python - 查找所有常见的、不重叠的子串

Question

给定两个字符串，我想从最长到最短的顺序识别所有常见的子字符串。

我想删除任何“子”子字符串。例如，“1234”的任何子字符串都不会包含在“12345”和“51234”之间的匹配中。

string1 = '51234'

string2 = '12345'

result = ['1234', '5']

我想找到 longest common substring ，然后递归地找到左边/右边最长的子串。但是，我不想在找到后删除公共(public)子字符串。例如，下面的结果共享中间的 6:

string1 = '12345623456'

string2 = '623456'

result = ['623456', '23456']

最后，我需要根据包含数千个字符串的固定列表检查一个字符串。我不确定我是否可以采取明智的步骤来散列这些字符串中的所有子字符串。

以前的回答:

在这个 thread 中，找到了一个动态规划解决方案，它需要 O(nm) 时间，其中 n 和 m 是字符串的长度。我对使用后缀树的更有效方法感兴趣。

背景:

我正在根据旋律片段创作歌曲旋律。有时，组合会设法生成与现有音符的一行中太多音符相匹配的旋律。

我可以使用字符串相似性度量，例如 Edit Distance，但相信与旋律差异非常小的曲调是独特且有趣的。不幸的是，这些曲调与连续复制许多旋律音符的歌曲具有相似的相似度。

Answer 1

让我们从树开始

from collections import defaultdict
def identity(x):
    return x


class TreeReprMixin(object):
    def __repr__(self):
        base = dict(self)
        return repr(base)


class PrefixTree(TreeReprMixin, defaultdict):
    '''
    A hash-based Prefix or Suffix Tree for testing for
    sequence inclusion. This implementation works for any
    slice-able sequence of hashable objects, not just strings.
    '''
    def __init__(self):
        defaultdict.__init__(self, PrefixTree)
        self.labels = set()

    def add(self, sequence, label=None):
        layer = self
        if label is None:
            label = sequence
        if label:
            layer.labels.add(label)
        for i in range(len(sequence)):
            layer = layer[sequence[i]]
            if label:
                layer.labels.add(label)

        return self

    def add_ngram(self, sequence, label=None):
        if label is None:
            label = sequence
        for i in range(1, len(sequence) + 1):
            self.add(sequence[:i], label)

    def __contains__(self, sequence):
        layer = self
        j = 0
        for i in sequence:
            j += 1
            if not dict.__contains__(layer, i):
                break
            layer = layer[i]
        return len(sequence) == j

    def depth_in(self, sequence):
        layer = self
        count = 0
        for i in sequence:
            if not dict.__contains__(layer, i):
                print "Breaking"
                break
            else:
                layer = layer[i]
            count += 1
        return count

    def subsequences_of(self, sequence):
        layer = self
        for i in sequence:
            layer = layer[i]
        return layer.labels

    def __iter__(self):
        return iter(self.labels)


class SuffixTree(PrefixTree):
    '''
    A hash-based Prefix or Suffix Tree for testing for
    sequence inclusion. This implementation works for any
    slice-able sequence of hashable objects, not just strings.
    '''
    def __init__(self):
        defaultdict.__init__(self, SuffixTree)
        self.labels = set()

    def add_ngram(self, sequence, label=None):
        if label is None:
            label = sequence
        for i in range(len(sequence)):
            self.add(sequence[i:], label=label)

要填充树，您将使用 .add_ngram 方法。

下一部分有点棘手，因为您要在跟踪树坐标的同时寻找字符串的并发遍历。为了完成这一切，我们需要一些对树和查询字符串进行操作的函数

def overlapping_substrings(string, tree, solved=None):
    if solved is None:
        solved = PrefixTree()
    i = 1
    last = 0
    matching = True
    solutions = []
    while i < len(string) + 1:
        if string[last:i] in tree:
            if not matching:
                matching = True
            else:
                i += 1
                continue
        else:
            if matching:
                matching = False
                solutions.append(string[last:i - 1])
                last = i - 1
                i -= 1
        i += 1
    if matching:
        solutions.append(string[last:i])
    for solution in solutions:
        if solution in solved:
            continue
        else:
            solved.add_ngram(solution)
            yield solution

def slide_start(string):
    for i in range(len(string)):
        yield string[i:]

def seek_subtree(tree, sequence):
    # Find the node of the search tree which
    # is found by this sequence of items
    node = tree
    for i in sequence:
        if i in node:
            node = node[i]
        else:
            raise KeyError(i)
    return node

def find_all_common_spans(string, tree):
    # We can keep track of solutions to avoid duplicates
    # and incomplete prefixes using a Prefix Tree
    seen = PrefixTree()
    for substring in slide_start(string):
        # Drive generator forward
        list(overlapping_substrings(substring, tree, seen))
    # Some substrings are suffixes of other substrings which you do not
    # want
    compress = SuffixTree()
    for solution in sorted(seen.labels, key=len, reverse=True):
        # A substrings may be a suffix of another substrings, but that substrings
        # is actually a repeating pattern. If a solution is
        # a repeating pattern, `not solution in seek_subtree(tree, solution)` will tell us.
        # Otherwise, discard the solution
        if solution in compress and not solution in seek_subtree(tree, solution):
            continue
        else:
            compress.add_ngram(solution)
    return compress.labels

def search(query, corpus):
    tree = SuffixTree()
    if isinstance(corpus, SuffixTree):
        tree = corpus
    else:
        for elem in corpus:
            tree.add_ngram(elem)
    return list(find_all_common_spans(query, tree))

所以现在要做你想做的事，这样做:

search("12345", ["51234"])
search("623456", ["12345623456"])

如果有什么不清楚的地方，请告诉我，我会尽力澄清。