如何过滤以任何元音开头的字符串？ [python]

Question

字符串是'banana'。我已经创建了所有子串，现在想过滤那些以元音开头或不以元音开头的子串。我想用什么操作不知道怎么用

l = ['', 'an', 'anan', 'na', 'ana', 'n', 'a', 'anana', 'ba', 'b', 'ban', 'nan', 'banan', 'banana', 'nana', 'bana']
vowels = 'aeiou'
for word in l:
    if word.startswith( any(vowels)): #this gives error
        print("this word starts with vowel")
        print(word)

Answer 1

l = ['', 'an', 'anan', 'na', 'ana', 'n', 'a', 'anana', 'ba', 'b', 'ban', 'nan', 'banan', 'banana', 'nana', 'bana']
vowels = 'aeiou'
for word in l:
    if any(word.startswith(v) for v in vowels):
        print("this word starts with vowel")
        print(word)

Answer 2

您可以使用集合来表示 vowels 而不是单个字符串，因此查找时间是 O(1) 而不是 O(n)，基本上它应该快一点：

words = ['', 'an', 'anan', 'na', 'ana', 'n', 'a', 'anana', 'ba', 'b', 'ban', 'nan', 'banan', 'banana', 'nana', 'bana']
vowels = {'a', 'e', 'i', 'o', 'u'}
for w in words:
  if len(w) > 0 and w[0] in vowels:
    print(f'"{w}" starts with a vowel')

输出：

"an" starts with a vowel
"anan" starts with a vowel
"ana" starts with a vowel
"a" starts with a vowel
"anana" starts with a vowel

Answer 3

您可以将 tuple 个允许的开始提供给 .startswith：

l = ['', 'an', 'anan', 'na', 'ana', 'n', 'a', 'anana', 'ba', 'b', 'ban', 'nan', 'banan', 'banana', 'nana', 'bana']
vowels = ('a','e','i','o','u')
for word in l:
    if word.startswith(vowels):
        print("this word starts with vowel")
        print(word)

输出：

this word starts with vowel
an
this word starts with vowel
anan
this word starts with vowel
ana
this word starts with vowel
a
this word starts with vowel
anana

如何过滤以任何元音开头的字符串？ [python]

How do I filter strings that start with any vowel? [python]

python

any