Log base 2 of large integers
Peter Otten
__peter__ at web.de
Wed Aug 13 10:17:47 EDT 2014
Steven D'Aprano wrote:
> 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
For base 2 there is also the bit_length() method:
>>> 511 .bit_length()
9
>>> 512 .bit_length()
10
>>> 513 .bit_length()
10
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