python - 如何在这里进行广义特征分解？

Question

我正在尝试实现一个拉普拉斯特征图算法，它包括:

1)构造一个图(我用的是kNN，说到k个最近邻有一条边)

2) 将每条边与权重相关联

3) 定义对角线(即对角线放置的行的总和)

4) 执行广义特征分解(应该是 Lv = lambdaDv，其中 L 和 D 在下面的代码中计算)

我认为这可以通过 scipy.linalg.eig(vals) 以某种方式解决，但我不明白如何正确输入我的两个矩阵。有人可以帮助我理解如何执行广义特征分解步骤吗？

import numpy as np
import random as r
from math import exp as exp
from scipy.spatial import distance

def rweights((vectors,features)):
    return 1 * np.random.random_sample((vectors,features)) - 0

def vEuclidean(v, m):
   return np.apply_along_axis(lambda x: distance.euclidean(v,x), 1, m)

def mEuclideans(m):
   return np.apply_along_axis(lambda v: vEuclidean(v,m), 1, m)

def neighbours(vector, neigh):
   size = (vector.shape[0] - neigh)
   for i in range(1,size):
      vector[np.argmax(vector)] = 0.0
   return vector

def kNN(m, k):
    me = mEuclideans(m)
    return np.array(map(lambda v: neighbours(v, k), me))

def diag(m):
    sums = np.sum(m,1)
    (vectors,features) = m.shape
    zeros = np.zeros(vectors*features).reshape((vectors,features))
    for i in range(features):
        zeros[i][i] = sums[i]
    return zeros

def vectorWeight(v, sigma):
      f = lambda x: exp((-(x)/(sigma**2)))
      size = v.shape[0]
      for i in range(size):
          v[i] = f(v[i])
      return v 

def weight(m):
    return np.array(np.apply_along_axis(lambda v: vectorWeight(v,0.5), 1, m))

if __name__ == "__main__":
    np.random.seed(666)
    m = rweights((5,3))
    w = weight(kNN(m, 2))
    D = diag(w)
    L = D-w

Answer 1

嗯， 这个答案得到了 Warrens 的帮助(所以他值得称赞)，但我找到了一个关于光谱聚类的视频 https://www.youtube.com/watch?v=Ln0mgyvXNQE ，他在图上使用拉普拉斯算子。我认为根据他的结果检查我的实现情况会很好。因此，我补充说:

from scipy.linalg import eig    

def distanceM():
    return np.array([[0.0,0.8,0.6, 0.1,0.0,0.0], 
    [0.8,0.0,0.9,0.0,0.0,0.0], [0.6,0.9,0.0,0.0,0.0,0.2], 
    [0.1,0.0,0.0,0.0,0.6,0.7],[0.0,0.0,0.0,0.6,0.0,0.8]
    [0.0,0.0,0.2,0.7,0.8,0.0]])

if __name__ == "__main__":
    w = distanceM()
    D = diag(w)
    L = D-w
    w,vr = eig(L)
    print vr

我发现我得到了相同的拉普拉斯矩阵，以及相同的特征向量(vr 的第二列)。