Indexing a list with an unique index

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You can do this in O(N) time using a defaultdict and a list comprehension:

>>> from itertools import count
>>> from collections import defaultdict
>>> lst = [10, 10, 20, 15, 10, 20]
>>> d = defaultdict(count(1).next)
>>> [d[k] for k in lst]
[1, 1, 2, 3, 1, 2]

In Python 3 use __next__ instead of next.

If you’re wondering how it works?

The default_factory(i.e count(1).next in this case) passed to defaultdict is called only when Python encounters a missing key, so for 10 the value is going to be 1, then for the next ten it is not a missing key anymore hence the previously calculated 1 is used, now 20 is again a missing key and Python will call the default_factory again to get its value and so on.

d at the end will look like this:

>>> d
defaultdict(<method-wrapper 'next' of itertools.count object at 0x1057c83b0>,
            {10: 1, 20: 2, 15: 3})

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