bit manipulation - Python How to get set bit -

say have number in memory looks (or similar, don't know how python stores numbers):

0000 0000 0000 1000 

i want bit set (thanks @muddyfish, clarify these numbers have 1 bit set), in case 3, because 2^3 = 8. use log2 achieve this, seems solved 'bit twiddling' techniques. loop through bits this:

def getsetbit(num):     if num == 0:         return -1      bit = 0     while num != 1:         bit += 1         num >>= 1      return bit  # getbitset(0b100000) == 5 # getbitset(1) == 0 

however me seems awkward , slower should be. if way this, guess live it, if wasn't.

to bit number of highest bit set, use

int.bit_length()-1 

this more efficient using math.log() or other function had posted

edit:

as requested, timeit results posted:

python -m timeit -s 'import math;x=100' 'int(math.log(x,2))'

1000000 loops, best of 3: 0.5 usec per loop

python -m timeit -s 'i = 100' 'i.bit_length()-1'

10000000 loops, best of 3: 0.106 usec per loop

also, using math.log yeild inaccurate results values higher 2**29.