From cb224fe3abf0e729770abfa7861dd34aa1c9c69b Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=A7=9C=E9=9B=AA=E8=99=B9?= <2496758638@qq.com> Date: Wed, 15 Jun 2022 12:04:43 +0800 Subject: [PATCH] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20'app/difficulty.py'?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- app/difficulty.py | 17 ++++++++--------- 1 file changed, 8 insertions(+), 9 deletions(-) diff --git a/app/difficulty.py b/app/difficulty.py index 50aa179..1c576ad 100644 --- a/app/difficulty.py +++ b/app/difficulty.py @@ -21,12 +21,11 @@ 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) - - level = min(level, 8) + else: + if(len(word)<5): #calculate the difficulty level based on length + level = math.log(len(word),math.pow(d[word],2)) + elseļ¼š + level = math.log(len[word],d[word]) return level @@ -44,8 +43,8 @@ def get_difficulty_level(d1, d2): 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]) + if k in d1: + d[k] = difficulty_level_from_frequency(k, d1)# BBC word rating 8 is too high,it is based on word fre elif k in d1: d[k] = difficulty_level_from_frequency(k, d1) @@ -86,7 +85,7 @@ def user_difficulty_level(d_user, d): if word in d: lst2.append((word, d[word])) - lst3 = sort_in_ascending_order(lst2) # easiest tuple first + lst3 = sort_in_descending_order(lst2) # easiest tuple first #print(lst3) for t in lst3: word = t[0]