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

开发者 https://www.devze.com 2023-02-06 17:48 出处:网络
I am dealing with arrays created via numpy.array(), and I need to draw points on a canvas simulating an image. Since there is a lot of zero values around the central part of the array which contains t

I am dealing with arrays created via numpy.array(), and I need to draw points on a canvas simulating an image. Since there is a lot of zero values around the central part of the array which contains the meaningful data, I would like to "trim" the array, erasing columns that only contain zeros and rows that only contain zeros.

So, I would like to know of some native numpy function or even a code snippet to "trim" or find a "bounding box" to slice only the data-containing part of the array.

(since it is a conceptual ques开发者_开发问答tion, I did not put any code, sorry if I should, I'm very fresh to posting at SO.)

Thanks for reading


This should do it:

from numpy import array, argwhere

A = 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'.

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