On 13/08/2014 14:46, Mok-Kong Shen wrote:
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:
<<< removed... see below >>>
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
Well, we can use Steven d'A's idea as a starting point:
import math
def log2_floor(n):
"""Return the floor of log2(n)."""
if n <= 0: raise ValueError
i = -1
while n:
n //= 2
i += 1
return i
def log2(n):
""" return log_2(n) by splitting the problem into the integer and
fractional parts"""
l2f = log2_floor(n)
if n == 2**l2f:
return l2f
else:
return l2f + math.log(n*2**-l2f, 2)
Andrew
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