bktree.py

class BKTree:
    """
    def __init__(self, distfn, words):
        # Create a new BK-tree from the given distance function and
        # words.
        #
        # Arguments:
        #
        # distfn: a binary function that returns the distance between
        # two words.  Return value is a non-negative integer.  the
        # distance function must be a metric space.
        #
        # words: an iterable.  produces values that can be passed to
        # distfn
        self.distfn = distfn
        it = iter(words)
        root = it.next()
        self.tree = (root, {})
        for i in it:
            self._add_word(self.tree, i)
    """

    def __init__(self, distfn, words):
        self.distfn = distfn
        root = words[0]
        self.tree = (root, {})
        for i in words[1:]:
            self._add_word(self.tree, i)

    def _add_word(self, parent, word):
        pword, children = parent
        d = self.distfn(word, pword)
        if d in children:
            self._add_word(children[d], word)
        else:
            children[d] = (word, {})

    def query(self, word, n):
        """
        Return all words in the tree that are within a distance of `n'
        from `word`.
        Arguments:
        word: a word to query on
        n: a non-negative integer that specifies the allowed distance
        from the query word.
        Return value is a list of tuples (distance, word), sorted in
        ascending order of distance.

        """

        def rec(parent):
            pword, children = parent
            d = self.distfn(word, pword)
            results = []
            if d <= n:
                results.append((d, pword))

            for i in range(d - n, d + n + 1):
                child = children.get(i)
                if child is not None:
                    results.extend(rec(child))
            return results

        # sort by distance
        return sorted(rec(self.tree))


def brute_query(word, words, distfn, n):
    """A brute force distance query
    Arguments:
    word: the word to query for
    words: a iterable that produces words to test
    distfn: a binary function that returns the distance between a
    `word' and an item in `words'.
    n: an integer that specifies the distance of a matching word
    """
    return [i for i in words
            if distfn(i, word) <= n]

def maxdepth(tree, c=0):
    _, children = tree
    if len(children):
        return max(maxdepth(i, c + 1) for i in children.values())
    else:
        return c

def levenshtein(s, t):
    m, n = len(s), len(t)
    d = [range(n + 1)]
    d += [[i] for i in range(1, m + 1)]
    for i in range(0, m):
        for j in range(0, n):
            cost = 1
            if s[i] == t[j]: cost = 0
            d[i + 1].append(min(d[i][j + 1] + 1,  # deletion
                                d[i + 1][j] + 1,  # insertion
                                d[i][j] + cost)  # substitution
                            )
    return d[m][n]

def list_words(dictfile):
    words = []
    with open(dictfile, "r", encoding="utf-8") as f:
        for line in f.readlines():
            line = line.strip()
            words.append(line)
    return words

def timeof(fn, *args):
    import time
    t = time.time()
    res = fn(*args)
    print("time: ", (time.time() - t))
    return res

if __name__ == "__main__":
    tree = BKTree(levenshtein, list_words('vocab.txt'))
    dist = 1
    for i in ["mil", "1mil", "1stqth", "houu",'1cle','11:11']:
        w = set(tree.query(i, dist))
        print(w)