Accessing items in an collections.OrderedDict by index

If its an OrderedDict() you can easily access the elements by indexing by getting the tuples of (key,value) pairs as follows

>>> import collections>>> d = collections.OrderedDict()>>> d['foo'] = 'python'>>> d['bar'] = 'spam'>>> d.items()[('foo', 'python'), ('bar', 'spam')]>>> d.items()[0]('foo', 'python')>>> d.items()[1]('bar', 'spam')

Note for 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

>>> items = list(d.items())>>> items[('foo', 'python'), ('bar', 'spam')]>>> items[0]('foo', 'python')>>> items[1]('bar', 'spam')

Do you have to use an OrderedDict or do you specifically want a map-like type that's ordered in some way with fast positional indexing? If the latter, then consider one of Python's many sorted dict types (which orders key-value pairs based on key sort order). Some implementations also support fast indexing. For example, the sortedcontainers project has a SortedDict type for just this purpose.

>>> from sortedcontainers import SortedDict>>> sd = SortedDict()>>> sd['foo'] = 'python'>>> sd['bar'] = 'spam'>>> print sd.iloc[0] # Note that 'bar' comes before 'foo' in sort order.'bar'>>> # If you want the value, then simple do a key lookup:>>> print sd[sd.iloc[1]]'python'

Here is a special case if you want the first entry (or close to it) in an OrderedDict, without creating a list. (This has been updated to Python 3):

>>> from collections import OrderedDict>>> >>> d = OrderedDict()>>> d["foo"] = "one">>> d["bar"] = "two">>> d["baz"] = "three">>> next(iter(d.items()))('foo', 'one')>>> next(iter(d.values()))'one'

(The first time you say "next()", it really means "first.")

In my informal test, next(iter(d.items())) with a small OrderedDict is only a tiny bit faster than items()[0]. With an OrderedDict of 10,000 entries, next(iter(d.items())) was about 200 times faster than items()[0].

BUT if you save the items() list once and then use the list a lot, that could be faster. Or if you repeatedly { create an items() iterator and step through it to to the position you want }, that could be slower.