Question

我有一个爱情计算器，它显示了有多少名1和名2恋爱（占100％），这取决于元音的数量是否匹配。到目前为止，这是我的代码：

name1 = input("Please type Name 1.\n")
name2 = input("Please type Name 2.\n")
vowels1 = 0
vowels2 = 0
VOWELS = 'aeiou'
def up_it(word):
    letters = []
    for letter in name1:
        if letter.lower() in VOWELS:
            vowels1 = 1+vowels1 
print("You have",vowels1,"vowels.")

它没有显示name1中有多少个元音。您如何制作它，以便确实显示name1中有多少个元音？

Answer 1

您应该创建一个直接返回解决方案的函数，而不是定义修改全局变量的函数。这样，它就不依赖于某些脆弱的全局状态。

def count_vowels(word):
    vowels = {'a', 'e', 'i', 'o', 'u'}  # I use a set here rather than a list because
                                        # sets have very fast membership checks.
    count = 0
    for letter in word:
        if letter in vowels:
            count += 1
    return count

然后您可以简单地做到：

num_vowels = count_vowels(name1)

可能值得注意的是，count_vowels可以简化为单层：

def count_vowels(word):
    return sum(lett in {'a', 'e', 'i', 'o', 'u'} for lett in word)

Answer 2

def up_it(word):
    vowel =['a','e','i','o','u']
    count =0
    for i in vowel:
        count+=word.count(i)
    return count

Answer 3

您定义了一个函数up_it(word)，但没有调用它。下面是该代码的简化版本：

name1 = input("Please type Name 1.\n")
name2 = input("Please type Name 2.\n")
vowels1 = 0
vowels2 = 0
VOWELS = 'aeiou'

def up_it(word):
    letters = []
    for letter in name1:
        if letter.lower() in VOWELS:
            vowels1 = 1+vowels1 

up_it(name1)  # call the function
print("You have", vowels1, "vowels.")

现在，该代码并未真正正确使用函数。更好的方法是使函数采用（并使用）参数，然后返回计数值。考虑到这一点，我们将有以下内容：

def count_vowels(word):
    VOWELS = 'aeiou'
    vowels_count = 0
    for letter in word:
        if letter.lower() in VOWELS:
            vowels_count += 1
    return vowels_count

name1 = input("Please type Name 1.\n")
name2 = input("Please type Name 2.\n")
print("You have", count_vowels(name1), "vowels.")
print("You have", count_vowels(name2), "vowels.")

Answer 4

您的代码看起来应该可以工作-无论如何对于name1-您只是忘记了调用up_it（）。所以试试这个：

def up_it(word):
    letters = []
    for letter in name1:
        if letter.lower() in VOWELS:
            vowels1 = 1+vowels1 

name1 = input("Please type Name 1.\n")
name2 = input("Please type Name 2.\n")
vowels1 = 0
vowels2 = 0
VOWELS = 'aeiou'

up_it("literally anything because you don't use the value passed")

print("You have",vowels1,"vowels.")

问题是您定义的up_it函数从不实际使用传入的可变字。它始终使用name1。另外，您还有一个称为来声明和初始化的字母的列表，但从未在方法中使用它。

我认为您真正想要的是以下代码：

def up_it(word):
    word = word.lower()
    return sum([word.count(vowel) for vowel in 'aeiou'])

name1 = input("Please type Name 1.\n")
name2 = input("Please type Name 2.\n")

vowels_name_1 = up_it(name1)
vowels_name_2 = up_it(name2)

print(name1,"has", vowels_name_1,"vowels.")
print(name2,"has", vowels_name_2,"vowels.")

您如何找出一个单词中有多少个元音？

5 个答案: