Iterate through True Bits of Binary Number (Fast)

Question

So I'm trying to make a function in Python to return all of the powers of two used in the binary of a number.

For example: 123 in binary is 1111011. I want my function to return the powers of two corresponding to the True bits of 123 (1, 2, 8, 16, 32 and 64) as fast as possible.

Right now the best I've found is:

def true_bits(num):
    while num:
        temp = num & -num
        num -= temp
        yield temp

Answer 1

Some (non-faster) alternatives:

Using numpy and assuming 8bits unsigned integers:

import numpy as np
def numpy_bits(num):
    bits = np.unpackbits(np.uint8(num), bitorder='little')
    return 2**np.arange(8)[bits.astype(bool)]

numpy_bits(123)
# array([ 1,  2,  8, 16, 32, 64])
# 6.8 µs ± 293 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

Using a python loop (decreasing bit order):

def python_bits(num):
    for i in range(7,-1,-1):
        if num >= (x:=2**i):
            yield x
            num -= x

list(python_bits(123))
# [64, 32, 16, 8, 2, 1]
# 2.26 µs ± 61.3 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

OP's approach:

list(true_bits(123))
# [1, 2, 8, 16, 32, 64]
# 1.14 µs ± 37.1 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)