使用有条件列表理解获取索引

发布于 2025-01-28 19:24:26 字数 346 浏览 0 评论 0原文

我有以下NP.array：

my_array=np.array([False, False, False, True, True, True, False, True, False, False, False, False, True])

如何使用与真实元素相对应的索引的列表理解列表。的输出将是[3,4,5,7,12]

我要寻找

cols = [index if feature_condition==True for index, feature_condition in enumerate(my_array)]

在这种情况下，

原文

I have the following np.array:

my_array=np.array([False, False, False, True, True, True, False, True, False, False, False, False, True])

How can I make a list using list comprehension of the indices corresponding to the True elements. In this case the output I'm looking for would be [3,4,5,7,12]

I've tried the following:

cols = [index if feature_condition==True for index, feature_condition in enumerate(my_array)]

But is not working

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

绿阴红影里的.如风往事 2025-02-04 19:24:26

为什么要特别是列表理解？

>>> np.where(my_array==True)
(array([ 3,  4,  5,  7, 12]),)

这可以完成工作，并且更快。列表解决方案将是：

>>> [index for index, feature_condition in enumerate(my_array) if feature_condition == True]
[3, 4, 5, 7, 12]

可接受的答案解释了订购的混乱：在列表中理解

我对差异感到好奇：

def np_time(array):
     np.where(my_array==True)
def list_time(array):
    [index for index, feature_condition in enumerate(my_array) if feature_condition == True]

timeit.timeit(lambda: list_time(my_array),number = 1000)
0.007574789000500459
timeit.timeit(lambda: np_time(my_array),number = 1000)
0.0010812399996211752

why specifically a list comprehension?

>>> np.where(my_array==True)
(array([ 3,  4,  5,  7, 12]),)

this does the job and is quicker. The list solution would be:

>>> [index for index, feature_condition in enumerate(my_array) if feature_condition == True]
[3, 4, 5, 7, 12]

Accepted answer of this explains the confusion of the ordering: if/else in a list comprehension

I was curious of the differences:

def np_time(array):
     np.where(my_array==True)
def list_time(array):
    [index for index, feature_condition in enumerate(my_array) if feature_condition == True]

timeit.timeit(lambda: list_time(my_array),number = 1000)
0.007574789000500459
timeit.timeit(lambda: np_time(my_array),number = 1000)
0.0010812399996211752

回复收藏 0 原文

献世佛 2025-02-04 19:24:26

如果不正确的顺序，则应在最后 -

$more numpy_1.py
import numpy as np
my_array=np.array([False, False, False, True, True, True, False, True, False, False, False, False, True])
print (my_array)

cols = [index for index, feature_condition in enumerate(my_array) if feature_condition]

print (cols)

$python numpy_1.py
[False False False  True  True  True False  True False False False False
  True]
[3, 4, 5, 7, 12]

The order of if is not correct, should be in last -

$more numpy_1.py
import numpy as np
my_array=np.array([False, False, False, True, True, True, False, True, False, False, False, False, True])
print (my_array)

cols = [index for index, feature_condition in enumerate(my_array) if feature_condition]

print (cols)

$python numpy_1.py
[False False False  True  True  True False  True False False False False
  True]
[3, 4, 5, 7, 12]

回复收藏 0 原文

~没有更多了~