问题描述
我想用数组中给定的标量划分稀疏矩阵的行.
I want to divide a sparse matrix's rows by scalars given in an array.
例如，我有一个 csr_matrix C :
For example, I have a csr_matrix C :
C = [[2,4,6], [5,10,15]]
D = [2,5]
我希望 C 除法后的结果是:
I want the result of C after division to be :
result = [[1, 2, 3], [1, 2, 3]]
我使用我们用于 numpy 数组的方法进行了尝试:
I have tried this using the method that we use for numpy arrays:
result = C / D[:,None]
但这似乎真的很慢.如何在稀疏矩阵中有效地做到这一点?
But this seems really slow. How to do this efficiently in sparse matrices?
推荐答案
方法 #1
这是使用手动复制和 indexing 的稀疏矩阵解决方案 -
Here's a sparse matrix solution using manual replication with indexing -
from scipy.sparse import csr_matrix
r,c = C.nonzero()
rD_sp = csr_matrix(((1.0/D)[r], (r,c)), shape=(C.shape))
out = C.multiply(rD_sp)
输出是一个稀疏矩阵，与创建完整矩阵的 C/D[:,None] 的输出相反.因此，所提出的方法节省了内存.
The output is a sparse matrix as well as opposed to the output from C / D[:,None] that creates a full matrix. As such, the proposed approach saves on memory.
使用 np.repeat 而不是索引复制可能会提高性能 -
Possible performance boost with replication using np.repeat instead of indexing -
val = np.repeat(1.0/D, C.getnnz(axis=1))
rD_sp = csr_matrix((val, (r,c)), shape=(C.shape))
方法#2
另一种方法可能涉及稀疏矩阵的 data 方法，它为我们提供了对 in-place 结果的稀疏矩阵的扁平化视图，并且还避免使用 非零，就像这样 -
Another approach could involve data method of the sparse matrix that gives us a flattened view into the sparse matrix for in-place results and also avoid the use of nonzero, like so -
val = np.repeat(D, C.getnnz(axis=1))
C.data /= val
这篇关于Scipy 稀疏矩阵中的行划分的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！