Numpy Routine（s）在2d数组中创建一个常规网格

Question

I am trying to write a function that would create a regular grid of 5 pixels by 5 pixels inside a 2d array. I was hoping some combination of numpy.arange and numpy.repeat might do it, but so far I haven't had any luck because numpy.repeat will just repeat along the same row.

Here is an example:

Let's say I want a 5x5 grid inside a 2d array of shape (20, 15) . It should look like:

array([[ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
       [ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
       [ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
       [ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
       [ 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2],
       [ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
       [ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
       [ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
       [ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
       [ 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5],
       [ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
       [ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
       [ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
       [ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
       [ 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8],
       [ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11],
       [ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11],
       [ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11],
       [ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11],
       [ 9, 9, 9, 9, 9,10,10,10,10,10,11,11,11,11,11]])

I realize I could simply use a loop and slicing to accomplish this, but I could be applying this to very large arrays and I worry that the performance of that would be too slow or impractical.

Can anyone recommend a method to accomplish this?

Thanks in advance.

UPDATE :

All the answers provided seem to work well. Can anyone tell me which will be the most efficient to use for large arrays? By large array I mean it could be 100000 x 100000 or more with 15 x 15 grid cell sizes.

Answer 1

Broadcasting is the answer here:

m, n, d = 20, 15, 5
arr = np.empty((m, n), dtype=np.int)
arr_view = arr.reshape(m // d, d, n // d, d)
vals = np.arange(m // d * n // d).reshape(m // d, 1, n // d, 1)
arr_view[:] = vals

>>> arr
array([[ 0,  0,  0,  0,  0,  1,  1,  1,  1,  1,  2,  2,  2,  2,  2],
       [ 0,  0,  0,  0,  0,  1,  1,  1,  1,  1,  2,  2,  2,  2,  2],
       [ 0,  0,  0,  0,  0,  1,  1,  1,  1,  1,  2,  2,  2,  2,  2],
       [ 0,  0,  0,  0,  0,  1,  1,  1,  1,  1,  2,  2,  2,  2,  2],
       [ 0,  0,  0,  0,  0,  1,  1,  1,  1,  1,  2,  2,  2,  2,  2],
       [ 3,  3,  3,  3,  3,  4,  4,  4,  4,  4,  5,  5,  5,  5,  5],
       [ 3,  3,  3,  3,  3,  4,  4,  4,  4,  4,  5,  5,  5,  5,  5],
       [ 3,  3,  3,  3,  3,  4,  4,  4,  4,  4,  5,  5,  5,  5,  5],
       [ 3,  3,  3,  3,  3,  4,  4,  4,  4,  4,  5,  5,  5,  5,  5],
       [ 3,  3,  3,  3,  3,  4,  4,  4,  4,  4,  5,  5,  5,  5,  5],
       [ 6,  6,  6,  6,  6,  7,  7,  7,  7,  7,  8,  8,  8,  8,  8],
       [ 6,  6,  6,  6,  6,  7,  7,  7,  7,  7,  8,  8,  8,  8,  8],
       [ 6,  6,  6,  6,  6,  7,  7,  7,  7,  7,  8,  8,  8,  8,  8],
       [ 6,  6,  6,  6,  6,  7,  7,  7,  7,  7,  8,  8,  8,  8,  8],
       [ 6,  6,  6,  6,  6,  7,  7,  7,  7,  7,  8,  8,  8,  8,  8],
       [ 9,  9,  9,  9,  9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11],
       [ 9,  9,  9,  9,  9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11],
       [ 9,  9,  9,  9,  9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11],
       [ 9,  9,  9,  9,  9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11],
       [ 9,  9,  9,  9,  9, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11]])

Answer 2

与Jaime的答案类似：

np.repeat(np.arange(0, 10, 3), 4)[..., None] + np.repeat(np.arange(3), 5)[None, ...]

Answer 3

kron will do this expansion (as Brionius also suggested in the comments):

xi, xj, ni, nj = 5, 5, 4, 3
r = np.kron(np.arange(ni*nj).reshape((ni,nj)), np.ones((xi, xj)))

Although I haven't tested it, I assume it's less efficient than the broadcasting approach, but a bit more concise and easier to understand (I hope). It's likely less efficient because: 1) it requires the array of ones, 2) it does xi*xj multiplications by 1, and 3) it does a bunch of concats.