两个矩形与 NumPy 的交集

Question

I have the following function to find the intersection of two rectangles. It's a little bit slow, I don't know if it's due to the OR condition or the >, < operators. I wonder if there's a way to improve the performance of the is_intersect() function. Maybe with NumPy? Or Cython?

import numpy as np

def is_intersect(rect1, rect2):
    xmin1, xmax1, ymin1, ymax1 = rect1
    xmin2, xmax2, ymin2, ymax2 = rect2
    if xmin1 > xmax2 or xmax1 < xmin2:
        return False
    if ymin1 > ymax2 or ymax1 < ymax2:
        return False
    return True

N_ELEMS = 100000000
rects1 = np.random.rand(N_ELEMS,4)
rects2 = np.random.rand(N_ELEMS,4)

temp_dct = dict()

for i in range(N_ELEMS):
    rect1 = rects1[i,:]
    rect2 = rects2[i,:]
    if is_intersect(rect1, rect2):
        temp_dct[i] = True

I can't profit from caching results as the points will be incremental, that is, one rectangle will move in space (never the same place). In this example, I used NumPy's random() function, but that's not the case for my real use. I will call the is_intersect() function 100 000 000 times or more.

Answer 1

You can improve performance by avoiding the for loop using vectorized comparison and np.any :

result = (1 - np.any([rects1[:,0] > rects2[:,1], 
                      rects1[:,1] < rects2[:,0], 
                      rects1[:,2] > rects2[:,3], 
                      rects1[:,3] < rects2[:,2]], 
                     axis=0)).astype(bool)

You don't get a dictionary, yet you can access result by index.

Performance with 100M elements:

import numpy as np
import timeit

N_ELEMS = 100_000_000
rects1 = np.random.rand(N_ELEMS,4)
rects2 = np.random.rand(N_ELEMS,4)

start_time = timeit.default_timer()
result = (1 - np.any([rects1[:,0] > rects2[:,1], 
                      rects1[:,1] < rects2[:,0], 
                      rects1[:,2] > rects2[:,3], 
                      rects1[:,3] < rects2[:,2]], 
                     axis=0)).astype(bool)

print(timeit.default_timer() - start_time)
2.9162093999999996