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goeunpark
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goeunpark
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Great job, Amel & Mitra! 🎉
Your tests pass, your commit messages are descriptive and frequent, and your code is crystal clear! 🔮 I'm impressed with the subtle styling in this submission, and it's clear to me that y'all are comfortable with working with lists, dictionaries, and tuples! This project is a Green. 🟢
Keep up the great work!
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| # Draw a random letter | ||
| while len(letters) < 10: | ||
| letter = random.choice(list(LETTER_POOL)) |
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Clever use of drawing out all the keys of LETTER_POOL as a list, here!
| def uses_available_letters(word, letter_bank): | ||
| pass | ||
| """ check each letter used in leter_bank_copy """ | ||
| letter_bank_copy = letter_bank[:] |
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Great use of the python slicing notation to make a copy! ✨
| word_and_score = [] | ||
| for word in word_list: | ||
| score = score_word(word) | ||
| score_tuple = (word, score) | ||
| word_and_score.append(score_tuple) | ||
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| # Create dictionary using word-score pair in each tuple as key and value | ||
| word_and_score_dict = {} | ||
| for word, score in word_and_score: | ||
| word_and_score_dict[word] = score |
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Since we're creating a word_and_score list of tuples only to form the word_and_score_dict on L108-110, I'm curious if we could have made this more efficient and tried to make word_and_score_dict from the get-go! 🤔
| if score > highest_score: | ||
| highest_score = score | ||
| best_word = word | ||
| # Break tie |
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Great tie breaker logic below! I see how you could have combined L121-122 and L127-128 if you wanted, but this reads very clearly!
| while len(letters) < 10: | ||
| letter = random.choice(list(LETTER_POOL)) | ||
| if LETTER_POOL[letter] == 0: # Check letter is in pool | ||
| continue | ||
| else: # Remove one letter from the letter pool | ||
| letters.append(letter) | ||
| LETTER_POOL[letter] -= 1 | ||
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| return letters |
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One thing to consider: the way this function is structured, this will prevent us from picking a letter that is not available. However, when we randomly pick on L65, it's a random pick from the 26 letters in an alphabet. That means A and Z both have 1/26th chance of being picked before we do the validation check, despite the letter bag distribution for A being 9/98 and Z being 1/98.
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