Ordered Sets Python 2.7

Question

I have a list that I'm attempting to remove duplicate items from. I'm using python 2.7.1 so I can simply use the set() function. However, this reorders my list. Which for my particular case is unacceptable.

Below is a function I wrote; which does this. However I'm wondering if there's a better/faster way. Also any comments on it would be appreciated.

    def ordered_set(list_):

        newlist = []
        lastitem = None
        for item in list_:

            if item != lastitem:
                newlist.append(item)
                lastitem = item

        return newlist

The above function assumes that none of the items will be None, and that the items are in order (ie, ['a', 'a', 'a', 'b', 'b', 'c', 'd'])

The above function returns ['a', 'a', 'a', 'b', 'b', 'c', 'd'] as ['a', 'b', 'c', 'd'].

Answer 1

I think this is perfectly OK. You get O(n) performance which is the best you could hope for.

If the list were unordered, then you'd need a helper set to contain the items you've already visited, but in your case that's not necessary.

Answer 2

I know this has already been answered, but here's a one-liner (plus import):

from collections import OrderedDict
def dedupe(_list):
    return OrderedDict((item,None) for item in _list).keys()

>>> dedupe(['q', 'w', 'e', 'r', 'q', 'w', 'y', 'u', 'i', 't', 'e', 'p', 't', 'y', 'e'])
['q', 'w', 'e', 'r', 'y', 'u', 'i', 't', 'p']