我正在寻找一种简洁的方法来提取大小为2x2的对角块，它位于（2N）x（2N）numpy数组的主对角线上（也就是说，会有N个这样的块）。这概括了numpy.diag，它沿着主对角线返回元素，人们可能会认为它是1x1块（当然numpy并不代表它们）。
为了更广泛地说明这一点，人们可能希望从（MN）x（MN）阵列中提取N个MxM块。这似乎是scipy.linalg.block_diag的补充，在How can I transform blocks into a blockdiagonal matrix (NumPy)中巧妙地讨论过，从block_diag将它们放置的地方拉出块。
借用the solution to that question中的代码，以下是如何设置代码：
>>> a1 = np.array([[1,1,1],[1,1,1],[1,1,1]])
>>> a2 = np.array([[2,2,2],[2,2,2],[2,2,2]])
>>> a3 = np.array([[3,3,3],[3,3,3],[3,3,3]])
>>> import scipy.linalg
>>> scipy.linalg.block_diag(a1, a2, a3)
array([[1, 1, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 3, 3, 3],
[0, 0, 0, 0, 0, 0, 3, 3, 3],
[0, 0, 0, 0, 0, 0, 3, 3, 3]])
然后，我们希望有一个像
这样的功能>>> A = scipy.linalg.block_diag(a1, a2, a3)
>>> extract_block_diag(A, M=3)
array([[[1, 1, 1],
[1, 1, 1],
[1, 1, 1]],
[[2, 2, 2],
[2, 2, 2],
[2, 2, 2]],
[[3, 3, 3],
[3, 3, 3],
[3, 3, 3]]])
为了继续使用numpy.diag进行类比，人们可能还希望提取非对角线块：它们在第k个块对角线上的N - k。 （顺便说一下，block_diag的扩展允许将块放在主对角线之外肯定会有用，但这不是这个问题的范围。）对于上面的数组，这可能会产生：
>>> extract_block_diag(A, M=3, k=1)
array([[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]],
[[0, 0, 0],
[0, 0, 0],
[0, 0, 0]]])
我看到使用this question中涵盖的stride_tricks旨在产生类似于此功能的东西，但我知道跨步操作在字节级别，这听起来不太合适。
通过动机，这源于我希望提取协方差矩阵（即方差）的对角元素的情况，其中元素本身不是标量而是2x2矩阵。
修改：根据the suggestion from Chris，我做了以下尝试：
def extract_block_diag(A,M,k=0):
"""Extracts blocks of size M from the kth diagonal
of square matrix A, whose size must be a multiple of M."""
# Check that the matrix can be block divided
if A.shape[0] != A.shape[1] or A.shape[0] % M != 0:
raise StandardError('Matrix must be square and a multiple of block size')
# Assign indices for offset from main diagonal
if abs(k) > M - 1:
raise StandardError('kth diagonal does not exist in matrix')
elif k > 0:
ro = 0
co = abs(k)*M
elif k < 0:
ro = abs(k)*M
co = 0
else:
ro = 0
co = 0
blocks = np.array([A[i+ro:i+ro+M,i+co:i+co+M]
for i in range(0,len(A)-abs(k)*M,M)])
return blocks
将为上述数据返回以下结果：
D = extract_block_diag(A,3)
[[[1 1 1]
[1 1 1]
[1 1 1]]
[[2 2 2]
[2 2 2]
[2 2 2]]
[[3 3 3]
[3 3 3]
[3 3 3]]]
D = extract_block_diag(A,3,-1)
[[[0 0 0]
[0 0 0]
[0 0 0]]
[[0 0 0]
[0 0 0]
[0 0 0]]]
答案 0 :(得分：3)
您也可以使用视图进行操作。这可能比索引方法更快。
import numpy as np
import scipy.linalg
a1 = np.array([[1,1,1],[1,1,1],[1,1,1]])
a2 = np.array([[2,2,2],[2,2,2],[2,2,2]])
a3 = np.array([[3,3,3],[3,3,3],[3,3,3]])
b = scipy.linalg.block_diag(a1, a2, a3)
b[1,4] = 4
b[1,7] = 5
b[4,1] = 6
b[4,7] = 7
b[7,1] = 8
b[7,4] = 9
print b
def extract_block_diag(a, n, k=0):
a = np.asarray(a)
if a.ndim != 2:
raise ValueError("Only 2-D arrays handled")
if not (n > 0):
raise ValueError("Must have n >= 0")
if k > 0:
a = a[:,n*k:]
else:
a = a[-n*k:]
n_blocks = min(a.shape[0]//n, a.shape[1]//n)
new_shape = (n_blocks, n, n)
new_strides = (n*a.strides[0] + n*a.strides[1],
a.strides[0], a.strides[1])
return np.lib.stride_tricks.as_strided(a, new_shape, new_strides)
print "-- Diagonal blocks:"
c = extract_block_diag(b, 3)
print c
print "-- They're views!"
c[0,1,2] = 9
print b[1,2]
print "-- 1st superdiagonal"
c = extract_block_diag(b, 3, 1)
print c
print "-- 2nd superdiagonal"
c = extract_block_diag(b, 3, 2)
print c
print "-- 3rd superdiagonal (empty)"
c = extract_block_diag(b, 3, 3)
print c
print "-- 1st subdiagonal"
c = extract_block_diag(b, 3, -1)
print c
print "-- 2nd subdiagonal"
c = extract_block_diag(b, 3, -2)
print c
答案 1 :(得分：2)
作为起点，您可以使用类似
的内容>>> a
array([[1, 1, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 0, 0, 0],
[1, 1, 1, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 2, 2, 2, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 3, 3, 3],
[0, 0, 0, 0, 0, 0, 3, 3, 3],
[0, 0, 0, 0, 0, 0, 3, 3, 3]])
>>> M = 3
>>> [a[i*M:(i+1)*M,i*M:(i+1)*M] for i in range(a.shape[0]/M)]
[array([[1, 1, 1],
[1, 1, 1],
[1, 1, 1]]), array([[2, 2, 2],
[2, 2, 2],
[2, 2, 2]]), array([[3, 3, 3],
[3, 3, 3],
[3, 3, 3]])]
答案 2 :(得分：0)
您不想使用简单方法的任何特殊原因？
>>> A=np.array([[1,1,1,1],[2,2,2,2],[3,3,3,3],[4,4,4,4]])
>>> M1s,M2s=0,2 # start from A[M1s,M2s]
>>> M=2 # extract an MxM block
>>> for a in A[M1s:M1s+M]:
... print a[M2s:M2s+M]
...
[1 1]
[2 2]
>>>
答案 3 :(得分：0)
根据unutbu对https://stackoverflow.com/a/8070716/219229的回答，我得到以下结论:( BTW在字节级操作有什么问题？）
from numpy.lib.stride_tricks import as_strided
...
def extract_block_diag(A, M=3, k=0):
ny, nx = A.shape
ndiags = min(map(lambda x: x//M, A.shape))
offsets = (nx*M+M, nx, 1)
strides = map(lambda x:x*A.itemsize, offsets)
if k > 0:
B = A[:,k*M]
ndiags = min(nx//M-k, ny//M)
else:
k = -k
B = A[k*M]
ndiags = min(nx//M, ny//M-k)
return as_strided(B, shape=(ndiags,M,M),
strides=((nx*M+M)*A.itemsize, nx*A.itemsize, A.itemsize))
使用示例：
# a1, a2, a3 from your example
>>> bigA = scipy.linalg.block_diag(a1, a2, a3)
>>> extract_block_diag ( bigA, 3 )
array([[[1, 1, 1],
[1, 1, 1],
[1, 1, 1]],
[[2, 2, 2],
[2, 2, 2],
[2, 2, 2]],
[[3, 3, 3],
[3, 3, 3],
[3, 3, 3]]])
>>> extract_block_diag ( bigA, 3 )[2]
array([[3, 3, 3],
[3, 3, 3],
[3, 3, 3]])
>>> extract_block_diag(np.arange(1,9*9+1).reshape(9,9),3,1)
[[[ 4 5 6]
[13 14 15]
[22 23 24]]
[[34 35 36]
[43 44 45]
[52 53 54]]]
请注意，上面的函数返回一个视图，如果更改返回的数组数组中的任何内容，原始视图也会受到影响。如有必要，请复印。
答案 4 :(得分：0)
将numpy导入为np
def extract_blocks（array）：
prev = -1
for i in xrange(len(array)-1):
if array[i+1][i] == 0 and array[i][i+1] == 0:
yield array[prev + 1: i + 1, prev + 1: i + 1]
print prev + 1, i
prev = i
yield array[prev + 1: len(array), prev + 1: len(array)]
d = np.array（[[4,4,0,0,0,0,0,0,0]，[4,4,0,0,0,0,0,0]， [0,0,1,0,0,0,0,0,0]，[0,0,0,2,2,2,0,0,0]，[0,0,0,2,2] ，2,0,0,0]，[0,0,0,2,2,0,0,0]，[0,0,0,0,0,0,3,3,3]， [0,0,0,0,0,0,3,3,3]，[0,0,0,0,0,0,3,3,3]]）
for extract_blocks（d）中的h：
print h
[[4 4] [4 4]]，[[1]]，[[2 2 2] [2 2 2] [2 2 2]]，[[3 3 3] [3 3 3] [ 3 3 3]]