NumPy - Selecting a submatrix from another 2d array

Question

This is my original 2d array A

[[0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 2, 2, 2, 2, 2, 2, 0, 0, 0],
 [0, 2, 2, 2, 2, 2, 2, 0, 0, 0],
 [0, 2, 2, 2, 2, 2, 2, 0, 0, 0],
 [0, 2, 2, 2, 2, 2, 2, 0, 0, 0],
 [0, 2, 2, 8, 8, 8, 2, 0, 0, 0],
 [0, 2, 2, 8, 8, 8, 2, 0, 0, 0],
 [0, 0, 0, 8, 8, 8, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]]

Let's say I want to return a 3x3 submatrix of 8's at the middle. I made a boolean mask with this expression A == 8 and it looks like this.

array([[False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False,  True,  True,  True, False, False, False,
        False],
       [False, False, False,  True,  True,  True, False, False, False,
        False],
       [False, False, False,  True,  True,  True, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False],
       [False, False, False, False, False, False, False, False, False,
        False]])

This is the point where I'm stuck. How can I return the submatrix with that boolean mask? If I do A[A == 8], it returns a flat array of 8s like this

array([8, 8, 8, 8, 8, 8, 8, 8, 8])

Another way is getting the row and column numbers with np.where(A == 8) which returns (array([5, 5, 5, 6, 6, 6, 7, 7, 7]), array([3, 4, 5, 3, 4, 5, 3, 4, 5])). How can I return the matrix using them?

Is there any better approach for this problem?

Boolean indexing like this returns a 1d result. In general such indexing does not a return a neat block of values. But if you know the results will be such block, just reshape them. — hpaulj
– hpaulj, Commented Feb 18, 2020 at 16:38
It works as intended, because you don't always have rectangular shape, eg. A[A==2]. Now if you are certain that 8 only appear in a block, you can detect the shape with argmin/argmax. — Quang Hoang
– Quang Hoang, Commented Feb 18, 2020 at 17:16

kuzand · Accepted Answer · 2020-02-18 21:45:54Z

1

As mentioned in comments, A[A==some_value] correctly returns a rank-1 array of values some_value. That makes sense because it's not always the case that the values some_value are organized in a block (take for example some_value = 2) or the number of values some_value is such that they cannot be put into a 2D array.

However If you are sure that there is such a block, you can do the following to get it:

import numpy as np

inds = np.where(A==8)
slice_x = slice(inds[0].min(), inds[0].max() + 1)  # slice(6, 9, None)
slice_y = slice(inds[1].min(), inds[1].max() + 1)  # slice(3, 6, None)

A[slice_x, slice_y]

# array([[8, 8, 8],
#        [8, 8, 8],
#        [8, 8, 8]])

As alternative, you can use scipy.ndimage.find_objects to obtain the slices:

from scipy import ndimage

slice_x, slice_y = ndimage.find_objects(A==8)[0]  # (slice(6, 9, None), slice(3, 6, None))

edited Feb 18, 2020 at 21:45

answered Feb 18, 2020 at 17:54

kuzand

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Comments

mathfux · Accepted Answer · 2020-02-18 19:27:53Z

You can capture row and column indexes and after that return submatrix from these indexes only using np.ix_ method:

x, y = np.where(A==8) #[5 5 5 6 6 6 7 7 7], [3 4 5 3 4 5 3 4 5]
x, y = np.unique(x), np.unique(y) # [5,6,7], [3,4,5]
print(A[np.ix_(x, y)]) #prints [[8 8 8], [8 8 8], [8 8 8]]

This should work even in more general cases, although not always as expected:

def submatrix(A):
    x, y = np.where(A==8)
    x, y = np.unique(x), np.unique(y)
    return A[np.ix_(x, y)]

>>> A = np.array([
[8, 2, 8, 2, 8, 0],
[2, 8, 8, 8, 2, 0],
[2, 8, 4, 8, 3, 0],
[0, 8, 8, 8, 0, 0],
[8, 0, 8, 0, 8, 0],
[0, 0, 0, 0, 0, 0]])
>>> submatrix(A)
array([[8, 2, 8, 2, 8],
       [2, 8, 8, 8, 2],
       [2, 8, 4, 8, 3],
       [0, 8, 8, 8, 0],
       [8, 0, 8, 0, 8]])
>>> A = np.array([
[8, 2, 8, 2, 8, 0],
[2, 2, 8, 8, 2, 0],
[2, 2, 4, 3, 3, 0],
[0, 2, 8, 8, 0, 0],
[8, 0, 8, 0, 8, 0],
[0, 0, 0, 0, 0, 0]])
>>> submatrix(A) # skipping empty rows and columns
array([[8, 8, 2, 8],
       [2, 8, 8, 2],
       [0, 8, 8, 0],
       [8, 8, 0, 8]])

Since np.unique returns sorted arrays, you can fill the gaps constucting x, y like so:

def submatrix(A):
    x, y = np.where(A==8)
    x, y = np.unique(x), np.unique(y)
    x, y = np.arange(x[0], x[-1]+1), np.arange(y[0], y[-1]+1)
    return A[np.ix_(x, y)]

Nik P · Accepted Answer · 2020-02-18 20:56:51Z

0

You can try the following

 data = A[np.any(A==8,axis=1)]
 data.T[np.all(data==8,axis=0)]

This should give,

 array([[8, 8, 8],
        [8, 8, 8],
        [8, 8, 8]])

answered Feb 18, 2020 at 20:56

Nik P

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