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EnglishPal
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更新 'app/difficulty.py'

Bug344-JiangXueHong
姜雪虹 2022-06-15 12:04:43 +08:00
parent e48008550a
commit cb224fe3ab
1 changed files with 8 additions and 9 deletions
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17
app/difficulty.py
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@ -21,12 +21,11 @@ def difficulty_level_from_frequency(word, d):
level = 1 level = 1
if not word in d: if not word in d:
return level return level
else:
if 'what' in d: if(len(word)<5): #calculate the difficulty level based on length
ratio = (d['what']+1)/(d[word]+1) # what is a frequent word level = math.log(len(word),math.pow(d[word],2))
level = math.log( max(ratio, 1), 2) else：
level = math.log(len[word],d[word])
level = min(level, 8)
return level return level
@ -44,8 +43,8 @@ def get_difficulty_level(d1, d2):
d[k] = 6 d[k] = 6
elif 'BBC' in d2[k]: elif 'BBC' in d2[k]:
d[k] = 8 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. if k in d1:
d[k] = min(difficulty_level_from_frequency(k, d1), d[k]) 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: elif k in d1:
d[k] = difficulty_level_from_frequency(k, d1) d[k] = difficulty_level_from_frequency(k, d1)
@ -86,7 +85,7 @@ def user_difficulty_level(d_user, d):
if word in d: if word in d:
lst2.append((word, d[word])) 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) #print(lst3)
for t in lst3: for t in lst3:
word = t[0] word = t[0]