I want to achieve this:
a = np.array ([[1, 2],
[2, 1]])
b = np.array ([[0.25, 0.25, 0.5, 0.5],
[0.25, 0.25, 0.5, 0.5],
[0.5, 0.5, 0.25, 0.25],
[0.5, 0.5, 0.25, 0.25])
Mathematically they r not the same matrices. But i think you get the idea what i want to do. I want to double the dimension of a matrix. But therefore i want to keep the information from the initial matrix a by take the quarter for the four corresponding cells.
Does some1 knows how to do this efficiently in numpy?
Here's one with np.broadcast_to that leverages broadcasting to avoid two stages of replications or tiling for performance benefits by doing it one -
# "Expand" array a by Height, H and Width, W
def expand_blockavg(a, H, W):
m,n = a.shape
return np.broadcast_to((a/float(H*W))[:,None,:,None],(m,H,n,W)).reshape(m*H,-1)
Sample runs -
In [93]: a
Out[93]:
array([[1, 2],
[2, 1]])
In [94]: expand_blockavg(a, H=2, W=2)
Out[94]:
array([[0.25, 0.25, 0.5 , 0.5 ],
[0.25, 0.25, 0.5 , 0.5 ],
[0.5 , 0.5 , 0.25, 0.25],
[0.5 , 0.5 , 0.25, 0.25]])
In [95]: expand_blockavg(a, H=2, W=3)
Out[95]:
array([[0.17, 0.17, 0.17, 0.33, 0.33, 0.33],
[0.17, 0.17, 0.17, 0.33, 0.33, 0.33],
[0.33, 0.33, 0.33, 0.17, 0.17, 0.17],
[0.33, 0.33, 0.33, 0.17, 0.17, 0.17]])
Runtime test on a large array -
In [2]: a = np.random.rand(200,200)
# Expand by (2 x 2)
# @Kasrâmvd's soln
In [85]: %timeit np.repeat(np.repeat(a, 2, 1), 2, 0)/4
1000 loops, best of 3: 492 µs per loop
In [86]: %timeit expand_blockavg(a, H=2, W=2)
1000 loops, best of 3: 382 µs per loop
# Expand by (20 x 20)
# @Kasrâmvd's soln
In [5]: %timeit np.repeat(np.repeat(a, 20, 1), 20, 0)/400
10 loops, best of 3: 32 ms per loop
In [6]: %timeit expand_blockavg(a, H=20, W=20)
10 loops, best of 3: 20.1 ms per loop
Larger array with (2 x 2) expansion -
In [87]: a = np.random.rand(2000,2000)
# @Kasrâmvd's soln
In [88]: %timeit np.repeat(np.repeat(a, 2, 1), 2, 0)/4
10 loops, best of 3: 70.2 ms per loop
In [89]: %timeit expand_blockavg(a, H=2, W=2)
10 loops, best of 3: 51.6 ms per loop