Python函数在二进制数组中查找1的索引和 [英] Python function to find indexes of 1s in binary array and

问题描述

我有一个看起来像这样的数组

I have an array which looks like this

[1, 0, 1 , 0 , 0, 1]

我想获取那些包含 1 的索引.所以在这里我会得到一个 [0, 2 , 5] 的数组然后基于它，我将创建一个新数组，该数组采用这些数字并用它们对 2 进行指数运算所以最后的数组是

And I want to get those indexes that have 1 in it. So here I would get an array of [0, 2 , 5] and then based on it I would create a new array that takes these numbers and exponantiate 2 with them So the end array is

[2**0, 2**2, 2**5]

有没有办法尽快写出来?

Is there a way to write it as shortly as possible?

推荐答案

您可以在列表推导式中使用枚举:

you could use enumerate in a list comprehension:

a = [1, 0, 1 , 0 , 0, 1]
b = [2**idx for idx, v in enumerate(a) if v]
b

输出:

[1, 4, 32]

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