Open alexott opened 11 months ago
The cell 27 fails with following stacktrace:
KeyError Traceback (most recent call last) Cell In[27], line 5 2 purchases = {'test1': 0, 'test3': 0} 3 for _ in range(0, NUM_USERS): ----> 5 model_name, purchase_made = a_or_b_model(query='transformers dvd', 6 a_model='test1', 7 b_model='test3') 8 if purchase_made: 9 purchases[model_name]+= 1 Cell In[17], line 20, in a_or_b_model(query, a_model, b_model) 17 else: 18 model_name=b_model ---> 20 purchase_made = live_user_query(query=query, 21 model_name=model_name, 22 desired=wants_to_purchase, 23 meh=might_purchase) 24 return (model_name, purchase_made) Cell In[16], line 15, in live_user_query(query, model_name, desired, meh, desired_prob, meh_prob, uninteresting_prob, quit_per_rank_prob) 1 def live_user_query(query, model_name, 2 desired, meh, 3 desired_prob=0.15, 4 meh_prob=0.03, 5 uninteresting_prob=0.01, 6 quit_per_rank_prob=0.2): 7 """Live user for 'query' where purchase probability depends on if 8 products upc is in one of three sets. 9 (...) 13 14 """ ---> 15 search_results = search(query, model_name, at=10) 17 results = pd.DataFrame(search_results).reset_index() 18 for doc in results.to_dict(orient="records"): Cell In[11], line 32, in search(query, model_name, at, log) 29 if log: 30 print(resp) ---> 32 search_results = resp['response']['docs'] 34 for rank, result in enumerate(search_results): 35 result['rank'] = rank KeyError: 'response'
Hey Alex I ran through this notebook, and didn't have this error.... are you still getting this?
@dcrouch26 - can you please confirm we're good here after your refactors an close this issue? Thanks!
The cell 27 fails with following stacktrace: