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'?=
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---
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))
+ elseļ¼
+ 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]