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

Question

There\'s an existing function that ends in the following, where d is a dictionary:

return d.iteritems()

that returns an unsort

Answer 1

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]