现在，我正在尝试创建一个oppish翻译器。也就是说，在连续一个辅音或多个辅音之后，您可以在这些字母后添加“ op”。举例来说，母牛会变成小矮人或街头小贩。这是我到目前为止的内容：
def oppish(phrase): #with this function I'm going to append 'op' or 'Op' to a string.
consonants = ['b','c','d','f','g','h','i','j','k','l','m','n','p','q','r','s','t','v','w','x','y','z']
vowels = ['a', 'e', 'i', 'o', 'u'] #this and the preceding line create a variable for the vowels we will be looking to append 'op' to.
if phrase == '': #error case. if the input is nothing then the program will return False.
return False
phrase_List = list(' ' + phrase) # turns the input phrase into a list which allows for an index to search for consonants later on.
new_phrase_List = list() #creates new list for oppish translation
for i in range(1, len(phrase_List)):
if phrase_List[i] == phrase_List[1]:
new_phrase_List.append(phrase_List[i])
elif phrase_List[i] in consonants:
new_phrase_List.append('op') #adds op to the end of a consonant within the list and then appends it to the newlist
new_phrase_List.append(phrase_List[i]) #if the indexed letter is not a consonant it is appended to the new_phrase_list.
print 'Translation: ' + ''.join(new_phrase_List)
oppish('street')
唯一的问题是上面的代码产生了
Translation: ssoptopreeopt
我不确定自己做错了什么，我尝试过使用展示台，但无济于事。感谢所有帮助！ :)
最佳答案
这非常适合itertools.groupby，它将使您可以使用键功能将项目归为一组。该组将累积，直到键函数的返回值更改为止，此时group by将产生键函数的返回值，并在累积的组上进行迭代。在这种情况下，如果字母是元音，我们希望我们的键函数返回True。这样，我们将从groupby中获得一组连续的辅音和一组连续的元音：
from itertools import groupby
vowels = {'a', 'e', 'i', 'o', 'u'} # set instead of list, because lookups are O(1)
def oppish(phrase):
if not phrase:
return False
out = []
for is_vowel, letters in groupby(phrase, lambda x: x in vowels):
out.append(''.join(list(letters)))
if not is_vowel:
out.append('op')
return ''.join(out)
print oppish('street')
print oppish('cow')
输出：
stropeetop
copowop