On Sun, 20 Nov 2005 22:03:34 +0100, Christoph Zwerschke <[EMAIL PROTECTED]>
wrote:
>> Ordering the keys isn't the normal case, and can be done easily when
>> needed.
>
>That depends. Maybe I do not want the keys to be sorted alphabetically,
>but according to some criteria which cannot be derived from the keys
>themselves.
You mean involving also the values? What's wrong with
sorted(plaindict.items(), key=your_ordering_function) ?
>>> def show(*a): print a
...
>>> sorted(dict((c,ord(c)) for c in 'abcd').items(), key=show)
(('a', 97),)
(('c', 99),)
(('b', 98),)
(('d', 100),)
[('a', 97), ('c', 99), ('b', 98), ('d', 100)]
What key function would you like, to generate the value that is actually used
to define the ordering?
>>> sorted(dict((c,ord(c)) for c in 'abcd').items(), key=lambda t:t[0])
[('a', 97), ('b', 98), ('c', 99), ('d', 100)]
>>> sorted(dict((c,ord(c)) for c in 'abcd').items(), key=lambda t:t[1])
[('a', 97), ('b', 98), ('c', 99), ('d', 100)]
>>> sorted(dict((c,ord(c)) for c in 'abcd').items(), key=lambda t:-t[1])
[('d', 100), ('c', 99), ('b', 98), ('a', 97)]
>>> sorted(dict((c,ord(c)) for c in 'abcd').items(), key=lambda t:t[1]&1)
[('b', 98), ('d', 100), ('a', 97), ('c', 99)]
>>> sorted(dict((c,ord(c)) for c in 'abcd').items(), key=lambda
>>> t:(t[1]&1,t[1]))
[('b', 98), ('d', 100), ('a', 97), ('c', 99)]
>>> sorted(dict((c,ord(c)) for c in 'abcd').items(), key=lambda
>>> t:(t[1]&1,-t[1]))
[('d', 100), ('b', 98), ('c', 99), ('a', 97)]
And being able to reverse the end result is handy
>>> sorted(dict((c,ord(c)) for c in 'abcd').items(), key=lambda
>>> t:(t[1]&1,-t[1]), reverse=True)
[('a', 97), ('c', 99), ('b', 98), ('d', 100)]
You may need to upgrade your Python though ;-)
Regards,
Bengt Richter
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