Re: Dictionary question.

[Brian van den Broek]

hawkesed said unto the world upon 2005-04-21 20:28:
Actually,
  I think I got it now. Here is what I did:
for num in alist:
...         if adict.has_key(num):
...             x = adict.get(num)
...             x = x + 1
...             adict.update({num:x})
...     else:
...             adict.update({num:1})
...
adict
{128: 2, 129: 2, 132: 1, 153: 1, 30: 1, 53: 1, 57: 1, 61: 1, 66: 2, 67:
1, 69: 2, 71: 1, 72: 1, 73: 2, 74: 1, 75: 2, 76: 2, 77: 2, 78: 2, 82:
1, 84: 1, 85: 4, 87: 2, 88: 3, 89: 1, 91: 1, 92: 1, 93: 2, 94: 2, 95:
3, 96: 1, 97: 3, 98: 2, 99: 1, 100: 6, 101: 4, 102: 2, 103: 1, 104: 1,
105: 1, 106: 2, 107: 2, 108: 2, 109: 2, 111: 3, 112: 1, 114: 3, 115: 3,
116: 3, 118: 1, 119: 2, 122: 2, 123: 1, 125: 1, 126: 1}
Thanks all for the help and best regards!
Ed

Hi Ed,
I think there is a more Pythonic way. Here's something I used in a 
similar context:

def get_frequency_dict(sequence):
    '''sequence of hashables -> dictionary of frequency of members
    Given a sequence of items, each of which is hashable (i.e. can
    serve as a key in a dictionary), the function returns a
    dictionary, using the items as keys, and the number of their
    occurrences as values.
    Usage:
    >>> get_frequency_dict([1,2,2,3,3,3,4,4,4,4])
    {1: 1, 2: 2, 3: 3, 4: 4}
    >>> get_frequency_dict([[], "will fail as list aren't hashable"])
    ...
    TypeError: list objects are unhashable
    '''
    frequency_dict = {}
    for element in sequence:
        frequency_dict[element] = frequency_dict.setdefault(
                                    element, 0) + 1
    return frequency_dict
HTH,
Brian vdB
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