In Python, how do I iterate over a dictionary in sorted key order?

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.

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]

Greg's answer is right. Note that in Python 3.0 you'll have to do

sorted(dict.items())

as iteritems will be gone.