b10hazard b10hazard - 6 months ago 46
Python Question

Find matching rows in 2 dimensional numpy array

I would like to get the index of a 2 dimensional Numpy array that matches a row. For example, my array is this:

vals = np.array([[0, 0],
[1, 0],
[2, 0],
[0, 1],
[1, 1],
[2, 1],
[0, 2],
[1, 2],
[2, 2],
[0, 3],
[1, 3],
[2, 3],
[0, 0],
[1, 0],
[2, 0],
[0, 1],
[1, 1],
[2, 1],
[0, 2],
[1, 2],
[2, 2],
[0, 3],
[1, 3],
[2, 3]])


I would like to get the index that matches the row [0, 1] which is index 3 and 15. When I do something like
numpy.where(vals == [0 ,1])
I get...

(array([ 0, 3, 3, 4, 5, 6, 9, 12, 15, 15, 16, 17, 18, 21]), array([0, 0, 1, 1, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0]))


I want index array([3, 15]).

Answer

You need the np.where function to get the indexes:

>>> np.where((vals == (0, 1)).all(axis=1))
(array([ 3, 15]),)

To dissassemble that:

>>> vals == (0, 1)
array([[ True, False],
       [False, False],
       ...
       [ True, False],
       [False, False],
       [False, False]], dtype=bool)

and calling the .all method on that array (with axis=1) gives you True where both are True:

>>> (vals == (0, 1)).all(axis=1)
array([False, False, False,  True, False, False, False, False, False,
       False, False, False, False, False, False,  True, False, False,
       False, False, False, False, False, False], dtype=bool)

and to get which indexes are True:

>>> np.where((vals == (0, 1)).all(axis=1))
(array([ 3, 15]),)

I find my solution a bit more readable, but as unutbu points out, the following may be faster, and returns the same value as (vals == (0, 1)).all(axis=1):

>>> (vals[:, 0] == 0) & (vals[:, 1] == 1)
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