问题
I need to convert a scipy sparse matrix to cvxopt's sparse matrix format, spmatrix, and haven't come across anything yet (the matrix is too big to be converted to dense, of course). Any ideas how to do this?
回答1:
The more robust answer is a combination of hpaulj's answer and OferHelman's answer.
def scipy_sparse_to_spmatrix(A):
coo = A.tocoo()
SP = spmatrix(coo.data.tolist(), coo.row.tolist(), coo.col.tolist(), size=A.shape)
return SP
Defining the shape variable preserves the dimensionality of A on SP. I found that any zero columns ending the scipy sparse matrix would be lost without this added step.
回答2:
taken from http://maggotroot.blogspot.co.il/2013/11/constrained-linear-least-squares-in.html
coo = A.tocoo()
SP = spmatrix(coo.data, coo.row.tolist(), coo.col.tolist())
回答3:
From http://cvxopt.org/userguide/matrices.html#sparse-matrices
cvxopt.spmatrix(x, I, J[, size[, tc]])
looks similar to the scipy.sparse
coo_matrix((data, (i, j)), [shape=(M, N)])
My guess is that if A
is a matrix in coo
format, that
cvxopt.spmatrix(A.data, A.row, A.col, A.shape)
would work. (I don't have cvxopt
installed to test this.)
来源:https://stackoverflow.com/questions/25314067/scipy-sparse-matrix-to-cvxopt-spmatrix