C18 Lions Carina Bauman #83
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checarina
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Oct 12, 2022
checarina
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alope107
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alope107
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Great job! Really like the helper functions you created here. In the future, please make more frequent and more descriptive commits. Nicely done, this project is green!
| def create_letter_list(): | ||
| #put the helper function here | ||
| LETTER_POOL = { | ||
| 'A': 9, |
| letter_pool = [] | ||
| for letter in LETTER_POOL: | ||
| for i in range(LETTER_POOL[letter]): | ||
| letter_pool.append(letter) | ||
| return letter_pool |
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Great logic! One stylistic note: in the inner for loop we never use the i variable. To make it clear that this variable is unneeded we can use an underscore instead: for _ in range(LETTER_POOL[letter]):. This is a common idiom in Python when we want to show we don't care about the value of a variable.
| def draw_letters(): | ||
| pass | ||
| #convert LETTER_POOL dict to list--helper function? | ||
| letter_pool = create_letter_list() | ||
| hand = [] | ||
| for i in range(10): | ||
| rand_index = random.randint(0, len(letter_pool) - 1) | ||
| hand.append(letter_pool.pop(rand_index)) | ||
| return hand |
| def uses_available_letters(word, letter_bank): | ||
| pass | ||
| word = word.upper() | ||
| hand_copy = letter_bank[:] | ||
| # print(hand_copy) | ||
| for letter in word: | ||
| if letter not in hand_copy: | ||
| hand_copy = letter_bank[:] | ||
| return False | ||
| hand_copy.remove(letter) | ||
| # print(hand_copy) | ||
| return True |
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This works well! Smart to copy the letter_bank to avoid mutating it. However, line 54 is unneeded.
| def get_highest_word_score(word_list): | ||
| pass | ||
| # print(word_list) | ||
| word_scores = [] | ||
| for word in word_list: | ||
| word_scores.append((word, score_word(word))) | ||
| high_score = 0 | ||
| for i in range(len(word_scores)): | ||
| if word_scores[i][1] > high_score: | ||
| winner = word_scores[i] | ||
| high_score = winner[1] | ||
| elif word_scores[i][1] == high_score: | ||
| if len(winner[0]) == 10: | ||
| break | ||
| elif len(word_scores[i][0]) < len(winner[0]) or len(word_scores[i][0]) == 10: | ||
| winner = word_scores[i] | ||
| high_score = winner[1] | ||
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| return winner |
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There's some great logic here, but I think it might not work in some edge cases due to the break on line 110. When we break, we exit the loop entirely and stop considering any new words. In this case, if we find words that are tied in points but one was already 10 letters, we assume that the 10 letter word must be the best of any of our words. This may not be the case. Think through the example below:
Word A: len 10, points 25
Word B: len 10, points 25
Word C: len 10, points 50
I believe your function would incorrectly return Word A, because it would break the loop when checking Word B for a tiebreak and never make it to Word C (the actual highest scoring word).
Also, consider unpacking your variables instead of using the index to make it more readable. For example, we could rewrite line 104 to be for word, score in word_scores:
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