Best way to turn word list into frequency dict

What's the best way to convert a list/tuple into a dict where the keys are the distinct values of the list and the values are the the frequencies of those distinct values?

In other words:

['a', 'b', 'b', 'a', 'b', 'c']
--> 
{'a': 2, 'b': 3, 'c': 1}

(I've had to do something like the above so many times, is there anything in the standard lib that does it for you?)

EDIT:

Jacob Gabrielson points out there is something coming in the standard lib for the 2.7/3.1 branch

From stackoverflow

• Kind of

  from collections import defaultdict
  fq= defaultdict( int )
  for w in words:
      fq[w] += 1
  

  That usually works nicely.

• I have to share an interesting but kind of ridiculous way of doing it that I just came up with:

  >>> class myfreq(dict):
  ...     def __init__(self, arr):
  ...         for k in arr:
  ...             self[k] = 1
  ...     def __setitem__(self, k, v):
  ...         dict.__setitem__(self, k, self.get(k, 0) + v)
  ... 
  >>> myfreq(['a', 'b', 'b', 'a', 'b', 'c'])
  {'a': 2, 'c': 1, 'b': 3}
  

  John Fouhy : (self.get(k) or 0) can be better written as self.get(k, 0)

• This is an abomination, but:

  from itertools import groupby
  dict((k, len(list(xs))) for k, xs in groupby(sorted(items)))
  

  I can't think of a reason one would choose this method over S.Lott's, but if someone's going to point it out, it might as well be me. :)

  ʞɔıu : points for cleverness

• I find that the easiest to understand (while might not be the most efficient) way is to do:

  {i:words.count(i) for i in set(words)}
  

  S.Lott : +1: Got to get me some of that Python 3.0 syntactic sugar.

  Trey Stout : That is pretty hot

  Nathan Ross Powell : Beautiful Python!

• Just a note that, starting with Python 2.7/3.1, this functionality will be built in to the collections module, see this bug for more information. Here's the example from the release notes:

  >>> from collections import Counter
  >>> c=Counter()
  >>> for letter in 'here is a sample of english text':
  ...   c[letter] += 1
  ...
  >>> c
  Counter({' ': 6, 'e': 5, 's': 3, 'a': 2, 'i': 2, 'h': 2,
  'l': 2, 't': 2, 'g': 1, 'f': 1, 'm': 1, 'o': 1, 'n': 1,
  'p': 1, 'r': 1, 'x': 1})
  >>> c['e']
  5
  >>> c['z']
  0
  

  ʞɔıu : looks even simpler than that, looks like you can just pass the string to the Counter constructor and it does it for you