#!/usr/bin/env/python
# -*- coding:utf-8 -*-

import re
from pprint import pprint


def resplit(item_res):
    """
    对题目的切分结果，判断是否还有题目切分到一起的情况，并进行继续拆分
    :param item_res:
    :return:
    """
    new_res = []
    for k, one_res in enumerate(item_res):
        p1 = re.search(r"((?<=\n)|(?<=</su[pb]>))\s*([1-9]|1[0-9])\s*[.．、､].+?([是为有]|等于)[(（]\s*[)）]", one_res['content'], re.S)
        # 大题出现2个相同序号如16A、16B
        p2 = re.compile(r"\n\s*([1-9]|1[0-9])\s*[ABC]\s*[.．、､].+?(求|试问|[是为等于]+多少)", re.S)

        if p1 and one_res['item_topic_name'].replace("题", "") in ['选择', '单选', '多选', '双向选择']:  # 主要针对2个题合在一起的情况
            con1 = one_res['content'][:p1.start()]
            con2 = one_res['content'][p1.start():]
            upletter_num1 = re.findall("[A-E]\s*[.．、､]|[A-E]\s*<imgsrc", con1)
            upletter_num1 = set([re.findall("[A-E]", i)[0] for i in upletter_num1])
            upletter_num2 = re.findall("[A-E]\s*[.．、､]|[A-E]\s*<imgsrc", con2)
            upletter_num2 = set([re.findall("[A-E]", i)[0] for i in upletter_num2])
            if len(upletter_num1)>=3 and len(upletter_num2)>=3:
                new_one = one_res.copy()
                new_one['item_id'] = int(re.search("^\n*\s*([1-9]|1[0-9])\s*[.．、､]", con2).group(1))
                one_res['content'] = con1
                new_one['content'] = con2
                new_res.append(one_res)
                new_res.append(new_one)
            else:
                new_res.append(one_res)
        elif re.search(p2, one_res['content']) and one_res['item_topic_name'].replace("题", "") not in ['选择', '单选', '多选', '双向选择', '填空']:
            temp = one_res['content']
            split_p = [i.start() for i in re.finditer(p2, one_res['content'])]
            one_res['content'] = temp[:split_p[0]]
            new_res.append(one_res)
            for i, j in zip(split_p, split_p[1:] + [None]):
                new_one = one_res.copy()
                new_one['content'] = temp[i:j]
                new_one['item_id'] = re.search("\n\s*([1-9]\s*[ABC]|1[0-9]\s*[ABC])", new_one['content'][:10]).group(1)
                new_res.append(new_one)
        else:
            new_res.append(one_res)

    return new_res