interleave string

[Andrea Crotti]

Just a curiosity not a real problem, I want to pass from a string like

xxaabbddee
to
xx:aa:bb:dd:ee

so every two characters insert a ":".
At the moment I have this ugly inliner
interleaved = ':'.join(orig[x:x+2] for x in range(0, len(orig), 2))

but also something like this would work
[''.join((x,y)) for x, y in zip(orig[0::2], orig[1::2])]

any other ideas?

 

[Wojciech Mu³a]

Just a curiosity not a real problem, I want to pass from a string like

xxaabbddee
to
xx:aa:bb:dd:ee

so every two characters insert a ":".
At the moment I have this ugly inliner
interleaved = ':'.join(orig[x:x+2] for x in range(0,
len(orig), 2))

but also something like this would work
[''.join((x,y)) for x, y in zip(orig[0::2], orig[1::2])]

any other ideas?

import re

s = 'xxaabbddee'
m = re.compile("(..)")
s1 = m.sub("\\1:", s)[:-1]

w.

 

[Alex Willmer]

import re

s = 'xxaabbddee'
m = re.compile("(..)")
s1 = m.sub("\\1:", s)[:-1]

One can modify this slightly:

s = 'xxaabbddee'
m = re.compile('..')
s1 = ':'.join(m.findall(s))

Depending on one's taste this could be clearer. The more general
answer, from the itertools docs:

from itertools import izip_longest

def grouper(n, iterable, fillvalue=None):
"grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)

s2 = ':'.join(''.join(pair) for pair in grouper(2, s, ''))

Note that this behaves differently to the previous solutions, for
sequences with an odd length.

 

[alex23]

At the moment I have this ugly inliner
        interleaved = ':'.join(orig[x:x+2] for x in range(0, len(orig), 2))

I actually prefer this over every other solution to date. If you feel
its too much behaviour in one line, I sometimes break it out into
separate values to provide some in-code documentation:

s = "xxaabbddee"
get_two_chars_at = lambda i: s[i:i+2]
string_index = xrange(0, len(s), 2)
':'.join(get_two_chars_at(i) for i in string_index)

'xx:aa:bb:dd:ee'

 

[Steven D'Aprano]

At the moment I have this ugly inliner
        interleaved = ':'.join(orig[x:x+2] for x in range(0,
        len(orig), 2))

I actually prefer this over every other solution to date.

Agreed. To me, it's the simplest, least contrived way to solve the
problem.