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

Assuming you are using CPython 2.x and have a large dictionary mydict, then using sorted(mydict) is going to be slow because sorted builds a sorted list of the keys of mydict.

In that case you might want to look at my ordereddict package which includes a C implementation of sorteddict in C. Especially if you have to go over the sorted list of keys multiple times at different stages (ie. number of elements) of the dictionaries lifetime.

http://anthon.home.xs4all.nl/Python/ordereddict/

Answer 2

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

sorted(dict.items())

as iteritems will be gone.

Answer 3

sorted returns a list, hence your error when you try to iterate over it, but because you can't order a dict you will have to deal with a list.

I have no idea what the larger context of your code is, but you could try adding an iterator to the resulting list. like this maybe?:

return iter(sorted(dict.iteritems()))

of course you will be getting back tuples now because sorted turned your dict into a list of tuples

ex: say your dict was: {'a':1,'c':3,'b':2} sorted turns it into a list:

[('a',1),('b',2),('c',3)]

so when you actually iterate over the list you get back (in this example) a tuple composed of a string and an integer, but at least you will be able to iterate over it.

Answer 4

>>> import heapq
>>> d = {"c": 2, "b": 9, "a": 4, "d": 8}
>>> def iter_sorted(d):
        keys = list(d)
        heapq.heapify(keys) # Transforms to heap in O(N) time
        while keys:
            k = heapq.heappop(keys) # takes O(log n) time
            yield (k, d[k])


>>> i = iter_sorted(d)
>>> for x in i:
        print x


('a', 4)
('b', 9)
('c', 2)
('d', 8)

This method still has an O(N log N) sort, however, after a short linear heapify, it yields the items in sorted order as it goes, making it theoretically more efficient when you do not always need the whole list.

Answer 5

In general, one may sort a dict like so:

for k in sorted(d):
    print k, d[k]

For the specific case in the question, having a "drop in replacement" for d.iteritems(), add a function like:

def sortdict(d, **opts):
    # **opts so any currently supported sorted() options can be passed
    for k in sorted(d, **opts):
        yield k, d[k]

and so the ending line changes from

return dict.iteritems()

to

return sortdict(dict)

or

return sortdict(dict, reverse = True)

Answer 6

If you want to sort by the order that items were inserted instead of of the order of the keys, you should have a look to Python's collections.OrderedDict. (Python 3 only)