如果相同，则删除多维列表中的元组

问题描述

我有一个元组列表:

[[[('p', 'u'), ('r', 'w')], [('t', 'q')]], [[('p', 'u'), ('r', 'w')], [('v', 'q')]], [[('p', 'u'), ('r', 'w')], [('t', 's')]], [[('p', 'u'), ('r', 'w')], [('v', 's')]], [[('p', 'w'), ('r', 'u')], [('t', 'q')]], [[('p', 'w'), ('r', 'u')], [('v', 'q')]], [[('p', 'w'), ('r', 'u')], [('t', 's')]], [[('p', 'w'), ('r', 'u')], [('v', 's')]], [[('r', 'u'), ('p', 'w')], [('t', 'q')]], [[('r', 'u'), ('p', 'w')], [('v', 'q')]], [[('r', 'u'), ('p', 'w')], [('t', 's')]], [[('r', 'u'), ('p', 'w')], [('v', 's')]], **[[('r', 'w'), ('p', 'u')], [('t', 'q')]]**, [[('r', 'w'), ('p', 'u')], [('v', 'q')]], [[('r', 'w'), ('p', 'u')], [('t', 's')]], [[('r', 'w'), ('p', 'u')], [('v', 's')]]]

但是现在例如元素[[('p','u'),('r','w')], [('t','q')]]

与[[('r','w'),('p','u')], [('t','q')]]相同，在列表中标记为 fat .

is the same as [[('r','w'),('p','u')], [('t','q')]], which are marked fat in the list.

因此，在列表中，我有16个元素，其中每个元素都是双精度的.

So in the list I have 16 elements, where every element is double.

现在，我要删除重复项，只剩下前八个元素.

Now, I want to delete the duplicates, that I have only the first eight elements left.

所以我很天真地尝试过

[[list(y) for y in set([tuple(set(x)) for x in doublegammas1])]]

但是在这里，他说:

TypeError: unhashable type: 'list'

所以我的问题:

如何扩展列表理解功能，使其适用于更多维的列表?

How can I extend the list comprehension, that it works for a more dimensional list?

推荐答案

可变对象(例如列表或集合)不能是集合的成员.您可以使用不可变的Frozenset.

A mutable object (such as a list or a set) cannot be a member of a set. You can use a frozenset, which is immutable.

main_list = [[[('p', 'u'), ('r', 'w')], [('t', 'q')]],
             [[('p', 'u'), ('r', 'w')], [('v', 'q')]],
             [[('p', 'u'), ('r', 'w')], [('t', 's')]],
             [[('p', 'u'), ('r', 'w')], [('v', 's')]],
             [[('p', 'w'), ('r', 'u')], [('t', 'q')]],
             [[('p', 'w'), ('r', 'u')], [('v', 'q')]],
             [[('p', 'w'), ('r', 'u')], [('t', 's')]],
             [[('p', 'w'), ('r', 'u')], [('v', 's')]],
             [[('r', 'u'), ('p', 'w')], [('t', 'q')]],
             [[('r', 'u'), ('p', 'w')], [('v', 'q')]],
             [[('r', 'u'), ('p', 'w')], [('t', 's')]],
             [[('r', 'u'), ('p', 'w')], [('v', 's')]],
             [[('r', 'w'), ('p', 'u')], [('t', 'q')]],
             [[('r', 'w'), ('p', 'u')], [('v', 'q')]],
             [[('r', 'w'), ('p', 'u')], [('t', 's')]],
             [[('r', 'w'), ('p', 'u')], [('v', 's')]]]

main_set = set(tuple(frozenset(innermost_list) for innermost_list in sublist) for sublist in main_list)

from pprint import pprint
pprint(main_set)

输出:

{(frozenset({('r', 'u'), ('p', 'w')}), frozenset({('t', 'q')})),
 (frozenset({('p', 'u'), ('r', 'w')}), frozenset({('v', 'q')})),
 (frozenset({('r', 'u'), ('p', 'w')}), frozenset({('v', 'q')})),
 (frozenset({('p', 'u'), ('r', 'w')}), frozenset({('t', 's')})),
 (frozenset({('r', 'u'), ('p', 'w')}), frozenset({('t', 's')})),
 (frozenset({('p', 'u'), ('r', 'w')}), frozenset({('v', 's')})),
 (frozenset({('r', 'u'), ('p', 'w')}), frozenset({('v', 's')})),
 (frozenset({('p', 'u'), ('r', 'w')}), frozenset({('t', 'q')}))}

要转换回嵌套列表的原始结构，请执行以下操作:

To convert back to the original structure of nested lists:

new_list = [[list(frozen) for frozen in subtuple] for subtuple in main_set]
pprint(new_list)

输出:

[[[('r', 'u'), ('p', 'w')], [('t', 'q')]],
 [[('p', 'u'), ('r', 'w')], [('v', 'q')]],
 [[('r', 'u'), ('p', 'w')], [('v', 'q')]],
 [[('p', 'u'), ('r', 'w')], [('t', 's')]],
 [[('r', 'u'), ('p', 'w')], [('t', 's')]],
 [[('p', 'u'), ('r', 'w')], [('v', 's')]],
 [[('r', 'u'), ('p', 'w')], [('v', 's')]],
 [[('p', 'u'), ('r', 'w')], [('t', 'q')]]]

更新:

一种从输入数据中就地删除重复项的解决方案.

A solution that removes the duplicate items in-place from the input data.

unique = []

for item in main_list[:]:
    frozen_item = frozenset(frozenset(innermost_list) for innermost_list in item)
    if frozen_item not in unique:
        unique.append(frozen_item)
    else:
        main_list.remove(item)

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