python, convert a dictionary to a sorted list by value instead of key

Question

I have a collections.defaultdict(int) that I\'m building to keep count of how many times a key shows up in a set of data. I later want to be able to sort it (obviously by turnin

Answer 1

Note: I'm putting this in as an answer so that it gets seen. I don't want upvotes. If you want to upvote anyone, upvote Nadia.

The currently accepted answer gives timing results which are based on a trivially small dataset (size == 6 - (-5) == 11). The differences in cost of the various methods are masked by the overhead. A use case like what are the most frequent words in a text or most frequent names in a membership list or census involves much larger datasets.

Repeating the experiment with range(-n,n+1) (Windows box, Python 2.6.4, all times in microseconds):

n=5: 11.5, 9.34, 11.3
n=50: 65.5, 46.2, 68.1
n=500: 612, 423, 614

These results are NOT "slightly" different. The itemgetter answer is a clear winner on speed.

There was also mention of "the simplicity of the get idiom". Putting them close together for ease of comparison:

[(k, adict[k]) for k in sorted(adict, key=adict.get, reverse=True)] sorted(adict.iteritems(), key=itemgetter(1), reverse=True)

The get idiom not only looks up the dict twice (as J. F. Sebastian has pointed out), it makes one list (result of sorted()) then iterates over that list to create a result list. I'd call that baroque, not simple. YMMV.