🔒 Making a function that detects duplicates in a give n list

[spring fiber]

Hi guys I need some hints how to make this function. The instructions say to make a function that detects duplicates in a given list with the parameter named "elements". It also says to use set(). Here's what I've written. Output gives nothing.

#// BEGIN_TODO [Duplicate_detection] Duplicate detection

def has_duplicates(elements) -> bool:
    '''
    This function tells us whether there's dupplicated items in the list or not.
    >>> emilys_list = ["tree", "cat", "dog"]
        return False
    >>> emilys_list = [1,2,3,1]
        return True
    '''
    emilys_list = list[any]
    set1 = set(emilys_list)
    for i in range(emilys_list):
        if i == i+1:
            print(set1)
            return True
        else:
            return False
            print(set1)
    set1 = set(emilys_list)
    return set1

    
#// END_TODO [Duplicate_detection]

Thanks in advance!

[half blaze]

@spring fiber

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[worthy lily]

are you calling the function?

actually nvm

most of this code makes no sense

[spring fiber]

how can I make it sense

[worthy lily]

do you know what a set does?

[spring fiber]

it takes a list and makes it unchangeable?

[worthy lily]

not really

in the context of this question

a set can be used to remove duplicates

!e

my_list = [1, 2, 3, 1]
my_set = set(my_list)

print(my_set)

[half blaze]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

{1, 2, 3}

[worthy lily]

notice how the duplicate 1 was removed

[spring fiber]

ohhh

I see because a set stores multiple variables into 1?

does it always do that?

[worthy lily]

it doesn't

sets have a property that they store only unique elements

[safe moon]

a set can only store one of any given thing
if you try to add something to a set which already exists in it, nothing will happen

[spring fiber]

interesting so thats why they put it in the instructions

[safe moon]

also, why did you do

emilys_list = list[any]

[spring fiber]

was also part of the instructions

they wanted me to use list[any]

[safe moon]

I doubt they meant it in this way... can you post the instructions?

[worthy lily]

I don't think they meant you need to assign list[any] to a variable

[spring fiber]

the instructions of the homework:

[normal verge]

!e

duplicate_list = [1, 1, 2, 3]
deduplicate_list = list(set(duplicate_list))
print(deduplicate_list)```

[half blaze]

:white_check_mark: Your 3.12 eval job has completed with return code 0.

[1, 2, 3]

[worthy lily]

yes, they want you to use list[Any] as a type hint, not as the value of a variable

[safe moon]

yep. the type hint is supposed to be applied to the input list you're taking, i.e. elements

[spring fiber]

ok

I changed the code a bit

#// BEGIN_TODO [Duplicate_detection] Duplicate detection

def has_duplicates(elements, list[any]) -> bool:
    '''
    This function tells us whether there's dupplicated items in the list or not.
    >>> emilys_list = ["tree", "cat", "dog"]
        return False
    >>> emilys_list = [1,2,3,1]
        return True
    '''
    emilys_list = []
    emilys_list_set = set(emilys_list)
    return emilys_list_set
print(has_duplicates(3))
    
#// END_TODO [Duplicate_detection]

[safe moon]

also, Any (the type denoting "anything") is not the same as any (the builtin function any), so you should use

from typing import Any
def has_duplicates(elements: list[Any]) -> bool:
    ...

[safe moon]

type hints are denoted with varname: typehint <- notice the use of :, not ,

[safe moon]

there are a number of problems

1. you're not using the input list elements inside the function
2. you're returning the set that you created, but you're actually supposed to return a True/False indicating whether there were any duplicate elements found or not
3. When calling, you're supposed to give it a list of items, not a number like 3

[spring fiber]

ok i will change it again

[maiden glacier]

lst: list[any] = []

[spring fiber]

!e

#// BEGIN_TODO [Duplicate_detection] Duplicate detection

def has_duplicates(elements, list[Any]) -> dupped_items bool:
    '''
    This function tells us whether there's dupplicated items in the list or not.
    >>> emilys_list = ["tree", "cat", "dog"]
        return False
    >>> emilys_list = [1,2,3,1]
        return True
    '''
    elements_set = set(elements)
    if elements_set > 0:
        return True
    elif elements_set == 0:
        return False
print(has_duplicates(["tree", "cat", "dog"]))
    
#// END_TODO [Duplicate_detection]

[half blaze]

:x: Your 3.12 eval job has completed with return code 1.

001 |   File "/home/main.py", line 3
002 |     def has_duplicates(elements, list[Any]) -> dupped_items bool:
003 |                                      ^
004 | SyntaxError: invalid syntax

[spring fiber]

huh says "(" was not closed

[normal verge]

(elements, list[Any])
Should be 
(elements: list[Any])```
More info: https://docs.python.org/3/library/typing.html

[safe moon]

^^

[acoustic patio]

your logic is wrong why should a list contain duplicates when its set has elements? only a empty list creates the empty set

return lst == list(set(lst))

[normal verge]

^^

[wicked pilot]

does set preserve the order tho? asking, not sure
if not, check the len instead

[safe moon]

also, -> dupped_items bool isn't a valid return typehint. there's no reason for dupped_items to be there

[spring fiber]

ok ill try to add this

[spring fiber]

I removed dupped_items and replaced , with :

[safe moon]

+1, len is the way to go. you can trivially construct an example where lst == list(set(lst)) fails

for example, in current CPython, [1, 24] == list(set([1, 24])) is False

[spring fiber]

should I make an if statement?

#// BEGIN_TODO [Duplicate_detection] Duplicate detection
from typing import Any, Optional
def has_duplicates(elements : list[Any]) -> bool:
    '''
    This function tells us whether there's dupplicated items in the list or not.
    >>> elements = ["tree", "cat", "dog"]
        return False
    >>> elements = [1,2,3,1]
        return True
    '''
    elements_set = set(elements)
    lst = list[Any]
    return lst == list(set(lst))
print(has_duplicates(["tree", "cat", "dog",]))
    
#// END_TODO [Duplicate_detection]

[wicked pilot]

    elements_set = set(elements)
    lst = list[Any]
    return lst == list(set(lst))

explain these lines you just wrote, like what do you think this does and why

[spring fiber]

i remove the duplicates from a list which is the set() function and then i make a list variable lst and turn it into a set

[safe moon]

what are the contents of lst?

[wicked pilot]

what does this list variable you mentioned contain?

[spring fiber]

it contains any list?

[safe moon]

but we're not supposed to be working with any list, are we? we're supposed to work on the list that was given to this function to check (i.e. elements)

[technical aside: lst = list[Any] just makes lst equal to the literal type hint list[Any]. it's not even an actual list]

[spring fiber]

ok ill change it

I added an if statement too. If the list of set(elements) is greater than 0 that means there's a duplicated element.

#// BEGIN_TODO [Duplicate_detection] Duplicate detection
from typing import Any, Optional
def has_duplicates(elements : list[Any]) -> bool:
    '''
    This function tells us whether there's dupplicated items in the list or not.
    >>> elements = ["tree", "cat", "dog"]
        return False
    >>> elements = [1,2,3,1]
        return True
    '''
    List_sets = list[Any]
    return List_sets == list(set(elements))
    if List_sets > 0:
        return True
    if List_sets == 0:
        return False
print(has_duplicates(["tree", "cat", "dog",]))
    
#// END_TODO [Duplicate_detection]

[safe moon]

You're comparing list(set(elements)) to an empty list... what do you think the value of list(set(elements)) is for, say, elements = [1, 2, 3] and for elements = [1, 2, 3, 1]?

[spring fiber]

for elements = [1,2,3] its 0 and for elements = [1,2,3,1] I think it's 1? because there's a dupped element

oh wait maybe it's 3 for both since set removes the extra 1 in the second list

[safe moon]

No. Please open a Python console and try to see what values it actually takes.

[spring fiber]

oh waitt... I think i'm getting an idea.

let me change the code again

[safe moon]

My advice would be to first understand what set actually does, by experimenting with it in a console as I said. Right now you have a wrong idea about what it does, so the code you produce like this is going to be wrong too.

[spring fiber]

ok I will check the wiki again

I will come back to this later. Thank you for your help 👍

!close

[half blaze]

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