1、删除了difficulty_level_from_frequency

2、修改了get_difficulty_level_for_user，按新的方式修改了单词的评级方式：CET4 等于 level 4， OXFORD3000 等于 level 5， CET6 等于 level 6， GRADUATE 等于 level 6, OXFORD5000 等于 level 7， BBC 等于 level 8，找不到等于 level 3

app/Article.py

@@ -62,8 +62,11 @@ def get_today_article(user_word_list, articleID):
                 d = reading
                 break
 
-    s = '<div class="alert alert-success" role="alert">According to your word list, your level is <span class="badge bg-success">%4.2f</span>  and we have chosen an article with a difficulty level of <span class="badge bg-success">%4.2f</span> for you.</div>' % (
+    s = '<div class="alert alert-success" role="alert">' \
-        user_level, text_level)
+        'According to your word list, your level is <span class="badge bg-success">%4.2f</span>  ' \
+        'and we have chosen an article with a difficulty level of ' \
+        '<span class="badge bg-success">%4.2f</span> for you.' \
+        '</div>' % (user_level, text_level)
     s += '<p class="text-muted">Article added on: %s</p>' % (d['date'])
     s += '<div class="p-3 mb-2 bg-light text-dark">'
     article_title = get_article_title(d['text'])

app/difficulty.py

@@ -17,26 +17,6 @@ def load_record(pickle_fname):
     f.close()
     return d
 
-
-def difficulty_level_from_frequency(word, d):
-    """
-    根据单词的频率进行难度的评级
-    :param word:
-    :param d:
-    :return:
-    """
-    level = 1
-    if not word in d:
-        return level
-    
-    if 'what' in d:
-        ratio = (d['what']+1)/(d[word]+1)   # what is a frequent word
-        level = math.log(max(ratio, 1), 2)
-
-    level = min(level, 8) 
-    return level
-
-
 def convert_test_type_to_difficulty_level(d):
     """
     对原本的单词库中的单词进行难度评级
@@ -49,14 +29,14 @@ def convert_test_type_to_difficulty_level(d):
     for k in L:
         if 'CET4' in d[k]:
             result[k] = 4  # CET4 word has level 4
+        elif 'OXFORD3000' in d[k]:
+            result[k] = 5
         elif 'CET6' in d[k] or 'GRADUATE' in d[k]:
             result[k] = 6
-        elif 'IELTS' in d[k]:   # 雅思
+        elif 'OXFORD5000' in d[k]:
             result[k] = 7
         elif 'BBC' in d[k]:
             result[k] = 8
-        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
-            result[k] = 3
 
     return result  # {'apple': 4, ...}
 
@@ -80,7 +60,7 @@ def simplify_the_words_dict(d):
 
 def get_difficulty_level_for_user(d1, d2):
     """
-    d2 来自于词库的27000个已标记单词
+    d2 来自于词库的35511个已标记单词
     d1 用户不会的词
     在d2的后面添加单词，没有新建一个新的字典
     """
@@ -89,12 +69,13 @@ def get_difficulty_level_for_user(d1, d2):
     stem = snowballstemmer.stemmer('english')
 
     for k in d1:        # 用户的词
-        for l in d2_simplified:        # l是词库的某个单词的词根
+        if k in d2:     # 如果用户的词以原型的形式存在于词库d2中
-            if stem.stemWord(k) == l:   # 两者相等则视为同一难度的词
+            continue    # 无需评级，跳过
-                d2[k] = d2_simplified[l]       # 给d2定级
+        elif stem.stemWord(k) in d2_simplified:     # 如果用户的词的词根存在于词库d2的词根库中
+            d2[k] = d2_simplified[k]        # 按照词根进行评级
             break
-            else:       # 不相等则表明词库中没这词，按照单词的频率定级
+        else:
-                d2[k] = difficulty_level_from_frequency(k, d1)
+            d2[k] = 3       # 如果k的词根都不在，那么就当认为是3级
     return d2