Question

所以我有一个功能，它会在一个句子中并在piglatin中返回。代码如下：

import re
from string import ascii_uppercase

def isvowel(ch):
    vowels = ("a", "e", "i", "o", "u", "A", "E", "I", "O", "U")
    for i in range(len(ch)):
        if ch[i] in vowels:
            return i
    return -1


def format(ch):
    if any(x in ascii_uppercase for x in ch): 
        ch = ch.capitalize()
    m = re.search(r'\W+', ch) 
    if m: 
        ch = ch.replace(m.group(0), '') + m.group(0)
    return ch 


def igpay(astring):
    vowels = ("a", "e", "i", "o", "u", "A", "E", "I", "O", "U")
    words =  astring.split()
    count = 0
    for ch in words:
        vowel = isvowel(ch)
        if vowel == 0:
            astring = astring.replace(ch,(format(ch + "way")))
        else:
            astring = astring.replace(ch,(format(ch[vowel:] + ch[:vowel] + "ay")))
    astring = astring.strip()        
    return astring

当我输入一个单词时输出正常，但是当我输入一个句子时，它会变得混乱。

实施例：我得到了什么：

>>>igpay("An apple a day keeps the doctor away")
>>>'Anwawaywayy awaywaypplewawaywayy awayway dawaywayy eepskay ethay octorday       awaywaywawaywayy'

我应该得到什么：

>>>igpay("An apple a day keeps the doctor away")
>>>'Anway appleway away ayday eepskay ethay octorday awayway'

如果我单独输入每个单词，我会得到正确的回报。发生了什么事？

Answer 1

这是因为您通过替换单词结尾来修改循环中的整个句子。

相反，您应该对列表中的每个单词执行此操作，将单词收集到列表中并返回按空格连接的单词：

def igpay(astring):
    words = astring.split()
    modified_words = []
    for ch in words:
        vowel = isvowel(ch)
        if vowel == 0:
            ch = ch.replace(ch, (format(ch + "way")))
        else:
            ch = ch.replace(ch, (format(ch[vowel:] + ch[:vowel] + "ay")))
        modified_words.append(ch.strip())
    return ' '.join(modified_words)

示例用法（正是您要查找的内容）：

>>> igpay("An apple a day keeps the doctor away")
'Anway appleway away ayday eepskay ethay octorday awayway'

将句子转换成piglatin

1 个答案: