Lion Calss-Kimia Hasani-C18 #69
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| def draw_letters(): | ||
| pass | ||
| #empty list to hold keys | ||
| all_letters =[] | ||
| #empty list to hole random letter | ||
| letter_bank =[] | ||
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| for key, value in LETTER_POOL.items(): | ||
| for i in range(value): | ||
| all_letters.append(key) | ||
| while len(letter_bank) < 10: | ||
| letter = random.choice(all_letters) | ||
| letter_bank.append(letter) | ||
| all_letters.remove(letter) | ||
| return letter_bank |
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Awesome logic here! In the future, try to use more descriptive variable names. Instead of key and value, you could say something like letter and repetitions.
| def draw_letters(): | ||
| pass | ||
| #empty list to hold keys | ||
| all_letters =[] |
| def uses_available_letters(word, letter_bank): | ||
| pass | ||
| #word is string describes some input word | ||
| #letter_bank is list of drawn leters in hand(10 string) | ||
| #if every letter in input is avilable in letter_bank return Tru | ||
| #if not letter in input in letter bank or has to much compare to letter bank return False | ||
| upper_letter = word.upper() | ||
| dict = {} | ||
| for letter in letter_bank: | ||
| # print(letter) | ||
| if letter in dict: | ||
| dict[letter]+=1 | ||
| dict[letter]=1 | ||
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| for letter in upper_letter: | ||
| if letter in dict and dict[letter] > 0: | ||
| dict[letter] -=1 | ||
| continue | ||
| else: | ||
| return False | ||
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| return True |
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I really like the idea here! However, there's a bug for words with repeated letters. The loop from lines 55-59 will always set each letter in the word to have a value of 1, even if it shows up multiple times. Line 59 needs to be enclosed in an else to avoid overwriting the updated value.
Smaller stylistic notes: line 64 is unneeded and try to use more descriptive variable names than dict.
| score = 0 | ||
| for letter in word.upper(): | ||
| score += SCORE_CHART[letter] | ||
| if len(word) > 6: | ||
| score += 8 | ||
| return score |
| score = score_word(word_list[word]) | ||
| word_and_score[word_list[word]] = score |
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Consider not using a range here. We could just do for word in word_list: and then we wouldn't need to index into the list inside the loop.
| if not best_word_and_score: | ||
| best_word_and_score.append(word) | ||
| best_word_and_score.append(score) | ||
| elif score > best_word_and_score[1]: | ||
| best_word_and_score[1] = score | ||
| best_word_and_score[0] = word | ||
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| elif score == best_word_and_score[1] and len(word)< len(best_word_and_score[0]) and len(best_word_and_score[0]) != 10: | ||
| best_word_and_score[1] = score | ||
| best_word_and_score[0] = word | ||
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| elif score == best_word_and_score[1] and len(word) == 10 and len(word) != len(best_word_and_score[0]): | ||
| best_word_and_score[1] = score | ||
| best_word_and_score[0] = word | ||
| return tuple(best_word_and_score) |
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Great logic! Instead of having best_word_and_score be a list of length 2, consider just having it be two different variables. Then, combine the two variables into a tuple at the end. That way we don't have to remember which is in index 0 and which is in index 1 in the list as we're performing out logic.
hi Auberon, I'm Kimia, As you know, Carina was my partner , She said, her feeling really unwell , and need to stay home We I work on my project by my self
wave 4 was difficult for me.
you are the best teacher ever
I learn a lot form you
Thank you so much