Repeat each element of a matrix in as a block into a new matrix

user_1951730 0
Viewed 541

In Matlab, there is the command repelem which works as follows (see https://www.mathworks.com/help/matlab/ref/repelem.html#buocbhj-2):

e.g.: Create a matrix and repeat each element into a 3-by-2 block of a new matrix.

A = [1 2; 3 4]

B = repelem(A,3,2)

A = (2×2)
     1     2
     3     4

B = (6×4)
     1     1     2     2
     1     1     2     2
     1     1     2     2
     3     3     4     4
     3     3     4     4
     3     3     4     4

What is the best way to do the same in Numpy?

A = np.arange(1,5).reshape((2,2))
B = ...
matlab numpy python
2 Answers

You can chain np.repeat specifying axis 

repelem = lambda a, x, y: np.repeat(np.repeat(a, x, axis=0), y, axis=1)
# same as repelem  = lambda a, x, y: a.repeat(x, 0).repeat(y, 1)

Just call

>>> a = np.array([[1,2], [3,4]])
>>> repelem(a, 3, 2)

array([[1, 1, 2, 2],
       [1, 1, 2, 2],
       [1, 1, 2, 2],
       [3, 3, 4, 4],
       [3, 3, 4, 4],
       [3, 3, 4, 4]])
user_25356110

Since you aren't changing any of the elements, numpy.broadcast_to is a perfect solution here. It allows us to return a view matching the desired number of repeats, in a very fast (and memory efficient) way:

def broadcast_tile(arr, h, w):
    x, y = a.shape
    m, n = x * h, y * w
    return np.broadcast_to(
        a.reshape(x, m//(h*x), y, n//(w*y)), (m//h, h, n//w, w)
    ).reshape(m, n)

broadcast_tile(a, 3, 2)


array([[1, 1, 2, 2],
       [1, 1, 2, 2],
       [1, 1, 2, 2],
       [3, 3, 4, 4],
       [3, 3, 4, 4],
       [3, 3, 4, 4]])

Timings

In [425]: %timeit repelem(a, 200, 200)
770 µs ± 35.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [426]: %timeit broadcast_tile(a, 200, 200)
57.5 µs ± 2.27 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
user_34832030
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