diff --git a/app/Article.py b/app/Article.py index e40717f..df9ac3a 100644 --- a/app/Article.py +++ b/app/Article.py @@ -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" diff --git a/app/difficulty.py b/app/difficulty.py index 50aa179..dfaf89b 100644 --- a/app/difficulty.py +++ b/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): @@ -17,41 +18,49 @@ def load_record(pickle_fname): return d -def difficulty_level_from_frequency(word, d): - level = 1 - if not word in d: - return level - - if 'what' in d: - ratio = (d['what']+1)/(d[word]+1) # what is a frequent word - level = math.log( max(ratio, 1), 2) +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 - level = min(level, 8) - return level + 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 + + 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) +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') - return d + 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)) - + diff --git a/app/static/words_and_tests.p b/app/static/words_and_tests.p index 62afd6d..5c46c5a 100644 Binary files a/app/static/words_and_tests.p and b/app/static/words_and_tests.p differ