将scipy稀疏行矩阵添加到另一个稀疏矩阵

I have a csr_matrix A of shape (70000, 80000) and another csr_matrix Bof shape (1, 80000). How can I efficiently add B to every row of A? One idea is to somehow create a sparse matrix B' which is rows of B repeated, but numpy.repeat does not work and using a matrix of ones to create B' is very memory inefficient.

我有一个形状的csr_matrix A(70000,80000)和另一个csr_matrix Bof形状(1,80000)。如何有效地将B添加到A的每一行?一种想法是以某种方式创建稀疏矩阵B',其是B行重复,但是numpy.repeat不起作用并且使用1的矩阵来创建B'是非常低效的内存。

I also tried iterating through every row of A and adding B to it, but that again is very time inefficient.

我也尝试迭代A的每一行并向其添加B,但这又是非常低效的时间。

Update: I tried something very simple which seems to be very efficient than the ideas I mentioned above. The idea is to use scipy.sparse.vstack:

更新:我尝试了一些非常简单的东西,它似乎比我上面提到的想法非常有效。想法是使用scipy.sparse.vstack:

C = sparse.vstack([B for x in range(A.shape[0])])
A + C

This performs well for my task! Few more realizations: I initially tried an iterative approach where I called vstackmultiple times, this approach is slower than calling it just once.

这对我的任务表现很好!几乎没有实现:我最初尝试了一种迭代方法,我称之为vstackmultiple次,这种方法比仅调用一次要慢。

1 个解决方案

#1

A + B[np.zeros(A.shape[0])] is another way to expand B to the same shape as A.

A + B [np.zeros(A.shape [0])]是将B扩展为与A相同形状的另一种方法。

It has about the same performance and memory footprint as Warren Weckesser's solution:

它与Warren Weckesser的解决方案具有相同的性能和内存占用:

import numpy as np
import scipy.sparse as sparse

N, M = 70000, 80000
A = sparse.rand(N, M, density=0.001).tocsr()
B = sparse.rand(1, M, density=0.001).tocsr()

In [185]: %timeit u = sparse.csr_matrix(np.ones((A.shape[0], 1), dtype=B.dtype)); Bp = u * B; A + Bp
1 loops, best of 3: 284 ms per loop

In [186]: %timeit A + B[np.zeros(A.shape[0])]
1 loops, best of 3: 280 ms per loop

and appears to be faster than using sparse.vstack:

并且似乎比使用sparse.vstack更快:

In [187]: %timeit A + sparse.vstack([B for x in range(A.shape[0])])
1 loops, best of 3: 606 ms per loop

#1