On Fri, 15 Apr 2011 13:58:22 +1000, Chris Angelico wrote:
> The dictionary is potentially a lot larger than this particular set of
> values (it's a mapping of header:value for a row of a user-provided CSV
> file). Does this make a difference to the best option? (Currently I'm
> looking at "likely" figures of 60+ keys in the dictionary and 3-8
> postage options to pick up, but both of those could increase
> significantly before production.)
SIXTY keys?
When you get to sixty thousand keys, it might take a few seconds to
process.
> Efficiency is important, though not king; this whole code will be inside
> a loop. But readability is important too.
>
> I can't just give all of dct.values() to parse_kwdlist; the function
> parses a particular format of text string, and it's entirely possible
> that other values would match that format (some of them are pure
> free-form text). This has to get only the ones starting with Keyword,
> and in order. Steven, the line you suggested:
>
> lst = [parse_kwdlist(dct["Keyword%d"%i]) for i in xrange(1, len(dct)+1)]
>
> will bomb with KeyError when it hits the first one that isn't present, I
> assume. Is there an easy way to say "and when you reach any exception,
> not just StopIteration, return the list"? (I could wrap the whole "lst =
> " in a try/except, but then it won't set lst at all.)
No.
You could use the list comprehension form at the cost of running over the
dict twice:
maxkey = 0
for key in dct:
if key.startswith("Keyword"):
maxkey = max(maxkey, int(key[7:]))
lst = [parse_kwdlist(dct["Keyword%d"%i]) for i in xrange(1, maxkey+1)]
but quite frankly, at this point I'd say, forget the one-liner, do it the
old-fashioned way with a loop. Or change your data structure: often you
can simplify a task drastically just by changing the way you store the
data.
--
Steven
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