如何删除列表中的部分重复项

Question

I have a list of lists where it looks like

[[1,'a',2],[1,'b',2],[1,'a',3]]

I want to remove the item from the list if the second element in the list of lists are the same (eg they are both a )

I want to create output that looks like:

[[1,'a',2],[1,'b',2]]

where it grabs the first one in the list of the duplicates.

Answer 1

that's a variant of How do you remove duplicates from a list whilst preserving order? .

You can use a marker set to track the already appended sublists since strings are immutable so hashable & storable in a set :

lst = [[1,'a',2],[1,'b',2],[1,'a',3]]

marker_set = set()

result = []

for sublist in lst:
    second_elt = sublist[1]
    if second_elt not in marker_set:
        result.append(sublist)
        marker_set.add(second_elt)

print(result)

prints:

[[1, 'a', 2], [1, 'b', 2]]

(using a marker set and not a list allows an average O(1) lookup instead of O(N) )

Answer 2

You can use a dictionary where the second element is the key, on the reverse of the list, to drop duplicates:

dct = {j: (i, k) for i, j, k in reversed(L)}

{'a': (1, 2), 'b': (1, 2)}

---

Getting the result back as a list:

[[i, j, k] for j, (i, k) in dct.items()]

[[1, 'a', 2], [1, 'b', 2]]

While this solution will always keep the first occurence of a duplicate, the relative order of elements is not guaranteed in the final result.

Answer 3

lst = [[1,'a',2],[1,'b',2],[1,'a',3]]
res = []
for i in lst:
    if not any(i[1] in j for j in res):
        res.append(i)

print(res)
# [[1, 'a', 2], [1, 'b', 2]]