sklearn.preprocessing.normalize中的norm ='l2'对于矩阵归一化有什么作用？

Question

I normalized scipy.sparse matrix sA here by using sklearn.preprocessing.normalize . I read the document but not understand about norm='l...' well, so I tested it.

norm='l1' went ok and I got the sum result as expected as 1 in all row.

A = np.array([[1,2,0],[0,0,3],[1,0,4]])
sA = sp.csr_matrix(A)   
normsA = normalize(sA, norm='l1', axis=0)
print normsA
print "---"
print sum(normsA)

>>(0, 0)    0.5
  (2, 0)    0.5
  (0, 1)    1.0
  (1, 2)    0.428571428571
  (2, 2)    0.571428571429
  ---
  (0, 0)    1.0
  (0, 1)    1.0
  (0, 2)    1.0

However, when I tried l2 , I can't find how it is normalizing the matrix. Sum of the matrix or the transposed matrix is not equal to one. What l2 does for normalizing here?

normsA2 = normalize(sA, norm='l2', axis=0)
print sum(normsA2)
print sum(normsA2.T)

>>(0, 0)    1.41421356237
  (0, 1)    1.0
  (0, 2)    1.4
  (0, 0)    1.70710678119
  (0, 1)    0.6
  (0, 2)    1.50710678119

Answer 1

That is using the l2 norm (or euclidean norm/distance), in other words the sum of the squares of the elements gives one.

The following outputs the expected vector of ones:

sum(normsA2 ** 2, axis=0)