Accessing items in an collections.OrderedDict by index

Question

Lets say I have the following code:

import collections
d = collections.OrderedDict()
d[\'foo\'] = \'python\'
d[\'bar\'] = \'spam\'

Is there

Answer 1

for OrderedDict() you can access the elements by indexing by getting the tuples of (key,value) pairs as follows or using '.values()'

>>> import collections
>>> d = collections.OrderedDict()
>>> d['foo'] = 'python'
>>> d['bar'] = 'spam'
>>> d.items()
[('foo', 'python'), ('bar', 'spam')]
>>>d.values()
odict_values(['python','spam'])
>>>list(d.values())
['python','spam']

Answer 2

It's a new era and with Python 3.6.1 dictionaries now retain their order. These semantics aren't explicit because that would require BDFL approval. But Raymond Hettinger is the next best thing (and funnier) and he makes a pretty strong case that dictionaries will be ordered for a very long time.

So now it's easy to create slices of a dictionary:

test_dict = {
                'first':  1,
                'second': 2,
                'third':  3,
                'fourth': 4
            }

list(test_dict.items())[:2]

Note: Dictonary insertion-order preservation is now official in Python 3.7.

Answer 3

It is dramatically more efficient to use IndexedOrderedDict from the indexed package.

Following Niklas's comment, I have done a benchmark on OrderedDict and IndexedOrderedDict with 1000 entries.

In [1]: from numpy import *
In [2]: from indexed import IndexedOrderedDict
In [3]: id=IndexedOrderedDict(zip(arange(1000),random.random(1000)))
In [4]: timeit id.keys()[56]
1000000 loops, best of 3: 969 ns per loop

In [8]: from collections import OrderedDict
In [9]: od=OrderedDict(zip(arange(1000),random.random(1000)))
In [10]: timeit od.keys()[56]
10000 loops, best of 3: 104 µs per loop

IndexedOrderedDict is ~100 times faster in indexing elements at specific position in this specific case.