Removing Duplicates in Lists
How Do I Get a New List Without Duplicates? [1, 2, 3, 1] → [1, 2, 3] How Do I Get a New List Where Items That Are Duplicated Are Entirely Removed? [1, 2, 3, 1]...
How do I get a new list without duplicates?
[1, 2, 3, 1] → [1, 2, 3]
How do I get a new list where items that are duplicated are entirely removed?
[1, 2, 3, 1] → [2, 3]
57 Answers
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.
In Python 2.7, the new way of removing duplicates from an iterable while keeping it in the original order is:
>>> from collections import OrderedDict
>>> list(OrderedDict.fromkeys('abracadabra'))
['a', 'b', 'r', 'c', 'd']
In Python 3.5, the OrderedDict has a C implementation. My timings show that this is now both the fastest and shortest of the various approaches for Python 3.5.
In Python 3.6, the regular dict became both ordered and compact. (This feature is holds for CPython and PyPy but may not present in other implementations). That gives us a new fastest way of deduping while retaining order:
>>> list(dict.fromkeys('abracadabra'))
['a', 'b', 'r', 'c', 'd']
In Python 3.7, the regular dict is guaranteed to both ordered across all implementations. So, the shortest and fastest solution is:
>>> list(dict.fromkeys('abracadabra'))
['a', 'b', 'r', 'c', 'd']
It's a one-liner: list(set(source_list)) will do the trick.
A set is something that can't possibly have duplicates.
Update: an order-preserving approach is two lines:
from collections import OrderedDict
OrderedDict((x, True) for x in source_list).keys()
Here we use the fact that OrderedDict remembers the insertion order of keys, and does not change it when a value at a particular key is updated. We insert True as values, but we could insert anything, values are just not used. (set works a lot like a dict with ignored values, too.)
>>> t = [1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> t
[1, 2, 3, 1, 2, 5, 6, 7, 8]
>>> s = []
>>> for i in t:
if i not in s:
s.append(i)
>>> s
[1, 2, 3, 5, 6, 7, 8]
If you don't care about the order, just do this:
def remove_duplicates(l):
return list(set(l))
A set is guaranteed to not have duplicates.
To make a new list retaining the order of first elements of duplicates in L:
newlist = [ii for n,ii in enumerate(L) if ii not in L[:n]]
For example: if L = [1, 2, 2, 3, 4, 2, 4, 3, 5], then newlist will be [1, 2, 3, 4, 5]
This checks each new element has not appeared previously in the list before adding it. Also it does not need imports.
There are also solutions using Pandas and Numpy. They both return numpy array so you have to use the function .tolist() if you want a list.
t=['a','a','b','b','b','c','c','c']
t2= ['c','c','b','b','b','a','a','a']
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Pandas solution
Using Pandas function unique():
import pandas as pd
pd.unique(t).tolist()
>>>['a','b','c']
pd.unique(t2).tolist()
>>>['c','b','a']
Maya is a hardware enthusiast who tests and reviews smart home devices, smartphones, wearables, and audio gear. She focuses on practical consumer value and build quality.
