list without repetitions

[nilamo]

You generate a new list, but then don't return it.

But also, you only increment i when a match is found. Which means once there isn't a match, the whole rest of the list would be included.

To demonstrate, I added a couple prints, along with a smaller sample size:

def lwr(liss):
    liss.sort(key=None, reverse=False)
    print(liss)
    i=1
    liss1=[]
    for n in liss:
        print(f"Index: {i}, Current Value: {n}, Next Value: {liss[i]}")
        if n!=liss[i]:
            liss1.append(n)
            i+=1
    return liss1

liss=[1, 1, 2, 3, 3, 4, 5, 5]
unique = lwr(liss)
print(unique)

Output:
[1, 1, 2, 3, 3, 4, 5, 5]
Index: 1, Current Value: 1, Next Value: 1
Index: 1, Current Value: 1, Next Value: 1
Index: 1, Current Value: 2, Next Value: 1
Index: 2, Current Value: 3, Next Value: 2
Index: 3, Current Value: 3, Next Value: 3
Index: 3, Current Value: 4, Next Value: 3
Index: 4, Current Value: 5, Next Value: 3
Index: 5, Current Value: 5, Next Value: 4
[2, 3, 4, 5, 5]

Unless you need to use indexes (for homework), I think you should just compare with the last value appended. ie: if n != liss1[-1]