当达到某个索引值时，如何停止将数字添加到列表中的 while 循环？

Question

I made a while loop that needs to stop when the number of things inside of it reaches the 'count' variable, but I'm not too sure on how to go about this. Here is the full code for context, but the part that I'm having trouble with is at the end. Also, when the user gives wrong input, the code gives an error, it's probably something that I could fix by myself after a few attempts, but I'm not exactly sure how. Anyways, the main problems here are the while loops. Thanks for the help in advance.

count = int(input("How many numbers should the sequence have? "))
question = input("AP or GP? ")

# Asking the user for input on how many numbers the sequence will have
if question == "AP":
    APdiff = int(input("Insert the common difference for the AP: "))
    while type(APdiff) != int:
        print("Insert a valid input.")
        APdiff = int(input("Insert the common difference for the AP: "))
elif question == "GP":
    GPdiff = int(input("Insert the common ratio for the GP: "))
    while type(GPdiff) != int:
        print("Insert a valid input.")
        GPdiff = int(input("Insert the common ratio for the GP: "))
while question != "AP" and question != "GP":
    print("Please enter a valid input.")
    question = input("AP or GP? ")


def sequence_generator():


    #Setting up the sequences
    sequence = []
    number = 1

    #Defining the AP
    if question == "AP":
        while number in range(1, 99999999999999999999999999999999):
            sequence.append(number)
            number += APdiff
            if sequence.index() == count:
                break
    #Defining the GP
    elif question == "GP":
        while number in range(1, 99999999999999999999999999999999):
            sequence.append(number)
            number *= GPdiff
            if sequence.index() == count:
                break
    return sequence



print(sequence_generator())

Answer 1

You could use while len(sequence)<count . You would end up with something like this

count = int(input("How many numbers should the sequence have? "))
question = input("AP or GP? ")

# Asking the user for input on how many numbers the sequence will have
if question == "AP":
    APdiff = int(input("Insert the common difference for the AP: "))
    while type(APdiff) != int:
        print("Insert a valid input.")
        APdiff = int(input("Insert the common difference for the AP: "))
elif question == "GP":
    GPdiff = int(input("Insert the common ratio for the GP: "))
    while type(GPdiff) != int:
        print("Insert a valid input.")
        GPdiff = int(input("Insert the common ratio for the GP: "))
while question != "AP" and question != "GP":
    print("Please enter a valid input.")
    question = input("AP or GP? ")


def sequence_generator():


    #Setting up the sequences
    sequence = []
    number = 1

    #Defining the AP
    if question == "AP":
        while len(sequence)<count:
            sequence.append(number)
            number += APdiff
    #Defining the GP
    elif question == "GP":
        while len(sequence)<count:
            sequence.append(number)
            number *= GPdiff
    return sequence



print(sequence_generator())

Answer 2

While loops will not break until the conditional statement is met. For example:

x = 0
while (x < 5):
  print(x)
  x += 1
print('All done')

will output

0
1
2
3
4
All done

The program will execute the block of code until the condition is met, and then break the loop and will not print 5.

To perform an action when a while loop is broken, simply put the code after the while block.