def to_pig_latin(s):
j = 0 # points to first character in word
i = 0
new_sentence_1 = '' # variable to store strings being changed
vowel_position = 0 # show the position of the first vowel
number_of_words = 0
number_of_spaces = s.count(" ")
number_of_words = number_of_spaces + 1
space_position = s.find(" ") # find the position of the first space
sent = s[:space_position] # slice the first word of the sentence
old_sent = s[len(sent)+1:] # stores the old sentence without the first word of s
while number_of_spaces >= 0:
if sent[j] in ["a", "e", "i", "o", "u"]: # checks if first character is a vowel
new_sentence = sent + "way" # adds 'way' to the first word
new_sentence_1 = new_sentence_1 + ' ' + new_sentence # adds the words
else: # if first character is not equal to a vowel
for i in range(len(sent)):
# check to see if first character in s is a vowel
if s[i] == 'a':
break
if s[i] == 'e':
break
if s[i] == 'i':
break
if s[i] == 'o':
break
if s[i] == 'u':
break
vowel_position = i # takes position of first vowel reached in word
consonant_sequence = sent[:vowel_position] # stores all the consonants up to the first vowel, but not the first vowel
sent = sent[vowel_position:] # slices the word from the first vowel to the end
new_sentence = sent + 'a' + consonant_sequence + 'ay' # adds strings
new_sentence_1 = new_sentence_1 + ' ' + new_sentence # adds the words
s = old_sent # takes the value of old_sent
space_position = s.find(" ") # find the position of the first space
如何更改下面的部分以便即使s中有一个单词也能进行检查？或者，如果字符串中的最后一个单词s以一个或多个辅音开头的单词结尾？
if space_position == -1:
space_position = len(s)
sent = s[:space_position]
if sent[j] in ["a", "e", "i", "o", "u"]:
new_sentence = sent + "way"
new_sentence_1 = new_sentence_1 + ' ' + new_sentence
break
else:
for i in range(len(sent)):
if s[i] == 'a':
break
if s[i] == 'e':
break
if s[i] == 'i':
break
if s[i] == 'o':
break
if s[i] == 'u':
break
vowel_position = i
consonant_sequence = sent[:vowel_position]
sent = sent[vowel_position:]
new_sentence = sent + 'a' + consonant_sequence + 'ay'
new_sentence_1 = new_sentence_1 + ' ' + new_sentence
sent = s[:space_position]
old_sent = s[len(sent)+1:]
number_of_spaces = s.count(" ")
number_of_words = number_of_spaces + 1
return new_sentence_1[1:]
英语/皮拉丁语翻译测试程序：
import piglatin
choice = input ("(E)nglish or (P)ig Latin?n")
action = choice[:1]
if action == 'E':
s = input("Enter an English sentence:n")
new_s = piglatin.to_pig_latin(s)
print("Pig-Latin:")
print(new_s)
elif action =='P':
s = input("Enter a Pig Latin sentence:n")
new_s = piglatin.to_english(s)
print("English:")
print(new_s)
输出：
(E)nglish or (P)ig Latin? E
Enter an English sentence: My friend next to me is wearing a shoe
Traceback (most recent call last):
File "/Applications/Wing101.app/Contents/Resources/src/debug/tserver/_sandbox.py",
line 9, in <module>
File "/Users/azhar/Desktop/Computer Science/Assignments/Assignment 4 (Functions & Strings)/piglatin.py", line 46, in to_pig_latin
if sent[j] in ["a", "e", "i", "o", "u"]: # checks if first value in j is equal to a vowel
builtins.IndexError: string index out of range
我认为
，与其解决该特定问题，不如简化代码更容易。
首先，您可以使用 s.split() 拆分句子，这将为您提供在空格上拆分的单词列表。 其次，您可以使用s.find()查找给定字符串的索引。 第三，您可以使用' '.join(sen)来连接使用空格的字符串列表。
因此，使用这些，您的代码可以简化为以下内容（我还添加了大写处理）：
def to_pig_latin(sen):
senlst = sen.lower().split() # split into list of words
for i, word in enumerate(senlst):
if word[0] in 'aeiou':
senlst[i] += 'way'
else:
for vow in 'aeiou':
ind = word.find(vow)
if ind >= 0:
break
if ind < 0: # no vowel in word
continue
senlst[i] = word[ind:]+'a'+word[:ind]
newsen = ' '.join(senlst) # combine list of words into sentence
if sen[0].isupper(): # make capitalization like original
newsen = newsen.capitalize()
return newsen
但是，如果您真的想检查是否有单词，则可以if s.strip():。 这将做的是去除和前导或尾随空格。 如果没有单词，这将给您留下一个空字符串。 在python中，空字符串（或空列表或元组）被认为是False，所以你可以在if测试中使用它，然后对这种情况进行任何你想要的处理。