使用矩阵乘法,您可以进行有效的切片,创建一个“切片器”矩阵,其中的矩阵位于正确的位置。切片矩阵将具有相同的type
作为“切片器”,因此您可以有效地控制输出类型。
下面您将看到一些比较,对您来说最有效的方法是询问.A
矩阵并对其进行切片。事实证明它比.toarray()
方法。当“切片器”创建为ndarray
,乘以csr
矩阵并对结果进行切片。
OBS:使用coo
稀疏矩阵A
导致时间稍微慢一些,保持相同的比例,并且sol3
不适用,我后来意识到在乘法中它被转换为csr
自动地。
import scipy
import scipy.sparse.csr as csr
test = csr.csr_matrix([
[11,12,13,14,15,16,17,18,19],
[21,22,23,24,25,26,27,28,29],
[31,32,33,34,35,36,37,38,39],
[41,42,43,44,45,46,47,48,49],
[51,52,53,54,55,56,57,58,59],
[61,62,63,64,65,66,67,68,69],
[71,72,73,74,75,76,77,78,79],
[81,82,83,84,85,86,88,88,89],
[91,92,93,94,95,96,99,98,99]])
def sol1():
B = test.A[2:5]
def sol2():
slicer = scipy.array([[0,0,0,0,0,0,0,0,0],
[0,0,0,0,0,0,0,0,0],
[0,0,1,0,0,0,0,0,0],
[0,0,0,1,0,0,0,0,0],
[0,0,0,0,1,0,0,0,0]])
B = (slicer*test)[2:]
return B
def sol3():
B = (test[2:5]).A
return B
def sol4():
slicer = csr.csr_matrix( ((1,1,1),((2,3,4),(2,3,4))), shape=(5,9) )
B = ((slicer*test).A)[2:] # just changing when we do the slicing
return B
def sol5():
slicer = csr.csr_matrix( ((1,1,1),((2,3,4),(2,3,4))), shape=(5,9) )
B = ((slicer*test)[2:]).A
return B
timeit sol1()
#10000 loops, best of 3: 60.4 us per loop
timeit sol2()
#10000 loops, best of 3: 91.4 us per loop
timeit sol3()
#10000 loops, best of 3: 111 us per loop
timeit sol4()
#1000 loops, best of 3: 310 us per loop
timeit sol5()
#1000 loops, best of 3: 363 us per loop
编辑:答案已更新替换.toarray()
by .A
,提供更快的结果,现在最好的解决方案被放置在顶部