新建了simplify_the_words_dict函数专门用于给评级后的词库单词清洗词根，词根的难度按照其变式的最低等级决定，

目前还未进行测试。

app/difficulty.py

@@ -8,6 +8,7 @@
 import pickle
 import math
 from wordfreqCMD import remove_punctuation, freq, sort_in_descending_order, sort_in_ascending_order
+import snowballstemmer
 
 
 def load_record(pickle_fname):
@@ -18,6 +19,12 @@ def load_record(pickle_fname):
 
 
 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
@@ -30,27 +37,62 @@ def difficulty_level_from_frequency(word, d):
     return level
 
 
-def get_difficulty_level(d1, d2):
+def get_difficulty_level_for_words_and_tests(dic):
+    """
+    对原本的单词库中的单词进行难度评级
+    :param dic: 存储了单词库pickle文件中的单词的字典
+    :return:
+    """
     d = {}
-    L = list(d1.keys())  # in d1, we have freuqence for each word
+    L = list(dic.keys())  # in dic, we have test types (e.g., CET4,CET6,BBC) for each word
-    L2 = list(d2.keys()) # in d2, we have test types (e.g., CET4,CET6,BBC) for each word
-    L.extend(L2)
-    L3 = list(set(L)) # L3 contains all words
-    for k in L3:
-        if k in d2:
-            if 'CET4' in d2[k]:
-                d[k] = 4 # CET4 word has level 4
-            elif 'CET6' in d2[k]:
-                d[k] = 6
-            elif 'BBC' in d2[k]:
-                d[k] = 8
-                if k in d1: # BBC could contain easy words that are not in CET4 or CET6.  So 4 is not reasonable.  Recompute difficulty level.
-                    d[k] = min(difficulty_level_from_frequency(k, d1), d[k])
-        elif k in d1:
-            d[k] = difficulty_level_from_frequency(k, d1)
 
-    return d
+    for k in L:
+        if 'CET4' in dic[k]:
+            d[k] = 4  # CET4 word has level 4
+        elif 'CET6' in dic[k]:
+            d[k] = 6
+        elif 'BBC' in dic[k]:
+            d[k] = 8
+        # print(k, d[k])
 
+    return d  # {'apple': 4, ...}
+
+def simplify_the_words_dict(dic):
+    """
+    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')
+
+    res = {}
+    for j in dic:   # j 在字典中
+        temp = stem.stemWord(j)     # 提取j得词根
+        if not temp in res:         # 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
+            res[temp] = dic[j]
+        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 res[temp] > dic[j]:
+                res[temp] = dic[j]
+
+    return res
+
+
+def get_difficulty_level(d1, d2):
+    """
+    d2 来自于词库的27000个已标记单词
+    d1 你个老六不会的词
+    在d2的后面添加单词，没有新建一个新的字典
+    """
+    d2 = get_difficulty_level_for_words_and_tests(d2)   # 根据d2的标记评级{'apple': 4, 'abandon': 4, ...}
+    d2_sim = 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
+        for l in d2_sim:        # l是词库的某个单词的词根
+            if stem.stemWord(k) == l:   # 两者相等则视为同一难度的词
+                d2[k] = d2_sim[l]       # 给d2定级
+                break
+            else:       # 不相等则表明词库中没这词，按照单词的频率定级
+                d2[k] = difficulty_level_from_frequency(k, d1)
+    return d2
         
 
 def revert_dict(d):