Python - Efficient Function With Scipy Sparse Matrices
Solution 1:
I revised your function and ran it in
import numpy as np
from scipy import sparse
def get_distances(X, Y):
ret=[]
for row in X:
sample = row.A
test = Y.getrow(0).A
dist = np.minimum(sample[0,:], test[0,:]).sum()
ret.append(dist)
return ret
X = [[0,0,2],
[1,0,0],
[3,1,0]]
Y = [[1,0,2]]
XM = sparse.csr_matrix(X)
YM = sparse.csr_matrix(Y)
print( get_distances(XM,YM))
print (np.minimum(XM.A, YM.A).sum(axis=1))
producing
1255:~/mypy$ python3 stack37056258.py
[2, 1, 1]
[211]
np.minimum
takes element wise minimum of two arrays (may be 2d), so I don't need to iterate on columns. I also don't need to use indexing.
minimum
is also implemented for sparse matrices, but I get a segmenation error when I try to apply it to your X
(with 3 rows) and Y
(with 1). If they are the same size this works:
Ys = sparse.vstack((YM,YM,YM))
print(Ys.shape)
print (XM.minimum(Ys).sum(axis=1))
Converting the single row matrix to an array also gets around the error - because it ends up using the dense version, np.minimum(XM.todense(), YM.A)
.
print (XM.minimum(YM.A).sum(axis=1))
When I try other element by element operations on these 2 matrices I get ValueError: inconsistent shapes
, e.g. XM+YM
, or XM<YM
. Looks like sparse does not implement broadcasting as numpy
arrays does.
=======================
Comparison of ways of replicating a 1 row sparse matrix many times
In [271]: A=sparse.csr_matrix([0,1,0,0,1])
In [272]: timeit sparse.vstack([A]*3000).A
10 loops, best of 3: 32.3 ms per loop
In [273]: timeit sparse.kron(A,np.ones((3000,1),int)).A
1000 loops, best of 3: 1.27 ms per loop
For many times, kron
is better than vstack
.
=======================
There's an overlap in issues with Scipy sparse matrix alternative for getrow()
Solution 2:
Try below code for sparse matrix:
from scipy.sparse import csr_matrix, vstack
X = csr_matrix([[0,0,2],[1,0,0],[3,1,0]])
Y = csr_matrix([[1,0,2]])
def matrix_dist(x,y):
y=vstack([y]*x.shape[1])
return (((x+y)-(x-y).multiply((x-y).sign())).sum(1)/2).A.ravel()
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