Question

我的代码的目的是从用户那里获取一个字符串并变成一个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: ')

Answer 1

要索引序列，您需要使用[]

if UserSentence[i] == UserWord:

如果您正在尝试找到哪个索引（按字词），则可以执行

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

或类似地

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

Answer 2

这里有一些错误：

如果您使用的是Python 2，请使用raw_input作为字符串。在Python 3中input没问题。
您的maketrans电话很奇怪
要查找单词是否在列表中，您不会进行任何循环或自己的比较。 Python可以为你做到这一点。
请坚持PEP0008。它告诉您如何格式化代码以便更容易阅读。

您重写和测试过的代码：

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.')

如何调试我的Python代码？

2 个答案: