Python 根据嵌套列表的概率删除嵌套列表中的项

所以我想在保留嵌套结构的同时删除嵌套列表中的重复项，删除哪个项是基于概率的决定。以下是我所拥有的：

import collections

new_sample = collections.defaultdict(list, {"dv1": [[3, 141], [13]], "dv2":[[3, 141], [157]], "dv3":[[0],[2]]})
new_sample_proba = collections.defaultdict(list, {"dv1": [[0.8, 0.9], [0.9]], "dv2": [[0.9, 0.8], [0.9]], "dv3": [[0.9], [0.95]]})

print("before fix: ",new_sample)
print("before fix: ",new_sample_proba)
full_views_list = list(new_sample.keys())
temp_views_list = list(new_sample.keys())
for curr_view in full_views_list:
    print("t1 ",new_sample[curr_view])
    print("t1 ",new_sample_proba[curr_view])
    temp_views_list.remove(curr_view)
    for i, curr_class_new in enumerate(new_sample[curr_view]):
        temp_pop_list = []
        for j, class_item in enumerate(curr_class_new):
            print(class_item)
            for other_view in temp_views_list:
                print("t2 org",new_sample[other_view])
                print("t2 org",new_sample_proba[other_view])
                for l, other_view_class_new in enumerate(new_sample[other_view]):
                    for m, other_class_item in enumerate(other_view_class_new):
                        # only perform delete if current view contain item, if it's deleted, move on
                        if new_sample[curr_view][i][j] == new_sample[other_view][l][m]:
                            if new_sample_proba[curr_view][i][j]>new_sample_proba[other_view][l][m]:
                                new_sample[other_view][l].pop(m)
                                new_sample_proba[other_view][l].pop(m)
                                print("t2 update",new_sample)
                                print("t2 update",new_sample_proba)
                            else:
                                temp_pop_list.append(j)
                            print(class_item, "check")
        temp_pop_list.sort(reverse=True)
        for idx in temp_pop_list:
            new_sample[curr_view][i].pop(idx)
            new_sample_proba[curr_view][i].pop(idx)

print("apply fix: ",new_sample)
print("apply fix: ",new_sample_proba)

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因此，上面的代码是有效的，我如何缩短它？

你能解释一下你的代码应该做什么吗？请提供3或4个示例的输入和预期输出（看看它们应该如何“冲突”，每个示例？只有“邻居”？）你能解释一下你的代码应该做什么吗？请提供3或4个示例的输入和预期输出（看看它们应该如何“冲突”，每个都是？只有“邻居”？…？）