Log base 2 of large integers
Steven D'Aprano
steve+comp.lang.python at pearwood.info
Wed Aug 13 09:32:51 EDT 2014
Mok-Kong Shen wrote:
>
> I like to compute log base 2 of a fairly large integer n but
> with math.log(n,2) I got:
>
> OverflowError: long int too large to convert to float.
>
> Is there any feasible work-around for that?
If you want the integer log2, that is, the floor of log2, the simplest way
is calculate it like this:
def log2(n):
"""Return the floor of log2(n)."""
if n <= 0: raise ValueError
i = -1
while n:
n //= 2
i += 1
return i
log2(511)
=> returns 8
log2(512)
=> returns 9
log2(513)
=> returns 9
Does that help?
--
Steven
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