反元音节目问题

Question

def anti_vowel(text):
    new = ''
    for char in text:
        if char in "aeiou" or char in "AEIOU":
            ala = text.replace(char, '')
    new = new + ala
    return new
print anti_vowel("Hey look Words!")

它返回'Hey lk Wrds!' 这意味着该功能以某种方式完全忽略了"e".

Answer 1

def anti_vowel(text):
    new = ''
    for char in text:
        if char in "aeiou" or char in "AEIOU":
            ala = text.replace(char, '')

让我们到此为止看看会发生什么:

对于消息中的每个字符,如果字符是元音,则删除该元音后复制原始文本.然后你把它分配给ala......但是？

    new = new + ala
    return new

缩进意味着new = new + ala 只运行一次 - 在for循环结束后.在你的样本数据上,"嘿看单词!",你看到的最后一个元音是'o',所以ala包含一个没有'o'的字符串(但是所有其他的元音都留下了):

print anti_vowel("Hey look Words!")   # => "Hey lk Wrds!"
print anti_vowel("aeiouAEIOU!")       # => "aeiouAEIO!"

(你的测试字符串只有两个元音字符,'e'和'o',这意味着你看不出"为什么不删除？"和"为什么只删除o？" 之间有任何区别,这将是一个更明显的问题指标!)

直接修复代码看起来像

def anti_vowel(text):
    result = text
    for char in text:
        if char in "aeiouAEIOU":
            result = result.replace(char, '')
    return result

但更多的Pythonic方法是:

# a set allows very fast look-ups
VOWELS = set("aeiouAEIOU")

def no_vowels(s):
    # keep every character that is not a vowel
    return "".join(ch for ch in s if ch not in VOWELS)

print(no_vowels("Look Ma, no vowels!"))  # => "Lk M, n vwls!"