Shortest way to get first item of `OrderedDict` in Python 3

Question

What's the shortest way to get first item of OrderedDict in Python 3?

My best:

list(ordered_dict.items())[0]

Quite long and ugly.

I can think of:

next(iter(ordered_dict.items()))       # Fixed, thanks Ashwini

But it's not very self-describing.

Any better suggestions?

Answer 1

Programming Practices for Readabililty

In general, if you feel like code is not self-describing, the usual solution is to factor it out into a well-named function:

def first(s):
    '''Return the first element from an ordered collection
       or an arbitrary element from an unordered collection.
       Raise StopIteration if the collection is empty.
    '''
    return next(iter(s))

With that helper function, the subsequent code becomes very readable:

>>> extension = {'xml', 'html', 'css', 'php', 'xhmtl'}
>>> one_extension = first(extension)

Patterns for Extracting a Single Value from Collection

The usual ways to get an element from a set, dict, OrderedDict, generator, or other non-indexable collection are:

for value in some_collection:
    break

and:

value = next(iter(some_collection))

The latter is nice because the next() function lets you specify a default value if collection is empty or you can choose to let it raise an exception. The next() function is also explicit that it is asking for the next item.

Alternative Approach

If you actually need indexing and slicing and other sequence behaviors (such as indexing multiple elements), it is a simple matter to convert to a list with list(some_collection) or to use [itertools.islice()][2]:

s = list(some_collection)
print(s[0], s[1])

s = list(islice(n, some_collection))
print(s)