小编典典

由多个较小的2D阵列组成一个大2D阵列

python

问题是这个问题的反面。我正在寻找一种通用方法,以从小数组中提取原始的大数组:

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]])

我目前正在开发一个解决方案,将在完成后发布,但是希望看到其他(更好)的方法。


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2020-12-20

共1个答案

小编典典

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))

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]]

注意,还有超级蝙蝠blockwise_view鱼的
。它以不同的格式(使用更多的轴)排列块,但是它的优点是(1)始终返回视图,并且(2)能够处理任何尺寸的数组。

2020-12-20