Intersection of 2-d numpy arrays

Question

I am looking for a way to get the intersection between two 2-dimensional numpy.array of shape (n_1, m) and (n_2, m). Note that n_1 and n_2 can differ but m is the same for both arrays. Here are two minimal examples with the expected results:

import numpy as np array1a = np.array([[2], [2], [5], [1]]) array1b = np.array([[5], [2]]) array_intersect(array1a, array1b) ## array([[2], ## [5]]) array2a = np.array([[1, 2], [3, 3], [2, 1], [1, 3], [2, 1]]) array2b = np.array([[2, 1], [1, 4], [3, 3]]) array_intersect(array2a, array2b) ## array([[2, 1], ## [3, 3]])

If someone have a clue on how I should implement the array_intersect function, I would be very grateful!

Sorry, I can't understand, how exactly you define the intersection. Is it a matrix consisting of all the rows of the second matrix present in the first one? Or vice versa? — aparpara
– aparpara, Commented Apr 15, 2019 at 20:12
I'm sorry if I was not clear! I just want that any row that exist in the two arrays to be returned. — J.P. Le Cavalier
– J.P. Le Cavalier, Commented Apr 15, 2019 at 20:14

cvanelteren · Accepted Answer · 2019-04-15 20:19:27Z

2

How about using sets?

import numpy as np array2a = np.array([[1, 2], [3, 3], [2, 1], [1, 3], [2, 1]]) array2b = np.array([[2, 1], [1, 4], [3, 3]]) a = set((tuple(i) for i in array2a)) b = set((tuple(i) for i in array2b)) a.intersection(b) # {(2, 1), (3, 3)}

answered Apr 15, 2019 at 20:19

cvanelteren

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2 Comments

J.P. Le Cavalier Over a year ago

This is my favorite one since it's easier to read. I implemented it in a function where I return np.array(list(a.intersection(b))) since I want the output to be a numpy.array. Thanks for your help!

Eelco Hoogendoorn Over a year ago

This may be acceptable in terms of performance for small arrays; but note that this foregoes numpy-like performance, and involves the creation of many seperate python objects with their associated overhead.

cvanelteren · Accepted Answer · 2019-04-15 20:36:42Z

Another approach would be to harness the broadcasting feature

import numpy as np array2a = np.array([[1, 2], [3, 3], [2, 1], [1, 3], [2, 1]]) array2b = np.array([[2, 1], [1, 4], [3, 3]]) test = array2a[:, None] == array2b print(array2b[np.all(test.mean(0) > 0, axis = 1)]) # [[2 1] # [3 3]]

but this is less readable imo. [edit]: or use the unique and set combination. In short, there are many options!

subnivean · Accepted Answer · 2019-04-15 20:46:06Z

Here's a way to do without any loops or list comprehensions, assuming you have scipy installed (I haven't tested for speed):

In [31]: from scipy.spatial.distance import cdist In [32]: np.unique(array1a[np.where(cdist(array1a, array1b) == 0)[0]], axis=0) Out[32]: array([[2], [5]]) In [33]: np.unique(array2a[np.where(cdist(array2a, array2b) == 0)[0]], axis=0) Out[33]: array([[2, 1], [3, 3]])

aparpara · Accepted Answer · 2019-04-15 20:52:58Z

Construct a set of tuples from the first array and test each line of the second array. Or vice versa.

def array_intersect(a, b): s = {tuple(x) for x in a} return np.unique([x for x in b if tuple(x) in s], axis=0)

Eelco Hoogendoorn · Accepted Answer · 2019-04-15 21:18:43Z

The numpy-indexed package (disclaimer: I am its author) was created with the exact purpose of providing such functionality in an expressive and efficient manner:

import numpy_indexed as npi npi.intersect(a, b)

Note that the implementation is fully vectorized; that is no loops over the arrays in python.

gives me the module 'numpy_indexed' has no attribute 'intersect' error

yazan sayed · Accepted Answer · 2022-09-09 11:03:50Z

arr1 = np.arange(20000).reshape(-1,2) arr2 = arr1.copy() np.random.shuffle(arr2) print(len(arr1)) #10000

%%timeit res= np.array([x for x in set(tuple(x) for x in arr1) & set(tuple(x) for x in arr2) ])

83.7 ms ± 16.1 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

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