Use numpy.argwhere to obtain the matching values in an np.array

Question

I\'d like to use np.argwhere() to obtain the values in an np.array.

For example:

z = np.arange(9).reshape(3,3)

[[0 1 2]
 [3 4          


        

Answer 1

Source for argwhere

def argwhere(a):
    """
    Find the indices of array elements that are non-zero, grouped by element.
    ...
    """
    return transpose(nonzero(a))

np.where is the same as np.nonzero.

In [902]: z=np.arange(9).reshape(3,3)
In [903]: z%3==0
Out[903]: 
array([[ True, False, False],
       [ True, False, False],
       [ True, False, False]], dtype=bool)
In [904]: np.nonzero(z%3==0)
Out[904]: (array([0, 1, 2], dtype=int32), array([0, 0, 0], dtype=int32))
In [905]: np.transpose(np.nonzero(z%3==0))
Out[905]: 
array([[0, 0],
       [1, 0],
       [2, 0]], dtype=int32)

In [906]: z[[0,1,2], [0,0,0]]
Out[906]: array([0, 3, 6])

z[np.nonzero(z%3==0)] is equivalent to using I,J as indexing arrays:

In [907]: I,J =np.nonzero(z%3==0)
In [908]: I
Out[908]: array([0, 1, 2], dtype=int32)
In [909]: J
Out[909]: array([0, 0, 0], dtype=int32)
In [910]: z[I,J]
Out[910]: array([0, 3, 6])