给定一个任意的numpy数组(ndarray
),是否有函数或简短的方法将其转换为scipy.sparse
矩阵?
我想要的东西就像:
A = numpy.array([0,1,0],[0,0,0],[1,0,0])
S = to_sparse(A, type="csr_matrix")
答案 0 :(得分:6)
我经常做类似
的事情>>> import numpy, scipy.sparse
>>> A = numpy.array([[0,1,0],[0,0,0],[1,0,0]])
>>> Asp = scipy.sparse.csr_matrix(A)
>>> Asp
<3x3 sparse matrix of type '<type 'numpy.int64'>'
with 2 stored elements in Compressed Sparse Row format>
答案 1 :(得分:1)
帮助中有一个非常有用且相关的例子!
import scipy.sparse as sp
help(sp)
这给出了:
Example 2
---------
Construct a matrix in COO format:
>>> from scipy import sparse
>>> from numpy import array
>>> I = array([0,3,1,0])
>>> J = array([0,3,1,2])
>>> V = array([4,5,7,9])
>>> A = sparse.coo_matrix((V,(I,J)),shape=(4,4))
值得注意的是各种构造函数(再次来自帮助):
1. csc_matrix: Compressed Sparse Column format
2. csr_matrix: Compressed Sparse Row format
3. bsr_matrix: Block Sparse Row format
4. lil_matrix: List of Lists format
5. dok_matrix: Dictionary of Keys format
6. coo_matrix: COOrdinate format (aka IJV, triplet format)
7. dia_matrix: DIAgonal format
To construct a matrix efficiently, use either lil_matrix (recommended) or
dok_matrix. The lil_matrix class supports basic slicing and fancy
indexing with a similar syntax to NumPy arrays.
你的例子很简单:
S = sp.csr_matrix(A)
答案 2 :(得分:0)
请参考以下答案:https://stackoverflow.com/a/65017153/9979257
在这个答案中,我解释了如何将二维NumPy矩阵转换为CSR或CSC格式。