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...
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:
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(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:
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.
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.
I was able to squeeze out a little more performance by replacing np.where with np.argmax and working on a boolean mask.
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
np.any, but this may require some tricky indexing that I wasn't able to get working efficiently with simple vectorized code.
