Skip Montanaro wrote:
...lotsa great stuff ...
You might want to sort your lists by the 'English' key. I don't know how to
use the new key arg to list.sort(), but you can still do it the
old-fashioned way:
oldl.sort(lambda a,b: cmp(a['English'], b['English']))
newl.sort(lambda a,b: cmp(a['English'], b['English']))
To complete the thought, for 2.4 and after the new-fashioned way is:
import operator
oldl.sort(key=operator.itemgetter('English'))
newl.sort(key=operator.itemgetter('English'))
Once sorted, you can then march through the lists in parallel, which should
give you an O(n) algorithm.
But overall you will have O(n log n) because of the sorts.
--Scott David Daniels
[EMAIL PROTECTED]
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