如何创建交互稀疏矩阵？

Question

假设我有两个稀疏矩阵：

from scipy.sparse import random
from scipy import stats

S0 = random(5000,100, density=0.01)
S1 = random(5000,100,density=0.01)

我想创建一个稀疏矩阵 S2，其中的形状是 (5000,100*100)。（在我的实际应用中，这个'5000'应该是2000万）。对于每一行，它是这两个 100 维向量中的某种交互。

S2 =  some_kind_of_tensor_multiplication(S0 ,S1 )

为了说明 S2[i,j] = S0[i,k0] * S1[i,k1]，我们从 [0,99] 遍历所有 k0,k1 以创建长度为 10000 的第 i 行。我找不到任何有效的方法来实现这一目标。有人能帮忙吗？

低效的方法看起来像，但我认为这将是非常低效的...:[=17=]

result=[]
for i in range(S0.shape[1]):
    for j in range(S1.shape[1]):
        result.append(S0[:,i]*S1[:,j])
result = np.vstack(result).T

类似问题： Special kind of row-by-row multiplication of 2 sparse matrices in Python

我试过了：

import numpy as np

from scipy.sparse import random
from scipy import stats
from scipy import sparse

S0 = random(20000000,100, density=0.01).tocsr()
S1 = random(20000000,100,density=0.01).tocsr()


def test_iter(A, B):
    m,n1 = A.shape
    n2 = B.shape[1]
    Cshape = (m, n1*n2)
    data = np.empty((m,),dtype=object)
    col =  np.empty((m,),dtype=object)
    row =  np.empty((m,),dtype=object)
    for i,(a,b) in enumerate(zip(A, B)):
        data[i] = np.outer(a.data, b.data).flatten()
        #col1 = a.indices * np.arange(1,a.nnz+1) # wrong when a isn't dense
        col1 = a.indices * n2   # correction
        col[i] = (col1[:,None]+b.indices).flatten()
        row[i] = np.full((a.nnz*b.nnz,), i)
    data = np.concatenate(data)
    col = np.concatenate(col)
    row = np.concatenate(row)
    return sparse.coo_matrix((data,(row,col)),shape=Cshape)

尝试：

%%time
S_result = test_iter(S0,S1)

Wall time: 53min 8s .我们有没有更快的方案，谢谢？

Answer 1

这是一个重写，直接使用 csr intptr。它通过直接切片 data 和 indices 来节省时间，而不是每行制作一个全新的 1 行 csr 矩阵：

def test_iter2(A, B): 
    m,n1 = A.shape 
    n2 = B.shape[1] 
    Cshape = (m, n1*n2) 
    data = [] 
    col =  [] 
    row =  [] 
    for i in range(A.shape[0]): 
        slc1 = slice(A.indptr[i],A.indptr[i+1]) 
        data1 = A.data[slc1]; ind1 = A.indices[slc1] 
        slc2 = slice(B.indptr[i],B.indptr[i+1])  
        data2 = B.data[slc2]; ind2 = B.indices[slc2]  
        data.append(np.outer(data1, data2).ravel()) 
        col.append(((ind1*n2)[:,None]+ind2).ravel()) 
        row.append(np.full(len(data1)*len(data2), i)) 
    data = np.concatenate(data) 
    col = np.concatenate(col) 
    row = np.concatenate(row) 
    return sparse.coo_matrix((data,(row,col)),shape=Cshape)

对于较小的测试用例，这可以节省相当多的时间：

In [536]: S0=sparse.random(200,200, 0.01, format='csr')                                                   
In [537]: S1=sparse.random(200,200, 0.01, format='csr')                                                   
In [538]: timeit test_iter(S0,S1)                                                                         
42.8 ms ± 1.7 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [539]: timeit test_iter2(S0,S1)                                                                        
6.94 ms ± 27 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

如何创建交互稀疏矩阵？

How do i create interacting sparse matrix?

python

scipy

sparse-matrix