Keep duplicates in a list in Python

问题

I know this is probably an easy answer but I can't figure it out. What is the best way in Python to keep the duplicates in a list:

x = [1,2,2,2,3,4,5,6,6,7]

The output should be:

[2,6]

I found this link: Find (and keep) duplicates of sublist in python, but I'm still relatively new to Python and I can't get it to work for a simple list.

回答1:

This is a short way to do it if the list is sorted already:

x = [1,2,2,2,3,4,5,6,6,7]

from itertools import groupby
print [key for key,group in groupby(x) if len(list(group)) > 1]

回答2:

I'd use a collections.Counter:

from collections import Counter
x = [1, 2, 2, 2, 3, 4, 5, 6, 6, 7]
counts = Counter(x)
output = [value for value, count in counts.items() if count > 1]

Here's another version which keeps the order of when the item was first duplicated that only assumes that the sequence passed in contains hashable items and it will work back to when set or yeild was introduced to the language (whenever that was).

def keep_dupes(iterable):
    seen = set()
    dupes = set()
    for x in iterable:
        if x in seen and x not in dupes:
            yield x
            dupes.add(x)
        else:
            seen.add(x)

print list(keep_dupes([1,2,2,2,3,4,5,6,6,7]))

回答3:

keepin' it simple:

array2 = []
aux = 0
aux2=0
for i in x:
    aux2 = i
    if(aux2==aux):
        array2.append(i)
    aux= i
list(set(array2))

That should work

来源：https://stackoverflow.com/questions/15812547/keep-duplicates-in-a-list-in-python