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

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')

Answer 2

If you're dealing with fixed number of keys that you know in advance, use Python's inbuilt namedtuples instead. A possible use-case is when you want to store some constant data and access it throughout the program by both indexing and specifying keys.

import collections
ordered_keys = ['foo', 'bar']
D = collections.namedtuple('D', ordered_keys)
d = D(foo='python', bar='spam')

Access by indexing:

d[0] # result: python
d[1] # result: spam

Access by specifying keys:

d.foo # result: python
d.bar # result: spam

Or better:

getattr(d, 'foo') # result: python
getattr(d, 'bar') # result: spam

Answer 3

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'

Answer 4

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.

Answer 5

This community wiki attempts to collect existing answers.

Python 2.7

In python 2, the keys(), values(), and items() functions of OrderedDict return lists. Using values as an example, the simplest way is

d.values()[0]  # "python"
d.values()[1]  # "spam"

For large collections where you only care about a single index, you can avoid creating the full list using the generator versions, iterkeys, itervalues and iteritems:

import itertools
next(itertools.islice(d.itervalues(), 0, 1))  # "python"
next(itertools.islice(d.itervalues(), 1, 2))  # "spam"

The indexed.py package provides IndexedOrderedDict, which is designed for this use case and will be the fastest option.

from indexed import IndexedOrderedDict
d = IndexedOrderedDict({'foo':'python','bar':'spam'})
d.values()[0]  # "python"
d.values()[1]  # "spam"

Using itervalues can be considerably faster for large dictionaries with random access:

$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 1000;   d = OrderedDict({i:i for i in range(size)})'  'i = randint(0, size-1); d.values()[i:i+1]'
1000 loops, best of 3: 259 usec per loop
$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 10000;  d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i:i+1]'
100 loops, best of 3: 2.3 msec per loop
$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 100000; d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i:i+1]'
10 loops, best of 3: 24.5 msec per loop

$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 1000;   d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); next(itertools.islice(d.itervalues(), i, i+1))'
10000 loops, best of 3: 118 usec per loop
$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 10000;  d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); next(itertools.islice(d.itervalues(), i, i+1))'
1000 loops, best of 3: 1.26 msec per loop
$ python2 -m timeit -s 'from collections import OrderedDict; from random import randint; size = 100000; d = OrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); next(itertools.islice(d.itervalues(), i, i+1))'
100 loops, best of 3: 10.9 msec per loop

$ python2 -m timeit -s 'from indexed import IndexedOrderedDict; from random import randint; size = 1000;   d = IndexedOrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i]'
100000 loops, best of 3: 2.19 usec per loop
$ python2 -m timeit -s 'from indexed import IndexedOrderedDict; from random import randint; size = 10000;  d = IndexedOrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i]'
100000 loops, best of 3: 2.24 usec per loop
$ python2 -m timeit -s 'from indexed import IndexedOrderedDict; from random import randint; size = 100000; d = IndexedOrderedDict({i:i for i in range(size)})' 'i = randint(0, size-1); d.values()[i]'
100000 loops, best of 3: 2.61 usec per loop

+--------+-----------+----------------+---------+
|  size  | list (ms) | generator (ms) | indexed |
+--------+-----------+----------------+---------+
|   1000 | .259      | .118           | .00219  |
|  10000 | 2.3       | 1.26           | .00224  |
| 100000 | 24.5      | 10.9           | .00261  |
+--------+-----------+----------------+---------+

Python 3.6

Python 3 has the same two basic options (list vs generator), but the dict methods return generators by default.

List method:

list(d.values())[0]  # "python"
list(d.values())[1]  # "spam"

Generator method:

import itertools
next(itertools.islice(d.values(), 0, 1))  # "python"
next(itertools.islice(d.values(), 1, 2))  # "spam"

Python 3 dictionaries are an order of magnitude faster than python 2 and have similar speedups for using generators.

+--------+-----------+----------------+---------+
|  size  | list (ms) | generator (ms) | indexed |
+--------+-----------+----------------+---------+
|   1000 | .0316     | .0165          | .00262  |
|  10000 | .288      | .166           | .00294  |
| 100000 | 3.53      | 1.48           | .00332  |
+--------+-----------+----------------+---------+

Answer 6

If you have pandas installed, you can convert the ordered dict to a pandas Series. This will allow random access to the dictionary elements.

>>> import collections
>>> import pandas as pd
>>> d = collections.OrderedDict()
>>> d['foo'] = 'python'
>>> d['bar'] = 'spam'

>>> s = pd.Series(d)

>>> s['bar']
spam
>>> s.iloc[1]
spam
>>> s.index[1]
bar