新建了get_difficulty_level_for_words_and_tests函数专门用于给词库单词评级，返回得到一个字典d2（{'apple': 4, 'banana': 4, ...}）;修改get_difficulty_level函数，给用户的单词评级， 分三种情况：1.以原型出现，则无需操作，因为已经出现在d2；2.词根与词库中某单词词根相同，视为一个难度的单词；3.以上两种情况之外的词视为不在词库中的词，按照difficulty_level_from_frequency函数方法评定难度。

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
@@ -30,26 +37,44 @@ 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
-    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)
+    L = list(dic.keys())  # in dic, we have test types (e.g., CET4,CET6,BBC) for each word
 
-    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 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
+    """
+    d2 = get_difficulty_level_for_words_and_tests(d2)  # 根据标记评级，仅适用于词库中的词
+    stem = snowballstemmer.stemmer('english')
+
+    for k in d1:        # k是用户不会的词
+        for l in d2:     # l是已经完成评级的词库的词
+            if k == l:  # k == l，这个用户也不会的词刚好以原型的形式出现在词库中，因为词库已经评过难度了，所以啥也不用干
+                break
+            elif stem.stemWord(k) in l:     # 这个词的词根与词库中的某个词一样，我们认为是同一难度的词
+                d1[k] = d2[l]
+            else:   # 这个词不在词库中，按频率来评定难度
+                d2[k] = difficulty_level_from_frequency(k, d1)
+
+    return d2