Fastest way to convert a byte of integer into a list

[Godzilla]

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?

[Paul McGuire]

On Jul 13, 3:46 pm, "mensana...@aol .com" <mensana...@aol .comwrote:

On Jul 13, 5:17 am, Paul McGuire <pt...@austin.r r.comwrote:

On Jul 12, 5:34 pm, Godzilla <godzillais...@ gmail.comwrote:

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(nu m)/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**177 149 - 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- Hide quoted text -

- Show quoted text -

Oof! Pre-calculating those byte bitmasks doesn't help at all! It
would seem it is faster to use a single list comp than to try to sum
together the precalcuated sublists.

I *would* say though that it is somewhat cheating to call the other
two algorithms with the hardcoded range length of 177149, when you
know this is the right range because this is tailored to fit the input
value 2**177149-1. This would be a horrible value to use if the input
number were something small, like 5. I think numBits still helps here
to handle integers of arbitrary length (and only adds a slight
performance penalty since it is called only once).

-- Paul

[mensanator]

On Jul 14, 5:49?pm, Paul McGuire <pt...@austin.r r.comwrote:

On Jul 13, 3:46 pm, "mensana...@aol .com" <mensana...@aol .comwrote:

On Jul 13, 5:17 am, Paul McGuire <pt...@austin.r r.comwrote:

On Jul 12, 5:34 pm, Godzilla <godzillais...@ gmail.comwrote:

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(nu m)/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**177 149 - 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

Oof! Pre-calculating those byte bitmasks doesn't help at all! It
would seem it is faster to use a single list comp than to try to sum
together the precalcuated sublists.

I *would* say though that it is somewhat cheating to call the other
two algorithms with the hardcoded range length of 177149, when you
know this is the right range because this is tailored to fit the input
value 2**177149-1. This would be a horrible value to use if the input
number were something small, like 5. I think numBits still helps here
to handle integers of arbitrary length (and only adds a slight
performance penalty since it is called only once).

I agree. But I didn't want to compare your numBits
against gmpy's numdigits() which I would normally use.
But since the one algorithm required a hardcoded number,
I thought it best to hardcode all three.

I originally coded this stupidly by converting
the number to a string and then converting to
a list of integers. But Matimus had already posted
his which looked a lot better than mine, so I didn't.

But then I got to wondering if Matimus' solution
requires (B**2+B)/2 shift operations for B bits.

My attempt to re-code his solution to only use B
shifts didn't work out and by then you had posted
yours. So I did the 3-way comparison using gmpy's
direct bit comparison. I was actually surprised at
the difference.

I find gmpy's suite of bit manipulation routines
extremely valuable and use them all the time.
That's another reason for my post, to promote gmpy.

It is also extremely important to keep coercion
out of loops. In other words, never use literals,
only pre-coerced constants. For example:

import gmpy
def collatz(n):
ONE = gmpy.mpz(1)
TWO = gmpy.mpz(2)
TWE = gmpy.mpz(3)
while n != ONE:
if n % TWO == ONE:
n = TWE*n + ONE
else:
n = n/TWO
collatz(gmpy.mp z(2**177149-1))

>
-- Paul

[jigloo]

On 7 13 , 6 34 , Godzilla <godzillais...@ gmail.comwrote:

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?

my clone *bin* function from python3000
def _bin(n, count=32):
"""returns the binary of integer n, using count number of digits"""
return ''.join([str((n >i) & 1) for i in range(count-1, -1, -1)])