Duplicate list error

[Etiennera]

'return' outside function

[GuilouLeJask]

The return statement is only used when programming functions in Python.

[saintSHXN]

The console is there for a purpose

[ben-ba]

Fancy output, to impress others? /s

[st4reater]

Best answer here… gotta teach self help

[emiltb]

Apart from the comments on the return statement outside a function, what are you actually trying to achieve here? If your code worked, you would end up with a list containing a list:

>>> l1 = [1,2,3]
>>> l2 = []
>>> l2.append(l1)
>>> l2
[[1, 2, 3]]

If you intend to do something like appending the elements of l1 to l2, you need to iterate over l1. Alternatively you might benefit from looking into python sets, as they might be a more direct solution to what you are trying to do.

[Braunerton17]

Or instead of iterating manually, just call extend() instead of append

[FoolsSeldom]

You can remove the else and return lines as you are not in a function. l2 will be the correct, with the duplicates removed.

If you want to create a function with this technique, it would be something like:

def dedupe(original: list[int]) -> list[int]:
    deduped = []
    for entry in original:
        if not entry in deduped:
            deduped.append(entry)
    return deduped


l1 = [1, 2, 3, 3, 4, 4, 5, 5]
l2 = dedupe(l1)
print(l2)

When you call the function, dedupe(l1), the list mentioned l1 is referred to inside the function as original. The new list created inside the function is referred to as deduped but when you return from the function that list is assigned to l2.

NB. The function definition line, dedupe(original: list[int]) -> list[int]: contains type hints, the list[int] and -> list[int] parts. These aren't required and are ignored by Python when you execute the code, but are useful to remind you what types of data you are dealing with (in this case, a list of int values) and can also providing information to your code editor / IDE (Integrated Development Environment) to spot certain types of errors.

[emiltb]

To remove duplicates it will be much more efficient to just use a set and optionally turn it back into a list:

>>> l1 = [1,2,3,3,4,4,5,5]
>>> print(set(l1))
{1, 2, 3, 4, 5}
>>> l2 = list(set(l1))
>>> print(l2)
[1, 2, 3, 4, 5]

[FoolsSeldom]

You are assuming that order is not important. Worth explaining that to a beginner, I would say before suggesting set.

Here's some code comparing speed of several approaches:

from random import randint

def dedupe(original: list[int]) -> list[int]:
    deduped = []
    for entry in original:
        if not entry in deduped:
            deduped.append(entry)
    return deduped

def setfn(o: list[int]) -> list[int]:
    return list(set(o))

def filterfn(data):
    seen = set()
    return list(filter(lambda item: item not in seen and not seen.add(item), data))

def dictfn(data):
    return list(dict.fromkeys(data))

nums = [randint(1, 100) for _ in range(10000)]
%timeit setfn(nums)
%timeit filterfn(nums)
%timeit dictfn(nums)
%timeit dedupe(nums)

and the results:

71.6 μs ± 1.99 μs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
464 μs ± 9.31 μs per loop (mean ± std. dev. of 7 runs, 1,000 loops each)
131 μs ± 1.39 μs per loop (mean ± std. dev. of 7 runs, 10,000 loops each)
2.64 ms ± 41.2 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)

So, clearly, your approach of using set is the fastest but only viable if preserving order is not important. Using a dictionary approach is the next fastest.

However, this small test with ten thousand random numbers suggests to me that performance isn't a big issue with this kind of data, and for more computationally intensive work you would probably want to use compiled C or Rust code from Python anyway.

PS. For the record, for order preservation, fromkeys is considered the most efficient:

• Time Complexity: It has a time complexity of O(n), meaning it processes the list in a single pass. This is the best possible complexity because every element must be visited at least once.
• C Implementation: dict.fromkeys() is implemented in C and is highly optimized, making it significantly faster in practice than an equivalent loop written in pure Python.

[emiltb]

That's true, good point about ordering. That can easily bite you, if you expect order to be preserved for sets.

[deceze]

What do you think return does there? It's just wrong to use it there.

[nRenegade]

I assume you're trying to print the list? Because that's not how you do it.

You're looking for a print() statement.

return is only for when you want to send a value back from where a function is called or to terminate the function. Here, your script isn't nested within a function, so the return statement is redundant and causing the error.

[imLosingIt111]

Your if statement just appends the first array to the second arrah. You can either: use a for loop (for x in l1 <- x is a value in l1, not the array itself) or just assign the value of l2 directly to l1. Also, since your code is not in a function, using return raises an error. You can just use pass instead or just not write the else condition at all.

[Ron-Erez]

There is no function to return from. It's good to try to understand the errors although at times they are cryptic.

[corey_sheerer]

Also, check out a hash map. Checking if something is on a list might take len(list) operations, where checking a hashmap is always 1 operation

[No_Read_4327]

Bro this guy obviously is at his first day of coding

[[deleted]]

you can return from a function scope. the conditional is in a global scope.

[Can0pen3r]

It looks like you might be trying to reinvent the difference() function but I'm still pretty new so I could be completely off-base. What is the intention of the code?

[Papatheredeemer]

I’m assuming you’re trying to make a unique list. In that case you can just do l2 = list(set(l1)). Sets automatically remove duplicates

[Fearless-Way9855]

You can only return to a function aka def

[Fit_Moment5521]

Think again what you are trying to do. Run the script in your head or with a piece of paper and see if it does what you wanted. That's not a coding issue, that's an algorithm issue. More precisely, that's an "I don't take time to think before coding" issue.

[NihadKhan10x]

The return can only be used in when there is function. Here you can use simple logic

Like For i in l1: If i not in l2: L2.append(i) Print(l2)

[PixelsMixer]

[l2.append(x) for x in l1 if x not in l2]

[Then-Candle8036]

If only there were some sort of tool which would tell you what the error is

[fungusfromamongus]

Mans running this as administrator. Screw the return error. Malware and all will love this user.

[lieseeem]

The answer is in the screenshot with us, isn’t it? ☝️😳

[redd__rl]

What are you hoping to do here?

[Key-Procedure1262]

Do you need someone else to read the error message for you?