Python: Remove Sublists from List if Same Including/Not Including Order

Question

Is there a way to remove duplicate sublists from a list of lists, even if they are not the same ordering?

So could I do something like make:

x = [[1,2],[3,4],[5,6],[2,1],[7,8]]

into

x = [[1,2],[3,4],[5,6],[7,8]]

Is there an itertools function or something with a for loop?

Thanks!

Will you ever have to distinguish [1,1,2] from [1,2,1]? And do you need to preserve the initial order of the sublists? — DSM
– DSM, Commented Dec 9, 2013 at 3:40
@DSM yes, the length of the sublists may increase. And no, the order is not important. — 5813
– 5813, Commented Dec 9, 2013 at 3:44
@DSM, but the order within the sublists is important, so [7,8] is not the same as [8,7]. — 5813
– 5813, Commented Dec 9, 2013 at 3:50
@DSM Yes, that is what I mean. If for a sublist there is a matching one elsewhere in the list, then one, preferably the second, should be removed. But, for a sublist which does not have a duplicate, it should remain the same. — 5813
– 5813, Commented Dec 9, 2013 at 3:55

behzad.nouri · Accepted Answer · 2015-02-27 00:49:42Z

4

this will preserve the order of list and sublists, with possible duplicates in sublists:

y, s = [], set()
for t in x:
    w = tuple(sorted(t))
    if not w in s:
        y.append(t)
        s.add(w)

if

x = [[1,2],[3,4],[5,6],[2,1,1],[2,1],[7,8],[4,3],[1,2,1]]

then y will be:

[[1, 2], [3, 4], [5, 6], [2, 1, 1], [7, 8]]

edited Feb 27, 2015 at 0:49

answered Dec 9, 2013 at 3:50

behzad.nouri

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1 Comment

DSM Over a year ago

+1: This preserves as much original information as possible (unlike my OrderedDict approach), and seems to satisfy the OP's desiderata.

score 1 · Accepted Answer · 2013-12-09 03:51:50Z

1

You can use frozenset:

>>> def remove_dups(L):
        return map(list, frozenset(map(frozenset, L)))

>>> x = [[1,2],[3,4],[5,6],[2,1],[7,8]]
>>> remove_dups(x)
[[5, 6], [1, 2], [8, 7], [3, 4]]
>>>

edited Dec 9, 2013 at 3:51

answered Dec 9, 2013 at 3:37

user2134086

Comments

Ashif Abdulrahman · Accepted Answer · 2013-12-09 06:10:14Z

0

try this:

a=[[1,2],[3,4],[5,6],[2,1],[7,8]]
y=[]
for i in a:
        if sorted(i) not in y:
                y.append(i)
print y

output is

[[1, 2], [3, 4], [5, 6], [7, 8]]

answered Dec 9, 2013 at 6:10

Ashif Abdulrahman

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