修复程序，检查字的出现，2个字符旁边的每个oth

Question

我试图让我的程序检查字符串中相邻的两个字符的实例，并返回一个不同的字符串来替换这两个字符

def main（）：

dubs = ["ai", "ae", "ao", "au", "ei", "eu", "iu", "oi", "ou", "ui"]
newdubs = [ "eye", "eye", "ow", "ow", "ay","eh-oo", "ew", "oy", "ow","ooey"]

word = input("ENTER WORD : " )
count = 0
fin = []
while count < len(word):

   if word[count:count+2] in dubs:

        if word[count:count+2] == dubs[0]:
            fin.append(newDubs[0] + "-")

        if word[count:count+2] == dubs[1]:
            fin.append(newDubs[1] + "-")

        if word[count:count+2] == dubs[2]:

            fin.append(newDubs[2] + "-")

        if word[count:count+2] == dubs[3]:
            fin.append(newDubs[3] + "-")

        if word[count:count+2] == dubs[4]:
            fin.append(newDubs[4] + "-")

        if word[count:count+2] == dubs[5]:
            fin.append(newDubs[5] + "-")

        if word[count:count+2] == dubs[6]:
            fin.append(newDubs[6] + "-")

        if word[count:count+2] == dubs[7]:
            fin.append(newDubs[7] + "-")

        if word[count:count+2] == dubs[8]:
            fin.append(newDubs[8] + "-")

       if word[count:count+2] == dubs[9]:
            fin.append(newDubs[9] + "-")

    if word[count:count+2] not in dubs:
        fin.append(word[count])

    count+=1
fin= "".join(fin)

print(fin)

像wanai这样的词，我期望wan-eye 结果是waneye-i
我还需要运行检查，看看dubs前面的字符是否是元音，但在正常工作之前不要担心这个问题

Answer 1

我会将您的代码重组为更加模块化：

dubs = ["ai", "ae", "ao", "au", "ei", "eu", "iu", "oi", "ou", "ui"]
newdubs = [ "eye", "eye", "ow", "ow", "ay","eh-oo", "ew", "oy", "ow","ooey"]

def dubbizer(word):
   for itter in range(len(dubs)):
       word = word.replace(dubs[itter], "-"+newdubs[itter])
   return word

print(dubbizer("wanai"))

这将为您提供wan-eye的输出

无需更换：

dubs = ["ai", "ae", "ao", "au", "ei", "eu", "iu", "oi", "ou", "ui"]
newdubs = [ "eye", "eye", "ow", "ow", "ay","eh-oo", "ew", "oy", "ow","ooey"]

def dubbizer(word):
   for itter in range(len(dubs)):
     while dubs[itter] in word:
       word = word[:word.find(dubs[itter])]+"-"+newdubs[itter]+word[word.find(dubs[itter])+len(dubs[itter]):]
   return word

print(dubbizer("wanai"))

Answer 2

使用zip()+replace()：

dubs = ["ai", "ae", "ao", "au", "ei", "eu", "iu", "oi", "ou", "ui"]
newdubs = [ "eye", "eye", "ow", "ow", "ay","eh-oo", "ew", "oy", "ow","ooey"]

s = 'wanai'
for x, y in zip(dubs, newdubs):
    s = s.replace(x, f'-{y}')

print(s)
# wan-eye