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。
请您参考如下方法:
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:
@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)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)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)
