Python removes equal lists

Question

I have a list of equals in my list a:

a = [[a],[a],[b],[b],[c],[c]]

How can I remove equal lists so that I have the following:

a = [[a],[b],[c]]

I tried to do it with a set, but it doesn't work:

my_set = set()
for elem in a:
    my_set.add(elem)
for x in my_set:
    print(x)

+3

Python123 Jul 17 17 at 18:07

6 answers

aquil.abdullah · Answer 1 · 2017-07-17T18:09:56+0000

You can try this:

a = [['a'],['a'],['b'],['b'],['c'],['c']]
b = list()
for item in a:
    if item not in b:
        b.append(item)

Mr_U4913 · Answer 2 · 2017-07-17T18:21:57+0000

numpy

is one of the options

import numpy as np
a = [['a'],['a'],['b'],['b'],['c'],['c']] 
np.unique(a).tolist() #  ['a', 'b', 'c']

DyZ · Answer 3 · 2017-07-17T18:11:21+0000

If you don't mind using itertools

:

[x for x,_ in itertools.groupby(sorted(a))]
#[['a'], ['b'], ['c']]

If you do, convert the lists to tuples, create a set of them, and then convert back to lists:

list(map(list, set(map(tuple, a))))
#[['b'], ['c'], ['a']]

AGN Gazer · Answer 4 · 2017-07-17T18:15:59+0000

A variation on @ DYZ's answer that works on lists containing identical elements regardless of order:

[x for x,_ in itertools.groupby(map(set, a))]

For example:

>>> a = [['a'],['a'],['b'],['b'],['c', 'a'],['a', 'c']]
>>> [list(x) for x,_ in itertools.groupby(map(set, a))]
[['a'], ['b'], ['a', 'c']]

If ['c', 'a']

both are ['a', 'c']

not considered equal (see comments below from @DYZ), you can do something like this with set()

:

map(list, set(map(tuple, a)))

AGN Gazer · Answer 5 · 2017-07-17T18:27:16+0000

If ['c', 'a']

and ['a', 'c']

are not considered to be equal, you can do something like this with the help of set()

:

map(list, set(map(tuple, a)))

Midhun c nair · Answer 6 · 2017-07-17T18:41:02+0000

You can use a list instead.

d = [it for i, it in enumerate(a) if it not in a[0:i]]
print(d)