查找所有常见的、不重叠的子字符串。

Question

给定两个字符串，我想识别所有公共子字符串从最长到最短。

我要删除任何“子”子字符串。例如，“12345”和“51234”之间的匹配将不包括任何'1234‘的子字符串。

string1 = '51234'

string2 = '12345'

result = ['1234', '5']

我正在考虑寻找最长公共子串，然后递归地查找左/右最长的子字符串。但是，在找到之后，我不想删除一个常见的子字符串。例如，下面的结果在中间部分共享一个6：

string1 = '12345623456'

string2 = '623456'

result = ['623456', '23456']

最后，我需要检查一个字符串与数千个字符串的固定列表。我不确定在散列这些字符串中的所有子字符串时是否有一个明智的步骤。

以前的答案：

在这个线程中，找到了一个动态规划解，它需要O(nm)时间，其中n和m是字符串的长度。我对一种更有效的方法感兴趣，它使用后缀树。

背景：

我正在用几段旋律谱写歌曲旋律。有时，一个组合设法产生一个旋律匹配太多的音符在一排现有的一个。

我可以用一个字符串相似性度量，如编辑距离，但相信与旋律差异很小的曲调是独特的和有趣的。不幸的是，这些曲调将有相似的水平相似的歌曲，复制了许多音符的旋律一条条。

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，可以使用.add_ngram方法。

下一部分比较棘手，因为您正在寻找并行遍历字符串，同时跟踪树坐标。为了实现这一切，我们需要一些在树上操作的函数和一个查询字符串。

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，做这个：

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

如果有什么不清楚的地方，请告诉我，我会尽力澄清的。