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问题描述

我想知道最好的方法是用scipy.sparse迭代稀疏矩阵的非零条目.例如,如果我执行以下操作:

I'm wondering what the best way is to iterate nonzero entries of sparse matrices with scipy.sparse. For example, if I do the following:

from scipy.sparse import lil_matrix

x = lil_matrix( (20,1) )
x[13,0] = 1
x[15,0] = 2

c = 0
for i in x:
  print c, i
  c = c+1

输出为

0
1
2
3
4
5
6
7
8
9
10
11
12
13   (0, 0) 1.0
14
15   (0, 0) 2.0
16
17
18
19

所以看起来迭代器正在接触每个元素,而不仅仅是非零条目.我看过了API

so it appears the iterator is touching every element, not just the nonzero entries. I've had a look at the API

http://docs.scipy.org/doc/scipy/reference/generation/scipy.sparse.lil_matrix.html

并进行了一些搜索,但是我似乎找不到有效的解决方案.

and searched around a bit, but I can't seem to find a solution that works.

推荐答案

(使用 coo_matrix )比我最初的建议要快得多. lil_matrix.nonzero.html#scipy-sparse-lil-matrix-nonzero"rel =" noreferrer>非零. Sven Marnach建议使用itertools.izip也可以提高速度.当前最快的是using_tocoo_izip:

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)

产生这些timeit结果:

% 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

这篇关于遍历scipy.sparse向量(或矩阵)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!

09-05 20:03