Python 如何获取列表中非零元素的索引列表？
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原文地址: http://stackoverflow.com/questions/4111412/
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How do I get a list of indices of non zero elements in a list?
提问by George Profenza
I have a list that will always contain only ones and zeroes. I need to get a list of the non-zero indices of the list:
我有一个始终只包含 1 和 0 的列表。我需要获取列表的非零索引列表：
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
What would be the 'pythonic' way of achieving this ?
实现这一目标的“pythonic”方式是什么？
采纳答案by Ignacio Vazquez-Abrams
[i for i, e in enumerate(a) if e != 0]
回答by John La Rooy
Since THC4k mentioned compress (available in python2.7+)
由于 THC4k 提到了压缩（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]
回答by Brian Larsen
回答by Lexa
Just wished to add explanation for 'funny' output from the previous asnwer. Result is a tuple that contains vectors of indexes for each dimension of the matrix. In this case user is processing what is considered a vector in numpy, so output is tuple with one element.
只是想为上一个答案的“有趣”输出添加解释。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，因此输出是具有一个元素的元组。
import numpy as np
a = [0, 1, 0, 1, 0, 0, 0, 0]
nonzeroind = np.nonzero(a)
print nonzeroind
(array([1, 3]),)
