Question

如何获得稀疏复矩阵的元素绝对值？通常，是否可以使用用户定义的函数将矩阵映射到另一个矩阵？

Answer 1

对我来说，以下工作正常

import numpy as np
from scipy import sparse

a = sparse.identity(5).tocsr() * 1j
a[2,4] = 1-1j
b = np.absolute(a)
b.A
# array([[ 1.        ,  0.        ,  0.        ,  0.        ,  0.        ],
#        [ 0.        ,  1.        ,  0.        ,  0.        ,  0.        ],
#        [ 0.        ,  0.        ,  1.        ,  0.        ,  1.41421356],
#        [ 0.        ,  0.        ,  0.        ,  1.        ,  0.        ],
#        [ 0.        ,  0.        ,  0.        ,  0.        ,  1.        ]])

# general functions: use the data attribute to access nonzeros

b.data = np.exp(b.data)
b.A
# array([[ 2.71828183,  0.        ,  0.        ,  0.        ,  0.        ],
#        [ 0.        ,  2.71828183,  0.        ,  0.        ,  0.        ],
#        [ 0.        ,  0.        ,  2.71828183,  0.        ,  4.11325038],
#        [ 0.        ,  0.        ,  0.        ,  2.71828183,  0.        ],
#        [ 0.        ,  0.        ,  0.        ,  0.        ,  2.71828183]])

Answer 2

scipy中的稀疏矩阵对象实现方法__abs__()。这意味着您可以使用Python的内置 abs()函数。它会将调用分派给__abs__()方法。例如，

In [16]: from scipy.sparse import csr_matrix

In [17]: a = csr_matrix([[1+1j, 0, 0, 0], [0, -2, 0, 0], [0, 0, 0, 0], [3-4j, 0, -5j, 0]])

In [18]: a.A
Out[18]: 
array([[ 1.+1.j,  0.+0.j,  0.+0.j,  0.+0.j],
       [ 0.+0.j, -2.+0.j,  0.+0.j,  0.+0.j],
       [ 0.+0.j,  0.+0.j,  0.+0.j,  0.+0.j],
       [ 3.-4.j,  0.+0.j,  0.-5.j,  0.+0.j]])

In [19]: b = abs(a)

In [20]: b.A
Out[20]: 
array([[ 1.41421356,  0.        ,  0.        ,  0.        ],
       [ 0.        ,  2.        ,  0.        ,  0.        ],
       [ 0.        ,  0.        ,  0.        ,  0.        ],
       [ 5.        ,  0.        ,  5.        ,  0.        ]])

稀疏矩阵的元素绝对值

2 个答案: