There\'s an existing function that ends in the following, where d
is a dictionary:
return d.iteritems()
that returns an unsort
You can now use OrderedDict
in Python 2.7 as well:
>>> from collections import OrderedDict
>>> d = OrderedDict([('first', 1),
... ('second', 2),
... ('third', 3)])
>>> d.items()
[('first', 1), ('second', 2), ('third', 3)]
Here you have the what's new page for 2.7 version and the OrderedDict API.
Use the sorted() function:
return sorted(dict.iteritems())
If you want an actual iterator over the sorted results, since sorted()
returns a list, use:
return iter(sorted(dict.iteritems()))
A dict's keys are stored in a hashtable so that is their 'natural order', i.e. psuedo-random. Any other ordering is a concept of the consumer of the dict.
sorted() always returns a list, not a dict. If you pass it a dict.items() (which produces a list of tuples), it will return a list of tuples [(k1,v1), (k2,v2), ...] which can be used in a loop in a way very much like a dict, but it is not in anyway a dict!
foo = {
'a': 1,
'b': 2,
'c': 3,
}
print foo
>>> {'a': 1, 'c': 3, 'b': 2}
print foo.items()
>>> [('a', 1), ('c', 3), ('b', 2)]
print sorted(foo.items())
>>> [('a', 1), ('b', 2), ('c', 3)]
The following feels like a dict in a loop, but it's not, it's a list of tuples being unpacked into k,v:
for k,v in sorted(foo.items()):
print k, v
Roughly equivalent to:
for k in sorted(foo.keys()):
print k, foo[k]
Haven't tested this very extensively, but works in Python 2.5.2.
>>> d = {"x":2, "h":15, "a":2222}
>>> it = iter(sorted(d.iteritems()))
>>> it.next()
('a', 2222)
>>> it.next()
('h', 15)
>>> it.next()
('x', 2)
>>>
If you are used to doing for key, value in d.iteritems(): ...
instead of iterators, this will still work with the solution above
>>> d = {"x":2, "h":15, "a":2222}
>>> for key, value in sorted(d.iteritems()):
>>> print(key, value)
('a', 2222)
('h', 15)
('x', 2)
>>>
With Python 3.x, use d.items()
instead of d.iteritems()
to return an iterator.