網格上的排序特徵值和特徵向量

Question

我在三維網格上的每個點上都有一個矩陣。我需要計算每個點的特徵值和特徵向量，並按照特徵值的升序對它們進行排序。我使用python編寫了下面的測試用例，我能夠對特徵值進行排序，但是相關的特徵向量具有較大的維度。

import numpy as np 
from numpy import linalg as LA 
n = 2 
a = np.zeros((3,3,n,n,n)) 
a[:,:,0,0,0] = [[5,0,0],[0,1,0],[0,0,3]] 
a[:,:,1,1,1] = [[2,0,0],[0,3,0],[0,0,1]] 
eigvals,eigvecs = LA.eig(a.swapaxes(0, -1).swapaxes(1,-2)) 
ev = eigvals.swapaxes(0,-1) 
evecs = eigvecs.swapaxes(0,-1).swapaxes(1,-2) 
evo = np.sort(ev,axis=0) 
print evo[:,0,0,0],evo[:,1,1,1] 
print evecs[:,:,0,0,0] 
print evecs[:,:,1,1,1] 
eveco = evecs[np.argsort(ev,axis=0)] 
print np.shape(eveco) 
print eveco[:,0,0,0,:,0,0,0] # decided after knowing the shape 
print eveco[:,1,1,1,:,0,0,0] # decided after knowing the shape 

它提供了正確的答案，但不是外形，eveco形狀應爲（3,3,2,2,2）：

[ 1. 3. 5.] [ 1. 2. 3.] 
[[ 1. 0. 0.] 
[ 0. 1. 0.] 
[ 0. 0. 1.]] 
[[ 1. 0. 0.] 
[ 0. 1. 0.] 
[ 0. 0. 1.]] 
(3, 2, 2, 2, 3, 2, 2, 2) 
[[ 0. 1. 0.] 
[ 0. 0. 1.] 
[ 1. 0. 0.]] 
[[ 0. 0. 1.] 
[ 1. 0. 0.] 
[ 0. 1. 0.]] 

Answer 1

這個怎麼樣，它把一個開放的網格在未排序的維度上防止它們被複制（用以下代碼替換代碼的最後四行）。

eveco = evecs[(np.argsort(ev,axis=0)[:, None, ...],) + tuple(np.ogrid[:3,:n,:n,:n])] 
print np.shape(eveco) 
print eveco[:,:,0,0,0] 
print eveco[:,:,1,1,1] 

輸出（新代碼的唯一）：

(3, 3, 2, 2, 2) 
[[ 0. 1. 0.] 
[ 0. 0. 1.] 
[ 1. 0. 0.]] 
[[ 0. 0. 1.] 
[ 1. 0. 0.] 
[ 0. 1. 0.]]