Is there a "bounding box" function (slice with non-zero values) for a ndarray in NumPy?

This should do it:

from numpy import array, argwhereA = array([[0, 0, 0, 0, 0, 0, 0],           [0, 0, 0, 0, 0, 0, 0],           [0, 0, 1, 0, 0, 0, 0],           [0, 0, 1, 1, 0, 0, 0],           [0, 0, 0, 0, 1, 0, 0],           [0, 0, 0, 0, 0, 0, 0],           [0, 0, 0, 0, 0, 0, 0]])B = argwhere(A)(ystart, xstart), (ystop, xstop) = B.min(0), B.max(0) + 1 Atrim = A[ystart:ystop, xstart:xstop]

The code below, from this answer runs fastest in my tests:

def bbox2(img):    rows = np.any(img, axis=1)    cols = np.any(img, axis=0)    ymin, ymax = np.where(rows)[0][[0, -1]]    xmin, xmax = np.where(cols)[0][[0, -1]]    return img[ymin:ymax+1, xmin:xmax+1]

The accepted answer using argwhere worked but ran slower. My guess is, it's because argwhere allocates a giant output array of indices. I tested on a large 2D array (a 1024 x 1024 image, with roughly a 50x100 nonzero region).

Something like:

empty_cols = sp.all(array == 0, axis=0)empty_rows = sp.all(array == 0, axis=1)

The resulting arrays will be 1D boolian arrays. Loop on them from both ends to find the 'bounding box'.