Question

我有一个总是只包含1和0的列表。我需要获得列表中的非零索引列表：

a = [0, 1, 0, 1, 0, 0, 0, 0]
b = []
for i in range(len(a)):
    if a[i] == 1:  b.append(i)
print b

实现这个目标的'pythonic'方法是什么？

Answer 1

[i for i, e in enumerate(a) if e != 0]

Answer 2

并非真正的“新”答案，但numpy也内置了此答案。

import numpy as np
a = [0, 1, 0, 1, 0, 0, 0, 0]
nonzeroind = np.nonzero(a)[0] # the return is a little funny so I use the [0]
print nonzeroind
[1 3]

Answer 3

由于THC4k提到了compress（在python2.7 +中可用）

>>> from itertools import compress, count
>>> x = [0, 1, 0, 1, 0, 0, 0, 0]
>>> compress(count(), x)
<itertools.compress object at 0x8c3666c>   
>>> list(_)
[1, 3]

Answer 4

只是希望为之前的asnwer添加“有趣”输出的解释。结果是一个元组，其中包含矩阵每个维度的索引向量。在这种情况下，用户正在处理numpy中被视为向量的内容，因此输出是带有一个元素的元组。

import numpy as np
a = [0, 1, 0, 1, 0, 0, 0, 0]
nonzeroind = np.nonzero(a) 
print nonzeroind
(array([1, 3]),)

Answer 5

两个答案的时间比较，长度为列表

a = [int(random.random()>0.5) for i in range(10)]
%timeit [i for i, e in enumerate(a) if e != 0]
683 ns ± 14 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
%timeit np.nonzero(a)[0]
4.43 µs ± 102 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

a = [int(random.random()>0.5) for i in range(1000)]
%timeit [i for i, e in enumerate(a) if e != 0]
53.1 µs ± 2.02 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit np.nonzero(a)[0]
73.8 µs ± 2.71 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

a = [int(random.random()>0.5) for i in range(100000)]
%timeit [i for i, e in enumerate(a) if e != 0]
5.86 ms ± 79.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit np.nonzero(a)[0]
6.61 ms ± 14.4 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

列表长度为100000，更改列表中的数字

a = [int(random.random()>0.1) for i in range(100000)]
%timeit [i for i, e in enumerate(a) if e != 0]
6.45 ms ± 28.8 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit np.nonzero(a)[0]
5.74 ms ± 9.25 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

a = [int(random.random()>0.9) for i in range(100000)]
%timeit [i for i, e in enumerate(a) if e != 0]
4.69 ms ± 12.6 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit np.nonzero(a)[0]
5.74 ms ± 6.3 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

1的数量仅影响第一个选项。 np.nonzero（）最好使用大量非零元素。如果长度小于10000，则第一个选项会更快。

如何获取列表中非零元素的索引列表？

5 个答案: