为系统添加一个全局变量记录数据库的单词及其等级，使得数据库单词等级只需在登录时进行一次评级，大致能将点击下一篇的时间缩减为原来的10^-15次以下，感谢章翊、赵煜涵、唐伟、宋江涛同学的建议，没有他们我懒得改的

app/difficulty.py

@@ -6,10 +6,14 @@
 # Purpose: compute difficulty level of a English text
 
 import pickle
-import math
-from wordfreqCMD import remove_punctuation, freq, sort_in_descending_order, sort_in_ascending_order
+import time
+
 import snowballstemmer
 
+from wordfreqCMD import remove_punctuation, freq, sort_in_descending_order, sort_in_ascending_order
+
+# 定义一个全局的res_d， 记录数据库单词评级之后的单词及其等级
+res_d = {}
 
 def load_record(pickle_fname):
     f = open(pickle_fname, 'rb')
@@ -24,6 +28,7 @@ def convert_test_type_to_difficulty_level(d):
     :param d: 存储了单词库pickle文件中的单词的字典
     :return:
     """
+    time_start = time.time()
     result = {}
     L = list(d.keys())  # in d, we have test types (e.g., CET4,CET6,BBC) for each word
 
@@ -38,7 +43,10 @@ def convert_test_type_to_difficulty_level(d):
             result[k] = 7
         elif 'BBC' in d[k]:
             result[k] = 8
-
+    time_end = time.time()
+    print('convert_test_type_to_difficulty_level totally cost', time_end - time_start)
+    global res_d
+    res_d = result
     return result  # {'apple': 4, ...}
 
 
@@ -48,8 +56,12 @@ def get_difficulty_level_for_user(d1, d2):
     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
     """
+    time_start = time.time()
     # TODO: convert_test_type_to_difficulty_level() should not be called every time.  Each word's difficulty level should be pre-computed.
-    d2 = convert_test_type_to_difficulty_level(d2)  # 根据d2的标记评级{'apple': 4, 'abandon': 4, ...}
+    if res_d == {}:
+        d2 = convert_test_type_to_difficulty_level(d2)  # 根据d2的标记评级{'apple': 4, 'abandon': 4, ...}
+    else:
+        d2 = res_d
     stemmer = snowballstemmer.stemmer('english')
 
     for k in d1:  # 用户的词
@@ -61,6 +73,8 @@ def get_difficulty_level_for_user(d1, d2):
                 d2[k] = d2[stem]  # 按照词根进行评级
             else:
                 d2[k] = 3  # 如果k的词根都不在，那么就当认为是3级
+    time_end = time.time()
+    print('get_difficulty_level_for_user totally cost', time_end - time_start)
     return d2