刪除python列表中的重復項但記住索引

Question

如何刪除列表中的重復項，保留項目的原始順序並記住列表中任何項目的第一個索引？

例如，從[1, 1, 2, 3] 1,1,2,3]中刪除重復項會產生[1, 2, 3]但我需要記住索引[0, 2, 3] 。

我使用的是Python 2.7。

Answer 1

我會稍微解決這個問題並使用OrderedDict並且列表index方法將返回項目的最低索引。

>>> 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)]

如果您需要列表（刪除重復項）和索引，您可以簡單地發出：

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

Answer 2

使用enumerate跟蹤索引和一組跟蹤元素：

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)

如果你想要兩個：

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

或者如果你想要兩個在不同的列表中：

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)

使用套裝是迄今為止最好的方法：

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

因此，對於100k元素10.3秒對22.6 ms ，如果你嘗試使用更少的欺騙，如[randint(1,100000) for _ in range(100000)]你將有時間閱讀一本書。 創建兩個列表的速度略慢，但仍比使用list.index快幾個數量級。

如果要一次獲取一個值，可以使用生成器函數：

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