擅长:python、mysql、java
<p>这是我的解决办法。</p>
<ul>
<li>转置A</li>
<li>计算每列的总和</li>
<li>求和倒数的对角矩阵B格式</li>
<li>A*B等于标准化</li>
<li><p>转置C</p>
<pre><code>import scipy.sparse as sp
import numpy as np
import math
minf = 0.0001
A = sp.lil_matrix((5,5))
b = np.arange(0,5)
A.setdiag(b[:-1], k=1)
A.setdiag(b)
print A.todense()
A = A.T
print A.todense()
sum_of_col = A.sum(0).tolist()
print sum_of_col
c = []
for i in sum_of_col:
for j in i:
if math.fabs(j)<minf:
c.append(0)
else:
c.append(1/j)
print c
B = sp.lil_matrix((5,5))
B.setdiag(c)
print B.todense()
C = A*B
print C.todense()
C = C.T
print C.todense()
</code></pre></li>
</ul>