How to transform numpy.matrix or array to scipy sparse matrix

You can pass a numpy array or matrix as an argument when initializing a sparse matrix. For a CSR matrix, for example, you can do the following.

>>> import numpy as np>>> from scipy import sparse>>> A = np.array([[1,2,0],[0,0,3],[1,0,4]])>>> B = np.matrix([[1,2,0],[0,0,3],[1,0,4]])>>> Aarray([[1, 2, 0],       [0, 0, 3],       [1, 0, 4]])>>> sA = sparse.csr_matrix(A)   # Here's the initialization of the sparse matrix.>>> sB = sparse.csr_matrix(B)>>> sA<3x3 sparse matrix of type '<type 'numpy.int32'>'        with 5 stored elements in Compressed Sparse Row format>>>> print sA  (0, 0)        1  (0, 1)        2  (1, 2)        3  (2, 0)        1  (2, 2)        4

There are several sparse matrix classes in scipy.

bsr_matrix(arg1[, shape, dtype, copy, blocksize]) Block Sparse Row matrix
coo_matrix(arg1[, shape, dtype, copy]) A sparse matrix in COOrdinate format.
csc_matrix(arg1[, shape, dtype, copy]) Compressed Sparse Column matrix
csr_matrix(arg1[, shape, dtype, copy]) Compressed Sparse Row matrix
dia_matrix(arg1[, shape, dtype, copy]) Sparse matrix with DIAgonal storage
dok_matrix(arg1[, shape, dtype, copy]) Dictionary Of Keys based sparse matrix.
lil_matrix(arg1[, shape, dtype, copy]) Row-based linked list sparse matrix

Any of them can do the conversion.

import numpy as npfrom scipy import sparsea=np.array([[1,0,1],[0,0,1]])b=sparse.csr_matrix(a)print(b)(0, 0)  1(0, 2)  1(1, 2)  1

See http://docs.scipy.org/doc/scipy/reference/sparse.html#usage-information .

As for the inverse, the function is inv(A), but I won't recommend using it, since for huge matrices it is very computationally costly and unstable. Instead, you should use an approximation to the inverse, or if you want to solve Ax = b you don't really need A^-1.