Find matching rows in 2 dimensional numpy array

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

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

Or, as the documentation states:

If only condition is given, return condition.nonzero()

You could directly call .nonzero() on the array returned by .all:

>>> (vals == (0, 1)).all(axis=1).nonzero()
(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]),)

or

>>> (vals == (0, 1)).all(axis=1).nonzero()
(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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