pythonalgorithmlistduplicatesintersection

# Removing duplicates in lists

How can I check if a list has any duplicates and return a new list without duplicates?

Solution

• The common approach to get a unique collection of items is to use a `set`. Sets are unordered collections of distinct objects. To create a set from any iterable, you can simply pass it to the built-in `set()` function. If you later need a real list again, you can similarly pass the set to the `list()` function.

The following example should cover whatever you are trying to do:

``````>>> t = [1, 2, 3, 1, 2, 3, 5, 6, 7, 8]
>>> list(set(t))
[1, 2, 3, 5, 6, 7, 8]
>>> s = [1, 2, 3]
>>> list(set(t) - set(s))
[8, 5, 6, 7]
``````

As you can see from the example result, the original order is not maintained. As mentioned above, sets themselves are unordered collections, so the order is lost. When converting a set back to a list, an arbitrary order is created.

### Maintaining order

If order is important to you, then you will have to use a different mechanism. A very common solution for this is to rely on `OrderedDict` to keep the order of keys during insertion:

``````>>> from collections import OrderedDict
>>> list(OrderedDict.fromkeys(t))
[1, 2, 3, 5, 6, 7, 8]
``````

Starting with Python 3.7, the built-in dictionary is guaranteed to maintain the insertion order as well, so you can also use that directly if you are on Python 3.7 or later (or CPython 3.6):

``````>>> list(dict.fromkeys(t))
[1, 2, 3, 5, 6, 7, 8]
``````

Note that this may have some overhead of creating a dictionary first, and then creating a list from it. If you don’t actually need to preserve the order, you’re often better off using a set, especially because it gives you a lot more operations to work with. Check out this question for more details and alternative ways to preserve the order when removing duplicates.

Finally note that both the `set` as well as the `OrderedDict`/`dict` solutions require your items to be hashable. This usually means that they have to be immutable. If you have to deal with items that are not hashable (e.g. list objects), then you will have to use a slow approach in which you will basically have to compare every item with every other item in a nested loop.