删除列表列表中的所有重复项（不包括原始列表）

Question

I have a list of lists and I want to remove all the duplicates so that the similar lists will not appear at all in the new list.

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

output == [[5,6,2], [3]]

So for example, [1,2] have a duplicate so it should not appear in the final list at all. This is different from what others have asked as they wanted to keep one copy of the duplicate in the final list.

Answer 1

You can use count to count the number of occurrences of a element.

Approach 1 Time complexity O(n^2)

final=[]
for i in k:
    if k.count(i)==1:
        final.append(i)
print(final)

---

[[5, 6, 2], [3]]

---

Pythonic way to write this is :

final=[i for i in k if k.count(i)==1]

---

Approach 2 Time complexity O(n^2)

Or you can search if ith element is present in rest of the list or not. If present don't add it to final .

for i,lst in enumerate(k):
    if lst not in k[:i]+k[i+1:]:
        final.append(lst)
print(final)

---

output

[[5, 6, 2], [3]]

pythonic way to write this:

final=[i for i,lst in enumerate(k) if lst not in k[:i]+k[i+1:]]

Approach 3 Time complexity is O(n)

You can achieve this in O(n) by using dictionaries.

dic={}
k=list(map(tuple,k))   #Since key values in dictionary should always be immutable.
for i in k:
    dic[i]=dic.setdefault(i,0)+1

final=[]

for k in dic:
    if dic[k]== 1:
        final.append(list(k))

Answer 2

For the general problem of dropping duplicates, you could use collections.Counter , although it requires hashable values.

import collections

def drop_all_duplicates(values):
    # As of CPython 3.7, Counter remembers insertion order.
    value_counts = collections.Counter(values)
    return (value for value, count in value_counts.items() if count == 1)

>>> list(drop_all_duplicates([1, 1, 2, 3, 3, 4, 5, 5, 6])
[2, 4, 6]

This could be expanded to cover non-hashable values by accepting a function that converts them to hashable (eg a function that converts lists to tuples).