Python 矩阵组合逻辑

我有以下矩阵：

,

及

其中每个元素为3x3矩阵（即a_01为3x3矩阵）。这意味着A和B张量是9x9矩阵。 如果不使用for命令，如何组合前面的方程式以获得：

---

目标是获得一个量子解，因为在实际情况下，a和B矩阵的大小可以是（N，N）。

您可以尝试以下方法

import numpy as np

def checker_index(n, m=3):
    """create an index vector that addresses m blocks of n consecutive
    elements with the blocks separated by gaps of n elements
    """
    # the next line creates the array
    # /   0    1 ...   n-1  \
    # |  2n 2n+1 ...   3n-1 |
    # \  4n 4n+1 ...   5n-1 /
    i = np.arange(m*n).reshape((m, n)) + n*np.arange(m)[:, None]
    # next line just puts the rows side by side
    # now observe the these are precisely the places where you want to
    # put your first row of of A's (with a leading id) in the first
    # row of the target structure, and similarly with columns
    # also, observe that one just needs to add n to get indices
    # suitable for placing the first row/column of B's
    return i.ravel()

def mingle(AA, BB, m=3):
    """combine AA and BB into the target structure

    here AA and BB are the full 3x3 block matrices you define in
    your question
    """
    n = len(AA) // m
    i1 = checker_index(n, m)
    # ix_ creates an "open grid" from its arguments
    # so indexing with y1, x1 below will select nm x nm elements
    # contrast this with ...
    y1, x1 = np.ix_(i1, i1)
    i2 = i1 + n
    y2, x2 = np.ix_(i2, i2)

    IAA = AA.copy()
    # ... the following line which only selects the diagonal,
    # thus just mn elements
    IAA[np.arange(m*n), np.arange(m*n)] = 1
    out = np.empty((2*m*n, 2*m*n))
    out[y1, x1] = IAA
    out[y1, x2] = BB
    out[y2, x1] = BB
    out[y2, x2] = IAA
    return out

Numpythonic够了吗？

这里有一个方法-

def matrix_combination(A,B):
    N = A.shape[0]//3  # Size of each block
    A4D = A.reshape(3,N,3,N)
    B4D = B.reshape(3,N,3,N)

    r,c = np.nonzero(~np.eye(3,dtype=bool))
    out = np.zeros((6,N,6,N),dtype=A.dtype)    

    idx0 = 2*np.arange(3)
    out[idx0[r],:,idx0[c]] = A4D[r,:,c]
    out[idx0[r]+1,:,idx0[c]+1] = A4D[r,:,c]

    out[idx0[r],:,idx0[c]+1] = B4D[r,:,c]
    out[idx0[r]+1,:,idx0[c]] = B4D[r,:,c]
    out = out.reshape(N*6,-1)
    np.fill_diagonal(out,1)
    return out

样本运行-

In [41]: A
Out[41]: 
array([[ 0,  0, 44, 98, 40, 69],
       [ 0,  0, 22, 55, 51, 19],
       [16, 58,  0,  0, 95, 95],
       [90, 88,  0,  0, 47, 91],
       [65, 96, 21, 50,  0,  0],
       [15, 91, 23, 91,  0,  0]])

In [42]: B
Out[42]: 
array([[ 0,  0, 20, 36, 85, 15],
       [ 0,  0, 17, 78, 56, 55],
       [86, 19,  0,  0, 60, 96],
       [76, 30,  0,  0, 34, 36],
       [73, 63, 28, 58,  0,  0],
       [40, 19, 22, 96,  0,  0]])

In [43]: matrix_combination(A,B)
Out[43]: 
array([[ 1,  0,  0,  0, 44, 98, 20, 36, 40, 69, 85, 15],
       [ 0,  1,  0,  0, 22, 55, 17, 78, 51, 19, 56, 55],
       [ 0,  0,  1,  0, 20, 36, 44, 98, 85, 15, 40, 69],
       [ 0,  0,  0,  1, 17, 78, 22, 55, 56, 55, 51, 19],
       [16, 58, 86, 19,  1,  0,  0,  0, 95, 95, 60, 96],
       [90, 88, 76, 30,  0,  1,  0,  0, 47, 91, 34, 36],
       [86, 19, 16, 58,  0,  0,  1,  0, 60, 96, 95, 95],
       [76, 30, 90, 88,  0,  0,  0,  1, 34, 36, 47, 91],
       [65, 96, 73, 63, 21, 50, 28, 58,  1,  0,  0,  0],
       [15, 91, 40, 19, 23, 91, 22, 96,  0,  1,  0,  0],
       [73, 63, 65, 96, 28, 58, 21, 50,  0,  0,  1,  0],
       [40, 19, 15, 91, 22, 96, 23, 91,  0,  0,  0,  1]])

为了好玩（对于一个Divakar来说），这里有一个非常紧凑的解决方案：

def mingle(AA, BB, m=3):
    n = len(AA) // m
    out = np.empty((m, 2, n, m, 2, n))
    out[:, [0, 1], ..., [0, 1], :] = AA.reshape((1, m, n, m, n))
    out[:, [0, 1], ..., [1, 0], :] = BB.reshape((1, m, n, m, n))
    out.shape = m * 2 * n, m * 2 * n
    out[np.arange(m * 2 * n), np.arange(m * 2 * n)] = 1
    return out

---

因此，如果A和B都是

（N，N）

，那么每个元素仍然是

3x3

？每个A和B都是3x3张量。这被称为块张量，其中每个元素是另一个张量。所以，每个a和B中总是有3 x 3=9个块？是的。例如A_01=[[A_xx，A_xy，A_xz]，[A_yx，A_yy，A_yz]，[A_zx，A_zy，A_zz]]是的（以更清楚的方式）：例如A_01=[[A_01^xx，A_01^xy，A_01^xz]，[A_01^yx，A_01^yy，A_01^yz]，[A_^zx，A_^zx，A divu^zy，A_^01]。@很显然，我无法按照你的思路完成这些问题，谢谢你的解决方案。