How to remove all duplicate items from a list [duplicate]

Answer 1

You can do this like that:

x = list(set(x))

Example: if you do something like that:

x = [1,2,3,4,5,6,7,8,9,10,2,1,6,31,20]
x = list(set(x))
x

you will see the following result:

[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 31]

There is only one thing you should think of: the resulting list will not be ordered as the original one (will lose the order in the process).

Answer 2

There is a faster way to fix this:

list = [1, 1.0, 1.41, 1.73, 2, 2, 2.0, 2.24, 3, 3, 4, 4, 4, 5, 6, 6, 8, 8, 9, 10]
list2=[]

for value in list:
    try:
        list2.index(value)
    except:
        list2.append(value)
list.clear()
for value in list2:
    list.append(value)
list2.clear()
print(list)
print(list2)

Answer 3

The modern way to do it that maintains the order is:

>>> from collections import OrderedDict
>>> list(OrderedDict.fromkeys(lseparatedOrbList))

as discussed by Raymond Hettinger (python core dev) in this answer. In python 3.5 and above this is also the fastest way - see the linked answer for details. However the keys must be hashable (as is the case in your list I think)

Answer 4

This should be faster and will preserve the original order:

seen = {}
new_list = [seen.setdefault(x, x) for x in my_list if x not in seen]

If you don't care about order, you can just:

new_list = list(set(my_list))

Answer 5

No, it's simply a typo, the "list" at the end must be capitalized. You can nest loops over the same variable just fine (although there's rarely a good reason to).

However, there are other problems with the code. For starters, you're iterating through lists, so i and j will be items not indices. Furthermore, you can't change a collection while iterating over it (well, you "can" in that it runs, but madness lies that way - for instance, you'll propably skip over items). And then there's the complexity problem, your code is O(n^2). Either convert the list into a set and back into a list (simple, but shuffles the remaining list items) or do something like this:

seen = set()
new_x = []
for x in xs:
    if x in seen:
        continue
    seen.add(x)
    new_xs.append(x)

Both solutions require the items to be hashable. If that's not possible, you'll probably have to stick with your current approach sans the mentioned problems.

Answer 6

In this way one can delete a particular item which is present multiple times in a list : Try deleting all 5

list1=[1,2,3,4,5,6,5,3,5,7,11,5,9,8,121,98,67,34,5,21]
print list1
n=input("item to be deleted : " )
for i in list1:
    if n in list1:
        list1.remove(n)
print list1