Get indices of intersecting rows of Numpy 2d Array

Question

I want to get the indices of the intersecting rows of a main numpy 2d array A, with another one B.

A=array([[1, 2],
         [3, 4],
         [5, 6],
                


        

Answer 1

With minimal changes, you can get your approach to work:

In [15]: A
Out[15]: 
array([[ 1,  2],
       [ 3,  4],
       [ 5,  6],
       [ 7,  8],
       [ 9, 10]])

In [16]: B
Out[16]: 
array([[1, 4],
       [1, 2],
       [5, 6],
       [6, 3]])

In [17]: np.in1d(A.view('i,i').reshape(-1), B.view('i,i').reshape(-1))
Out[17]: array([ True, False,  True, False, False], dtype=bool)

In [18]: np.nonzero(np.in1d(A.view('i,i').reshape(-1), B.view('i,i').reshape(-1)))
Out[18]: (array([0, 2], dtype=int64),)

In [19]: np.nonzero(np.in1d(A.view('i,i').reshape(-1), B.view('i,i').reshape(-1)))[0]
Out[19]: array([0, 2], dtype=int64)

If your arrays are not floats, and are both contiguous, then the following will be faster:

In [21]: dt = np.dtype((np.void, A.dtype.itemsize * A.shape[1]))

In [22]: np.nonzero(np.in1d(A.view(dt).reshape(-1), B.view(dt).reshape(-1)))[0]
Out[22]: array([0, 2], dtype=int64)

And a quick timing:

In [24]: %timeit np.nonzero(np.in1d(A.view('i,i').reshape(-1), B.view('i,i').reshape(-1)))[0]
10000 loops, best of 3: 75 µs per loop

In [25]: %timeit np.nonzero(np.in1d(A.view(dt).reshape(-1), B.view(dt).reshape(-1)))[0]
10000 loops, best of 3: 29.8 µs per loop

Answer 2

You can use np.char.array() objects to do this comparison using np.in1d():

s1 = np.char.array(A[:,0]) + '-' + np.char.array(A[:,1])
s2 = np.char.array(B[:,0]) + '-' + np.char.array(B[:,1])

np.where(np.in1d(s1, s2))[0]
#array([0, 2], dtype=int64)

NOTE: A and B must be of the same data type (int, float, etc) for this to work.