Sphinx - Christelle Nkera#27
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| def draw_letters(): | ||
| pass | ||
| LETTER_POOL = { |
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Nice job on using the constant variable naming convention for Python here!
| for letter, original_letter_count in LETTER_POOL.items(): | ||
| for letter_index in range(original_letter_count): | ||
| all_available_let.append(letter) |
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Yes, this makes sure that the letters have their weighted distribution and ensures letters like Z and E don't have the same chances of being chosen! ⭐️
| current_hand.append(all_available_let[available_let_list_index]) | ||
| all_available_let.pop(available_let_list_index) | ||
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| return current_hand |
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| def uses_available_letters(word, letter_bank): | ||
| pass | ||
| word = word.upper() # capitalizes input word |
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Nice work on ensuring that all the letters where of the same casing. You could also use casefold, which is used for caseless matching. You can find more information on the method here.
Also, be mindful of re-assigning the parameter of a function. That could cause issues for the caller of the function if the change is not expected. Usually we want to avoid this unless specifically told to.
| for input_letter in word: | ||
| letter_freq_in_letter_bank = letter_bank.count(input_letter) | ||
| letter_freq_in_word = word.count(input_letter) |
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Under the hood, .count() is essentially looping over the iterable that provided it and counts the instances of its provided elements (input_letter). Essentially you are looping over word twice but this same functionality could be provided without looping over word with count. How do you think this could be done?
| points_dict = { | ||
| '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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Love this data structure, this allows us to avoid making a nested loop later on in the function.
| if len(word) >= 7: | ||
| total_points += BONUS_POINTS | ||
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| return total_points |
| for word in word_list: | ||
| word_score = score_word(word) | ||
| word_length = len(word) | ||
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| # selects the highest scoring word in the word_list | ||
| if word_score > best_score: | ||
| best_word = word | ||
| best_score = word_score | ||
| best_word_length = word_length | ||
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| # selects best word according to the adagrams tie breaking rules | ||
| elif word_score == best_score: | ||
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| # chooses the 10 letter word as the best word | ||
| if word_length == 10 and best_word_length != 10: | ||
| best_word = word | ||
| best_word_length = word_length | ||
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| # chooses the first 10 letter word | ||
| elif word_length == 10 and best_word_length == 10: | ||
| continue | ||
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| # selects the shortest word if not word length is fewer than 10 letters | ||
| elif word_length < best_word_length and best_word_length != 10: | ||
| best_word = word | ||
| best_word_length = word_length |
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Great job on this loop and breaking down the tie-breaking logic. There is a way that you re-arrange your if/else statements to where they can be un-nested and be more concise. Two of them will only have pass and break in them as well.
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