How to get the “next” item in an OrderedDict?

Question

I\'m using an OrderedDict to random access a list, but now want the next item in the list from the one that I have:

foo = OrderedDict([(\'apple\', 4         


        

Answer 1

I shudder to think how slow this will be on a list of size, but the only way I've come up with so far...

>>> foo.items()[foo.keys().index('apple') + 1]
('banana', 3)

Edit:

The example was slightly contrived; my actual collection is keyed by dates. If I need the entry after today; found a solution using dropwhile...

>>> foo = OrderedDict([(datetime.date(2000,1,1), 4), (datetime.date(2000,5,23), 3), datetime.date(2000,10,1), 2), (datetime.date(2000,12,31), 1)])
>>> today = datetime.date(2000,1,30)
>>> foo.items()[foo.keys().index((itertools.dropwhile(lambda d: d

Quite a mouthful.