Python字符串模式识别/压缩

Question

I can do basic regex alright, but this is slightly different, namely I don't know what the pattern is going to be.

For example, I have a list of similar strings:

lst = ['asometxt0moretxt', 'bsometxt1moretxt', 'aasometxt10moretxt', 'zzsometxt999moretxt']

In this case the common pattern is two segments of common text: 'sometxt' and 'moretxt' , starting and separated by something else that is variable in length.

The common string and variable string can of course occur at any order and at any number of occasions.

What would be a good way to condense/compress the list of strings into their common parts and individual variations?

An example output might be:

c = ['sometxt', 'moretxt']

v = [('a','0'), ('b','1'), ('aa','10'), ('zz','999')]

Answer 1

This solution finds the two longest common substrings and uses them to delimit the input strings:

def an_answer_to_stackoverflow_question_1914394(lst):
    """
    >>> lst = ['asometxt0moretxt', 'bsometxt1moretxt', 'aasometxt10moretxt', 'zzsometxt999moretxt']
    >>> an_answer_to_stackoverflow_question_1914394(lst)
    (['sometxt', 'moretxt'], [('a', '0'), ('b', '1'), ('aa', '10'), ('zz', '999')])
    """
    delimiters = find_delimiters(lst)
    return delimiters, list(split_strings(lst, delimiters))

find_delimiters and friends finds the delimiters:

import itertools

def find_delimiters(lst):
    """
    >>> lst = ['asometxt0moretxt', 'bsometxt1moretxt', 'aasometxt10moretxt', 'zzsometxt999moretxt']
    >>> find_delimiters(lst)
    ['sometxt', 'moretxt']
    """
    candidates = list(itertools.islice(find_longest_common_substrings(lst), 3))
    if len(candidates) == 3 and len(candidates[1]) == len(candidates[2]):
        raise ValueError("Unable to find useful delimiters")
    if candidates[1] in candidates[0]:
        raise ValueError("Unable to find useful delimiters")
    return candidates[0:2]

def find_longest_common_substrings(lst):
    """
    >>> lst = ['asometxt0moretxt', 'bsometxt1moretxt', 'aasometxt10moretxt', 'zzsometxt999moretxt']
    >>> list(itertools.islice(find_longest_common_substrings(lst), 3))
    ['sometxt', 'moretxt', 'sometx']
    """
    for i in xrange(min_length(lst), 0, -1):
        for substring in common_substrings(lst, i):
            yield substring


def min_length(lst):
    return min(len(item) for item in lst)

def common_substrings(lst, length):
    """
    >>> list(common_substrings(["hello", "world"], 2))
    []
    >>> list(common_substrings(["aabbcc", "dbbrra"], 2))
    ['bb']
    """
    assert length <= min_length(lst)
    returned = set()
    for i, item in enumerate(lst):
        for substring in all_substrings(item, length):
            in_all_others = True
            for j, other_item in enumerate(lst):
                if j == i:
                    continue
                if substring not in other_item:
                    in_all_others = False
            if in_all_others:
                if substring not in returned:
                    returned.add(substring)
                    yield substring

def all_substrings(item, length):
    """
    >>> list(all_substrings("hello", 2))
    ['he', 'el', 'll', 'lo']
    """
    for i in range(len(item) - length + 1):
        yield item[i:i+length]

split_strings splits the strings using the delimiters:

import re

def split_strings(lst, delimiters):
    """
    >>> lst = ['asometxt0moretxt', 'bsometxt1moretxt', 'aasometxt10moretxt', 'zzsometxt999moretxt']
    >>> list(split_strings(lst, find_delimiters(lst)))
    [('a', '0'), ('b', '1'), ('aa', '10'), ('zz', '999')]
    """
    for item in lst:
        parts = re.split("|".join(delimiters), item)
        yield tuple(part for part in parts if part != '')

Answer 2

Here is a scary one to get the ball rolling.

>>> import re
>>> makere = lambda n: ''.join(['(.*?)(.+)(.*?)(.+)(.*?)'] + ['(.*)(\\2)(.*)(\\4)(.*)'] * (n - 1))
>>> inp = ['asometxt0moretxt', 'bsometxt1moretxt', 'aasometxt10moretxt', 'zzsometxt999moretxt']
>>> re.match(makere(len(inp)), ''.join(inp)).groups()
('a', 'sometxt', '0', 'moretxt', '', 'b', 'sometxt', '1', 'moretxt', 'aa', '', 'sometxt', '10', 'moretxt', 'zz', '', 'sometxt', '999', 'moretxt', '')

I hope its sheer ugliness will inspire better solutions :)

Answer 3

This seems to be an example of the longest common subsequence problem . One way could be to look at how diffs are generated. The Hunt-McIlroy algorithm seems to have been the first, and is such the simplest, especially since it apparently is non-heuristic.

The first link contains detailed discussion and (pseudo) code examples. Assuming, of course, Im not completely of the track here.

Answer 4

I guess you should start by identifying substrings (patterns) that frequently occur in the strings. Since naively counting substrings in a set of strings is rather computationally expensive, you'll need to come up with something smart.

I've done substring counting on a large amount of data using generalized suffix trees (example here) . Once you know the most frequent substrings/patterns in the data, you can take it from there.

Answer 5

This look much like the LZW algorithm for data (text) compression. There should be python implementations out there, which you may be able to adapt to your need.

I assume you have no a priori knowledge of these sub strings that repeat often.

Answer 6

How about subbing out the known text, and then splitting?

import re
[re.sub('(sometxt|moretxt)', ',', x).split(',') for x in lst]
# results in
[['a', '0', ''], ['b', '1', ''], ['aa', '10', ''], ['zz', '999', '']]