On Jul 13, 5:17 am, Paul McGuire <[EMAIL PROTECTED]> wrote: > On Jul 12, 5:34 pm, Godzilla <[EMAIL PROTECTED]> 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 -- http://mail.python.org/mailman/listinfo/python-list
