1、添加误删的IELTS单词标签

2、简化评级流程（快很多，几乎与网站相同）
3、删除了原先用于提取词根的函数

app/difficulty.py

@@ -24,7 +24,7 @@ def convert_test_type_to_difficulty_level(d):
     :return:
     """
     result = {}
-    L = list(d.keys())  # in dic, we have test types (e.g., CET4,CET6,BBC) for each word
+    L = list(d.keys())  # in d, we have test types (e.g., CET4,CET6,BBC) for each word
 
     for k in L:
         if 'CET4' in d[k]:
@@ -33,30 +33,13 @@ def convert_test_type_to_difficulty_level(d):
             result[k] = 5
         elif 'CET6' in d[k] or 'GRADUATE' in d[k]:
             result[k] = 6
-        elif 'OXFORD5000' in d[k]:
+        elif 'OXFORD5000' in d[k] or 'IELTS' in d[k]:
             result[k] = 7
         elif 'BBC' in d[k]:
             result[k] = 8
 
     return result  # {'apple': 4, ...}
 
-def simplify_the_words_dict(d):
-    """
-    用于把保存了词库中评级后的词新建一个以词根为键、以同词根的最低等级为值
-    """
-    stem = snowballstemmer.stemmer('english')
-
-    result = {}
-    for k in d:   # j 在字典中
-        temp = stem.stemWord(k)     # 提取j得词根
-        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
-            result[temp] = d[k]
-        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
-            if result[temp] > d[k]:
-                result[temp] = d[k]
-
-    return result
-
 
 def get_difficulty_level_for_user(d1, d2):
     """
@@ -65,14 +48,13 @@ def get_difficulty_level_for_user(d1, d2):
     在d2的后面添加单词，没有新建一个新的字典
     """
     d2 = convert_test_type_to_difficulty_level(d2)   # 根据d2的标记评级{'apple': 4, 'abandon': 4, ...}
-    d2_simplified = simplify_the_words_dict(d2)                # 提取d2的词根   {'appl': 4, 'abandon': 4, ...}
     stem = snowballstemmer.stemmer('english')
 
     for k in d1:        # 用户的词
         if k in d2:     # 如果用户的词以原型的形式存在于词库d2中
             continue    # 无需评级，跳过
-        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
+        elif stem.stemWord(k) in d2:     # 如果用户的词的词根存在于词库d2的词根库中
+            d2[k] = d2[stem.stemWord(k)]        # 按照词根进行评级
             break
         else:
             d2[k] = 3       # 如果k的词根都不在，那么就当认为是3级
@@ -108,7 +90,7 @@ def user_difficulty_level(d_user, d):
     for date in sorted(d_user2.keys(), reverse=True): # most recently added words are more important while determining user's level
         lst = d_user2[date] # a list of words
         lst2 = [] # a list of tuples, (word, difficulty level)
-        for  word in lst:
+        for word in lst:
             if word in d:
                 lst2.append((word, d[word]))