用户输出分析器将以元音打印的程序

Question

我想打印一个用户输入，我的输出将以元音形式出现。结果将是：输入一句话：我叫迈克你想输入更多吗：是输入一句话：我来自中国是否要输入更多：否[a,e,i,i,e,i,a,o,i,a]

Answer 1

class PrintVowels():

    def __init__(self):

        self.vowels = []

    def get_vowels(self):

        while True:

            print("Enter a sentence: ")

            inp = input()

            self.vowels.extend([i for i in inp.lower() if i in 'aeiou'])

            print("Do you want to input more?")

            inp2 = input()

            if inp2.lower() == 'no':

                break

        return self.vowels

ob1 = PrintVowels()

print (ob1.get_vowels())

#output['e', 'o', 'o', 'e', 'o', 'o']

Answer 2

既然你提到用户输入应该只是yes/noor YES/NO，我还添加了一些使用输入和验证它的方法。

def validate_input(inp2):

    if inp2 == 'yes' or inp2 == 'YES' or inp2 == 'no' or inp2 == 'NO':

        return True

    else:

        return False

def take_input(self):

    print("Enter a sentence: ")

    inp = input()

    self.vowels.extend([i for i in inp.lower() if i in 'aeiou'])

class PrintVowels():

    def __init__(self):

        self.vowels = []

    def get_vowels(self):

        take_input(self)

        while True:

            print("Do you want to input more?")

            inp2 = input()

            valid_inp = validate_input(inp2)

            if valid_inp:

                if inp2.lower() == 'no':

                    break

                else:

                    take_input(self)

            else:

                print("Error: Please try again with a valid input!")

                continue

        return self.vowels

ob1 = PrintVowels()

print (ob1.get_vowels())

输出：

Enter a sentence: 

my name is Mike

Do you want to input more?

blahhhhWHAAT

Error: Please try again with a valid input!

Do you want to input more?

oops

Error: Please try again with a valid input!

Do you want to input more?

yes

Enter a sentence: 

I am from China

Do you want to input more?

Huhhhh

Error: Please try again with a valid input!

Do you want to input more?

NO

['a', 'e', 'i', 'i', 'e', 'i', 'a', 'o', 'i', 'a']