Log base 2 of large integers

[Mok-Kong Shen]

Am 13.08.2014 15:32, schrieb Steven D'Aprano:
> 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?

That is too inaccurate (e.g. for 513 above) for me, I would like
to get accuracy around 0.01 and that for very large n.

M. K. Shen