如何有效地找到一组点的边界框？

Question

I have several points stored in an array. I need to find bounds of that points ie. the rectangle which bounds all the points. I know how to solve this in plain Python.

I would like to know is there a better way than the naive max, min over the array or built-in method to solve the problem.

points = [[1, 3], [2, 4], [4, 1], [3, 3], [1, 6]]
b = bounds(points) # the function I am looking for
# now b = [[1, 1], [4, 6]]

Answer 1

My approach to getting performance is to push things down to C level whenever possible:

def bounding_box(points):
    x_coordinates, y_coordinates = zip(*points)

    return [(min(x_coordinates), min(y_coordinates)), (max(x_coordinates), max(y_coordinates))]

By my (crude) measure, this runs about 1.5 times faster than @ReblochonMasque's bounding_box_naive() . And is clearly more elegant. ;-)

Answer 2

You cannot do better than O(n) , because you must traverse all the points to determine the max and min for x and y .

But, you can reduce the constant factor, and traverse the list only once; however, it is unclear if that would give you a better execution time, and if it does, it would be for large collections of points.

[EDIT]: in fact it does not, the "naive" approach is the most efficient.

Here is the "naive" approach: (it is the fastest of the two)

def bounding_box_naive(points):
    """returns a list containing the bottom left and the top right 
    points in the sequence
    Here, we use min and max four times over the collection of points
    """
    bot_left_x = min(point[0] for point in points)
    bot_left_y = min(point[1] for point in points)
    top_right_x = max(point[0] for point in points)
    top_right_y = max(point[1] for point in points)

    return [(bot_left_x, bot_left_y), (top_right_x, top_right_y)]

and the (maybe?) less naive:

def bounding_box(points):
    """returns a list containing the bottom left and the top right 
    points in the sequence
    Here, we traverse the collection of points only once, 
    to find the min and max for x and y
    """
    bot_left_x, bot_left_y = float('inf'), float('inf')
    top_right_x, top_right_y = float('-inf'), float('-inf')
    for x, y in points:
        bot_left_x = min(bot_left_x, x)
        bot_left_y = min(bot_left_y, y)
        top_right_x = max(top_right_x, x)
        top_right_y = max(top_right_y, y)

    return [(bot_left_x, bot_left_y), (top_right_x, top_right_y)]

profiling results:

import random
points = [(random.randrange(-1000, 1000), random.randrange(-1000, 1000))  for _ in range(1000000)]

%timeit bounding_box_naive(points)
%timeit bounding_box(points)

size = 1,000 points

1000 loops, best of 3: 573 µs per loop
1000 loops, best of 3: 1.46 ms per loop

size = 10,000 points

100 loops, best of 3: 5.7 ms per loop
100 loops, best of 3: 14.7 ms per loop

size 100,000 points

10 loops, best of 3: 66.8 ms per loop
10 loops, best of 3: 141 ms per loop

size 1,000,000 points

1 loop, best of 3: 664 ms per loop
1 loop, best of 3: 1.47 s per loop

Clearly, the first "not so naive" approach is faster by a factor 2.5 - 3