python - 查找公共(public)列表序列

Question

我有一本列表字典。每个列表都是一个数字序列。没有两个列表是相等的，但两个或更多列表可能以相同的数字序列开头(请参阅下面的示例输入)。我想做的是找到这些常见的序列，并使它们成为字典中的新元素。

示例输入:

sequences = {
    18: [1, 3, 5, 6, 8, 12, 15, 17, 18],
    19: [1, 3, 5, 6, 9, 13, 14, 16, 19],
    25: [1, 3, 5, 6, 9, 13, 14, 20, 25],
    11: [0, 2, 4, 7, 11],
    20: [0, 2, 4, 10, 20],
    26: [21, 23, 26],
}

示例输出:

expected_output = {
    6: [1, 3, 5, 6],
    18: [8, 12, 15, 17, 18],
    14: [9, 13, 14],
    19: [16, 19],
    25: [20, 25],
    4: [0, 2, 4],
    11: [7, 11],
    20: [10, 20],
    26: [21, 23, 26],
}

每个列表的键是它的最后一个元素。顺序无关紧要。

我有一个工作代码。但是，它非常困惑。有人可以建议更简单/更清洁的解决方案吗？

from collections import Counter

def split_lists(sequences):
    # get first elem from each sequence
    firsts = list(map(lambda s: s[0], sequences))

    # get non-duplicate first elements
    not_duplicates = list(map(lambda c: c[0], filter(lambda c: c[1] == 1, Counter(firsts).items())))

    # start the new_sequences with the non-duplicate lists
    new_sequences = dict(map(lambda s: (s[-1], s), filter(lambda s: s[0] in not_duplicates, sequences)))

    # get duplicate first elements
    duplicates = list(map(lambda c: c[0], filter(lambda c: c[1] > 1, Counter(firsts).items())))
    for duplicate in duplicates:
        # get all lists that start with the duplicate element
        duplicate_lists = list(filter(lambda s: s[0] == duplicate, sequences))

        # get the common elements from the duplicate lists and make it a new
        # list to add to our new_sequences dict
        repeated_sequence = sorted(list(set.intersection(*list(map(set, duplicate_lists)))))
        new_sequences[repeated_sequence[-1]] = repeated_sequence

        # get lists from where I left of
        i = len(repeated_sequence)
        sub_lists = list(filter(lambda s: len(s) > 0, map(lambda s: s[i:], duplicate_lists)))

        # recursively split them and store them in new_sequences
        new_sequences.update(split_lists(sub_lists))

    return new_sequences

另外，你能帮我弄清楚我的算法的复杂性吗？递归让我头晕。我最好的猜测是 O(n*m)，其中 n 是列表的数量，m 是最长列表的长度。

Answer 1

将其分解为逻辑函数:

• 找出哪些序列以相同元素开头
• 找到共同的元素

同样的开始:

可以使用 defaultdict 轻松完成

from collections import defaultdict
def same_start(sequences):
    same_start = defaultdict(list)
    for seq in sequences:
        same_start[seq[0]].append(seq)
    return same_start.values()

list(same_start(sequences.values()))

[[[1, 3, 5, 6, 8, 12, 15, 17, 18],
  [1, 3, 5, 6, 9, 13, 14, 16, 19],
  [1, 3, 5, 6, 9, 13, 14, 20, 25]],
 [[0, 2, 4, 7, 11], [0, 2, 4, 10, 20]],
 [[21, 23, 26]]]

找到共同的元素:

一个简单的生成器，只要它们都相同就生成值

def get_beginning(sequences):
    for values in zip(*sequences):
        v0 = values[0]
        if not all(i == v0 for i in values):
            return
        yield v0

聚合

def aggregate(same_start):
    for seq in same_start:
        if len(seq) < 2:
            yield  seq[0]
            continue
        start = list(get_beginning(seq))
        yield start
        yield from (i[len(start):] for i in seq)

list(aggregate(same_start(sequences.values())))

[[1, 3, 5, 6],
 [8, 12, 15, 17, 18],
 [9, 13, 14, 16, 19],
 [9, 13, 14, 20, 25],
 [0, 2, 4],
 [7, 11],
 [10, 20],
 [21, 23, 26]]

进一步

如果你想组合序列 18 和 25，那么你可以这样做

def combine(sequences):
    while True:
        s = same_start(sequences)
        if all(len(i) == 1 for i in s):
            return sequences
        sequences = tuple(aggregate(s))

{i[-1]: i for i in combine(sequences.values())}

{4: [0, 2, 4],
 6: [1, 3, 5, 6],
 11: [7, 11],
 14: [9, 13, 14],
 18: [8, 12, 15, 17, 18],
 19: [16, 19],
 20: [10, 20],
 25: [20, 25],
 26: [21, 23, 26]}