Get intersecting rows across two 2D numpy arrays

Question

I want to get the intersecting (common) rows across two 2D numpy arrays. E.g., if the following arrays are passed as inputs:

array([[1, 4],        [2, 5],        [3, 6]])  array([[1, 4],        [3, 6],        [7, 8]]) 

the output should be:

array([[1, 4],        [3, 6]) 

I know how to do this with loops. I'm looking at a Pythonic/Numpy way to do this.

Answer 1

For short arrays, using sets is probably the clearest and most readable way to do it.

Another way is to use numpy.intersect1d. You'll have to trick it into treating the rows as a single value, though... This makes things a bit less readable...

import numpy as np  A = np.array([[1,4],[2,5],[3,6]]) B = np.array([[1,4],[3,6],[7,8]])  nrows, ncols = A.shape dtype={'names':['f{}'.format(i) for i in range(ncols)],        'formats':ncols * [A.dtype]}  C = np.intersect1d(A.view(dtype), B.view(dtype))  # This last bit is optional if you're okay with "C" being a structured array... C = C.view(A.dtype).reshape(-1, ncols) 

For large arrays, this should be considerably faster than using sets.

Answer 2

You could use Python's sets:

>>> import numpy as np >>> A = np.array([[1,4],[2,5],[3,6]]) >>> B = np.array([[1,4],[3,6],[7,8]]) >>> aset = set([tuple(x) for x in A]) >>> bset = set([tuple(x) for x in B]) >>> np.array([x for x in aset & bset]) array([[1, 4],        [3, 6]]) 

As Rob Cowie points out, this can be done more concisely as

np.array([x for x in set(tuple(x) for x in A) & set(tuple(x) for x in B)]) 

There's probably a way to do this without all the going back and forth from arrays to tuples, but it's not coming to me right now.