratch'N'Purr · Accepted Answer · 2016-06-17 16:38:23Z

1

Try this:

import numpy as np
myarray = np.array([[0,3,4,5],[9,4,0,4],[1,2,3,4]])
ix = np.in1d(myarray.ravel(), 0).reshape(myarray.shape)

Output of ix:

array([[ True, False, False, False],
       [False, False,  True, False],
       [False, False, False, False]], dtype=bool)

answered Jun 17, 2016 at 16:38

Scratch'N'Purr

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