Algorithmic way to search a list of tuples for a matching substring?

问题

I have a list of tuples, about 100k entries. Each tuple consists of an id and a string, my goal is to list the ids of the tuples, whose strings contain a substring from a given list of substrings. My current solution is through set comprehension, ids can repeat.

tuples = [(id1, 'cheese trees'), (id2, 'freezy breeze'),...]
vals = ['cheese', 'flees']
ids = {i[0] for i in tuples if any(val in i[1] for val in vals)}

output: {id1}

Is there an algorithm that would allow doing that quicker? I'm interested in both exact substring matches and also possibly in the approximate ones. The main thing I'm after here is an algorithm that would offer speed advantage over the comprehension.

回答1:

DISCLAIMER I'm the author of trrex

For the case of the exact matching, one approach for solving this, is to use a Trie, as mentioned in the comments. trrex is a library that makes a Trie-Regex (a Trie in regex format) that can be used in conjunction with the regular expression engine of Python:

import random
import pandas as pd
import trrex as tx
import re

df = pd.read_csv('jeopardy-small.csv')
with open('words-sample') as infile:
    words = [line.strip() for line in infile]


tuples = [(random.randint(1, 250), sentence) for sentence in df['question']]


def fun_kislyuk(ws, ts):
    return {t[0] for t in ts if any(w in t[1] for w in ws)}


def fun_trrex(ws, ts):
    pattern = re.compile(tx.make(ws, left='', right=''))
    return {i for i, s in ts if pattern.search(s)}


if __name__ == "__main__":
    print(fun_trrex(words, tuples) == fun_kislyuk(words, tuples))

Output

True

The timings for the above functions are:

%timeit fun_trrex(words, tuples)
11.3 ms ± 34.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit fun_kislyuk(words, tuples)
67.5 ms ± 1.75 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

The data is a list of around 2K questions from jeopardy, and 500 randomly chosen words. You can find here the resources for reproducing the experiments.

UPDATE

If you add the grouping strategy mentioned in the comments the time improvements increases, below is the function:

def fun_grouping_trrex(ws, ts):
    pattern = re.compile(tx.make(ws, left='', right=''))
    groups = defaultdict(list)
    for i, s in ts:
        groups[i].append(s)

    return {i for i, vs in groups.items() if any(pattern.search(v) for v in vs)}

and the timings:

%timeit fun_trrex(words, tuples)
11.2 ms ± 61.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit fun_grouping_trrex(words, tuples)
4.96 ms ± 320 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
%timeit fun_kislyuk(words, tuples)
67.4 ms ± 1.47 ms per loop (mean ± std. dev. of 7 runs, 10 loops each)

The approach of grouping + trrex gives you an approximated 10 times improvement on performance. But take this last result with a grain of salt because it's very dependent on the dataset.

来源：https://stackoverflow.com/questions/65003988/algorithmic-way-to-search-a-list-of-tuples-for-a-matching-substring