删除python列表中的重复项但记住索引

Question

How can I remove duplicates in a list, keep the original order of the items and remember the first index of any item in the list?

For example, removing the duplicates from [1, 1, 2, 3] yields [1, 2, 3] but I need to remember the indices [0, 2, 3] .

I am using Python 2.7.

Answer 1

I'd tackle this a little differently and use an OrderedDict and the fact that a lists index method will return the lowest index of an item.

>>> from collections import OrderedDict
>>> lst = [1, 1, 2, 3]
>>> d = OrderedDict((x, lst.index(x)) for x in lst)
>>> d
OrderedDict([(1, 0), (2, 2), (3, 3)]

If you need the list (with its duplicates removed) and the indices separately, you can simply issue:

>>> d.keys()
[1, 2, 3]
>>> d.values()
[0, 2, 3]

Answer 2

Use enumerate to keep track of the index and a set to keep track of element seen:

l = [1, 1, 2, 3]
inds = []
seen = set()
for i, ele in enumerate(l):
    if ele not in seen:
        inds.append(i)
    seen.add(ele)

If you want both:

inds = []
seen = set()
for i, ele in enumerate(l):
    if ele not in seen:
        inds.append((i,ele))
    seen.add(ele)

Or if you want both in different lists:

l = [1, 1, 2, 3]
inds, unq = [],[]
seen = set()
for i, ele in enumerate(l):
    if ele not in seen:
        inds.append(i)
        unq.append(ele)
    seen.add(ele)

Using a set is by far the best approach:

In [13]: l = [randint(1,10000) for _ in range(10000)]     

In [14]: %%timeit                                         
inds = []
seen = set()
for i, ele in enumerate(l):
    if ele not in seen:
        inds.append((i,ele))
    seen.add(ele)
   ....: 
100 loops, best of 3: 3.08 ms per loop

In [15]: timeit  OrderedDict((x, l.index(x)) for x in l)
1 loops, best of 3: 442 ms per loop

In [16]: l = [randint(1,10000) for _ in range(100000)]      
In [17]: timeit  OrderedDict((x, l.index(x)) for x in l)
1 loops, best of 3: 10.3 s per loop

In [18]: %%timeit                                       
inds = []
seen = set()
for i, ele in enumerate(l):
    if ele not in seen:
        inds.append((i,ele))
    seen.add(ele)
   ....: 
10 loops, best of 3: 22.6 ms per loop

So for 100k elements 10.3 seconds vs 22.6 ms , if you try with anything larger with less dupes like [randint(1,100000) for _ in range(100000)] you will have time to read a book. Creating two lists is marginally slower but still orders of magnitude faster than using list.index.

If you want to get a value at a time you can use a generator function:

def yield_un(l):
    seen = set()
    for i, ele in enumerate(l):
        if ele not in seen:
            yield (i,ele)
        seen.add(ele)