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 once in mylist, so that my desired output would be:

 [A,D]

Not sure how to even being this, but my though process is to first append a count of each item, then remove anything equal to 1. Then dedupe, but this seems like a really roundabout, inefficient way to do it, so I am looking for advice.

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

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

Answer 3

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 4

Try some thing like this:

a = ['A','A','B','C','D','E','D']

import collections
print [x for x, y in collections.Counter(a).items() if y > 1]
 ['A', 'D']

Reference: How to find duplicate elements in array using for loop in Python?

OR

def list_has_duplicate_items( mylist ):
    return len(mylist) > len(set(mylist))
def get_duplicate_items( mylist ):
    return [item for item in set(mylist) if mylist.count(item) > 1]
mylist = [ 'oranges' , 'apples' , 'oranges' , 'grapes' ]
print 'List: ' , mylist
print 'Does list have duplicate item(s)? ' , list_has_duplicate_items( mylist )
print 'Redundant item(s) in list: ' , get_duplicate_items( mylist )

Reference https://www.daniweb.com/software-development/python/threads/286996/get-redundant-items-in-list

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])))

Answer 7

Use set to help you do that, like this maybe :

X = ['A','A','B','C','D','E','D']
Y = set(X)
Z = []

for val in Y :
    occurrences = X.count(val)
    if(occurrences > 1) :
        #print(val,'occurs',occurrences,'times')
        Z.append(val)

print(Z)

The list Z will save the list item which occur more than once. And the part I gave comment (#), that will show the number of occurrences of each list item which occur more than once

Answer 8

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 9

another solution what's written:

def delete_rep(list_):
new_list = []
for i in list_:
    if i not in list_[i:]:
        new_list.append(i)

return new_list

Answer 10

This is my approach without using packages

result = []
for e in listy:
    if listy.count(e) > 1:
        result.append(e)
    else:
        pass
print(list(set(result)))