Print a dict sorted by values

One can take advantage of the fact that sort works on tuples by considering the first element as more important than the second etc:

d = { "a":4, "c":3, "b":12 }
d_view = [ (v,k) for k,v in d.iteritems() ]
d_view.sort(reverse=True) # natively sort tuples by first element
for v,k in d_view:
    print "%s: %d" % (k,v)

Output:

b: 12
a: 4
c: 3

EDIT: one-liner, generator expression:

sorted( ((v,k) for k,v in d.iteritems()), reverse=True)

Output:

[(12, 'b'), (4, 'a'), (3, 'c')]

>>> d = { "a":4, "c":3, "b":12 }
>>> from operator import itemgetter
>>> lst = sorted(d.iteritems(), key=itemgetter(1))
>>> for t in lst: print '%s : %d' % (t[0], t[1])
... 
c : 3
a : 4
b : 12

You can use the key parameter of sorted to sort by the 2nd item:

>>> d = { "a":4, "c":3, "b":12 }
>>> from operator import itemgetter
>>> for k, v in sorted(d.items(), key=itemgetter(1)):
    print k, v


c 3
a 4
b 12
>>> 

EDIT: one-liner, in opposite order:

>>> d = {"a": 4, "c": 3, "b": 12}
>>> [(k, v) for k, v in sorted(d.items(), key=lambda x: x[1], reverse=True)]
[('b', 12), ('a', 4), ('c', 3)]
>>> 

Here's a solution with Python 3:

d = { "a":4, "c":3, "b":12 }
sorted(((v, k) for k, v in d.items()), reverse=True)