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添加测试convert_test_type_to_difficulty_level和get_difficulty_level_for_user函数运行所需时间的几行代码

Bug476-Yuhuangtao
俞黄焘 2023-06-18 13:21:49 +08:00
parent 4d2535a6e8
commit 3361e4ba79
1 changed files with 8 additions and 0 deletions

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@ -7,6 +7,8 @@
import pickle import pickle
import math import math
import time
from wordfreqCMD import remove_punctuation, freq, sort_in_descending_order, sort_in_ascending_order from wordfreqCMD import remove_punctuation, freq, sort_in_descending_order, sort_in_ascending_order
import snowballstemmer import snowballstemmer
@ -24,6 +26,7 @@ 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 :param d: 存储了单词库pickle文件中的单词的字典
:return: :return:
""" """
time_start = time.time()
result = {} result = {}
L = list(d.keys()) # in d, 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
@ -38,6 +41,8 @@ def convert_test_type_to_difficulty_level(d):
result[k] = 7 result[k] = 7
elif 'BBC' in d[k]: elif 'BBC' in d[k]:
result[k] = 8 result[k] = 8
time_end = time.time()
print('convert_test_type_to_difficulty_level totally cost', time_end - time_start)
return result # {'apple': 4, ...} return result # {'apple': 4, ...}
@ -48,6 +53,7 @@ def get_difficulty_level_for_user(d1, d2):
d1 用户不会的词 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 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. # 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, ...} d2 = convert_test_type_to_difficulty_level(d2) # 根据d2的标记评级{'apple': 4, 'abandon': 4, ...}
stemmer = snowballstemmer.stemmer('english') stemmer = snowballstemmer.stemmer('english')
@ -61,6 +67,8 @@ def get_difficulty_level_for_user(d1, d2):
d2[k] = d2[stem] # 按照词根进行评级 d2[k] = d2[stem] # 按照词根进行评级
else: else:
d2[k] = 3 # 如果k的词根都不在那么就当认为是3级 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 return d2