Question

现在我有一个包含两个词典的列表的JSON。此代码允许用户通过索引号搜索列表，检索字典。目前我得到两个打印结果，每个字典一个，例如，如果我输入索引号1，我得到：没有索引号，然后打印出一个字典。而不是这个我想只得到一个结果打印找到的字典或1错误。我不应该使用枚举吗？

这是我的JSON（问题），其中包含2个dicts列表。

[{
    "wrong3": "Nope, also wrong",
    "question": "Example Question 1",
    "wrong1": "Incorrect answer",
    "wrong2": "Another wrong one",
    "answer": "Correct answer"
}, {
    "wrong3": "0",
    "question": "How many good Matrix movies are there?",
    "wrong1": "2",
    "wrong2": "3",
    "answer": "1"
}]

这是我的代码

f = open('question.txt', 'r')
questions = json.load(f)
f.close()


value = inputSomething('Enter Index number: ')



for index, question_dict in enumerate(questions):

   if index == int(value):
      print(index, ') ', question_dict['question'],
         '\nCorrect:', question_dict['answer'],
         '\nIncorrect:', question_dict['wrong1'],
         '\nIncorrect:', question_dict['wrong2'],
         '\nIncorrect:', question_dict['wrong3'])
      break

   if not index == int(value):
      print('No index exists')

Answer 1

恕我直言，我认为你不需要使用enumerate。我个人认为你不应该绕过questions。

为什么不这样做：

# assuming the user enters integers from 1 to N.
user_index = int(value) - 1
if -1 < user_index < len(questions):
    # print the question
else:
    print('No index exists')

虽然我们在其中，但为什么不使用with关键字：

with open('question.txt', 'r') as f:
    questions = json.load(f)

而不是close：

f = open('question.txt', 'r')
questions = json.load(f)
f.close()

Answer 2

# Use `with` to make sure the file is closed properly
with open('question.txt', 'r') as f:
    questions = json.load(f)

value = inputSomething('Enter Index number: ')
queried_index = int(value)

# Since `questions` is a list, why don't you just *index` it with the input value
if 0 <= queried_index < len(questions):
    question_dict = questions[queried_index]
    # Let `print` take care of the `\n` for you
    print(index, ') ', question_dict['question'])
    print('Correct:', question_dict['answer'])
    print('Incorrect:', question_dict['wrong1'])
    print('Incorrect:', question_dict['wrong2'])
    print('Incorrect:', question_dict['wrong3'])
else:
    print('No index exists')

对于循环枚举。蟒蛇

2 个答案: