我想实现这一目标：

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])

从数学上讲,它们不是相同的矩阵.但是我认为您知道我想要做什么.我想将矩阵的尺寸加倍.但是因此我想通过取四个对应单元格的四分之一来保留来自初始矩阵a的信息.

some1知道如何在numpy中有效地做到这一点吗？

解决方法:

这是一个np.broadcast_to,利用广播来避免复制的两个阶段,或者通过这样做来获得性能优势,这两个阶段-

# "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)

样品运行-

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]])

在大型阵列上进行运行时测试-

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

具有(2 x 2)扩展的更大阵列-

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