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 turning it into a list first) in a descending fashion, ordered with the highest values first. I created my dictionary like the following:

adict = defaultdict(int)

later on I do a bunch of:

adict['someval'] += 1
adict['anotherval'] +=1
adict['someval'] += 1

Ideally after that I'd like to get a print out of:

someval => 2
anotherval => 1

Answer 1

A dict's keys, reverse-sorted by the corresponding values, can best be gotten as

sorted(adict, key=adict.get, reverse=True)

since you want key/value pairs, you could work on the items as all other answers suggest, or (to use the nifty adict.get bound method instead of itemgetters or weird lambdas;-),

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

Edit: in terms of performance, there isn't much into it either way:

$ python -mtimeit -s'adict=dict((x,x**2) for x in range(-5,6))' '[(k, adict[k]) for k in sorted(adict, key=adict.get, reverse=True)]'
100000 loops, best of 3: 10.8 usec per loop
$ python -mtimeit -s'adict=dict((x,x**2) for x in range(-5,6)); from operator import itemgetter' 'sorted(adict.iteritems(), key=itemgetter(1), reverse=True)'
100000 loops, best of 3: 9.66 usec per loop
$ python -mtimeit -s'adict=dict((x,x**2) for x in range(-5,6))' 'sorted(adict.iteritems(), key=lambda (k,v): v, reverse=True)'
100000 loops, best of 3: 11.5 usec per loop

So, the .get-based solution is smack midway in performance between the two items-based ones -- slightly slower than the itemgetter, slightly faster than the lambda. In "bottleneck" cases, where those microsecond fractions are crucial to you, by all means do focus on that. In normal cases, where this operation is only one step within some bigger task and a microsecond more or less matters little, focusing on the simplicity of the get idiom is, however, also a reasonable alternative.