主页 / coding / 问题 / 77340701
Accepted
Samvel Safaryan
Asked: 2023-10-22 23:58:43 +0800 CST

如何沿轴连接张量的内部矩阵？

  • 772

假设我们有一个形状为 $n\times d\times h\times w\times p\times p$ 的张量，我们想要将形状为 $p\time p$ 的内网矩阵连接起来，这样我们就制作了一个形状为 $ 的矩阵n\times d\times ph\times pw$。我该怎么做？

array([[[[[[ 0,  1,  2],
           [ 3,  4,  5],
           [ 6,  7,  8]],

          [[ 9, 10, 11],
           [12, 13, 14],
           [15, 16, 17]]],


         [[[18, 19, 20],
           [21, 22, 23],
           [24, 25, 26]],

          [[27, 28, 29],
           [30, 31, 32],
           [33, 34, 35]]]]]])

连接后

array([[[[0,   1,  2,  9, 10, 11],
         [3,   4,  5, 12, 13, 14],
         [6,   7,  8, 15, 16, 17],
         [18, 19, 20, 27, 28, 29],
         [21, 22, 23, 30, 31, 32],
         [24, 25, 26, 33, 34, 35]]]])

我使用重塑做了很多实验，但没有成功。我的实验之一

a.reshape(n, d, p*h, p*w)

我可以使用 for 循环来做到这一点，但我认为没有这个也是可能的。请帮我。使用for循环的代码

p = 3
arr = np.arange(1*1*2*2*p*p).reshape(1, 1, 2, 2, p, p)
answer = np.zeros(shape=(1, 1, 2*p, 2*p))

for (n, d, h, w) in np.ndindex(*arr.shape[:4]):
    answer[n, d, h:h+p, w:w+p] = arr[n, d, h, w]
numpy

1 个回答

  1. Best Answer
    In [15]: arr=np.arange(0,36).reshape(2,2,3,3)

    reshape无法对数组的元素重新排序。我从 [0,1,...35] 开始，并reshape保留：

    In [18]: arr.reshape(2,3,6)
Out[18]: 
array([[[ 0,  1,  2,  3,  4,  5],
        [ 6,  7,  8,  9, 10, 11],
        [12, 13, 14, 15, 16, 17]],

       [[18, 19, 20, 21, 22, 23],
        [24, 25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34, 35]]])

    我们必须以某种方式重新排序元素，将 [9,10,11] 块放在 [0,1,2] 附近。 transpose就是这样一种工具：

    In [19]: arr.transpose(0,2,1,3)
Out[19]: 
array([[[[ 0,  1,  2],
         [ 9, 10, 11]],

        [[ 3,  4,  5],
         [12, 13, 14]],

        [[ 6,  7,  8],
         [15, 16, 17]]],


       [[[18, 19, 20],
         [27, 28, 29]],

        [[21, 22, 23],
         [30, 31, 32]],

        [[24, 25, 26],
         [33, 34, 35]]]])

In [20]: arr.transpose(0,2,1,3).reshape(6,6)
Out[20]: 
array([[ 0,  1,  2,  9, 10, 11],
       [ 3,  4,  5, 12, 13, 14],
       [ 6,  7,  8, 15, 16, 17],
       [18, 19, 20, 27, 28, 29],
       [21, 22, 23, 30, 31, 32],
       [24, 25, 26, 33, 34, 35]])

    要通过分配给“空白”来做同样的事情，我们需要类似的东西：

    In [32]: res=np.zeros((6,6),int)
In [33]: res[:,:3] = arr[:,::2,:].reshape(6,3)

In [34]: res
Out[34]: 
array([[ 0,  1,  2,  0,  0,  0],
       [ 3,  4,  5,  0,  0,  0],
       [ 6,  7,  8,  0,  0,  0],
       [18, 19, 20,  0,  0,  0],
       [21, 22, 23,  0,  0,  0],
       [24, 25, 26,  0,  0,  0]])

In [35]: res[:,3:] = arr[:,1::2,:].reshape(6,3)

In [36]: res
Out[36]: 
array([[ 0,  1,  2,  9, 10, 11],
       [ 3,  4,  5, 12, 13, 14],
       [ 6,  7,  8, 15, 16, 17],
       [18, 19, 20, 27, 28, 29],
       [21, 22, 23, 30, 31, 32],
       [24, 25, 26, 33, 34, 35]])

    同一块连接的串联版本：

    In [41]: np.concatenate((arr[:,::2], arr[:,1::2]), axis=3)
Out[41]: 
array([[[[ 0,  1,  2,  9, 10, 11],
         [ 3,  4,  5, 12, 13, 14],
         [ 6,  7,  8, 15, 16, 17]]],


       [[[18, 19, 20, 27, 28, 29],
         [21, 22, 23, 30, 31, 32],
         [24, 25, 26, 33, 34, 35]]]])
    • 1

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