在不导入库和使用集合的情况下删除列表中重复项的最快方法

Question

I was trying to remove duplicates from a list using the following code:

a = [1,2,3,4,2,6,1,1,5,2]
res = []
[res.append(i) for i in a if i not in res]

But I would like to do this without defining the list I want as an empty list (ie, omit the line res = [] ) like:

a = [1,2,3,4,2,6,1,1,5,2]
#Either:
res = [i for i in a if i not in res]
#Or:
[i for i in a if i not in 'this list'] # this list is not a string. I meant it as the list being comprehensed

I want to avoid library imports and set()

Answer 1

I think may work for you. It removes duplicates from the list while keeping the order.

newlist=[i for n,i in enumerate(L) if i not in L[:n]]

Answer 2

For Python3.6+, you can use dict.fromkeys() :

>>> a = [1, 2, 3, 4, 2, 6, 1, 1, 5, 2]
>>> list(dict.fromkeys(a))
[1, 2, 3, 4, 6, 5]

From the docs :

Create a new dictionary with keys from iterable and values set to value.

If you are using a lower Python version, you will need to use collections.OrderedDict to maintain order:

>>> from collections import OrderedDict
>>> a = [1, 2, 3, 4, 2, 6, 1, 1, 5, 2]
>>> list(OrderedDict.fromkeys(a))
[1, 2, 3, 4, 6, 5]

Answer 3

here is a simple benchmark with the proposed solutions,

it shows that dict.fromkeys will perform the best

from simple_benchmark import BenchmarkBuilder
import random


b = BenchmarkBuilder()

@b.add_function()
def AmitDavidson(a):
    return [i for n,i in enumerate(a) if i not in a[:n]]

@b.add_function()
def RoadRunner(a):
    return list(dict.fromkeys(a))

@b.add_function()
def DaniMesejo(a):
    return  list({k: '' for k in a})


@b.add_function()
def rdas(a):
    return  sorted(list(set(a)), key=lambda x: a.index(x))


@b.add_function()
def unwanted_set(a):
    return  list(set(a))


@b.add_arguments('List lenght')
def argument_provider():
    for exp in range(2, 18):
        size = 2**exp
        yield size, [random.randint(0, 10) for _ in range(size)]

r = b.run()
r.plot()

Answer 4

Here is a solution using set that does preserve the order:

a = [1,2,3,4,2,6,1,1,5,2]
a_uniq = sorted(list(set(a)), key=lambda x: a.index(x))
print(a_uniq)

Answer 5

One-liner, comprehension, O(n) , that preserves order in Python 3.6+:

a = [1, 2, 3, 4, 2, 6, 1, 1, 5, 2]

res = list({k: '' for k in a})
print(res)