How to remove multiples in a list

[jasper100125]

How can remove something in a list if it appears more than once?
examples:

list1 = [1, 2, 3, 4, 1]
>>>>>
list1 = [1, 2, 3, 4]


or


list2 = ['a', 'b', 'a', 'a']
>>>>>
list2 = ['a', 'b']

[perfringo]

What have you tried? Is order important or not?

[jasper100125]

I have tried the following:

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old_list = ['a', 'b', 'a', 'a', 'b'] new_list = [] for i in range(len(old_list)):     if old_list[i] not in new_list:           new_list.append(old_list[i]) print(new_list) >>>>> #This prints: ['a', 'b']

This works ofcourse, but i was wondering if there was an easier way of doing this.
Or atleast something less than 5 lines of code in order to make my code look a little bit cleaner.
I was thinking that maybe there was a command for it like

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list1 = ['a', 'b', 'a', 'a', 'b'] remove_multiples(list1) print(list1) >>>>> #So it will print:['a', 'b']

Order is not important.

[ichabod801]

If order is not important, you can use a set. A set can only have one of each value, so it can very efficiently strip out duplicates:

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list1 = ['a', 'b', 'a', 'a', 'b'] uniques = set(list1)               # set(['a', 'b']) uniques2 = list(set(list1))        # ['a', 'b']

Order is generally lost when you convert to a set, however.

[jasper100125]

Thanks!

[Gribouillis]

This works ofcourse, but i was wondering if there was an easier way of doing this.

If you don't mind using a third party library, there is

1 2	from more_itertools import unique_everseen new_list = list(unique_everseen(old_list))

[buran]

if order is important, in 3.7+ you can use order preserving feature of dicts

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>>> spam = ['a', 'b', 'a', 'c', 'b', 'a'] >>> eggs = list(dict(zip(spam, spam)).keys()) >>> eggs ['a', 'b', 'c'] >>>

before 3.7 you need to use collections.OrderedDict instead.

also it's worth mentioning that this solution as well using set() will work only if list elements are hashable.