通过使用
np.any将包含非零值的行和列减少到一维向量，您可以大致减少执行时间，而无需使用
np.where以下命令查找所有非零值的索引：
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 bboxdef 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
将这种方法扩展到3D情况仅涉及沿每对轴执行缩减：
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
通过迭代每个轴的唯一组合来执行归约，很容易将其概括为 N个 维度
itertools.combinations：
import itertoolsdef 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角的坐标，旋转角度和旋转中心，则可以通过计算相应的仿射变换矩阵并将其与输入点相乘，直接获得已变换边界框角的坐标。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
与使用
np.any小型示例数组相比，这要快得多：
%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，它的速度可能要快得多。
将变换方法扩展到3D会稍微复杂一点，因为旋转现在具有三个不同的分量（一个绕x轴，一个绕y轴，一个绕z轴），但是基本方法是相同的：
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
我从这里基本上使用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，因此请谨慎使用。
