Count the multiple occurrences in a set

if group not in g:
    g[group] = set()
g[group].add(name)

goes through a list of groups with this structure:

Group: 开发者_如何学GoA
Name: Bob

and adds to the set the names of persons belonging to a specific group. The names in the set are unique and we don't know how many similar names there are in a group. So e.g. if there are two 'Bob' names or 5 'Mike' names, how can I count the multiple occurrences of the names as well to have something like this:

Group A: Bob 2, Mike 5
Group B: Jane 4

and so on. Thanks in advance.

---

Looks like you might be better off with a Counter:

>>> from collections import Counter
>>> mylist = ["Bob", "Mike", "Bob", "Mike", "Mike", "Mike", "Bob"]
>>> Counter(mylist)
Counter({'Mike': 4, 'Bob': 3})

---

use a dict of dicts to count, e.g. as follows:

tralala = dict()

for group, name in [('A', 'Bob'), ('B', 'Jane'), ('A', 'Bob')]:
    tralala.setdefault(group, dict()).setdefault(name, 0) 
    tralala[group][name] += 1

print tralala

This results in

{'A': {'Bob': 2}, 'B': {'Jane': 1}}

---

from collections import Counter, defaultdict


lst = [('B', 'Bob'), ('A', 'Andy'), ('C', 'Charles'), ('A', 'Adam'), ('B', 'Abraham')]

# assumes people can appear in more than one group
def groups(lst):
    counter = Counter(lst)
    result = defaultdict(dict)
    for (group, name), value in counter.iteritems():
        result[group][name] = value
    return result