checkio = [1, 2, 3, 2, 1]
checkio1 = []
for i in checkio:
    if i not in checkio1:
        checkio1.append(i)
checkio = []
for j in checkio1:
    if j not in checkio:
        checkio.append(j)
print(checkio)

>>> a = [1, 2, 3, 2, 3, 4]
>>> a
[1, 2, 3, 2, 3, 4]
>>> b = set(a)
>>> b
{1, 2, 3, 4}
>>> c = list(b)
>>> c
[1, 2, 3, 4]
>>>

>>> a = [1, 2, 3, 2, 3, 4]
>>> b = set(a)    # get clear set without dublicates
>>> c = set([x for x in a if a.count(x) > 1])    # check
>>> d = list(b - c)    # use subtraction operator on sets
>>> d
[1, 4]
>>>

a = [1, 2, 3, 2, 3, 4]

 a =  [2, 3, 2, 3]

for elem in a:
   эту строчку додумай сам 
        a.remove(elem)

a = [1, 2, 3, 2, 3, 4]
for elem in a:
   if a.count(elem) == 1:
        a.remove(elem)
print(a)

>>> a=range(10)
>>> a
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> for i in a:
...     a.remove(i)
...
>>> a
[1, 3, 5, 7, 9]

>>> def f(lst):
...     s = set()
...     return [i for i in lst
...             if i not in s and (s.add(i) or True)]
... 
>>> checkio = [1, 2, 3, 2, 1]
>>> checkio = f(checkio)
>>> checkio
[1, 2, 3]
>>>

 for elem in a[:]: