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pull最新的snapshot-20230511，后更新了difficulty.py和Article.py的部分代码，提交了新的pickle文件

Bug476-YuHuangtao
俞黄焘 2023-05-18 23:29:38 +08:00
parent d9f6df7fbe
commit 39d96014d9
3 changed files with 67 additions and 68 deletions
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4
app/Article.py
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@ -7,7 +7,7 @@ import random, glob
import hashlib
from datetime import datetime
from flask import Flask, request, redirect, render_template, url_for, session, abort, flash, get_flashed_messages
from difficulty import get_difficulty_level, text_difficulty_level, user_difficulty_level
from difficulty import get_difficulty_level_for_user, text_difficulty_level, user_difficulty_level
path_prefix = '/var/www/wordfreq/wordfreq/'
@ -53,7 +53,7 @@ def get_today_article(user_word_list, visited_articles):
# Choose article according to reader's level
d1 = load_freq_history(path_prefix + 'static/frequency/frequency.p')
d2 = load_freq_history(path_prefix + 'static/words_and_tests.p')
d3 = get_difficulty_level(d1, d2)
d3 = get_difficulty_level_for_user(d1, d2)
d = None
result_of_generate_article = "not found"

127
app/difficulty.py
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@ -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):
@ -17,40 +18,48 @@ def load_record(pickle_fname):
return d
def difficulty_level_from_frequency(word, d):
level = 1
if not word in d:
return level
def convert_test_type_to_difficulty_level(d):
"""
对原本的单词库中的单词进行难度评级
:param d: 存储了单词库pickle文件中的单词的字典
:return:
"""
result = {}
L = list(d.keys()) # in d, we have test types (e.g., CET4,CET6,BBC) for each word
if 'what' in d:
ratio = (d['what']+1)/(d[word]+1) # what is a frequent word
level = math.log( max(ratio, 1), 2)
for k in L:
if 'CET4' in d[k]:
result[k] = 4 # CET4 word has level 4
elif 'OXFORD3000' in d[k]:
result[k] = 5
elif 'CET6' in d[k] or 'GRADUATE' in d[k]:
result[k] = 6
elif 'OXFORD5000' in d[k] or 'IELTS' in d[k]:
result[k] = 7
elif 'BBC' in d[k]:
result[k] = 8
level = min(level, 8)
return level
return result # {'apple': 4, ...}
def get_difficulty_level(d1, d2):
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)
return d
def get_difficulty_level_for_user(d1, d2):
"""
d2 来自于词库的35511个已标记单词
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
"""
d2 = convert_test_type_to_difficulty_level(d2) # 根据d2的标记评级{'apple': 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: # 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
return d2
def revert_dict(d):
@ -62,12 +71,13 @@ def revert_dict(d):
for k in d:
if type(d[k]) is list: # d[k] is a list of dates.
lst = d[k]
elif type(d[k]) is int: # for backward compatibility. d was sth like {'word':1}. The value d[k] is not a list of dates, but a number representing how frequent this word had been added to the new word book.
elif type(d[
k]) is int: # for backward compatibility. d was sth like {'word':1}. The value d[k] is not a list of dates, but a number representing how frequent this word had been added to the new word book.
freq = d[k]
lst = freq*['2021082019'] # why choose this date? No particular reasons. I fix the bug in this date.
lst = freq * ['2021082019'] # why choose this date? No particular reasons. I fix the bug in this date.
for time_info in lst:
date = time_info[:10] # until hour
date = time_info[:10] # until hour
if not date in d2:
d2[date] = [k]
else:
@ -76,42 +86,43 @@ def revert_dict(d):
def user_difficulty_level(d_user, d):
d_user2 = revert_dict(d_user) # key is date, and value is a list of words added in that date
d_user2 = revert_dict(d_user) # key is date, and value is a list of words added in that date
count = 0
geometric = 1
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 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:
if word in d:
lst2.append((word, d[word]))
lst3 = sort_in_ascending_order(lst2) # easiest tuple first
#print(lst3)
lst3 = sort_in_ascending_order(lst2) # easiest tuple first
# print(lst3)
for t in lst3:
word = t[0]
hard = t[1]
#print('WORD %s HARD %4.2f' % (word, hard))
# print('WORD %s HARD %4.2f' % (word, hard))
geometric = geometric * (hard)
count += 1
if count >= 10:
return geometric**(1/count)
return geometric ** (1 / count)
return geometric**(1/max(count,1))
return geometric ** (1 / max(count, 1))
def text_difficulty_level(s, d):
s = remove_punctuation(s)
L = freq(s)
lst = [] # a list of tuples, each tuple being (word, difficulty level)
lst = [] # a list of tuples, each tuple being (word, difficulty level)
for x in L:
word = x[0]
if word in d:
lst.append((word, d[word]))
lst2 = sort_in_descending_order(lst) # most difficult words on top
#print(lst2)
lst2 = sort_in_descending_order(lst) # most difficult words on top
# print(lst2)
count = 0
geometric = 1
for t in lst2:
@ -119,24 +130,20 @@ def text_difficulty_level(s, d):
hard = t[1]
geometric = geometric * (hard)
count += 1
if count >= 20: # we look for n most difficult words
return geometric**(1/count)
return geometric**(1/max(count,1))
if count >= 20: # we look for n most difficult words
return geometric ** (1 / count)
return geometric ** (1 / max(count, 1))
if __name__ == '__main__':
d1 = load_record('frequency.p')
#print(d1)
# print(d1)
d2 = load_record('words_and_tests.p')
#print(d2)
# print(d2)
d3 = get_difficulty_level(d1, d2)
d3 = get_difficulty_level_for_user(d1, d2)
s = '''
South Lawn
@ -197,7 +204,6 @@ Amidst the aftermath of this shocking referendum vote, there is great uncertaint
'''
s = '''
British Prime Minister Boris Johnson walks towards a voting station during the Brexit referendum in Britain, June 23, 2016. (Photo: EPA-EFE)
@ -218,7 +224,6 @@ The prime minister was forced to ask for an extension to Britain's EU departure
Johnson has repeatedly pledged to finalize the first stage, a transition deal, of Britain's EU divorce battle by Oct. 31. A second stage will involve negotiating its future relationship with the EU on trade, security and other salient issues.
'''
s = '''
Thank you very much. We have a Cabinet meeting. We’ll have a few questions after grace. And, if you would, Ben, please do the honors.
@ -233,17 +238,11 @@ We need — for our farmers, our manufacturers, for, frankly, unions and non-uni
'''
#f = open('bbc-fulltext/bbc/entertainment/001.txt')
# f = open('bbc-fulltext/bbc/entertainment/001.txt')
f = open('wordlist.txt')
s = f.read()
f.close()
print(text_difficulty_level(s, d3))

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app/static/words_and_tests.p
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