如何调试我的Python代码？

我的代码的目的是从用户那里获取字符串，并将其转换为不区分大小写的列表。然后我需要从用户那里获取第二个字符串，然后输出第二个给定字符串的位置。这是我的代码：

UserSentence = input('Enter your chosen sentence: ') #this is where the user inputs their sentence
from string import punctuation #this 'fetches' the punctuation from the string
tbl=str.maketrans({ord(ch):" " for ch in punctuation})
UserSentence = UserSentence.lower().translate(tbl).split()#.split() turns the input sentence into a list,...
#...this will help to identify where a word appears...
#...in the sentence. The .lower() also turns the...
#...string into lowercase so it is not case sensitive.
UserWord = input('Enter a word from the sentence: ')#this is where the user inputs their word from the sentence
UserWord = UserWord.lower()#.lower() is used to make UserWord not case sensitive
for i in range(len(UserSentence)):
     if UserSentence (i) == UserWord:
         print ('Your chosen word appears in: ')

---

要为序列编制索引，您需要使用

[]

if UserSentence[i] == UserWord:

如果你想找到他们的单词是哪个索引，你可以这样做

if UserWord in UserSentence:
    print('Your word is located at {}'.format(UserSentence.index(UserWord)))

或者类似地

try:
    print('Your word is located at {}'.format(UserSentence.index(UserWord)))
except ValueError:
    print('Your word is not in the sentence')

这里有几个错误：

如果您使用的是Python 2，请对字符串使用

raw\u input

。在Python3中，输入是可以的

你的

maketrans

呼叫很奇怪

如果一个单词在一个列表中，你不需要任何循环或自己的比较。Python可以为您做到这一点

请坚持。它告诉您应该如何格式化代码，以便更易于阅读

---

您的重写和测试代码：

from string import punctuation, maketrans

user_sentence = raw_input('Enter a sentence: ')
trans = maketrans("", "")
user_sentence = user_sentence.lower().translate(trans, punctuation).split()
user_word = raw_input('Enter a word from the sentence: ')
user_word = user_word.lower()
if user_word in user_sentence:
    print ('Your chosen word appears in the sentence.')

---

“需要调试我的代码的帮助吗”-告诉我们哪里出了问题。投票结束时没有MCVE。“使用字符串的原始输入。”从Python 3.x开始，这在2.x中才是正确的