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…_in_english_dictionary() to ignore input word capitalization.
jbieniosek
reviewed
Oct 6, 2021
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| for letter in letter_bank: | ||
| if letter_bank.count(letter) > LETTER_POOL.count(letter): | ||
| is_freq_correct = False | ||
| #Letter frequency is incorrect. Breaks the loop and creates new hand of letter. | ||
| break |
|
|
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| def uses_available_letters(word, letter_bank): | ||
| pass | ||
| word = word.upper() |
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| if letter in letter_bank and word.count(letter) <= letter_bank.count(letter): | ||
| continue | ||
| else: | ||
| #Will return False if input word uses letters not in letter_bank, | ||
| # and/or over the quantity of the letter available in letter_bank. | ||
| return False |
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Another way to write this conditional is:
Suggested change
| if letter in letter_bank and word.count(letter) <= letter_bank.count(letter): | |
| continue | |
| else: | |
| #Will return False if input word uses letters not in letter_bank, | |
| # and/or over the quantity of the letter available in letter_bank. | |
| return False | |
| if not letter in letter_bank or word.count(letter) > letter_bank.count(letter): | |
| return False |
| score += SCORE_CHART[letter] | ||
| if len(word) >= 7: | ||
| score += 8 | ||
| return score |
| highest_words = [word] | ||
| elif highest_score == score: | ||
| highest_words.append(word) |
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| word = word.lower() | ||
| if word in english_dict.keys(): | ||
| return True | ||
| else: | ||
| return False |
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| if len(word) == 10: | ||
| return(word, highest_score) | ||
| return(min(highest_words, key=len), highest_score) |
| is_valid = word_in_english_dictionary(word) | ||
|
|
||
| # Assert | ||
| assert is_valid == False |
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We did an extension on the project by creating a function to check if the word is in the English dictionary. We imported the package english-dictionary 1.0.24 (pip install english-dictionary).