What I'm trying to achieve:
l = []
n = 100
while (n):
l.append(n//2)
n //= 2
print(l)
# [50, 25, 12, 6, 3, 1, 0]
What I've tried:
>>> from itertools import takewhile
>>> n = 100
>>> [(n := n//2) for _ in takewhile(lambda x: x > 0, [n] * n)]
[50, 25, 12, 6, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, ... 0, 0, 0]
This clearly doesn't work, and I also don't like the idea of creating an array of size n...
from math import log2, floor
n = 100
l = [n//(2**x) for x in range(1, floor(log2(n))+2)]
This will leave l as [50, 25, 12, 6, 3, 1, 0].
If you don't want the 0, change the +2 to +1.
You can use numpy to get vectorized solution:
import numpy as np
n= 100
arr = np.array(n * np.ones(int(np.log(n)/np.log(2))))
arr = np.floor(arr / 2**(np.arange(len(arr)) + 1))
Outputs:
[50. 25. 12. 6. 3. 1.]
Brief explanation:
Size of array you're looking for is: log_2(n), which I calculate from: int(np.log(n)/np.log(2)) which in essence returns highest integer exponent satisfying the inequality: 2^x < n
Then you can just restate the problem as sum[i=1 to x](n//2^i)