"""
词汇难度评估系统
功能：根据单词在不同考试中的出现情况评估其难度级别，并计算用户或文章的词汇水平
"""

import re
import pickle
from typing import Dict, List, Tuple, Union

# 预编译正则表达式提高性能
WORD_PATTERN = re.compile(r'\b[\w-]+\b')


def load_record(pickle_fname: str) -> Dict[str, List[str]]:
    """
    加载pickle格式的单词-考试类型数据

    参数:
    pickle_fname: pickle文件名

    返回:
    字典格式的单词到考试类型列表的映射

    异常:
    FileNotFoundError: 当文件不存在时抛出
    ValueError: 当pickle文件损坏时抛出
    """
    try:
        with open(pickle_fname, 'rb') as f:
            return pickle.load(f)
    except FileNotFoundError:
        raise FileNotFoundError(f"Pickle文件 {pickle_fname} 未找到")
    except pickle.PickleError:
        raise ValueError(f"Pickle文件 {pickle_fname} 损坏或格式不正确")


def convert_test_type_to_difficulty_level(d: Dict[str, List[str]]) -> Dict[str, int]:
    """
    将考试类型映射为难度级别

    难度级别定义:
    0: 未知/未分类
    4: CET4
    5: OXFORD3000
    6: CET6或GRADUATE
    7: IELTS或OXFORD5000
    8: BBC

    参数:
    d: 单词到考试类型列表的映射

    返回:
    单词到难度级别的映射
    """
    result = {}
    for word, test_types in d.items():
        if 'CET4' in test_types:
            result[word] = 4
        elif 'OXFORD3000' in test_types:
            result[word] = 5
        elif 'CET6' in test_types or 'GRADUATE' in test_types:
            result[word] = 6
        elif 'IELTS' in test_types:
            result[word] = 7
        elif 'OXFORD5000' in test_types:
            result[word] = 7
        elif 'BBC' in test_types:
            result[word] = 8
        else:
            result[word] = 0
    return result


class VocabularyLevelEstimator:
    """
    词汇难度评估基类
    使用预定义的单词-考试类型数据评估单词难度级别

    类属性:
    _test_raw: 原始单词-考试类型数据
    _difficulty_dict: 转换后的单词-难度级别映射
    """
    _test_raw = None
    _difficulty_dict = None

    @classmethod
    def _load_data(cls):
        """延迟加载数据，避免不必要的文件操作"""
        if cls._test_raw is None:
            cls._test_raw = load_record('words_and_tests.p')
            cls._difficulty_dict = convert_test_type_to_difficulty_level(cls._test_raw)

    @classmethod
    def get_word_level(cls, word: str) -> int:
        """
        获取单词难度级别

        参数:
        word: 要查询的单词

        返回:
        单词的难度级别(0-8)
        """
        cls._load_data()
        return cls._difficulty_dict.get(word, 0)


class UserVocabularyLevel(VocabularyLevelEstimator):
    """
    用户词汇水平评估
    根据用户最近查询的单词评估其词汇水平
    """

    def __init__(self, d: Dict[str, List[int]]):
        """
        初始化用户词汇数据

        参数:
        d: 单词到时间戳列表的映射
        """
        self.d = d
        # 获取每个单词的最新查询时间并排序
        word_time = [(word, max(times)) for word, times in d.items() if times]
        sorted_words = sorted(word_time, key=lambda x: x[1], reverse=True)
        self.recent_words = [word for word, _ in sorted_words[:3]]

    @property
    def level(self) -> float:
        """
        计算用户词汇水平

        返回:
        最近查询的有效单词的平均难度级别
        如果没有有效单词则返回0
        """
        levels = [self.get_word_level(word) for word in self.recent_words]
        valid_levels = [lvl for lvl in levels if lvl > 0]
        return sum(valid_levels) / len(valid_levels) if valid_levels else 0


class ArticleVocabularyLevel(VocabularyLevelEstimator):
    """
    文章词汇水平评估
    根据文章中出现的最高难度单词评估文章词汇水平
    """

    def __init__(self, content: str):
        """
        初始化文章内容

        参数:
        content: 文章内容字符串

        异常:
        ValueError: 当内容为空或不是字符串时抛出
        """
        if not content or not isinstance(content, str):
            raise ValueError("文章内容必须是非空字符串")

        self.content = content
        # 提取所有单词并计算难度
        words = WORD_PATTERN.findall(content.lower())
        word_levels = [self.get_word_level(word) for word in words]
        # 筛选有效难度并排序
        valid_levels = sorted([lvl for lvl in word_levels if lvl > 0], reverse=True)
        self.top_levels = valid_levels[:5] if valid_levels else []

    @property
    def level(self) -> float:
        """
        计算文章词汇水平

        返回:
        文章中最难5个单词的平均难度级别
        如果没有有效单词则返回0
        """
        if not self.top_levels:
            return 0
        return sum(self.top_levels) / len(self.top_levels)