如何使用Python有效地选择子矩阵？

Question

我有一个大小为nxn的邻接矩阵（所以矩阵是对称的），我想选择一个大小为mxm的子矩阵，然后得到其上三角。 目前，我正在执行以下操作：

from numpy import *
am = array([array([0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0], dtype=uint8),
 array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0], dtype=uint8)])

w= (0, 1, 2, 5, 22) # select the window that consist of these nodes only
window = am[ix_(list(w),list(w))]
upper = [el for arr in [window[i][i+1:] for i in range(len(window)-1)] for el in arr]

但是我需要多次执行此操作以获取子矩阵+获取其上三角，目前这是我的代码的瓶颈。

有什么办法可以使速度更快？ 如果有办法直接获得子矩阵的上三角，我也将不胜感激，因为我仍然不需要完整的子矩阵。

Answer 1

这有效：

w = np.array((0, 1, 2, 5, 22))
n = len(w)
rang = np.arange(n, 0, -1)
rows = np.repeat(w, rang)
col_idx  = np.arange(n * (n + 1) // 2)
delta = np.repeat(np.concatenate(([0], np.cumsum(rang[1:]))),
                  rang)
col_idx -= delta
cols = np.take(w, col_idx)

现在：

>>> rows
array([ 0,  0,  0,  0,  0,  1,  1,  1,  1,  2,  2,  2,  5,  5, 22])
>>> cols
array([ 0,  1,  2,  5, 22,  1,  2,  5, 22,  2,  5, 22,  5, 22, 22])
>>> am[rows, cols]
array([0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0], dtype=uint8)

Answer 2

尝试numpy.triu （上三角）和numpy.tril（下三角）。