Ada C19 Amethyst Kaliane V. #125
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Awesome job Kaliane ✨
Your submission covers all the learning goals and passes all the tests! You have earned a 🟢 grade!
In your PR, you'll find comments on ways to refactor and DRY up your code. I see that you only made 3 commits. For future projects, please make your commits small, frequent, and with detailed commit messages! Your commit messages should describe what changes you've added to your code. For example, "added score_word logic and passed the tests". Detailed commit messages are super helpful for future bug hunting!
Otherwise, keep up the great work Kaliane! 💜✨
| LETTER_POOL = { | ||
| 'A': 9, | ||
| 'B': 2, | ||
| 'C': 2, | ||
| 'D': 4, | ||
| 'E': 12, | ||
| 'F': 2, | ||
| 'G': 3, | ||
| 'H': 2, | ||
| 'I': 9, | ||
| 'J': 1, | ||
| 'K': 1, | ||
| 'L': 4, | ||
| 'M': 2, | ||
| 'N': 6, | ||
| 'O': 8, | ||
| 'P': 2, | ||
| 'Q': 1, | ||
| 'R': 6, | ||
| 'S': 4, | ||
| 'T': 6, | ||
| 'U': 4, | ||
| 'V': 2, | ||
| 'W': 2, | ||
| 'X': 1, | ||
| 'Y': 2, | ||
| 'Z': 1 | ||
| } | ||
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| SCORE_CHART = { | ||
| 'A': 1, | ||
| 'B': 3, | ||
| 'C': 3, | ||
| 'D': 2, | ||
| 'E': 1, | ||
| 'F': 4, | ||
| 'G': 2, | ||
| 'H': 4, | ||
| 'I': 1, | ||
| 'J': 8, | ||
| 'K': 5, | ||
| 'L': 1, | ||
| 'M': 3, | ||
| 'N': 1, | ||
| 'O': 1, | ||
| 'P': 3, | ||
| 'Q': 10, | ||
| 'R': 1, | ||
| 'S': 1, | ||
| 'T': 1, | ||
| 'U': 1, | ||
| 'V': 4, | ||
| 'W': 4, | ||
| 'X': 8, | ||
| 'Y': 4, | ||
| 'Z': 10 | ||
| } |
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Nice work using the uppercase naming convention to indicate that this is a constant variable ✨
Since LETTER_POOL is accessed by one function, it would be reasonable to place them inside the functions rather than as a constant. There are tradeoffs, the structures would clutter the function some, but it keeps the data as close as possible to where it's being used, and would mean other functions couldn't access it to accidentally alter the contents.
| def draw_letters(): | ||
| pass | ||
| letters = [] | ||
| pool = LETTER_POOL.copy() | ||
| while len(letters) < 10: | ||
| hand = random.choice(list(pool)) | ||
| if pool[hand] <= 0: | ||
| continue | ||
| pool[hand] -= 1 | ||
| letters.append(hand) | ||
| return letters |
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Looks like you made ✨Hard Mode✨! Using list on a dictionary ( LETTER_POOL) creates a list of the dictionary's keys. In this case, you're making a list of all 26 letters with an equal probability of being randomly chosen via random.choice. So in other words, (from the README.md):
Since there are 12
Es but only 1Z, it should be 12 times as likely for the user to draw anEas aZ.
The tests do not focus on the statistical accuracy of your code. Instead, the tests check for outcomes that should NOT be possible such as having more than 1 Z in our hand.
Hard Mode is a totally valid way of making this project so don't fret about resubmitting or redoing this (if you have time, you definitely can but it's low-priority compared to the other Learn, PSE, and projects you also need to focus on).
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| def uses_available_letters(word, letter_bank): | ||
| pass | ||
| bank = letter_bank.copy() |
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Nice use of a copy to avoid side effects on letter_bank.
| for character in word: | ||
| character = character.upper() | ||
| if character in bank: | ||
| bank.remove(character) | ||
| else: | ||
| return False | ||
| return True |
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We can refactor this slightly by using upper on the word and checking for falsiness first to exit the function early:
for character in word.upper():
if character not in bank:
return False
bank.remove(character)
return True | def score_word(word): | ||
| pass | ||
| score = 0 | ||
| for character in word: | ||
| character = character.upper() | ||
| score += SCORE_CHART[character] | ||
| if len(word) >= 7: | ||
| score += 8 | ||
| return score |
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👍 Nice work! This is a really clean solution ✨
| @@ -0,0 +1,287 @@ | |||
| { | |||
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Did you mean to add this file to your PR? If not, definitely be wary of what you're adding to your commits!
| candidates = [(w, score_word(w)) for w in word_list] |
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🔥 Beautiful use of a list comprehension to build a list of tuples!!
| ties = [c for c in candidates if c[1] == highest_score[1]] | ||
| if len(ties) == 1: | ||
| return ties[0] | ||
| for c in candidates: | ||
| if len(c[0]) == 10: | ||
| return c | ||
| return min(ties, key=lambda t: len(t[0])) |
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🔥 Great use of lambda, list comprehension, and min !
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