3.5 Python 交互式测验应用程序答案检查不起作用

Question

我试图为此只定义一个函数

欢迎参加考试

Word 1/5: Potato 这个词包含多少辅音？

3

正确！

单词 2/5：土豆 单词包含多少个元音？

1

正确！

单词 3/5：名称 单词包含多少个元音

5

不正确！正确答案 4

Word 4/5: YES 单词包含多少个字母？ 3 正确！

单词 5/5：天

单词的第3个字母是什么？

是的

正确！

游戏结束。你的分数是 4/5

@Niemmi 喜欢这个？ 随机导入 导入字符串

def consonant_count(word):
    word = word.lower()
    return len([x for x in word if x in consonants])

def vowel_count(word):
    word = word.lower()
    return len([x for x in word if x in vowels])

def prompt_letter_count(word):
    correct = word_map[word]['letters']
    ans = input('How many letters does "{}" contain?'.format(word))
    return check(int(ans), correct)

def prompt_vowel_count(word):
    correct = word_map[word]['vowels']
    ans = input('How many vowels does "{}" contain?'.format(word))
    return check(int(ans), correct)

def prompt_consonant_count(word):
    correct = word_map[word]['consonants']
    ans = input('How many consonants does "{}" contain?'.format(word))
    return check(int(ans), correct)

def prompt_random_letter(word):
    n = random.randint(0, len(word))
    correct = word[n-1]
    ans = input('What is letter {} of "{}"?'.format(n, word))
    return check(int(ans).lower(), correct.lower())

def check(ans, correct):
    if ans == correct:
        return prompt_correct()
    return prompt_incorrect()


def prompt_correct():
    print('That is correct! :)')
    return 1

def prompt_incorrect():
    print('That is incorrect :(')
    return 0

def next_question(word):
    q_type = input_map[random.randint(1, 4)]
    return q_type(word)

vowels = ['a', 'e', 'i', 'o', 'u']
consonants = [x for x in string.ascii_lowercase if x not in vowels]
quizWords = ['WOMBAT', 'COMPUTER', 'BOOKS', 'DAY', 'NIGHT', 'HOUR', 'POTATO', 'HUGE', 'TINY', 'GOOD', 'BAD', 'YES', 'NO', 'HOUSE', 'QUESTION', 'BALLOON', 'CAT', 'DUCK', 'KOOKARBURRA', 'POSTER', 'TELEVISION', 'PRINCE', 'RHYTHM', 'SUDDENLY', 'SNOW', 'MAGNET', 'TOWEL', 'WALKING', 'SPEAKER', 'UNCHARACTERISTICALLY']
word_map = {x:{'consonants':consonant_count(x), 'vowels':vowel_count(x), 'letters':len(x)} for x in quizWords}
input_map = {1:prompt_letter_count, 2:prompt_vowel_count, 3:prompt_consonant_count, 4:prompt_random_letter}

def start_quiz(number_questions):
    current_question = 0
    correct_questions = 0
    if number_questions > len(quizWords):
        number_questions = len(quizWords)
    sample_questions = random.sample(quizWords, number_questions)
    print('WELCOME TO YOUR QUIZ')
    print ('---------------------')
    for x in sample_questions:
        print ("Question {}/{}:".format(current_question+1,  number_questions))
        correct_questions += next_question(x)
        print ('---------------------')
        current_question += 1
    print ('Congragulations on completing your quiz!')
    print ("    Score {}/{}:".format(correct_questions , number_questions))
    try_again = input('Would you like to try again? (y/n)').lower()
    if try_again == 'y' or try_again == 'yes':
        start_quiz(number_questions)

start_quiz(4)

Answer 1

问题是您希望 input 向您返回一个 int，而它在 Python 3 上没有这样做。Python 3 input 的工作方式与 Python 2 上的 raw_input 相同，返回一个字符串您需要自己转换为其他类型。您应该能够通过在所需位置进行转换并将所有 raw_input 调用切换到 input 来修复您的代码，因为 Python 3 上没有 raw_input。

例子：

def prompt_letter_count(word):
    correct = word_map[word]['letters']
    ans = input('How many letters does "{}" contain?'.format(word))
    return check(int(ans), correct) # instead of check(ans, correct)