Thanks for the responses, all! In its strictest sense, itertools.count() seems to be what I'm after, but may not be what I need.
On Fri, Apr 15, 2011 at 12:33 PM, Steven D'Aprano <steve+comp.lang.pyt...@pearwood.info> wrote: > No. But you can use an itertools.count([start=0]) object, and then catch > the KeyError when you pass the end of the dict. But assuming keys are > consecutive, better to do this: > > lst = [parse_kwdlist(dct["Keyword%d"%i]) for i in xrange(1, len(dct)+1)] Ah, didn't think to use len(dct) as the top! And yes, the keys will be consecutive (or rather, if someone omits Keyword5 then I can justifiably ignore Keyword6). > If you don't care about the order of the results: > > lst = [parse_kwdlist(value) for value in dct.values()] No, order will matter (I have to pick up the first N that fit certain conditions, after parsing). 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.) 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.) If not, I think I'll go with: for i in xrange(1,len(dct)+1): and otherwise as per OP. Having a check for "if key%d in dct" going all the way up seems like an odd waste of effort (or maybe I'm wrong there). Chris Angelico -- http://mail.python.org/mailman/listinfo/python-list
