# 14. remove_duplicates

#1
``````def remove_duplicates(lst):
lst2 == []
for i in lst:
lst2.append(i)
lst2 == lst
return lst2``````

i got this error :Oops, try again. remove_duplicates([4, 5, 5, 4]) resulted in an error: global name 'lst2' is not defined

#2

this line:

``lst2 == []``

two equals sings means compare. to declare lst2 and give it an empty list as value, use a single equal sign

#3
``````def remove_duplicates(lst):
lst2 = []
for i in lst:
lst2.append(i)
lst2 == lst
return lst2``````

Oops, try again. remove_duplicates([4, 5, 5, 4]) returned [4, 5, 5, 4] instead of [4, 5]

still does not work

#4

you just append all items from lst to lst2 without any check for duplicate values

#5

sorry, i dont really get you

#6

the purpose of this exercise is to remove duplicates from the list. so if you have the list:

``[4,5,5,4]``

your function remove_duplicates should return a new list with the values:

``[4,5]``

so you should only append items to the new list (`lst2`) if the number (stored in `i`, thanks to your for loop) if the number isn't in lst2 yet ( `if i not in lst2`)

#7
``if i not in lst2:``

this gives syntax error!

#8

can i see how you implanted it?

you do still need the for loop, i hope you gave it a good thought where to implant this

#9
``````def remove_duplicates(lst):
lst2 = []
for i not in lst2:
lst2.append(i)
lst2 == lst
return lst2``````

#10

here it is
...........

#11

what i gave you was a if condition which you should have nested in the for loop, this way, each run of the loop, the if condition checks if the number is not yet in lst2

#12
``````def remove_duplicates(lst):
lst2 = []
for i in lst:
if i not in lst2:
lst2.append(i)
lst2 == lst
return lst2``````

@stetim94
thanks a lot i got it

#13

do you also understand why this works?

#14

Yes, thanks a lot ...

#15