<[EMAIL PROTECTED]> wrote in message news:[EMAIL PROTECTED] >I do this: > > def unique(keys): > unique = [] > for i in keys: > if i not in unique:unique.append(i) > return unique > > I don't know what is faster at the moment.
This is quadratic, O(n^2), in the length n of the list if all keys are unique. Conversion to a set just might use a better sorting algorithm than this (i.e. n*log(n)) and throwing out duplicates (which, after sorting, are positioned next to each other) is O(n). If conversion to a set should turn out to be slower than O(n*log(n)) [depending on the implementation], then you are well advised to sort the list first (n*log(n)) and then throw out the duplicate keys with a single walk over the list. In this case you know at least what to expect for large n... Regards, Christian -- http://mail.python.org/mailman/listinfo/python-list
