Python3 中的错误处理 - 在单词列表中查找特定字母的位置

Question

我正在运行解决通过单词列表将我的代码运行并列出字母在列表中的位置的问题。它可以很好地列出前两个单词的位置，但是当遇到没有指定字母的单词时，代码会跳过。我将在下面粘贴问题和我当前的代码以及当前的输出。

words = ["dog", "sparrow", "cat", "frog"]

#You may modify the lines of code above, but don't move them! #When you Submit your code, we'll change these lines to #assign different values to the variables.

#This program is supposed to print the location of the 'o' #in each word in the list. However, line 22 causes an #error if 'o' is not in the word. Add try/except blocks #to print "Not found" when the word does not have an 'o'. #However, when the current word does not have an 'o', the #program should continue to behave as it does now.

#Hint: don't worry that you don't know how line 18 works. #All you need to know is that it may cause an error.

#You may not use any conditionals.

#Fix this code!

我的代码

for word in words:
print(word.index("o"))

输出

1
5
Traceback (most recent call last):
  File "FindingO.py", line 22, in <module>
    print(word.index("o"))
ValueError: substring not found
Command exited with non-zero status 1

Answer 1

你只需要像这样添加 try-except 块：

words = ["dog", "sparrow", "cat", "frog"]
for word in words:
    try:
        print(word.index('o'))
    except:
        print("Not found")
        pass

Answer 2

使用 try 和 except 块来获取字母在单词中的位置。

for letter in words:
try:
    print(letter.index("o"))
except:
    print("not found")
    pass

Python3 中的错误处理 - 在单词列表中查找特定字母的位置

Error Handling in Python3 - Finding location of specific letters in a list of words

python

error-handling

list

letter

python-3.x