Iterate over a string 2 (or n) characters at a time in Python

Question

Earlier today I needed to iterate over a string 2 characters at a time for parsing a string formatted like \"+c-R+D-E\" (there are a few extra letters).

Answer 1

from itertools import izip_longest
def grouper(iterable, n, fillvalue=None):
    args = [iter(iterable)] * n
    return izip_longest(*args, fillvalue=fillvalue)
def main():
    s = "+c-R+D-e"
    for item in grouper(s, 2):
        print ' '.join(item)
if __name__ == "__main__":
    main()
##output
##+ c
##- R
##+ D
##- e

izip_longest requires Python 2.6( or higher). If on Python 2.4 or 2.5, use the definition for izip_longest from the document or change the grouper function to:

from itertools import izip, chain, repeat
def grouper(iterable, n, padvalue=None):
    return izip(*[chain(iterable, repeat(padvalue, n-1))]*n)

Answer 2

>>> s = "+c-R+D-e"
>>> s
'+c-R+D-e'
>>> s[::2]
'+-+-'
>>>

Answer 3

Great opportunity for a generator. For larger lists, this will be much more efficient than zipping every other elemnent. Note that this version also handles strings with dangling ops

def opcodes(s):
    while True:
        try:
            op   = s[0]
            code = s[1]
            s    = s[2:]
        except IndexError:
            return
        yield op,code        


for op,code in opcodes("+c-R+D-e"):
   print op,code

edit: minor rewrite to avoid ValueError exceptions.

Answer 4

Maybe this would be cleaner?

s = "+c-R+D-e"
for i in xrange(0, len(s), 2):
    op, code = s[i:i+2]
    print op, code

You could perhaps write a generator to do what you want, maybe that would be more pythonic :)

Answer 5

This approach support an arbitrary number of elements per result, evaluates lazily, and the input iterable can be a generator (no indexing is attempted):

import itertools

def groups_of_n(n, iterable):
    c = itertools.count()
    for _, gen in itertools.groupby(iterable, lambda x: c.next() / n):
        yield gen

Any left-over elements are returned in a shorter list.

Example usage:

for g in groups_of_n(4, xrange(21)):
    print list(g)

[0, 1, 2, 3]
[4, 5, 6, 7]
[8, 9, 10, 11]
[12, 13, 14, 15]
[16, 17, 18, 19]
[20]

Answer 6

Here's my answer, a little bit cleaner to my eyes:

for i in range(0, len(string) - 1):
    if i % 2 == 0:
        print string[i:i+2]