How To Get The "next" Item In An Ordereddict?

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), ('banana', 3), ('orange', 2), ('

Solution 1:

If you are OK with accessing those parts of the OrderedDict implementation that are intentionally kept private:

>>>classMyOrderedDict(OrderedDict):...defnext_key(self, key):...next = self._OrderedDict__map[key][1]...ifnextis self._OrderedDict__root:...raise ValueError("{!r} is the last key".format(key))...returnnext[2]...deffirst_key(self):...for key in self: return key...raise ValueError("OrderedDict() is empty")...>>>od = MyOrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])>>>od.next_key("apple")
'banana'
>>>od.next_key("banana")
'orange'
>>>od.next_key("orange")
'pear'
>>>od.next_key("pear")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 5, in next_key
ValueError: 'pear' is the last key
>>>od.first_key()
'apple'

Solution 2:

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<today, foo)).next())]
(datetime.date(2000, 5, 23), 3)

Quite a mouthful.

Solution 3:

Python 3.X

dict.items would return an iterable dict view object rather than a list. We need to wrap the call onto a list in order to make the indexing possible:

>>>from collections import OrderedDict>>>>>>foo = OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])>>>>>>defnext_item(odic, key):...returnlist(odic)[list(odic.keys()).index(key) + 1]...>>>next = next_item(foo, 'apple')>>>print(next, foo[next])
banana 3

Solution 4:

Reworked from your code, this way I guess would be a little better:

import collections as co
import datetime as dt
import itertools as it

foo = co.OrderedDict([
    (dt.date(2000,1,1), 4),
    (dt.date(2000,5,23), 3),
    (dt.date(2000,10,1), 2),
    (dt.date(2000,12,31), 1)
])
today = dt.date(2000,1,30)

fooiter = it.dropwhile(lambda d: d <= today, foo)
printnext(fooiter)
printlist(fooiter)

Basically having iterator at the right place is already enough.

Would be cool to start iteration from any position, but not sure if-how possible. Needs some thought.