diff options
| -rw-r--r-- | app/difficulty.py | 15 |
1 files changed, 9 insertions, 6 deletions
diff --git a/app/difficulty.py b/app/difficulty.py index 58740f6..1c8a08f 100644 --- a/app/difficulty.py +++ b/app/difficulty.py @@ -20,29 +20,32 @@ def load_record(pickle_fname): def difficulty_level_from_frequency(word, d): level = 0 + if not word in d: + return level + if 'what' in d: - ratio = (d['what']+1)/(d[word]+1) + ratio = (d['what']+1)/(d[word]+1) # what is a frequent word level = math.log( max(ratio, 1), 10) - level = min(level+1, 4) + level = min(level, 4) return level def get_difficulty_level(d1, d2): d = {} - L = list(d1.keys()) # in d1, we have freuqence for each word + 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 = list(set(L)) # L3 contains all words for k in L3: if k in d2: if 'CET4' in d2[k]: - d[k] = 1 + d[k] = 1 # CET4 word has level 1 elif 'CET6' in d2[k]: d[k] = 2 elif 'BBC' in d2[k]: d[k] = 4 - if k in d1: # BBC could contain easy words not in CET4 or CET6. So 4 is not reasonable. Recompute difficulty level. + 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) |