Most efficient way to remove duplicates from Python list while preserving order and removing the oldest element

I've seen a bunch of solutions on the site to remove duplicates while preserving the oldest element. I'm interested in the opposite: removing duplicates while preserving the newest element, for example:

list = ['1234','2345','3456','1234']
list.append('1234')
>>> ['1234','2345','3456','1234','1234']
list = unique(list)
>>开发者_运维百科> ['2345','3456','1234']

How would something like this work?

Thanks.

Requires the items (or keys) to be hashable, works in-place on list-likes:

def inplace_unique_latest(L, key=None):
  if key is None:
    def key(x):
      return x
  seen = set()
  n = iter(xrange(len(L) - 1, -2, -1))
  for x in xrange(len(L) - 1, -1, -1):
    item = L[x]
    k = key(item)
    if k not in seen:
      seen.add(k)
      L[next(n)] = item
  L[:next(n) + 1] = []