'remove duplicates from 2d lists regardless of order [duplicate]
I have a 2d list
a = [[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]]
How can I get the result:
result = [[1,2],[1,3],[2,3]]
Where duplicates are removed regardless of their order of the inner lists.
Solution 1:[1]
In [3]: b = []
In [4]: for aa in a:
...: if not any([set(aa) == set(bb) for bb in b if len(aa) == len(bb)]):
...: b.append(aa)
In [5]: b
Out[5]: [[1, 2], [1, 3], [2, 3]]
Solution 2:[2]
Try using a set to keep track of what lists you have seen:
from collections import Counter
a = [[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2], [1, 2, 1]]
seen = set()
result = []
for lst in a:
current = frozenset(Counter(lst).items())
if current not in seen:
result.append(lst)
seen.add(current)
print(result)
Which outputs:
[[1, 2], [1, 3], [2, 3], [1, 2, 1]]
Note: Since lists are not hash able, you can store frozensets of Counter
objects to detect order less duplicates. This removes the need to sort at all.
Solution 3:[3]
You can try
a = [[1, 2], [1, 3], [2, 1], [2, 3], [3, 1], [3, 2]]
aa = [tuple(sorted(elem)) for elem in a]
set(aa)
output
{(1, 2), (1, 3), (2, 3)}
Solution 4:[4]
The 'Set' concept would come in handy here. The list you have (which contains duplicates) can be converted to a Set (which will never contain a duplicate). Find more about Sets here : Set
Example :
l = ['foo', 'foo', 'bar', 'hello']
A set can be created directly:
s = set(l)
now if you check the contents of the list
print(s)
>>> {'foo', 'bar', 'hello'}
Set will work this way with any iterable object! Hope it helps!
Sources
This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.
Source: Stack Overflow
Solution | Source |
---|---|
Solution 1 | |
Solution 2 | |
Solution 3 | difegam3 |
Solution 4 | Joseph Biju Cadavil |