CODEWITHSUNDEEP
pythonlist
user4307777
user4307777

Reputation:

Creating a function that removes duplicates in list

I'm trying to manually make a function that removes duplicates from a list. I know there is a Python function that does something similar (set()), but I want to create my own. This is what I have:

def remove(lst):

for i in range(len(lst)):
    aux = lst[0:i] + lst[i+1:len(lst)]
    if lst[i] in aux:
        del(lst[i])

return lst

I was trying something like creating a sub-list with all the items except the one the for is currently on, and then check if the item is still in the list. If it is, remove it.

The problem is that it gives me an index out of range error. Does the for i in range(len(lst)): line not update every time it starts over? Since I'm removing items from the list, the list will be shorter, so for a list that has 10 items and 2 duplicates, it will go up to index 9 instead of stopping on the 7th.

Is there anyway to fix this, or should I just try doing this is another way?

Upvotes: 2

Views: 123

Answers (5)

Joe T. Boka
Joe T. Boka

Reputation: 6585

You can also try this:

lst = [1,2,3,3,4,4,5,6]
lst2 = []
for i in lst:
    if i not in lst2:
        lst2.append(i)
print(lst2)
[1, 2, 3, 4, 5, 6]

Upvotes: 0

Arthur Zopellaro
Arthur Zopellaro

Reputation: 158

def remove(lst):
    new_list = []
    for i in lst:
        if i not in new_list:
            new_list.append(i)
    return new_list

You should append the values to a secondary list. As Bobbyrogers said, it's not a good idea to iterate over a list that is changing.

Upvotes: 0

Prune
Prune

Reputation: 77837

You got it the first time: len(lst) is evaluated only when you enter the loop. If you want it re-evaluated, try the while version:

i = 0
while i < len(lst):
    ...
    i += 1

Next, you get to worry about another problem: you increment i only when you don't delete an item. When you do delete, shortening the list gets you to the next element. 

i = 0
while i < len(lst):
    aux = lst[0:i] + lst[i+1:len(lst)]
    if lst[i] in aux:
        del(lst[i])
    else:
        i += 1

I think that should solve your problem ... using the logic you intended.

Upvotes: 0

LMc
LMc

Reputation: 18632

I know this does not fix your current script, but would something like this work?

def remove(lst):
    unique=[]
    for i in lst:
        if i not in unique: unique.append(i)
    return unique

Just simply looping through, creating another list and checking for membership?

Upvotes: 1

Bobbyrogers
Bobbyrogers

Reputation: 302

The problem is you are manipulating the list as you are iterating over it. This means that when you reach the end of the list, it is now shorter because you're removed elements. You should (generally) avoid removing elements while you are looping over lists.

Upvotes: 0

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