如何在Python迭代器中匹配对象模式？

Question

It is very easy to use the Python re functions to match and manipulate patterns in text, for example:

re.match('a[efg]*c', 'aggggc')

How do I do the same thing on a list or other python iterator? For example, I may have a list that looks like this:

>>> list = ['foo', 'bar', 3, (1, 2, 3), 'a', 'b', {5, 6, 7}, 'apple']

And, following the regex idiom, I may want to match a pattern like so:

>>> pattern = ['a', '[', {7, 6, 5}, 'b', 'c', ']', '*', 'apple']

and I want to find a match inside this list. If it were regex, I'd write it like this:

>>> match = re.search(pattern, list)
>>> match.group(0)
['a', 'b', {5, 6, 7}, 'apple']

But, of course, it doesn't work because Python regex expects to see a string.

How do I do this?

Note: it's the ability to match patterns that I'm looking for, not this exact syntax. I guess, the ideal answer would be a module or library (or succinct function) that provided a variety of regex style pattern matching tools that worked on lists.

Explanation for why I want this: I'm working on scripts to process text from SE-Asian languages which use complex scripts. The program I'm working on now will intelligently correct typing mistakes (this language has characters which can go above, below, in front, around, etc., and have specific rules about which order they can occur in). The fist pass of my program uses a state machine to assign each character to a class, such as consonants, vowel, tone, number, etc. The second pass will try to correct invalid syllables and other kinds of mistakes. There's no analogy in English as far as the syllable bit goes, but in the numbers, suppose I saw the pattern ['number', 'o', 'number'] , then I would presume that the typist meant 'zero' rather than 'oh' and make the proper corrections.

Answer 1

You can do something like this and check if the item is a str before trying to match it.

import re
from collections import Iterable

pattern = re.compile('a[efg]*')
items = ['foo', 'bar', 3, (1, 2, 3), 'a', 'b', {5, 6, 7}, 'apple']

def _find_matches(it, pattern):
    matches = []
    for i in it:
        if isinstance(i, str):
            m = pattern.match(i)
            if m:
                matches.append(m)
        elif isinstance(i, Iterable):
            m = _find_matches(i, pattern)
            matches.extend(m)
        else:
            print "Could not process: {}".format(i)
    return matches

results = _find_matches(items, pattern)

Answer 2

Mostly you would need to write a function for checking this. some thing like this.

import sys


my_list =  ['foo', 'bar', 3, (1, 2, 3), 'a', 'b', {5, 6, 7}, 'apple']
pattern = ['fo', 'bar', 3, (1, 2, 3), 'a', '*', {5, 6, 7}, 'apple']


if len(my_list) != len(pattern):
    print('List length dose not match with the pattern')
    sys.exit(1)

for offset,value in enumerate(my_list):
    if pattern[offset] != value and pattern[offset] != '*':
        print('Pattern matching failed at offset {} with value {}'.format(offset, my_list[offset]))
        break;
else:
    print('Pattern matched perfectly..');