问题
Say I would like to remove the diagonal from a scipy.sparse.csr_matrix
. Is there an efficient way of doing so? I saw that in the sparsetools
module there are C
functions to return the diagonal.
Based on other SO answers here and here my current approach is the following:
def csr_setdiag_val(csr, value=0):
"""Set all diagonal nonzero elements
(elements currently in the sparsity pattern)
to the given value. Useful to set to 0 mostly.
"""
if csr.format != "csr":
raise ValueError('Matrix given must be of CSR format.')
csr.sort_indices()
pointer = csr.indptr
indices = csr.indices
data = csr.data
for i in range(min(csr.shape)):
ind = indices[pointer[i]: pointer[i + 1]]
j = ind.searchsorted(i)
# matrix has only elements up until diagonal (in row i)
if j == len(ind):
continue
j += pointer[i]
# in case matrix has only elements after diagonal (in row i)
if indices[j] == i:
data[j] = value
which I then follow with
csr.eliminate_zeros()
Is that the best I can do without writing my own Cython
code?
回答1:
Based on @hpaulj's comment, I created an IPython Notebook which can be seen on nbviewer. This shows that out of all methods mentioned the following is the fastest (assume that mat
is a sparse CSR matrix):
mat - scipy.sparse.dia_matrix((mat.diagonal()[scipy.newaxis, :], [0]), shape=(one_dim, one_dim))
来源:https://stackoverflow.com/questions/22660374/remove-set-the-non-zero-diagonal-elements-of-a-sparse-matrix-in-scipy