bounding box of numpy array

Question

Suppose you have a 2D numpy array with some random values and surrounding zeros.

Example "tilted rectangle":

import numpy as np from skimage import transform  img1 = np.zeros((100,100)) img1[25:75,25:75] = 1. img2 = transform.rotate(img1, 45) 

Now I want to find the smallest bounding rectangle for all the nonzero data. For example:

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] 

What would be the fastest way to achieve this result? I am sure there is a better way since the np.where function takes quite a time if I am e.g. using 1000x1000 data sets.

Edit: Should also work in 3D...

Answer 1

Here is an algorithm to calculate the bounding box for N dimensional arrays,

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 

which can be used with 2D, 3D, etc arrays as follows,

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]] 

Although replacing the np.diff and np.nonzero calls by one np.where might be better.

Answer 2

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.