> Pythons 2.7 and later have dictionary comprehensions. So you can do
> this:
>
> >>> {item: s.count(item) for item in set(s)}
>
> {'a': 1, 'b': 1, '1': 2, '3': 1, '2': 2, '4': 1}
>
> Which gives counts for all letters. To filter out the digit-counts
> only:
>
> >>> digits="0123456789"
> >>> {item: s.count(item) for item in set(s) if item in dig}
>
> {'1': 2, '3': 1, '2': 2, '4': 1}
>
> You can then print out the values in d in any which way you want.
> [Starting with printing is usually a bad idea]Sorry cut-paste slip-up. 1. use dig or digits (or none, just inline the "0123456789") 2. I assumed >>> s = "12ab3412" -- http://mail.python.org/mailman/listinfo/python-list
