Delete unique elements from a list

Question

I faced some problem with solving the next problem:

We have a list of elements (integers), and we should return a list consisting of only the non-unique elements in

Answer 1

Try this:

>>> a=[1,2,3,3,4,5,6,6,7,8,9,2,0,0]
>>> a=[i for i in a if a.count(i)>1]
>>> a
[2, 3, 3, 6, 6, 2, 0, 0]
>>> a=[1, 2, 3, 1, 3]
>>> a=[i for i in a if a.count(i)>1]
>>> a
[1, 3, 1, 3]
>>> a=[1, 2, 3, 4, 5]
>>> a=[i for i in a if a.count(i)>1]
a
[]

Answer 2

I used an integer and bool to check every time the list was modified within a while loop.

rechecks = 1
runscan = True
while runscan == True:
    for i in data:
         if data.count(i) <2:
             data.remove(i)
             rechecks+=1
            #need to double check now
    if rechecks >0:
        runscan = True
        rechecks-=1
    else:
        runscan = False       
return data

Answer 3

Following what you have started, iterating on the list of integers, but not counting or deleting elements, try just testing if the element has already been seen, append it to a list of duplicated elements:

def checkio(data):
    elements = []
    duplicates = []
    for i in data:
        if i not in elements:
            elements.append(i)
        else:
            if i not in duplicates:
                duplicates.append(i)
    return duplicates

d = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]

print (checkio(d))
#[1, 3, 5, 2]

Answer 4

You can implement a OrderedCounter, eg:

from collections import OrderedDict, Counter

class OrderedCounter(Counter, OrderedDict): 
    pass

data = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]

duplicates = [k for k, v in OrderedCounter(data).items() if v > 1]
# [1, 3, 2, 5]

So you count the occurrence of each value, then filter on if it has a frequency of more than one. Inheriting from OrderedDict means the order of the original elements is preserved.

---

Going by comments, you want all duplicated elements reserved, so you can pre-build a set of the duplicate entries, then re-iterate your original list, eg:

from collections import Counter

data = [1, 3, 1, 2, 3, 5, 8, 1, 5, 2]
duplicates = {k for k, v in Counter(data).items() if v > 1}
result = [el for el in data if el in duplicates]
# [1, 3, 1, 2, 3, 5, 1, 5, 2]

Answer 5

def checkio(data):
    lis = []
    for i in data:
        if data.count(i)>1:
            lis.append(i)
    print(lis)
checkio([1,2,3,3,2,1])

Yeah it's a bit late to contribute to this thread but just wanted to put it there on the net for anyone else use.

Answer 6

 numbers = [1, 1, 1, 1, 3, 4, 9, 0, 1, 1, 1]
 x=set(numbers)
 print(x)

You can use the set key word too to get the desired solution.