# \$Id: snutils.py,v 1.7 2012/07/03 14:44:33 samn Exp \$ # # got this function off the internet - sn def unique(s): """Return a list of the elements in s, but without duplicates. For example, unique([1,2,3,1,2,3]) is some permutation of [1,2,3], unique("abcabc") some permutation of ["a", "b", "c"], and unique(([1, 2], [2, 3], [1, 2])) some permutation of [[2, 3], [1, 2]]. For best speed, all sequence elements should be hashable. Then unique() will usually work in linear time. If not possible, the sequence elements should enjoy a total ordering, and if list(s).sort() doesn't raise TypeError it's assumed that they do enjoy a total ordering. Then unique() will usually work in O(N*log2(N)) time. If that's not possible either, the sequence elements must support equality-testing. Then unique() will usually work in quadratic time. """ n = len(s) if n == 0: return [] # Try using a dict first, as that's the fastest and will usually # work. If it doesn't work, it will usually fail quickly, so it # usually doesn't cost much to *try* it. It requires that all the # sequence elements be hashable, and support equality comparison. u = {} try: for x in s: u[x] = 1 except TypeError: del u # move on to the next method else: return u.keys() # We can't hash all the elements. Second fastest is to sort, # which brings the equal elements together; then duplicates are # easy to weed out in a single pass. # NOTE: Python's list.sort() was designed to be efficient in the # presence of many duplicate elements. This isn't true of all # sort functions in all languages or libraries, so this approach # is more effective in Python than it may be elsewhere. try: t = list(s) t.sort() except TypeError: del t # move on to the next method else: assert n > 0 last = t[0] lasti = i = 1 while i < n: if t[i] != last: t[lasti] = last = t[i] lasti += 1 i += 1 return t[:lasti] # Brute force is all that's left. u = [] for x in s: if x not in u: u.append(x) return u