Find different pair in a list by python

Question

I have a list and I want to find different pair in list. I implement a function --> different()

import numpy as np


def different(array):
    res = []
    for (x1, y1), (x2, y2) in array:
        if (x1, y1) != (x2, y2):
            res.append([(x1, y1), (x2, y2)])
    return res


a = np.array([[[1, 2], [3, 4]],
              [[1, 2], [1, 2]],
              [[7, 9], [6, 3]],
              [[3, 3], [3, 3]]])

out = different(a)  # get [[(1, 2), (3, 4)],
                    #      [(7, 9), (6, 3)]]

Is there any other better way to do it? I want to improve my function different. List size may be greater than 100,000.

Answer 1

The numpy way to do it is

import numpy as np

a = np.array([[[1, 2], [3, 4]],
              [[1, 2], [1, 2]],
              [[7, 9], [6, 3]],
              [[3, 3], [3, 3]]])

b = np.logical_or(a[:,0,0] != a[:,1,0],  a[:,0,1] != a[:,1,1])

print(a[b])

Answer 2

Vectorized Comparison

a[~(a[:, 0] == a[:, 1]).all(1)]

array([[[1, 2],
        [3, 4]],

       [[7, 9],
        [6, 3]]])

This works by taking the first pair of each subarray and comparing each one with the second pair. All subarrays for which entries which are not identical only are selected. Consider,

a[:, 0] == a[:, 1]

array([[False, False],
       [ True,  True],
       [False, False],
       [ True,  True]])

From this, we want those rows which do not have True at each column. So, on this result, use all and then negate the result.

~(a[:, 0] == a[:, 1]).all(1)
array([ True, False,  True, False])

This gives you a mask you can then use to select subarrays from a.

---

np.logical_or.reduce

Similar to the first option above, but approaches this problem from the other end (see DeMorgan's Law).

a[np.logical_or.reduce(a[:, 0] != a[:, 1], axis=1)]