Log base 2 of large integers
Mok-Kong Shen
mok-kong.shen at t-online.de
Wed Aug 13 09:46:09 EDT 2014
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
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