Otters - Roshni #111
Otters - Roshni #111
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| if title == None or genre == None or rating == None: | ||
| new_movie = None | ||
| return new_movie |
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This is a good implementation of this function, but something to think about now that we have discussed time complexity is how much work is being done - Does your function have varying outcomes? If so, what's the least amount of work that can be done for each scenario?
In this case, we have 2 scenarios: The inputs are all valid and we create a dictionary or we need to return None. In the case that the input is not valid, we do not need to create a dictionary. Think about how you can refactor this code so that there is no unnecessary work being done.
| "rating": rating | ||
| } | ||
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| if title == None or genre == None or rating == None: |
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Small note - when checking if a value is none, it's better to use var is None rather than var == None :)
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| def watch_movie(user_data, title): | ||
| for movie in user_data["watchlist"]: |
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This is a super clean solution! Only note is to make sure you're paying attention to how many times your loop will iterate and when it exits. Right now, your loop will keep going until it reaches the end of watchlist even if it already found the title. This is a suuuuper easy mistake to make. Now, that we have chatted about Big O, pay attention to if your code is doing unnecessary work and think about how you could refactor this code to stop looping once the movie is found.
| length = len(user_data["watched"]) | ||
| rating = 0.0 | ||
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| if length == 0: |
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Love that you get this edge case out the way early so that there is no unnecessary work being done. Beautiful solution ✨
| def get_most_watched_genre(user_data): | ||
| most_watched = None | ||
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| if len(user_data["watched"]) == 0: |
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Small note but the Pythonic way to write this would be if not user_data["watched"] since Python evaluates empty lists to False
| # Consider the movies that the user has watched, and consider the movies that their friends have watched. | ||
| # Determine which movies at least one of the user's friends have watched, but the user has not watched. | ||
| # Return a list of dictionaries, that represents a list of movies | ||
| def get_friends_unique_watched(user_data): |
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Same comments from get_unique_watched() apply here too. See if you can get this solution down to 2 for loops instead of 4.
| # At least one of the user's friends has watched | ||
| # The "host" of the movie is a service that is in the user's "subscriptions" | ||
| # Return the list of recommended movies | ||
| def get_available_recs(user_data): |
| # The movie is in the user's "favorites" | ||
| # None of the user's friends have watched it | ||
| # Return the list of recommended movies | ||
| def get_rec_from_favorites(user_data): |
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This is a great solution! Now I challenge you to see if you can refactor this down to 2 for loops instead of 3. 😄
| # else: | ||
| # genres[movie["genre"]] += 1 | ||
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| if not movie["genre"] in genres: |
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Another if-else block that would be a great place to implement a try-except statement to adhere to EAFP (Easier to Ask Forgiveness than Permission). Oooor you could look into the get method to handle keys that don't exist. Here's a resource for that here.
| else: | ||
| genres[movie["genre"]] += 1 | ||
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| most_frequently_watched_genre = max(genres, key=genres.get, default=0) |
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Hmmmm seems like there might be an existing function that could do this work for us (hint hint get_most_watched_genre()) :)
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