179 lines
5.0 KiB
Python
179 lines
5.0 KiB
Python
"""
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词汇难度评估系统
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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
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"""
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import re
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import pickle
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from typing import Dict, List, Tuple, Union
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# 预编译正则表达式提高性能
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WORD_PATTERN = re.compile(r'\b[\w-]+\b')
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def load_record(pickle_fname: str) -> Dict[str, List[str]]:
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"""
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加载pickle格式的单词-考试类型数据
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参数:
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pickle_fname: pickle文件名
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返回:
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字典格式的单词到考试类型列表的映射
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异常:
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FileNotFoundError: 当文件不存在时抛出
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ValueError: 当pickle文件损坏时抛出
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"""
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try:
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with open(pickle_fname, 'rb') as f:
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return pickle.load(f)
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except FileNotFoundError:
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raise FileNotFoundError(f"Pickle文件 {pickle_fname} 未找到")
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except pickle.PickleError:
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raise ValueError(f"Pickle文件 {pickle_fname} 损坏或格式不正确")
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def convert_test_type_to_difficulty_level(d: Dict[str, List[str]]) -> Dict[str, int]:
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"""
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将考试类型映射为难度级别
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难度级别定义:
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0: 未知/未分类
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4: CET4
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5: OXFORD3000
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6: CET6或GRADUATE
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7: IELTS或OXFORD5000
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8: BBC
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参数:
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d: 单词到考试类型列表的映射
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返回:
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单词到难度级别的映射
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"""
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result = {}
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for word, test_types in d.items():
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if 'CET4' in test_types:
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result[word] = 4
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elif 'OXFORD3000' in test_types:
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result[word] = 5
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elif 'CET6' in test_types or 'GRADUATE' in test_types:
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result[word] = 6
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elif 'IELTS' in test_types:
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result[word] = 7
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elif 'OXFORD5000' in test_types:
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result[word] = 7
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elif 'BBC' in test_types:
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result[word] = 8
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else:
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result[word] = 0
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return result
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class VocabularyLevelEstimator:
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"""
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词汇难度评估基类
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使用预定义的单词-考试类型数据评估单词难度级别
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类属性:
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_test_raw: 原始单词-考试类型数据
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_difficulty_dict: 转换后的单词-难度级别映射
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"""
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_test_raw = None
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_difficulty_dict = None
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@classmethod
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def _load_data(cls):
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"""延迟加载数据,避免不必要的文件操作"""
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if cls._test_raw is None:
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cls._test_raw = load_record('words_and_tests.p')
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cls._difficulty_dict = convert_test_type_to_difficulty_level(cls._test_raw)
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@classmethod
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def get_word_level(cls, word: str) -> int:
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"""
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获取单词难度级别
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参数:
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word: 要查询的单词
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返回:
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单词的难度级别(0-8)
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"""
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cls._load_data()
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return cls._difficulty_dict.get(word, 0)
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class UserVocabularyLevel(VocabularyLevelEstimator):
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"""
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用户词汇水平评估
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根据用户最近查询的单词评估其词汇水平
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"""
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def __init__(self, d: Dict[str, List[int]]):
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"""
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初始化用户词汇数据
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参数:
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d: 单词到时间戳列表的映射
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"""
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self.d = d
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# 获取每个单词的最新查询时间并排序
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word_time = [(word, max(times)) for word, times in d.items() if times]
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sorted_words = sorted(word_time, key=lambda x: x[1], reverse=True)
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self.recent_words = [word for word, _ in sorted_words[:3]]
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@property
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def level(self) -> float:
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"""
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计算用户词汇水平
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返回:
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最近查询的有效单词的平均难度级别
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如果没有有效单词则返回0
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"""
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levels = [self.get_word_level(word) for word in self.recent_words]
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valid_levels = [lvl for lvl in levels if lvl > 0]
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return sum(valid_levels) / len(valid_levels) if valid_levels else 0
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class ArticleVocabularyLevel(VocabularyLevelEstimator):
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"""
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文章词汇水平评估
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根据文章中出现的最高难度单词评估文章词汇水平
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"""
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def __init__(self, content: str):
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"""
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初始化文章内容
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参数:
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content: 文章内容字符串
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异常:
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ValueError: 当内容为空或不是字符串时抛出
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"""
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if not content or not isinstance(content, str):
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raise ValueError("文章内容必须是非空字符串")
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self.content = content
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# 提取所有单词并计算难度
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words = WORD_PATTERN.findall(content.lower())
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word_levels = [self.get_word_level(word) for word in words]
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# 筛选有效难度并排序
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valid_levels = sorted([lvl for lvl in word_levels if lvl > 0], reverse=True)
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self.top_levels = valid_levels[:5] if valid_levels else []
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@property
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def level(self) -> float:
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"""
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计算文章词汇水平
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返回:
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文章中最难5个单词的平均难度级别
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如果没有有效单词则返回0
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"""
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if not self.top_levels:
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return 0
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return sum(self.top_levels) / len(self.top_levels) |