python - NumPy 数组的反向堆叠操作

Question

我有一行代码可以有效地将 numpy 数组从 400x8x8 数组 reshape 为 160x160 数组，我需要反转该过程但无法弄清楚该行的反转。

我已经可以执行此过程，但它非常低效并且需要嵌套循环，出于性能目的我想避免这种情况。

这是我目前必须反转该过程的代码 (160x160 > 400x8x8):

       previousRow = 0
    for rowBlock in range(noBlocksOn1Axis):
        previousRow = rowBlock * blockSize  
        previousColumn = 0
        for columnBlock in range(noBlocksOn1Axis):
            previousColumn = columnBlock * blockSize
            block = 
            arrayY[previousRow:previousRow+blockSize, 
            previousColumn:previousColumn + blockSize]
            blocksList.append(block) 

下面是 reshape 400x8x8 > 160x160 的代码行:

    xy = np.zeros((160,160), dtype = np.uint8)

    xy = np.vstack(np.hstack(overDone[20*i:20+20*i]) for i in 
    range(overDone.shape[0]//20))

关于如何反向执行这行代码有什么想法吗？

谢谢:D

Answer 1

reshape 、交换轴(或转置轴)并 reshape 以获得 overDone -

xy.reshape(20,8,20,8).swapaxes(1,2).reshape(400,8,8)

有关 intuition behind nd-to-nd array transformation 的更多信息.

使其通用以处理通用形状 -

m,n = xy.shape
M,N = 20,20 # block size used to get xy
overDone_ = xy.reshape(M,m//M,N,n//N).swapaxes(1,2).reshape(-1,m//M,n//N)

sample 运行-

# Original input
In [21]: overDone = np.random.rand(400,8,8)

# Perform forward step to get xy
In [22]: xy = np.vstack(np.hstack(overDone[20*i:20+20*i]) for i in range(overDone.shape[0]//20))

# Use proposed approach to get back overDone
In [23]: out = xy.reshape(20,8,20,8).swapaxes(1,2).reshape(400,8,8)

# Verify output to be same as overDone
In [42]: np.array_equal(out,overDone)
Out[42]: True

奖励:

我们可以使用那些相同的向量化 reshape+permute-axes 步骤来为正向过程创建 xy -

xy = overDone.reshape(20,20,8,8).swapaxes(1,2).reshape(160,160)