scipy: Adding a sparse vector to a specific row of a sparse matrix - numpy

Question

In python, what is the best way to add a CSR vector to a specific row of a CSR matrix? I found one workaround here, but wondering if there is a better/more efficient way to do this. Would appreciate any help.
Given an NxM CSR matrix A and a 1xM CSR matrix B, and a row index i, the goal is to add B to the i-th row of A efficiently.

Answer 1

The obvious indexed addition does work. It gives a efficiency warning, but that doesn't mean it is the slowest way, just that you shouldn't count of doing this repeatedly. It suggests working with the lil format, but conversion to that and back probably takes more time than performing the addition to the csr matrix.
In [1049]: B.A
Out[1049]:
array([[0, 9, 0, 0, 1, 0],
[2, 0, 5, 0, 0, 9],
[0, 2, 0, 0, 0, 0],
[2, 0, 0, 0, 0, 0],
[0, 9, 5, 3, 0, 7],
[1, 0, 0, 8, 9, 0]], dtype=int32)
In [1051]: B[1,:] += np.array([1,0,1,0,0,0])
/usr/local/lib/python3.5/dist-packages/scipy/sparse/compressed.py:730: SparseEfficiencyWarning: Changing the sparsity structure of a csr_matrix is expensive. lil_matrix is more efficient.
SparseEfficiencyWarning)
In [1052]: B
Out[1052]:
<6x6 sparse matrix of type '<class 'numpy.int32'>'
with 17 stored elements in Compressed Sparse Row format>
In [1053]: B.A
Out[1053]:
array([[0, 9, 0, 0, 1, 0],
[3, 0, 6, 0, 0, 9],
[0, 2, 0, 0, 0, 0],
[2, 0, 0, 0, 0, 0],
[0, 9, 5, 3, 0, 7],
[1, 0, 0, 8, 9, 0]])
As your linked question shows, it is possible to act directly on the attributes of the sparse matrix. His code shows why there's an efficiency warning - in the general case it has to rebuild the matrix attributes.
lil is more efficient for row replacement because it just has to change a sublist in the matrix .data and .rows attributes. A change in one row doesn't change the attributes of any of the others.
That said, IF your addition has the same sparsity as the original row, it is possible change specific elements of the data attribute without reworking .indices or .indptr. Drawing on the linked code
A.data[:idx_start_row : idx_end_row]
is the slice of A.data that will be changed. You need of course the corresponding slice from the 'vector'.
Starting with the In [1049] B
In [1085]: B.indptr
Out[1085]: array([ 0, 2, 5, 6, 7, 11, 14], dtype=int32)
In [1086]: B.data
Out[1086]: array([9, 1, 2, 5, 9, 2, 2, 9, 5, 3, 7, 1, 8, 9], dtype=int32)
In [1087]: B.indptr[[1,2]] # row 1
Out[1087]: array([2, 5], dtype=int32)
In [1088]: B.data[2:5]
Out[1088]: array([2, 5, 9], dtype=int32)
In [1089]: B.indices[2:5] # row 1 column indices
Out[1089]: array([0, 2, 5], dtype=int32)
In [1090]: B.data[2:5] += np.array([1,2,3])
In [1091]: B.A
Out[1091]:
array([[ 0, 9, 0, 0, 1, 0],
[ 3, 0, 7, 0, 0, 12],
[ 0, 2, 0, 0, 0, 0],
[ 2, 0, 0, 0, 0, 0],
[ 0, 9, 5, 3, 0, 7],
[ 1, 0, 0, 8, 9, 0]], dtype=int32)
Notice where the changed values, [3,7,12], are in the lil format:
In [1092]: B.tolil().data
Out[1092]: array([[9, 1], [3, 7, 12], [2], [2], [9, 5, 3, 7], [1, 8, 9]], dtype=object)

Answer 2

csr / csc matrices are efficient for most operations including addition (O(nnz)). However, little changes that affect the sparsity structure such as your example or even switching a single position from 0 to 1 are not because they require a O(nnz) reorganisation of the representation. Values and indices are packed; inserting one, all above need to move.
If you do just a single such operation, my guess would be that you can't easily beat scipy's implementation. However, if you are adding multiple rows for example it may be worthwile first making a sparse matrix of them and then adding that in one go.
Creating a csr matrix by hand from rows, say, is not that difficult. For example if your rows are dense and in order:
row_numbers, indices = np.where(rows)
data = rows[row_numbers, indices]
indptr = np.searchsorted(np.r_[true_row_numbers[row_numbers], N], np.arange(N+1))
If you have a collection of sparse rows and their row numbers:
data = np.r_[tuple([r.data for r in rows])]
indices = np.r_[tuple(r.indices for r in rows])]
jumps = np.add.accumulate([0] + [len(r) for r in rows])
indptr = np.repeat(jumps, np.diff(np.r_[-1, true_row_numbers, N]))