如何在一个句子中的每个元音后检测并打印第一个字母？

Question

I wrote this code to detect and print the first letter after each vowel. It works except when I input a word with two consecutive vowels. It ignores the second vowel. For example, if I input 'school year' the result should be:

o
l
a
r

but I only get

o
a

So what am I doing wrong? I am fairly new to python and I am still learning.

def find_after_vowel(word):
for match in re.findall(r"[ouieaOUIEA](\w{1}|\s)", word):
      print (match)

Answer 1

You can use finditer and the start method of the resulting match objects to find the index of each match, and then use that to get the letter after each vowel:

import re

def find_after_vowel(word):
    for match in re.finditer(r"[ouieaOUIEA]", word):
          print word[match.start()+1] 

find_after_vowel("school year")

Which will output:

o
l
a
r

If you want it to return a list instead of printing the results, use:

import re

def find_after_vowel(word):
    after_vowels = []
    for match in re.finditer(r"[ouieaOUIEA]", word):
          after_vowels.append(word[match.start()+1])
    return after_vowels

after_vowels = find_after_vowel("school year")
print after_vowels

Which will output:

['o', 'l', 'a', 'r']

Answer 2

试试以下正则表达式： [ouieaOUIEA]([^ouieaOUIEA]|\\s)

Answer 3

If I understand you right, you can try this:

import re

def find_after_vowel(word):
    index = re.search(r'[ouieaOUIEA]', word).span()[0] + 1
    return word[index]

print(find_after_vowel('test')) # s
print(find_after_vowel('abc')) # b
print(find_after_vowel('def')) # f
print(find_after_vowel('abcdef')) # b