# Remove_duplicates

#1

So I'm trying to see if appending the numbers to new_list then going through new_list for any duplicates and removing them, is easier or a right way of doing this. Is their an easy function that looks over a list and leaves one of each?

``````def remove_duplicates(lits):
new_list = []
for x in lits:
new_list.append(x)
for y in new_list:

Replace this line with your code.``````

#2

you can use `in` keyword to check if something is present in a list

#3

I came up with the soultion of the problem by mistake, but then I relized why it was so simple.

def remove_duplicates(x): # this is the defined function
dup_list=[] # this is the list that will keep all the charachers from x
for i in x: # this is looping into X
if i not in dup_list: # basically this says, if the number i is not in the dup_list add it, if it is do not add it.
dup_list.append(i)# this actually adds the number that is not already in the list.

``return dup_list # this says ok at the end of the fuction return me the list that we created``

EXAMPLE LIST:
n=[4,44,44,4,33,33,2,1]
print remove_duplicates(n)
[4, 44, 33, 2, 1]

I hope this helps to clarify the excersie, by the way I am learning so I do not know for sure if what I said make sense for you or anyone, so feedback is highly appreciated.

#4

def remove_duplicates(lits):
new_list = []
for x in lits:
if new_list==[]:
new_list.append(x)
continue
for y in new_list:
if x!=y:
new_list.append(x)

``return new_list``

#5

please stop hijacking other topics, go here and press the make new topic button and start making your own topics

#6

I'm a little confused here. I originally tried to return new_list but it wouldn't work. Replaced it with x and now it works. It also prints just the 2.

``````def remove_duplicates(lits):
new_list = []
for x in lits:
if x not in lits:
new_list.append(x)
return x

print remove_duplicates([1,1,2,2])``````

#7

here:

``if x not in lits:``

why would you want to check if `x` is not in lits? all values are in lits, Please rethink this. You want to check if `x` is not yet in `new_list`, and then return new_list

#9