numpy数组的边界框

问题描述 投票:21回答:3

假设您有一个2D numpy数组,其中包含一些随机值和周围的零。

示例“倾斜矩形”:

import numpy as np
from skimage import transform

img1 = np.zeros((100,100))
img1[25:75,25:75] = 1.
img2 = transform.rotate(img1, 45)

现在我想找到所有非零数据的最小边界矩形。例如:

a = np.where(img2 != 0)
bbox = img2[np.min(a[0]):np.max(a[0])+1, np.min(a[1]):np.max(a[1])+1]

实现这一结果的最快方法是什么?我确信有更好的方法,因为np.where函数需要相当长的时间,如果我是使用1000x1000数据集。

编辑:还应该在3D中工作......

python arrays numpy transformation
3个回答
43
投票

通过使用np.any将包含非零值的行和列减少到1D向量,而不是使用np.where查找所有非零值的索引,可以大致将执行时间减半:

def bbox1(img):
    a = np.where(img != 0)
    bbox = np.min(a[0]), np.max(a[0]), np.min(a[1]), np.max(a[1])
    return bbox

def bbox2(img):
    rows = np.any(img, axis=1)
    cols = np.any(img, axis=0)
    rmin, rmax = np.where(rows)[0][[0, -1]]
    cmin, cmax = np.where(cols)[0][[0, -1]]

    return rmin, rmax, cmin, cmax

一些基准:

%timeit bbox1(img2)
10000 loops, best of 3: 63.5 µs per loop

%timeit bbox2(img2)
10000 loops, best of 3: 37.1 µs per loop

将此方法扩展到3D案例只需沿每对轴执行缩减:

def bbox2_3D(img):

    r = np.any(img, axis=(1, 2))
    c = np.any(img, axis=(0, 2))
    z = np.any(img, axis=(0, 1))

    rmin, rmax = np.where(r)[0][[0, -1]]
    cmin, cmax = np.where(c)[0][[0, -1]]
    zmin, zmax = np.where(z)[0][[0, -1]]

    return rmin, rmax, cmin, cmax, zmin, zmax

通过使用itertools.combinations迭代每个独特的轴组合来执行减少,可以很容易地将其推广到N维:

import itertools

def bbox2_ND(img):
    N = img.ndim
    out = []
    for ax in itertools.combinations(reversed(range(N)), N - 1):
        nonzero = np.any(img, axis=ax)
        out.extend(np.where(nonzero)[0][[0, -1]])
    return tuple(out)

如果您知道原始边界框的角坐标,旋转角度和旋转中心,则可以通过计算相应的affine transformation matrix并使用输入坐标对其进行点缀来直接获取变换的边界框角的坐标:

def bbox_rotate(bbox_in, angle, centre):

    rmin, rmax, cmin, cmax = bbox_in

    # bounding box corners in homogeneous coordinates
    xyz_in = np.array(([[cmin, cmin, cmax, cmax],
                        [rmin, rmax, rmin, rmax],
                        [   1,    1,    1,    1]]))

    # translate centre to origin
    cr, cc = centre
    cent2ori = np.eye(3)
    cent2ori[:2, 2] = -cr, -cc

    # rotate about the origin
    theta = np.deg2rad(angle)
    rmat = np.eye(3)
    rmat[:2, :2] = np.array([[ np.cos(theta),-np.sin(theta)],
                             [ np.sin(theta), np.cos(theta)]])

    # translate from origin back to centre
    ori2cent = np.eye(3)
    ori2cent[:2, 2] = cr, cc

    # combine transformations (rightmost matrix is applied first)
    xyz_out = ori2cent.dot(rmat).dot(cent2ori).dot(xyz_in)

    r, c = xyz_out[:2]

    rmin = int(r.min())
    rmax = int(r.max())
    cmin = int(c.min())
    cmax = int(c.max())

    return rmin, rmax, cmin, cmax

这比使用np.any作为小示例数组要快得多:

%timeit bbox_rotate([25, 75, 25, 75], 45, (50, 50))
10000 loops, best of 3: 33 µs per loop

但是,由于此方法的速度与输入数组的大小无关,因此对于较大的数组,它可以快得多。

将转换方法扩展到3D稍微复杂一点,因为旋转现在有三个不同的组件(一个围绕x轴,一个围绕y轴,一个围绕z轴),但基本方法是相同的:

def bbox_rotate_3d(bbox_in, angle_x, angle_y, angle_z, centre):

    rmin, rmax, cmin, cmax, zmin, zmax = bbox_in

    # bounding box corners in homogeneous coordinates
    xyzu_in = np.array(([[cmin, cmin, cmin, cmin, cmax, cmax, cmax, cmax],
                         [rmin, rmin, rmax, rmax, rmin, rmin, rmax, rmax],
                         [zmin, zmax, zmin, zmax, zmin, zmax, zmin, zmax],
                         [   1,    1,    1,    1,    1,    1,    1,    1]]))

    # translate centre to origin
    cr, cc, cz = centre
    cent2ori = np.eye(4)
    cent2ori[:3, 3] = -cr, -cc -cz

    # rotation about the x-axis
    theta = np.deg2rad(angle_x)
    rmat_x = np.eye(4)
    rmat_x[1:3, 1:3] = np.array([[ np.cos(theta),-np.sin(theta)],
                                 [ np.sin(theta), np.cos(theta)]])

    # rotation about the y-axis
    theta = np.deg2rad(angle_y)
    rmat_y = np.eye(4)
    rmat_y[[0, 0, 2, 2], [0, 2, 0, 2]] = (
        np.cos(theta), np.sin(theta), -np.sin(theta), np.cos(theta))

    # rotation about the z-axis
    theta = np.deg2rad(angle_z)
    rmat_z = np.eye(4)
    rmat_z[:2, :2] = np.array([[ np.cos(theta),-np.sin(theta)],
                               [ np.sin(theta), np.cos(theta)]])

    # translate from origin back to centre
    ori2cent = np.eye(4)
    ori2cent[:3, 3] = cr, cc, cz

    # combine transformations (rightmost matrix is applied first)
    tform = ori2cent.dot(rmat_z).dot(rmat_y).dot(rmat_x).dot(cent2ori)
    xyzu_out = tform.dot(xyzu_in)

    r, c, z = xyzu_out[:3]

    rmin = int(r.min())
    rmax = int(r.max())
    cmin = int(c.min())
    cmax = int(c.max())
    zmin = int(z.min())
    zmax = int(z.max())

    return rmin, rmax, cmin, cmax, zmin, zmax

我基本上只使用here中的旋转矩阵表达式修改了上面的函数 - 我还没有时间编写测试用例,所以请谨慎使用。


4
投票

这是一个计算N维数组边界框的算法,

def get_bounding_box(x):
    """ Calculates the bounding box of a ndarray"""
    mask = x == 0
    bbox = []
    all_axis = np.arange(x.ndim)
    for kdim in all_axis:
        nk_dim = np.delete(all_axis, kdim)
        mask_i = mask.all(axis=tuple(nk_dim))
        dmask_i = np.diff(mask_i)
        idx_i = np.nonzero(dmask_i)[0]
        if len(idx_i) != 2:
            raise ValueError('Algorithm failed, {} does not have 2 elements!'.format(idx_i))
        bbox.append(slice(idx_i[0]+1, idx_i[1]+1))
    return bbox

可用于2D,3D等阵列,如下所示,

In [1]: print((img2!=0).astype(int))
   ...: bbox = get_bounding_box(img2)
   ...: print((img2[bbox]!=0).astype(int))
   ...: 
[[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0]
 [0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 0 0 0 0 0]
 [0 0 0 0 0 0 1 1 1 1 1 1 1 1 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 1 1 1 1 1 1 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
[[0 0 0 0 0 0 1 1 0 0 0 0 0 0]
 [0 0 0 0 0 1 1 1 1 0 0 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 1 1 1 1 1 1 1 1 1 1 0 0]
 [0 1 1 1 1 1 1 1 1 1 1 1 1 0]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [1 1 1 1 1 1 1 1 1 1 1 1 1 1]
 [0 1 1 1 1 1 1 1 1 1 1 1 1 0]
 [0 0 1 1 1 1 1 1 1 1 1 1 0 0]
 [0 0 0 1 1 1 1 1 1 1 1 0 0 0]
 [0 0 0 0 1 1 1 1 1 1 0 0 0 0]
 [0 0 0 0 0 1 1 1 1 0 0 0 0 0]
 [0 0 0 0 0 0 1 1 0 0 0 0 0 0]]

虽然用一个np.diff取代np.nonzeronp.where电话可能会更好。


0
投票

通过用np.where替换np.argmax并使用布尔掩码,我能够挤出更多的性能。

def bbox(img):
    img = (img > 0)
    rows = np.any(img, axis=1)
    cols = np.any(img, axis=0)
    rmin, rmax = np.argmax(rows), img.shape[0] - 1 - np.argmax(np.flipud(rows))
    cmin, cmax = np.argmax(cols), img.shape[1] - 1 - np.argmax(np.flipud(cols))
    return rmin, rmax, cmin, cmax

对于我来说,这比同一基准测试中的bbox2解决方案快10μs。还应该有一种方法可以使用argmax的结果来找到非零的行和列,避免使用np.any完成额外的搜索,但这可能需要一些棘手的索引,我无法有效地使用简单的矢量化代码。

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