我应该如何获得列表中重复子列表的列表？

Question

我正在尝试创建允许我获取列表的唯一子列表的列表的函数。这些函数适用于某些列表列表，但不适用于其他列表，我不确定为什么。

获取重复子列表的索引然后构建它们的列表的可靠有效方法是什么？

以下最小工作示例说明了该功能。已找到列表 a 的重复项，但未正确找到列表 b.

的重复项

def indices_of_list_element_duplicates(x):
    seen = set()
    for index, element in enumerate(x):
        if isinstance(element, list):
            element = tuple(element)
        if element not in seen:
            seen.add(element)
        else:
            yield index

def list_element_duplicates(x):
    indices = list(indices_of_list_element_duplicates(x))
    return [x[index] for index in indices]

a = [[1, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [5, 2]]

print(list_element_duplicates(a))

print("--------------------------------------------------------------------------------")

b = [[10], [15], [20], [10, 10], [10, 15], [10, 20], [15, 10], [15, 15], [15, 20], [20, 10], [20, 15], [20, 20], [10, 10, 10], [10, 10, 15], [10, 10, 20], [10, 15, 10], [10, 15, 15], [10, 15, 20], [10, 20, 10], [10, 20, 15], [10, 20, 20], [15, 10, 10], [15, 10, 15], [15, 10, 20], [15, 15, 10], [15, 15, 15], [15, 15, 20], [15, 20, 10], [15, 20, 15], [15, 20, 20], [20, 10, 10], [20, 10, 15], [20, 10, 20], [20, 15, 10], [20, 15, 15], [20, 15, 20], [20, 20, 10], [20, 20, 15], [20, 20, 20], [10], [15], [20], [10, 10], [10, 15], [10, 20], [15, 10], [15, 15], [15, 20], [20, 10], [20, 15], [20, 20], [10, 10, 10], [10, 10, 15], [10, 10, 20], [10, 15, 10], [10, 15, 15], [10, 15, 20], [10, 20, 10], [10, 20, 15], [10, 20, 20], [15, 10, 10], [15, 10, 15], [15, 10, 20], [15, 15, 10], [15, 15, 15], [15, 15, 20], [15, 20, 10], [15, 20, 15], [15, 20, 20], [20, 10, 10], [20, 10, 15], [20, 10, 20], [20, 15, 10], [20, 15, 15], [20, 15, 20], [20, 20, 10], [20, 20, 15], [20, 20, 20], [10], [15], [20], [10, 10], [10, 15], [10, 20], [15, 10], [15, 15], [15, 20], [20, 10], [20, 15], [20, 20], [10, 10, 10], [10, 10, 15], [10, 10, 20], [10, 15, 10], [10, 15, 15], [10, 15, 20], [10, 20, 10], [10, 20, 15], [10, 20, 20], [15, 10, 10], [15, 10, 15], [15, 10, 20], [15, 15, 10], [15, 15, 15], [15, 15, 20], [15, 20, 10], [15, 20, 15], [15, 20, 20], [20, 10, 10], [20, 10, 15], [20, 10, 20], [20, 15, 10], [20, 15, 15], [20, 15, 20], [20, 20, 10], [20, 20, 15], [20, 20, 20], [10], [15], [20], [10, 10], [10, 15], [10, 20], [15, 10], [15, 15], [15, 20], [20, 10], [20, 15], [20, 20], [10, 10, 10], [10, 10, 15], [10, 10, 20], [10, 15, 10], [10, 15, 15], [10, 15, 20], [10, 20, 10], [10, 20, 15], [10, 20, 20], [15, 10, 10], [15, 10, 15], [15, 10, 20], [15, 15, 10], [15, 15, 15], [15, 15, 20], [15, 20, 10], [15, 20, 15], [15, 20, 20], [20, 10, 10], [20, 10, 15], [20, 10, 20], [20, 15, 10], [20, 15, 15], [20, 15, 20], [20, 20, 10], [20, 20, 15], [20, 20, 20]]

print(list_element_duplicates(b))

Answer 1

Python 列表有一个漂亮的内置函数，叫做 count。使用这个你可以做：

a = [[1, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [5, 2]]
dups = list()

for e in a:
    if a.count(e) > 1:
        dups.append(e)

这将为您提供一个名为 dups 的列表，其中包含 [[1,2],[1,2],[5,2],[5,2]]

Answer 2

您可以使用 Counter dict 将子列表映射到元组并获取计数，仅保留计数大于 1 的子列表：

from collections import Counter
a = [[1, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [5, 2]]


cn = Counter(map(tuple,a))

print([sub for sub in a if cn[tuple(sub)] > 1])

适用于混合类型并获得独特性 returns:

from collections import Counter    

a = [[1, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [5, 2], "foo", 123, 123]

def counts(x):
    for ele in x:
        if isinstance(ele, Hashable):
            yield ele
        else:
            yield tuple(ele)


def unique_dupes(x):
    cnts = Counter(counts(x))
    for ele in x:
        t = ele
        if not isinstance(ele, Hashable):
            t = tuple(ele)
        if cnts[t] > 1:
            yield ele
            del cnts[t]

print(list(unique_dupes(a)))

输出：

 [[1, 2], [5, 2], 123]

Answer 3

问题必须来自这些行：

if isinstance(element, list):
    element = tuple(element)
if element not in seen:
    seen.add(element)

所以如果你已经在 seen 中说例如 [10,15] 然后你想检查 [15,10] 中看到的，它会 return FALSE.

当您认为 [x,y] 与 [y,x] 相同时，解决此问题的方法是对您检查的每个元素进行排序，这样：

if isinstance(element, list):
    element = tuple(sorted(element))
if element not in seen:
    seen.add(element)

Answer 4

列表理解很容易

list_a = [[1, 2], [1, 2], [2, 2], [3, 2], [4, 2], [5, 2], [5, 2]]

    unique_list=[]
    duplicate_list=[]
    sorted_list=[sorted(item) for item in list_a]

    final_list=[unique_list.append(item) if item not in unique_list else duplicate_list.append(item) for item in sorted_list]
    print(unique_list)
    print(duplicate_list)

我应该如何获得列表中重复子列表的列表？

How should I get a list of duplicate sublists in a list?

python

list

duplicates

sublist