bounding box of numpy array

You can roughly halve the execution time by using np.any to reduce the rows and columns that contain non-zero values to 1D vectors, rather than finding the indices of all non-zero values using np.where:

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

Some benchmarks:

%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

Extending this approach to the 3D case just involves performing the reduction along each pair of axes:

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

It’s easy to generalize this to N dimensions by using itertools.combinations to iterate over each unique combination of axes to perform the reduction over:

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)

---

If you know the coordinates of the corners of the original bounding box, the angle of rotation, and the centre of rotation, you could get the coordinates of the transformed bounding box corners directly by computing the corresponding affine transformation matrix and dotting it with the input coordinates:

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

This works out to be very slightly faster than using np.any for your small example array:

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

However, since the speed of this method is independent of the size of the input array, it can be quite a lot faster for larger arrays.

Extending the transformation approach to 3D is slightly more complicated, in that the rotation now has three different components (one about the x-axis, one about the y-axis and one about the z-axis), but the basic method is the same:

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

I’ve essentially just modified the function above using the rotation matrix expressions from here – I haven’t had time to write a test-case yet, so use with caution.