If element is repeated in list remove element from that list

One option is to use collections.Counter, and then filter out items of multiple appearances:

[v for v, c in collections.Counter([1, 2, 3, 4, 2, 3]).items() if c == 1] 

The Counter will create a dictionary with the items: Counter({1: 1, 2: 2, 3: 2, 4: 1}).

You can use the .count() method of the lists, along with a list comprehension.

For example:

my_list = [1,2,3,4,2,3]
new_list = [x for x in my_list if l.count(x) == 1]
print(new_list) # Prints [1,4]

all approaches are good I want to add a different approach

def duplicate_deleter(list_object):
    set_obj=set(list_object)
    result=[]
    for i in set_obj:
        if list_object.count(i)==1:
            result.append(i)
    return result

Without collections you can use a dictionary to count the frequency of the items in the list then keep only those with low enough frequency. This example runs in O(n+m) where n is the number of elements in the original list and m is the number of unique elements.

lst = [1,2,2,3,4,5,4]
freq = dict()
for el in lst:
    if el in freq:
        freq[el] += 1
    else:
        freq[el] = 1
print([el for el, f in freq.items() if f < 2])

You can create a program to pop the list x number of times for each duplicate element, where x is equals to the count of a duplicate element in the list.

my_list = [1,2,3,4,2,3]
#Sort the list so the index matches the count arrays index
my_list.sort()
#Create the count array using set to count each unique element
count = [my_list.count(x) for x in set(my_list)]
#Create another count to cycle through the count array
count1 = 0
#Create a popped variable to keep record of how many times the list has been altered
popped = 0
while (count1 < len(count)):
    if count[count1] > 1:
        #Create a variable to pop r number of times depending on the count
        r = 0
        while (r < count[count1]):
            #If the list has already been popped, the index to be examined has to be the previous one.
            #For example, after the 2's have been popped we want it to check the first 3
            #which is now in the count1-1 index.
            my_list.pop((count1 - popped))
            r = r+1
        popped = popped + 1
        count1 += 1
        print(my_list)
    else:
        count1 += 1