forked from mrlan/EnglishPal
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
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d9f6df7fbe
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@ -7,7 +7,7 @@ import random, glob
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import hashlib
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from datetime import datetime
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from flask import Flask, request, redirect, render_template, url_for, session, abort, flash, get_flashed_messages
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from difficulty import get_difficulty_level, text_difficulty_level, user_difficulty_level
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from difficulty import get_difficulty_level_for_user, text_difficulty_level, user_difficulty_level
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path_prefix = '/var/www/wordfreq/wordfreq/'
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@ -53,7 +53,7 @@ def get_today_article(user_word_list, visited_articles):
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# Choose article according to reader's level
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d1 = load_freq_history(path_prefix + 'static/frequency/frequency.p')
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d2 = load_freq_history(path_prefix + 'static/words_and_tests.p')
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d3 = get_difficulty_level(d1, d2)
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d3 = get_difficulty_level_for_user(d1, d2)
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d = None
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result_of_generate_article = "not found"
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@ -8,6 +8,7 @@
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import pickle
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import math
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from wordfreqCMD import remove_punctuation, freq, sort_in_descending_order, sort_in_ascending_order
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import snowballstemmer
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def load_record(pickle_fname):
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@ -17,40 +18,48 @@ def load_record(pickle_fname):
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return d
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def difficulty_level_from_frequency(word, d):
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level = 1
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if not word in d:
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return level
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def convert_test_type_to_difficulty_level(d):
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"""
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对原本的单词库中的单词进行难度评级
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:param d: 存储了单词库pickle文件中的单词的字典
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:return:
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"""
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result = {}
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L = list(d.keys()) # in d, we have test types (e.g., CET4,CET6,BBC) for each word
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if 'what' in d:
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ratio = (d['what']+1)/(d[word]+1) # what is a frequent word
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level = math.log( max(ratio, 1), 2)
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for k in L:
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if 'CET4' in d[k]:
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result[k] = 4 # CET4 word has level 4
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elif 'OXFORD3000' in d[k]:
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result[k] = 5
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elif 'CET6' in d[k] or 'GRADUATE' in d[k]:
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result[k] = 6
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elif 'OXFORD5000' in d[k] or 'IELTS' in d[k]:
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result[k] = 7
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elif 'BBC' in d[k]:
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result[k] = 8
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level = min(level, 8)
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return level
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return result # {'apple': 4, ...}
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def get_difficulty_level(d1, d2):
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d = {}
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L = list(d1.keys()) # in d1, we have freuqence for each word
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L2 = list(d2.keys()) # in d2, we have test types (e.g., CET4,CET6,BBC) for each word
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L.extend(L2)
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L3 = list(set(L)) # L3 contains all words
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for k in L3:
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if k in d2:
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if 'CET4' in d2[k]:
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d[k] = 4 # CET4 word has level 4
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elif 'CET6' in d2[k]:
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d[k] = 6
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elif 'BBC' in d2[k]:
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d[k] = 8
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if k in d1: # BBC could contain easy words that are not in CET4 or CET6. So 4 is not reasonable. Recompute difficulty level.
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d[k] = min(difficulty_level_from_frequency(k, d1), d[k])
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elif k in d1:
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d[k] = difficulty_level_from_frequency(k, d1)
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return d
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def get_difficulty_level_for_user(d1, d2):
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"""
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d2 来自于词库的35511个已标记单词
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d1 用户不会的词
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在d2的后面添加单词，没有新建一个新的字典
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"""
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d2 = convert_test_type_to_difficulty_level(d2) # 根据d2的标记评级{'apple': 4, 'abandon': 4, ...}
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stem = snowballstemmer.stemmer('english')
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for k in d1: # 用户的词
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if k in d2: # 如果用户的词以原型的形式存在于词库d2中
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continue # 无需评级，跳过
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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
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d2[k] = d2[stem.stemWord(k)] # 按照词根进行评级
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break
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else:
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d2[k] = 3 # 如果k的词根都不在，那么就当认为是3级
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return d2
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def revert_dict(d):
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@ -62,9 +71,10 @@ def revert_dict(d):
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for k in d:
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if type(d[k]) is list: # d[k] is a list of dates.
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lst = d[k]
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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.
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elif type(d[
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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.
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freq = d[k]
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lst = freq*['2021082019'] # why choose this date? No particular reasons. I fix the bug in this date.
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lst = freq * ['2021082019'] # why choose this date? No particular reasons. I fix the bug in this date.
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for time_info in lst:
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date = time_info[:10] # until hour
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@ -79,7 +89,8 @@ def user_difficulty_level(d_user, d):
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d_user2 = revert_dict(d_user) # key is date, and value is a list of words added in that date
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count = 0
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geometric = 1
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for date in sorted(d_user2.keys(), reverse=True): # most recently added words are more important while determining user's level
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for date in sorted(d_user2.keys(),
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reverse=True): # most recently added words are more important while determining user's level
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lst = d_user2[date] # a list of words
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lst2 = [] # a list of tuples, (word, difficulty level)
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for word in lst:
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@ -87,17 +98,17 @@ def user_difficulty_level(d_user, d):
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lst2.append((word, d[word]))
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lst3 = sort_in_ascending_order(lst2) # easiest tuple first
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#print(lst3)
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# print(lst3)
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for t in lst3:
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word = t[0]
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hard = t[1]
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#print('WORD %s HARD %4.2f' % (word, hard))
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# print('WORD %s HARD %4.2f' % (word, hard))
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geometric = geometric * (hard)
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count += 1
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if count >= 10:
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return geometric**(1/count)
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return geometric ** (1 / count)
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return geometric**(1/max(count,1))
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return geometric ** (1 / max(count, 1))
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def text_difficulty_level(s, d):
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@ -111,7 +122,7 @@ def text_difficulty_level(s, d):
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lst.append((word, d[word]))
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lst2 = sort_in_descending_order(lst) # most difficult words on top
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#print(lst2)
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# print(lst2)
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count = 0
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geometric = 1
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for t in lst2:
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@ -120,23 +131,19 @@ def text_difficulty_level(s, d):
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geometric = geometric * (hard)
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count += 1
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if count >= 20: # we look for n most difficult words
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return geometric**(1/count)
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return geometric**(1/max(count,1))
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return geometric ** (1 / count)
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return geometric ** (1 / max(count, 1))
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if __name__ == '__main__':
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d1 = load_record('frequency.p')
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#print(d1)
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# print(d1)
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d2 = load_record('words_and_tests.p')
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#print(d2)
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# print(d2)
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d3 = get_difficulty_level(d1, d2)
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d3 = get_difficulty_level_for_user(d1, d2)
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s = '''
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South Lawn
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@ -197,7 +204,6 @@ Amidst the aftermath of this shocking referendum vote, there is great uncertaint
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'''
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s = '''
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British Prime Minister Boris Johnson walks towards a voting station during the Brexit referendum in Britain, June 23, 2016. (Photo: EPA-EFE)
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@ -218,7 +224,6 @@ The prime minister was forced to ask for an extension to Britain's EU departure
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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.
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'''
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s = '''
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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.
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@ -233,17 +238,11 @@ We need — for our farmers, our manufacturers, for, frankly, unions and non-uni
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'''
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#f = open('bbc-fulltext/bbc/entertainment/001.txt')
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# f = open('bbc-fulltext/bbc/entertainment/001.txt')
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f = open('wordlist.txt')
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s = f.read()
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f.close()
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print(text_difficulty_level(s, d3))
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