Numpy 块重塑
Numpy blocks reshaping
我正在寻找一种方法来重塑以下 1d-numpy 数组:
# dimensions
n = 2 # int : 1 ... N
h = 2 # int : 1 ... N
m = n*(2*h+1)
input_data = np.arange(0,(n*(2*h+1))**2)
预期输出应重塑为 (2*h+1)**2
个形状为 (n,n)
的块,例如:
input_data.reshape(((2*h+1)**2,n,n))
>>> array([[[ 0 1]
[ 2 3]]
[[ 4 5]
[ 6 7]]
...
[[92 93]
[94 95]]
[[96 97]
[98 99]]]
这些块最终需要重新整形为 (m,m)
矩阵,以便它们以 2*h+1
块的行堆叠:
>>> array([[ 0, 1, 4, 5, 8, 9, 12, 13, 16, 17],
[ 2, 3, 6, 7, 10, 11, 14, 15, 18, 19],
...
[80, 81, 84, 85, 88, 89, 92, 93, 96, 97],
[82, 83, 86, 87, 90, 91, 94, 95, 98, 99]])
我的问题是,在第一次重塑 (n,n)
块后,我似乎无法找到合适的轴排列。我已经看过几个答案,例如 但徒劳无功。
由于实际尺寸 n
和 h
相当大,并且此操作在迭代过程中进行,我正在寻找一种有效的重塑操作。
我不认为你可以单独使用 reshape
和 transpose
来做到这一点(尽管我很想被证明是错误的)。使用 np.block
可以,但有点乱:
np.block([list(i) for i in input_data.reshape( (2*h+1), (2*h+1), n, n )])
array([[ 0, 1, 4, 5, 8, 9, 12, 13, 16, 17],
[ 2, 3, 6, 7, 10, 11, 14, 15, 18, 19],
[20, 21, 24, 25, 28, 29, 32, 33, 36, 37],
[22, 23, 26, 27, 30, 31, 34, 35, 38, 39],
[40, 41, 44, 45, 48, 49, 52, 53, 56, 57],
[42, 43, 46, 47, 50, 51, 54, 55, 58, 59],
[60, 61, 64, 65, 68, 69, 72, 73, 76, 77],
[62, 63, 66, 67, 70, 71, 74, 75, 78, 79],
[80, 81, 84, 85, 88, 89, 92, 93, 96, 97],
[82, 83, 86, 87, 90, 91, 94, 95, 98, 99]])
编辑:没关系,你可以不用 np.block
:
input_data.reshape( (2*h+1), (2*h+1), n, n).transpose(0, 2, 1, 3).reshape(10, 10)
array([[ 0, 1, 4, 5, 8, 9, 12, 13, 16, 17],
[ 2, 3, 6, 7, 10, 11, 14, 15, 18, 19],
[20, 21, 24, 25, 28, 29, 32, 33, 36, 37],
[22, 23, 26, 27, 30, 31, 34, 35, 38, 39],
[40, 41, 44, 45, 48, 49, 52, 53, 56, 57],
[42, 43, 46, 47, 50, 51, 54, 55, 58, 59],
[60, 61, 64, 65, 68, 69, 72, 73, 76, 77],
[62, 63, 66, 67, 70, 71, 74, 75, 78, 79],
[80, 81, 84, 85, 88, 89, 92, 93, 96, 97],
[82, 83, 86, 87, 90, 91, 94, 95, 98, 99]])
我正在寻找一种方法来重塑以下 1d-numpy 数组:
# dimensions
n = 2 # int : 1 ... N
h = 2 # int : 1 ... N
m = n*(2*h+1)
input_data = np.arange(0,(n*(2*h+1))**2)
预期输出应重塑为 (2*h+1)**2
个形状为 (n,n)
的块,例如:
input_data.reshape(((2*h+1)**2,n,n))
>>> array([[[ 0 1]
[ 2 3]]
[[ 4 5]
[ 6 7]]
...
[[92 93]
[94 95]]
[[96 97]
[98 99]]]
这些块最终需要重新整形为 (m,m)
矩阵,以便它们以 2*h+1
块的行堆叠:
>>> array([[ 0, 1, 4, 5, 8, 9, 12, 13, 16, 17],
[ 2, 3, 6, 7, 10, 11, 14, 15, 18, 19],
...
[80, 81, 84, 85, 88, 89, 92, 93, 96, 97],
[82, 83, 86, 87, 90, 91, 94, 95, 98, 99]])
我的问题是,在第一次重塑 (n,n)
块后,我似乎无法找到合适的轴排列。我已经看过几个答案,例如
由于实际尺寸 n
和 h
相当大,并且此操作在迭代过程中进行,我正在寻找一种有效的重塑操作。
我不认为你可以单独使用 reshape
和 transpose
来做到这一点(尽管我很想被证明是错误的)。使用 np.block
可以,但有点乱:
np.block([list(i) for i in input_data.reshape( (2*h+1), (2*h+1), n, n )])
array([[ 0, 1, 4, 5, 8, 9, 12, 13, 16, 17],
[ 2, 3, 6, 7, 10, 11, 14, 15, 18, 19],
[20, 21, 24, 25, 28, 29, 32, 33, 36, 37],
[22, 23, 26, 27, 30, 31, 34, 35, 38, 39],
[40, 41, 44, 45, 48, 49, 52, 53, 56, 57],
[42, 43, 46, 47, 50, 51, 54, 55, 58, 59],
[60, 61, 64, 65, 68, 69, 72, 73, 76, 77],
[62, 63, 66, 67, 70, 71, 74, 75, 78, 79],
[80, 81, 84, 85, 88, 89, 92, 93, 96, 97],
[82, 83, 86, 87, 90, 91, 94, 95, 98, 99]])
编辑:没关系,你可以不用 np.block
:
input_data.reshape( (2*h+1), (2*h+1), n, n).transpose(0, 2, 1, 3).reshape(10, 10)
array([[ 0, 1, 4, 5, 8, 9, 12, 13, 16, 17],
[ 2, 3, 6, 7, 10, 11, 14, 15, 18, 19],
[20, 21, 24, 25, 28, 29, 32, 33, 36, 37],
[22, 23, 26, 27, 30, 31, 34, 35, 38, 39],
[40, 41, 44, 45, 48, 49, 52, 53, 56, 57],
[42, 43, 46, 47, 50, 51, 54, 55, 58, 59],
[60, 61, 64, 65, 68, 69, 72, 73, 76, 77],
[62, 63, 66, 67, 70, 71, 74, 75, 78, 79],
[80, 81, 84, 85, 88, 89, 92, 93, 96, 97],
[82, 83, 86, 87, 90, 91, 94, 95, 98, 99]])