TypeError: string indices must be integers

[JonWayn]

given the code snippet:

i = 0

for key, value in quiz.items():
    i += 1
    print(f"Question {str(i)} : {key['question']}")

Line 5 fails with the error code,

Error:
TypeError: string indices must be integers

The variable named 'i' is clearly an integer. Can someone explain why it fails with this error?

Yoriz write Aug-30-2022, 02:04 PM:
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[ndc85430]

Why do you assume the problem is with i?

The error message talks about an index - i isn't being used as an index. However, 'question' is, so it looks like

Output:
key

is a string.

In cases like this, if it isn't at first glance obvious where the error is coming from, break down the expression into several parts. That way, the error will come from a more specific place.

[snippsat]

There is not need need(a bad way) to make a varibale i at start an use it at index later.
Just loop over the dictionary like this.

quiz = {
    'What is the fastest land animal?': 'Cheetah',
    'What is Pi?': 3.14
    }

for key, value in quiz.items():
    #print(key, value)
    print(f"Question {key} answer is: {value}")

Output:
Question What is the fastest land animal? answer is: Cheetah
Question What is Pi? answer is: 3.14

[JonWayn]

Why do you assume the problem is with i?

The error message talks about an index - i isn't being used as an index. However, 'question' is, so it looks like

Output:
key

is a string.

In cases like this, if it isn't at first glance obvious where the error is coming from, break down the expression into several parts. That way, the error will come from a more specific place.

You were absolutely right. It wasn't referring to the i but to 'question'. The fact is, there is no key in the dictionary by the name of 'question', but rather a value does. So all I needed to do was change key to value and it worked beautifully.

[deanhystad]

for i, value in enumerate(quiz.values()):
    print(f"Question {i+1} : {value}")

[JonWayn]

given the code snippet:

i = 0

for key, value in quiz.items():
    i += 1
    print(f"Question {str(i)} : {key['question']}")

Line 5 fails with the error code,

Error:
TypeError: string indices must be integers

The variable named 'i' is clearly an integer. Can someone explain why it fails with this error?

full code:

quiz = {
    "question1": {
        "question": "What is the capital of France?",
        "answer": "Paris"
    },
    "question2": {
        "question": "What is the capital of Germany?",
        "answer": "Berlin"
    },
    "question3": {
        "question": "What is the capital of Italy?",
        "answer": "Rome"
    },
    "question4": {
        "question": "What is the capital of England?",
        "answer": "London"
    },
    "question5": {
         "question": "What is the capital of Norway?",
        "answer": "Oslo"
    },
    "question6": {
        "question": "What is the capital of Switzerland?",
        "answer": "Bern"
    },
    "question7": {
        "question": "What is the capital of Austria?",
        "answer": "Vienna"
    },
    "question8": {
        "question": "What is the capital of Spain?",
        "answer": "Madrid"
    },
    "question9": {
        "question": "What is the capital of Portugal?",
        "answer": "Lisbon"
    },
    "question10": {
        "question": "What is the capital of Belgium?",
        "answer": "Brussels"
    },
    "question11": {
        "question": "What is the capital of Sweden?",
        "answer": "Stockholm"
    },
    "question12": {
        "question": "What is the capital of Denmark?",
        "answer": "Copenhagen"
    },
    "question6": {
        "question": "What is the capital of Holland?",
        "answer": "Amsterdam"
    },
}

score = 0
i = 0

for key, value in quiz.items():
    i += 1
    print(f"Question {str(i)} : {value['question']}")
    answer = input("Answer? ")

    if answer.lower() == value['answer'].lower():
        score += 1
        print('Correct!')
    else:
        print("Incorrect :{")
        print("The correct answer is: " + value['answer'])

    print(f"Your score is {score}")

Error:
TypeError: string indices must be integers

I did get it resolved with the help of another reply but thank you for pointing me to the proper usage of the interface. It's pretty awesome

[snippsat]

Some advice can simplify the dictionary be using enumerate() then no need to hardcode Question1,Question2....,
then also can remove i = 0.
Also if using function it look cleaner.
Example.

def questions():
   return {
       "What is the capital of France?": 'Paris',
       "What is the capital of Germany? ": 'Berlin',
     }

def quiz(questions):
    score = 0
    for quiz_numb, question in enumerate(questions.keys(), 1):
        answer = input(f'Question.{quiz_numb}\n{question} ')
        if questions[question] == answer:
            print('Correct\n')
            score += 1
        else:
            print(f'Incorrect the answer was: {questions[question]}\n')
    print(f'Total score for this round: {score}')

if __name__ == '__main__':
   quiz(questions())

Output:
Question.1
What is the capital of France? Paris
Correct

Question.2
What is the capital of Germany?  Bern
Incorrect the answer was: Berlin

Total score for this round: 1

[deanhystad]

Dictionaries are not a good choice for your quiz. I can accept this as a dictionary:

{
        "question": "What is the capital of France?",
        "answer": "Paris"
    }

But this should be a list.

quiz = {
    "question1": {},
    "question2": {},
    ...
}

Using "questionN" keys just makes it more difficult to access the questions, and forced you to use items().

I would not use dictionaries for either the questions or the quiz. Since neither the question or quiz change I used tuples.

quiz = (
    ("What is the capital of France?", "PARIS")
    ("What is the capital of Germany?", "BERLIN")
)
 
score = 0
for index, question in (quiz):
    q, a = question
    print(f"Question {index+1} : {q}")
    if input("Answer? ").upper() == a:
        score += 1
        print('Correct!')
    else:
        print(f"Incorrect :(\nThe correct answer is: {a}")
 
print(f"Your score is {score}")

[JonWayn]

Dictionaries are not a good choice for your quiz. I can accept this as a dictionary:

{
        "question": "What is the capital of France?",
        "answer": "Paris"
    }

But this should be a list.

quiz = {
    "question1": {},
    "question2": {},
    ...
}

Using "questionN" keys just makes it more difficult to access the questions, and forced you to use items().

I would not use dictionaries for either the questions or the quiz. Since neither the question or quiz change I used tuples.

quiz = (
    ("What is the capital of France?", "PARIS")
    ("What is the capital of Germany?", "BERLIN")
)
 
score = 0
for index, question in (quiz):
    q, a = question
    print(f"Question {index+1} : {q}")
    if input("Answer? ").upper() == a:
        score += 1
        print('Correct!')
    else:
        print(f"Incorrect :(\nThe correct answer is: {a}")
 
print(f"Your score is {score}")

Thank you. I like the brevity of code that you implement but I will have to wrap my mind around the logic of its elegance and simplicity

[rob101]

This code:

quiz = (
    ("What is the capital of France?", "PARIS")
    ("What is the capital of Germany?", "BERLIN")
)
  
score = 0
for index, question in (quiz):
    q, a = question
    print(f"Question {index+1} : {q}")
    if input("Answer? ").upper() == a:
        score += 1
        print('Correct!')
    else:
        print(f"Incorrect :(\nThe correct answer is: {a}")
  
print(f"Your score is {score}")

...even with a correction to the syntax (a missing comma in the nested tuple), I can't see how it's going to work.

If I change the code to this:

quiz = (
    ("What is the capital of France?", "PARIS"),
    ("What is the capital of Germany?", "BERLIN")
)
  
score = 0
for index, question in enumerate(quiz):
    q, a = question
    print(f"Question {index+1} : {q}")
    if input("Answer? ").upper() == a:
        score += 1
        print('Correct!')
    else:
        print(f"Incorrect :(\nThe correct answer is: {a}")
  
print(f"Your score is {score}"))

... then yes, it works fine, and is a nice, simple way to do this.