Fastest way to convert a byte of integer into a list

[mensanator at aol.com]

On Jul 13, 5:17 am, Paul McGuire <pt... at austin.rr.com> wrote:
> On Jul 12, 5:34 pm, Godzilla <godzillais... at gmail.com> wrote:
>
> > Hello,
>
> > I'm trying to find a way to convert an integer (8-bits long for
> > starters) and converting them to a list, e.g.:
>
> > num = 255
> > numList = [1,1,1,1,1,1,1,1]
>
> > with the first element of the list being the least significant, so
> > that i can keep appending to that list without having to worry about
> > the size of the integer. I need to do this because some of the
> > function call can return a 2 lots of 32-bit numbers. I have to find a
> > way to transport this in a list... or is there a better way?
>
> Standing on the shoulders of previous posters, I put this together.
>
> -- Paul

But aren't we moving backwards? The OP did ask for the fastest way.

I put this together (from other posters and my own):

import gmpy
import time

y = 2**177149 - 1

# init list of tuples by byte
bytebits = lambda num : [num >> i & 1 for i in range(8)]
bytes = [ tuple(bytebits(i)) for i in range(256) ]
# use bytes lookup to get bits in a 32-bit integer
bits = lambda num : sum((bytes[num >> i & 255] for i in range(0,32,8)),
())
# use base-2 log to find how many bits in an integer of arbitrary
length
from math import log,ceil
log_of_2 = log(2)
numBits = lambda num : int(ceil(log(num)/log_of_2))
# expand bits to integers of arbitrary length
arbBits = lambda num : sum((bytes[num >> i & 255] for i in
range(0,numBits(num),8)),())
t0 = time.time()
L = arbBits(y)
t1 = time.time()
print 'Paul McGuire algorithm:',t1-t0

t0 = time.time()
L = [y >> i & 1 for i in range(177149)]
t1 = time.time()
print '     Matimus algorithm:',t1-t0

x = gmpy.mpz(2**177149 - 1)
t0 = time.time()
L = [gmpy.getbit(x,i) for i in range(177149)]
t1 = time.time()
print '  Mensanator algorithm:',t1-t0

##    Paul McGuire algorithm: 17.4839999676
##         Matimus algorithm: 3.28100013733
##      Mensanator algorithm: 0.125