如何通过函数实现从用户输入中去除元音字母？

Hey there, let's get that vowel-stripping code working properly! Let's break down what's going wrong and fix it step by step.

First, let's spot the key issues in your original code:

• You defined your function d_vowel with a word parameter, but then immediately overwrite that parameter inside the function using input(). When you call d_vowel() without passing any arguments, that parameter is totally unused—this creates unnecessary confusion.
• The word[:] slice is redundant; you can iterate directly over the string word since strings are natively iterable in Python.

Fix 1: Make the function reusable (separate input from logic) #

If you want the function to work with any string (not just user input), split the input handling from the vowel-removing logic. This makes the function way more flexible:

def remove_vowels(word):
    # Using a set for vowels makes lookups faster (O(1) vs O(n) for lists)
    vowels = {'a', 'e', 'i', 'o', 'u'}
    new_word = []
    for letter in word:
        # Check if lowercase version of the letter isn't a vowel
        if letter.lower() not in vowels:
            new_word.append(letter)
    # Return the result instead of printing it—better for reusability
    return ''.join(new_word)

# Handle input and output outside the function
user_word = input('Give me a word: ')
stripped_word = remove_vowels(user_word)
print(stripped_word)

Fix 2: Keep everything self-contained (if you prefer that workflow) #

If you want the function to handle input, processing, and output all in one go, just remove the unused word parameter entirely. We can also use a list comprehension to make the code cleaner:

def remove_vowels():
    word = input('Give me a word: ')
    vowels = {'a', 'e', 'i', 'o', 'u'}
    # List comprehension is a concise way to build the filtered list
    new_word = [letter for letter in word if letter.lower() not in vowels]
    print(''.join(new_word))

# Call the function directly
remove_vowels()

Quick optimizations to note: #

• Using a set for vowels instead of a list: Sets have faster membership checks, which doesn't matter much for short words, but it's a good habit for larger inputs.
• Returning the result instead of printing it (in Fix 1): This lets you use the stripped word elsewhere in your code, instead of just printing it immediately.
• List comprehensions: They're more readable and often faster than manually appending to a list in a for loop.

内容的提问来源于stack exchange，提问作者Josh Bennett