问题是这个问题的反面.我正在寻找一个从小数组原始大数组的泛型方法:
array([[[ 0, 1, 2],
[ 6, 7, 8]],
[[ 3, 4, 5],
[ 9, 10, 11]],
[[12, 13, 14],
[18, 19, 20]],
[[15, 16, 17],
[21, 22, 23]]])
->
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]])
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我目前正在开发一种解决方案,它会在完成后发布,但是我希望看到其他(更好)的方法.
unu*_*tbu 27
import numpy as np
def blockshaped(arr, nrows, ncols):
"""
Return an array of shape (n, nrows, ncols) where
n * nrows * ncols = arr.size
If arr is a 2D array, the returned array looks like n subblocks with
each subblock preserving the "physical" layout of arr.
"""
h, w = arr.shape
return (arr.reshape(h//nrows, nrows, -1, ncols)
.swapaxes(1,2)
.reshape(-1, nrows, ncols))
def unblockshaped(arr, h, w):
"""
Return an array of shape (h, w) where
h * w = arr.size
If arr is of shape (n, nrows, ncols), n sublocks of shape (nrows, ncols),
then the returned array preserves the "physical" layout of the sublocks.
"""
n, nrows, ncols = arr.shape
return (arr.reshape(h//nrows, -1, nrows, ncols)
.swapaxes(1,2)
.reshape(h, w))
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例如,
c = np.arange(24).reshape((4,6))
print(c)
# [[ 0 1 2 3 4 5]
# [ 6 7 8 9 10 11]
# [12 13 14 15 16 17]
# [18 19 20 21 22 23]]
print(blockshaped(c, 2, 3))
# [[[ 0 1 2]
# [ 6 7 8]]
# [[ 3 4 5]
# [ 9 10 11]]
# [[12 13 14]
# [18 19 20]]
# [[15 16 17]
# [21 22 23]]]
print(unblockshaped(blockshaped(c, 2, 3), 4, 6))
# [[ 0 1 2 3 4 5]
# [ 6 7 8 9 10 11]
# [12 13 14 15 16 17]
# [18 19 20 21 22 23]]
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请注意,还有超级鱼类
blockwise_view
.它以不同的格式排列块(使用更多的轴),但它的优点是(1)总是返回视图,(2)能够处理任何维度的数组.