python - 重新排序一个numpy数组python

Question

我有一个像这样的大二维数组:

array([[ 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],
       [33,34,35,36,37,38,39,40],
       [41,42,43,44,45,46,47,48],
       ....])

我需要将它转换成:

array([ 1, 9,17, 2,10,18, 3,11,19, 4,12,20, 5,13,21, 6,14,22, 7,15,23, 8,16,24],
      [25,33,41,26,34,42,27,35,43,28,36,44,29,37,45,30,38,46,31,39,47,32,40,48],
      ...

请注意，这应该只是演示它应该做什么。
原始数组仅包含 bool 值，大小为 512x8。在我的示例中，我只将 3 行 8 个元素排列成一行，但我真正需要的分别是 32 行 8 个元素。

我真的很抱歉，但是在写了 30 分钟之后，这是我对我的问题的唯一描述。我希望这足够了。

Answer 1

我认为您可以使用两个 reshape 来达到您想要的结果操作和 transpose :

x = np.array([[ 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],
              [33,34,35,36,37,38,39,40],
              [41,42,43,44,45,46,47,48]])

y = x.reshape(2, 3, 8).transpose(0, 2, 1).reshape(2, -1)

print(repr(y))
# array([[ 1,  9, 17,  2, 10, 18,  3, 11, 19,  4, 12, 20,  5, 13, 21,  6, 14,
#         22,  7, 15, 23,  8, 16, 24],
#        [25, 33, 41, 26, 34, 42, 27, 35, 43, 28, 36, 44, 29, 37, 45, 30, 38,
#         46, 31, 39, 47, 32, 40, 48]])

稍微分解一下:

1. @hpaulj 的第一个 reshape 操作为我们提供了这个:

   x1 = x.reshape(2, 3, 8)
   print(repr(x1))
   # array([[[ 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, 36, 37, 38, 39, 40],
   #         [41, 42, 43, 44, 45, 46, 47, 48]]])
   print(x1.shape)
   # (2, 3, 8)
   

2. 为了获得所需的输出，我们需要沿第二个维度(大小为 3)“折叠”此数组，然后沿第三个维度(大小为 8)“折叠”。实现这种事情的最简单方法是首先 transpose数组，以便您要折叠的维度从头到尾排序:

   x2 = x1.transpose(0, 2, 1)  # you could also use `x2 = np.rollaxis(x1, 1, 3)`
   print(repr(x2))
   # array([[[ 1,  9, 17],
   #     [ 2, 10, 18],
   #     [ 3, 11, 19],
   #     [ 4, 12, 20],
   #     [ 5, 13, 21],
   #     [ 6, 14, 22],
   #     [ 7, 15, 23],
   #     [ 8, 16, 24]],
   
   #    [[25, 33, 41],
   #     [26, 34, 42],
   #     [27, 35, 43],
   #     [28, 36, 44],
   #     [29, 37, 45],
   #     [30, 38, 46],
   #     [31, 39, 47],
   #     [32, 40, 48]]])
   print(x2.shape)
   # (2, 8, 3)
   

3. 最后，我可以使用 reshape(2, -1) 在最后两个维度上折叠数组。 -1 导致 numpy 根据 x 中的元素数量推断最后一个维度中的适当大小。

   y = x2.reshape(2, -2)