forked from mrlan/EnglishPal
				
			评审后的修改(words_and_tests.p文件没有提交,为防止版本的控制)
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			@ -37,26 +37,28 @@ def difficulty_level_from_frequency(word, d):
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    return level
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def get_difficulty_level_for_words_and_tests(d_in):
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def get_difficulty_level_for_words_and_tests(d):
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    """
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    对原本的单词库中的单词进行难度评级
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    :param d_in: 存储了单词库pickle文件中的单词的字典
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    :param d: 存储了单词库pickle文件中的单词的字典
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    :return:
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    """
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    d = {}
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    L = list(d_in.keys())  # in dic, we have test types (e.g., CET4,CET6,BBC) for each word
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    result = {}
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    L = list(d.keys())  # in dic, we have test types (e.g., CET4,CET6,BBC) for each word
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    for k in L:
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        if 'CET4' in d_in[k]:
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            d[k] = 4  # CET4 word has level 4
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        elif 'CET6' in d_in[k]:
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            d[k] = 6
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        elif 'IELTS' in d_in[k] or 'GRADUATE' in d_in[k]:   # 雅思或研究生英语
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            d[k] = 8
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        elif 'EnWords' in d_in[k]:      # 除基础词汇外的绝大多数词,包括一些犄角旮旯的专业词汇,近九万个,绝大多数我是真不认识
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            d[k] = 7
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        if 'CET4' in d[k]:
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            result[k] = 4  # CET4 word has level 4
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        elif 'CET6' in d[k] or 'GRADUATE' in d[k]:
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            result[k] = 6
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        elif 'IELTS' in d[k]:   # 雅思或研究生英语
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            result[k] = 7
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        elif 'BBC' in d[k]:
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            result[k] = 8
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        # elif 'EnWords' in d[k]:      # 除基础词汇外的绝大多数词,包括一些犄角旮旯的专业词汇,近九万个,定级不太好处理,绝大多数我是真不认识
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        #     result[k] = 7
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    return d  # {'apple': 4, ...}
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    return result  # {'apple': 4, ...}
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def simplify_the_words_dict(d):
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    """
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			@ -65,13 +67,13 @@ def simplify_the_words_dict(d):
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    stem = snowballstemmer.stemmer('english')
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    result = {}
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    for j in d:   # j 在字典中
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        temp = stem.stemWord(j)     # 提取j得词根
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        if not temp in result:         # 如果这个词根不在结果字典中,则以词根为键、以dic中的等级作为值添加
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            result[temp] = d[j]
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    for k in d:   # j 在字典中
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        temp = stem.stemWord(k)     # 提取j得词根
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        if temp not in result:         # 如果这个词根不在结果字典中,则以词根为键、以dic中的等级作为值添加
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            result[temp] = d[k]
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        else:                   # 如果这个词在结果词典中,则比较一下单词的难度等级是否最小
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            if result[temp] > d[j]:
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                result[temp] = d[j]
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            if result[temp] > d[k]:
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                result[temp] = d[k]
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    return result
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