How can I maintain the consistency of my list with set?

Question

In [1]: l1 = ['a',2,3,0,9.0,0,2,6,'b','a']

In [2]: l2 = list(set(l1))

In [3]: l2
Out[3]: ['a', 0, 2, 3, 6, 9.0, 'b']

Here you can see the list l2 falling with a different sequence, and then the original l1, I need to remove duplicate elements from my list without changing the sequence / order of the elements of the list ....

Answer 1

If you are not interested in efficiency, this is O (n * m)

>>> sorted(set(l1), key=l1.index)
['a', 2, 3, 0, 9.0, 6, 'b']

Using an intermediate dict is more complicated, but O (n + m * logm)

where n is the number of elements in l1 and m is the number of unique elements in l1

>>> l1 = ['a',2,3,0,9.0,0,2,6,'b','a']
>>> d1=dict((k,v) for v,k in enumerate(reversed(l1)))
>>> sorted(d1, key=d1.get, reverse=True)
['a', 2, 3, 0, 9.0, 6, 'b']

In Python3.1, you have an OrderedDict, so it’s very easy

>>> l1 = ['a',2,3,0,9.0,0,2,6,'b','a'] 
>>> list(OrderedDict.fromkeys(l1))
['a', 2, 3, 0, 9.0, 6, 'b']

Answer 2

, ​​:

def dedupe(items):
    seen = set()
    for item in items:
        if item not in seen:
            yield item
            seen.add(item)

:

>>> l1 = ['a',2,3,0,9.0,0,2,6,'b','a']
>>> l2 = list(dedupe(l1))
>>> l2
['a', 2, 3, 0, 9.0, 6, 'b']

Answer 3

( dicts):

l1 = ['a',2,3,0,9.0,0,2,6,'b','a']
l2 = []
s = {}
for i in l1:
    if not i in s:
        l2.append(i)
        s[i] = None

# l2 contains ['a', 2, 3, 0, 9.0, 6, 'b', 'a']

: ( ):

l1 = ['a',2,3,0,9.0,0,2,6,'b','a']
l2 = []
s = set()
for i in l1:
   if not i in s:
       l2.append(i)
       s.add(i)