将NumPy矩阵（二维数组）在垂直和水平方向分成相等的块的最pythonic方法是什么？
例如：
aa = np.reshape(np.arange(270),(18,15)) # a 18x15 matrix
然后是“功能”，如
ab = np.split2d(aa,(2,3))
将产生每个形状为（9,5）的6个矩阵的列表。第一个猜测是组合hsplit，map和vsplit，但是如果有两个参数要为它定义，则必须如何应用mar，例如：
map(np.vsplit(@,3),np.hsplit(aa,2))
答案 0 :(得分：4)
这是保持在NumPy环境中的一种方法 -
def view_as_blocks(arr, BSZ):
# arr is input array, BSZ is block-size
m,n = arr.shape
M,N = BSZ
return arr.reshape(m//M, M, n//N, N).swapaxes(1,2).reshape(-1,M,N)
示例运行
1）验证形状的实际大案例：
In [41]: aa = np.reshape(np.arange(270),(18,15))
In [42]: view_as_blocks(aa, (9,5)).shape
Out[42]: (6, 9, 5)
2）手动验证值的小案例：
In [43]: aa = np.reshape(np.arange(36),(6,6))
In [44]: aa
Out[44]:
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],
[24, 25, 26, 27, 28, 29],
[30, 31, 32, 33, 34, 35]])
In [45]: view_as_blocks(aa, (2,3)) # Blocks of shape (2,3)
Out[45]:
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]],
[[24, 25, 26],
[30, 31, 32]],
[[27, 28, 29],
[33, 34, 35]]])
如果您愿意与其他图书馆合作，scikit-image可能会在这里使用，就像这样 -
from skimage.util import view_as_blocks as viewB
out = viewB(aa, tuple(BSZ)).reshape(-1,*BSZ)
运行时测试 -
In [103]: aa = np.reshape(np.arange(270),(18,15))
# @EFT's soln
In [99]: %timeit split_2d(aa, (2,3))
10000 loops, best of 3: 23.3 µs per loop
# @glegoux's soln-1
In [100]: %timeit list(get_chunks(aa, 2,3))
100000 loops, best of 3: 3.7 µs per loop
# @glegoux's soln-2
In [111]: %timeit list(get_chunks2(aa, 9, 5))
100000 loops, best of 3: 3.39 µs per loop
# Proposed in this post
In [101]: %timeit view_as_blocks(aa, (9,5))
1000000 loops, best of 3: 1.86 µs per loop
请注意，我已使用(2,3) split_2d和get_chunks作为其定义，他们将其用作块数。在view_as_blocks的情况下，我有参数BSZ表示块大小。所以，我有(9,5)。 get_chunks2遵循与view_as_blocks相同的格式。输出应该代表相同的。
答案 1 :(得分：2)
切割，这个矩阵（18,15）：
+-+-+-+
+ +
+-+-+-+
在2x3块（9,5）中喜欢它：
+-+-+-+
+-+-+-+
+-+-+-+
执行：
from pprint import pprint
import numpy as np
M = np.reshape(np.arange(18*15),(18,15))
def get_chunks(M, n, p):
n = len(M)//n
p = len(M[0])//p
for i in range(0, len(M), n):
for j in range(0, len(M[0]), p):
yield M[i:i+n,j:j+p]
def get_chunks2(M, n, p):
for i in range(0, len(M), n):
for j in range(0, len(M[0]), p):
yield M[i:i+n,j:j+p]
# list(get_chunks2(M, 9, 5)) same result more faster
chunks = list(get_chunks(M, 2, 3))
pprint(chunks)
输出：
[array([[ 0, 1, 2, 3, 4],
[ 15, 16, 17, 18, 19],
[ 30, 31, 32, 33, 34],
[ 45, 46, 47, 48, 49],
[ 60, 61, 62, 63, 64],
[ 75, 76, 77, 78, 79],
[ 90, 91, 92, 93, 94],
[105, 106, 107, 108, 109],
[120, 121, 122, 123, 124]]),
array([[ 5, 6, 7, 8, 9],
[ 20, 21, 22, 23, 24],
[ 35, 36, 37, 38, 39],
[ 50, 51, 52, 53, 54],
[ 65, 66, 67, 68, 69],
[ 80, 81, 82, 83, 84],
[ 95, 96, 97, 98, 99],
[110, 111, 112, 113, 114],
[125, 126, 127, 128, 129]]),
array([[ 10, 11, 12, 13, 14],
[ 25, 26, 27, 28, 29],
[ 40, 41, 42, 43, 44],
[ 55, 56, 57, 58, 59],
[ 70, 71, 72, 73, 74],
[ 85, 86, 87, 88, 89],
[100, 101, 102, 103, 104],
[115, 116, 117, 118, 119],
[130, 131, 132, 133, 134]]),
array([[135, 136, 137, 138, 139],
[150, 151, 152, 153, 154],
[165, 166, 167, 168, 169],
[180, 181, 182, 183, 184],
[195, 196, 197, 198, 199],
[210, 211, 212, 213, 214],
[225, 226, 227, 228, 229],
[240, 241, 242, 243, 244],
[255, 256, 257, 258, 259]]),
array([[140, 141, 142, 143, 144],
[155, 156, 157, 158, 159],
[170, 171, 172, 173, 174],
[185, 186, 187, 188, 189],
[200, 201, 202, 203, 204],
[215, 216, 217, 218, 219],
[230, 231, 232, 233, 234],
[245, 246, 247, 248, 249],
[260, 261, 262, 263, 264]]),
array([[145, 146, 147, 148, 149],
[160, 161, 162, 163, 164],
[175, 176, 177, 178, 179],
[190, 191, 192, 193, 194],
[205, 206, 207, 208, 209],
[220, 221, 222, 223, 224],
[235, 236, 237, 238, 239],
[250, 251, 252, 253, 254],
[265, 266, 267, 268, 269]])]
答案 2 :(得分：2)
您可以使用np.split＆amp; np.concatenate，后者允许第二次拆分一步完成：
def split_2d(array, splits):
x, y = splits
return np.split(np.concatenate(np.split(array, y, axis=1)), x*y)
ab = split_2d(aa,(2,3))
ab[0].shape
Out[95]: (9, 5)
len(ab)
Out[96]: 6
这似乎也应该相对简单地概括为n-dim情况，尽管我还没有完全遵循这个想法。
修改
对于单个数组作为输出，只需添加np.stack：
np.stack(ab).shape
Out[99]: (6, 9, 5)
答案 3 :(得分：0)
对于一个更简单的解决方案，我将np.array_split与转换矩阵一起使用。因此，假设我希望将其垂直拆分为3个相等的块，水平拆分为2个相等的块，然后：
# Create your matrix
matrix = np.reshape(np.arange(270),(18,15)) # a 18x15 matrix
# Container for your final matrices
final_matrices = []
# Then split into 3 equal chunks vertically
vertically_split_matrices = np.array_split(matrix)
for v_m in vertically_split_matrices:
# Then split the transformed matrices equally
m1, m2 = np.array_split(v_m.T, 2)
# And transform the matrices back
final_matrices.append(m1.T)
final_matrices.append(m2.T)
因此，我最终得到了6个块，这些块都具有相同的高度和宽度。