Question

I would like to split a string according to the title in a single call. I’m looking for a simple syntax using list comprehension, but i don’t got it yet:

s = "123456"

And the result would be:

["12", "34", "56"]

What i don’t want:

re.split('(?i)([0-9a-f]{2})', s)
s[0:2], s[2:4], s[4:6]
[s[i*2:i*2+2] for i in len(s) / 2]

Edit:

Ok, i wanted to parse a hex RGB[A] color (and possible other color/component format), to extract all the component.
It seem that the fastest approach would be the last from sven-marnach:

1. sven-marnach xrange: 0.883 usec per loop

   python -m timeit -s 's="aabbcc";' '[int(s[i:i+2], 16) / 255. for i in xrange(0, len(s), 2)]'
   

2. pair/iter: 1.38 usec per loop

   python -m timeit -s 's="aabbcc"' '["%c%c" % pair for pair in zip(* 2 * [iter(s)])]'
   

3. Regex: 2.55 usec per loop

   python -m timeit -s 'import re; s="aabbcc"; c=re.compile("(?i)([0-9a-f]{2})"); 
   split=re.split' '[int(x, 16) / 255. for x in split(c, s) if x != ""]'
   

Answer 1

Reading through the comments, it turns out the actual question is: What is the fastest way to parse a color definition string in hexadecimal RRGGBBAA format. Here are some options:

def rgba1(s, unpack=struct.unpack):
    return unpack("BBBB", s.decode("hex"))

def rgba2(s, int=int, xrange=xrange):
    return [int(s[i:i+2], 16) for i in xrange(0, 8, 2)]

def rgba3(s, int=int, xrange=xrange):
    x = int(s, 16)
    return [(x >> i) & 255 for i in xrange(0, 32, 8)]

As I expected, the first version turns out to be fastest:

In [6]: timeit rgba1("aabbccdd")
1000000 loops, best of 3: 1.44 us per loop

In [7]: timeit rgba2("aabbccdd")
100000 loops, best of 3: 2.43 us per loop

In [8]: timeit rgba3("aabbccdd")
100000 loops, best of 3: 2.44 us per loop