Identifying the values which have the maximum number of keys

Question

I have a dict which goes something like this:

 ip = { "1" : ['a','b'],
      "2" : ['a','c'],
      "3" : ['a','b','c','d'],
      "4" : ['a','b','d','e']}

I need to find which of the items in the value sets have maximum number of keys against them and also have the items listed in descending order. The output will be something like :

op = {"a":4,"b":3,"c":2,"d":2,"e":1}

But I read somewhere that the dict cannot be in a sorted fashion, so the output can be a tuple too:

op = [('a', 4), ('b', 3), ('c', 2), ('d', 2), ('e', 1)]

We can iterate through the dict and for each of the items in the value set increment the result in a defaultdict for that item.

op = defaultdict(int)
for k,v in ip.iteritems():
    for item in v:
        op[item]+=1
op = sorted(op.items(), key=lambda x: x[1], reverse=True)

Is there a quicker/better way of doing this than the nested for?

Answer 1

Simply use Counter and chain.from_iterable

In [9]: from collections import Counter

In [10]: from itertools import chain

In [11]: ip = { "1" : ['a','b'],
    ...:       "2" : ['a','c'],
    ...:       "3" : ['a','b','c','d'],
    ...:       "4" : ['a','b','d','e']}

In [12]: Counter(chain.from_iterable(ip.values()))
Out[12]: Counter({'a': 4, 'b': 3, 'c': 2, 'd': 2, 'e': 1})

To remove duplicate value you can always do something like this:

>>> from operator import itemgetter
>>> sorted(Counter(chain.from_iterable(map(set, ip.values()))).items(), key=itemgetter(1), reverse=True)
[('a', 4), ('b', 3), ('c', 2), ('d', 2), ('e', 1)]