Get a unique list of items that occur more than once in a list

Question

I have a list of items:

mylist = [\'A\',\'A\',\'B\',\'C\',\'D\',\'E\',\'D\']

I want to return a unique list of items that appear more than

Answer 1

You can use collections.Counter to do what you have described easily:

from collections import Counter
mylist = ['A','A','B','C','D','E','D']
cnt = Counter(mylist)
print [k for k, v in cnt.iteritems() if v > 1]
# ['A', 'D']

Answer 2

Might not be as fast as internal implementations, but takes (almost) linear time (since set lookup is logarithmic)

mylist = ['A','A','B','C','D','E','D']
myset = set()
dups = set()
for x in mylist:
    if x in myset:
        dups.add(x)
    else:
        myset.add(x)
dups = list(dups)
print dups

Answer 3

>>> mylist = ['A','A','B','C','D','E','D']
>>> set([i for i in mylist if mylist.count(i)>1])
set(['A', 'D'])

Answer 4

import collections
cc = collections.Counter(mylist) # Counter({'A': 2, 'D': 2, 'C': 1, 'B': 1, 'E': 1})
cc.subtract(cc.keys())           # Counter({'A': 1, 'D': 1, 'C': 0, 'B': 0, 'E': 0})
cc += collections.Counter()      # remove zeros (trick from the docs)
print cc.keys()                  # ['A', 'D']

Answer 5

Using a similar approach to others here, heres my attempt:

from collections import Counter

    def return_more_then_one(myList):
         counts = Counter(my_list)
         out_list = [i for i in counts if counts[i]>1]
         return out_list

Answer 6

It can be as simple as ...

print(list(set([i for i in mylist if mylist.count(i) > 1])))