Edit: bbtrb's method (using coo_matrix) is much faster than my original suggestion, using nonzero. Sven Marnach's suggestion to use itertools.izip
also improves the speed. Current fastest is using_tocoo_izip
:
import scipy.sparse
import random
import itertools
def using_nonzero(x):
rows,cols = x.nonzero()
for row,col in zip(rows,cols):
((row,col), x[row,col])
def using_coo(x):
cx = scipy.sparse.coo_matrix(x)
for i,j,v in zip(cx.row, cx.col, cx.data):
(i,j,v)
def using_tocoo(x):
cx = x.tocoo()
for i,j,v in zip(cx.row, cx.col, cx.data):
(i,j,v)
def using_tocoo_izip(x):
cx = x.tocoo()
for i,j,v in itertools.izip(cx.row, cx.col, cx.data):
(i,j,v)
N=200
x = scipy.sparse.lil_matrix( (N,N) )
for _ in xrange(N):
x[random.randint(0,N-1),random.randint(0,N-1)]=random.randint(1,100)
yields these timeit
results:
% python -mtimeit -s'import test' 'test.using_tocoo_izip(test.x)'
1000 loops, best of 3: 670 usec per loop
% python -mtimeit -s'import test' 'test.using_tocoo(test.x)'
1000 loops, best of 3: 706 usec per loop
% python -mtimeit -s'import test' 'test.using_coo(test.x)'
1000 loops, best of 3: 802 usec per loop
% python -mtimeit -s'import test' 'test.using_nonzero(test.x)'
100 loops, best of 3: 5.25 msec per loop
与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…