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From: Ian Kelly <ian.g.kelly@gmail.com>
Date: Wed, 20 Jul 2011 01:29:10 -0600
Subject: =?UTF-8?Q?Re=3A_a_little_parsing_challenge_=E2=98=BA?=
To: jmfauth <wxjmfauth@gmail.com>
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On Wed, Jul 20, 2011 at 12:29 AM, jmfauth <wxjmfauth@gmail.com> wrote:
>> Then it is hard to code precisely.
>>
>
> Not really. The trick is to count the different opener/closer
> separately.
> That is what I am doing to check balanced brackets in
> chemical formulas. The rules are howerver not the same
> as in math.

I think the difficulty is not in the algorithm, but in adhering to the
desired output when it is ambiguously described.

> But, if I want to parse a string from right to left,
> what's the trick?
> The best I found so far:
>
>>>> s =3D 'abcd'
>>>> for i, c in enumerate(reversed(s)):
> ... =A0 =A0 print len(s) - 1 - i, c

That violates DRY, since you have reversal logic in the iterator
algebra and then again in the loop body.  I prefer to keep all such
logic in the iterator algebra, if possible.  This is one possibility,
if you don't mind it building an intermediate list:

>>> for i, c in reversed(list(enumerate(s))):
...

Otherwise, here's another non-DRY solution:

>>> from itertools import izip
>>> for i, c in izip(reversed(xrange(len(s))), reversed(s)):
...

Unfortunately, this is one space where there just doesn't seem to be a
single obvious way to do it.