查找列表Python中以元音开头的第一个字符串

Question

我是蟒蛇的新手-

我必须构建一个名为'first__vowel'的函数。接受字符串列表作为输入，并返回以小写元音("a","e","i","o", or "u")开头的第一个字符串。如果没有以元音开头的字符串，则返回空字符串("")。

你能帮我构建这个功能吗。

谢谢

Answer 1

你需要：

def first_vowel(lst):
    # iterate over list using for loop
    for i in lst:
        # check if first letter is vowel
        if i and i[0] in ['a','e','i','o','u']:
            return i 
    return ""

k = ['sad','dad','mad','asd','eas']
print(first_vowel(k))

或者您也可以使用regex

import re

def first_vow(lst):
    pat = re.compile(r'^[aeiou][a-zA-Z]*')
    for i in lst:
        match = re.match(pat, lst)
        if match:
            return i
    return ""

k = ['sad','Aad','mad','','asd','eas']
first_vow(k)

Answer 2

检查一下这个：

vowels = ["a", "e", "i", "o", "u"]

def first_vowel(ss):
    for s in ss:
        if s and s[0] in vowels:
            return s
    return ""

测试：

first_vowel(["Drere", "fdff", "", "aBD", "eDFF"])
'aBD'

Answer 3

您也可以始终使用str.startswith()函数：

def first_vowel(lst):
    for elem in lst:
        if elem.startswith(('a','e','i','o','u')):
            return elem
    return ''
k = ['', 'b','sad','dad','mad','asd','eas']
print(first_vowel(k))