Question

我有来自多个点云文件的数百万个xyz坐标，我将其存储在一个二维的numpy数组中：[[x1, y1, z1], [x2, y2, z2],..., [xn, yn, zn]]。

我想过滤4个坐标[[x1, y1], [x2, y2]]描述的特定边界框内的所有点，即矩形的左下和右上坐标。

我已经找到了以下代码来过滤numpy的坐标，它几乎就是我想要的。唯一的区别是（如果我说得对）我的二维数组也有z坐标。

import random
import numpy as np

points = [(random.random(), random.random()) for i in range(100)]

bx1, bx2 = sorted([random.random(), random.random()])
by1, by2 = sorted([random.random(), random.random()])

pts = np.array(points)
ll = np.array([bx1, by1])  # lower-left
ur = np.array([bx2, by2])  # upper-right

inidx = np.all(np.logical_and(ll <= pts, pts <= ur), axis=1)
inbox = pts[inidx]
outbox = pts[np.logical_not(inidx)]

我如何修改上面的代码，使其与xyz坐标一起使用，用两个xy坐标描述的边界框进行过滤？

Answer 1

选择点的X和Y坐标：

xy_pts = pts[:,[0,1]]

现在，只需在比较中使用xy_pts代替pts：

inidx = np.all((ll <= xy_pts) & (xy_pts <= ur), axis=1)

Answer 2

我正在写一个Python library for working with point clouds我有这个功能，我认为这应该适合你：

def bounding_box(points, min_x=-np.inf, max_x=np.inf, min_y=-np.inf,
                        max_y=np.inf, min_z=-np.inf, max_z=np.inf):
    """ Compute a bounding_box filter on the given points

    Parameters
    ----------                        
    points: (n,3) array
        The array containing all the points's coordinates. Expected format:
            array([
                [x1,y1,z1],
                ...,
                [xn,yn,zn]])

    min_i, max_i: float
        The bounding box limits for each coordinate. If some limits are missing,
        the default values are -infinite for the min_i and infinite for the max_i.

    Returns
    -------
    bb_filter : boolean array
        The boolean mask indicating wherever a point should be keeped or not.
        The size of the boolean mask will be the same as the number of given points.

    """

    bound_x = np.logical_and(points[:, 0] > min_x, points[:, 0] < max_x)
    bound_y = np.logical_and(points[:, 1] > min_y, points[:, 1] < max_y)
    bound_z = np.logical_and(points[:, 2] > min_z, points[:, 2] < max_z)

    bb_filter = np.logical_and(bound_x, bound_y, bound_z)

    return bb_filter

以下是您要问的一个示例：

1000万点：

points = np.random.rand(10000000, 3)

您指定格式的矩形：

rectangle = np.array([[0.2, 0.2],
                     [0.4, 0.4]])

打开矩形包装：

min_x = rectangle[:,0].min()
max_x = rectangle[:,0].max()
min_y = rectangle[:,1].min()
max_y = rectangle[:,1].max()

在框内获取布尔数组标记点：

%%timeit
inside_box = bounding_box(points, min_x=min_x, max_x=max_x, min_y=min_y, max_y=max_y)
1 loop, best of 3: 247 ms per loop

这样您可以按如下方式使用数组：

points_inside_box = points[inside_box]
points_outside_box = points[~inside_box]

用numpy在一个边界框内找到点

2 个答案: