Python function to find indexes of 1s in binary array and

Question

I have an array which looks like this

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

And I want to get those indexes that have 1 in it. So here I would get an array of <

Answer 1

Here's one compact way -

2**np.where(a)[0]

Sample run -

In [83]: a = np.array([1, 0, 1 , 0 , 0, 1])

In [84]: 2**np.where(a)[0]
Out[84]: array([ 1,  4, 32])

Answer 2

There is np.where, np.argwhere, np.nonzero and np.flatnonzero. Take your pick.

np.where and np.nonzero are as far as I know equivalent, only there is also a three argument version of where that does something different. np.argwhere is the transpose and np.flatnonzero gives flat indices.

Answer 3

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

output:

[1, 4, 32]