Python: Replace duplicate values in a list with the same values

This code automatically replace all duplicate items with 3

my_list = [1, 2, 2, 5, 4, 8, 6]
new = []
for item in my_list:
   if my_list.count(item) > 1:
    new.append(3)
   else:
    new.append(item)
print(new)

Among the fastest

fr = {x:0 for x in X}
for n in X:
    fr[n] += 1
Y = [3 if fr[n]>1 else n for n in X]

You iterate over the list and add one to the dictionary counter for that number. Then you create a list with list comprehension: you change the value to 3 if it is repeated more than once.

Little bit slower and little bit shorter

xi = {x:i for i, x in enumerate(X)}
dp = {x: xi[x]>i for i, x in reversed(list(enumerate(X)))}
Y = [3 if dp[x] else x for x in X]

You iterate over X and keep the lastest index of each value. Then you iterate again but in reverse order, and ask if there is another index for that value. Then with that info you create the desired output. All using list/dict comprehension. This is more of a functional approach.

You should be able to use a for loop for this

my_list = [1, 2, 2, 5, 4, 8, 6]
new_list = []

for i in range(len(my_list)):
    if my_list[i] in new_list:
        new_list.append(3)
    else:
        new_list.append(my_list[i])
print(new_list)

Output:

 [1, 3, 3, 5, 4, 8, 6]

Maybe something like this:

X = [t if t not in X[:i] + X[i+1:] else 3 for i, t in enumerate(X)]