Develop Reference

Python While loop breakout issues

The question I have is about the flag I have here for the while loop. This works but not like I think it should. I assume I'm not understanding something so if someone is able to explain, that would be great.
From my understanding this should break out of the loop as soon as one of my conditionals is met. So if I input 'q' it should break out and stop the loop. But what happens is it keeps going through the loop and then it breaks out. so it goes through the last prompt and prints the exception.
(Python version is 3.8.5)
# Statement that tells the user what we need.
print("Enter two numbers and I will tell you the sum of the numbers.")
# Lets the user know they can press 'q' to exit the program.
print("Press 'q' at anytime to exit.")
keep_going = True
# Loop to make the program keep going until its told to stop.
while keep_going:
# Prompt for user to input first number and store it in a variable.
first_number = input("First number: ")
# Create a break when entering the first number.
if first_number == 'q':
keep_going = False
# Prompt for user to input second number and store it in a variable.
second_number = input("Second number: ")
# Create a break when entering the second number.
if second_number == 'q':
keep_going = False
# Exception for non integers being input "ValueError"
try:
# Convert input to integers and add them.
# storing the answer in a variable.
answer = int(first_number) + int(second_number)
except ValueError:
# Tell the user what they did wrong.
print("Please enter a number!")
else:
# Print the sum of the numbers
print(f"\nThe answer is: {answer}")
Using this code it breaks out right away like I expect it to.
while True:
first_number = input("First number: ")
if first_number == 'q':
break
second_number = input("Second number: ")
if second_number == 'q':
break
I just would like to understand what the difference is and if thats how it should work. I feel like I'm missing something or misunderstanding something.

The condition of the while loop is only checked between iterations of the loop body, so if you change the condition in the middle of the loop, the current iteration will finish before the loop terminates. If you want to break a loop immediately, you need to either break (which automatically breaks the loop regardless of the condition) or continue (which jumps to the next iteration, and will therefore terminate the loop if the condition is no longer true).
Using while True: with a break when you want to stop the loop is generally much more straightforward than trying to control the loop by setting and unsetting a flag.
FWIW, rather than copying and pasting the code to input the two numbers, and have two different ways to break out of the loop, I might put that all into a function and break the loop with an Exception, like this:
print("Enter two numbers and I will tell you the sum of the numbers.")
print("Press 'q' at anytime to exit.")
def input_number(prompt: str) -> int:
"""Ask the user to input a number, re-prompting on invalid input.
Exception: raise EOFError if the user enters 'q'."""
while True:
try:
number = input(f"{prompt} number: ")
if number == 'q':
raise EOFError
return int(number)
except ValueError:
print("Please enter a number!")
while True:
try:
numbers = (input_number(n) for n in ("First", "Second"))
print(f"The answer is: {sum(numbers)}")
except EOFError:
break
This makes it easier to extend the program to handle more than two inputs; try adding a "Third" after where it says "First" and "Second"! :)

Once you run the program and type "q", Yes indeed keep_going will be set to False but it DOES NOT MEAN it will break the loop already, it will just make the keep_going be equal to False thus on the NEXT ITERATION will stop the loop. Why is that? because it would be like this while keep_going: -> while False: so since it is not True thus not executing the program anymore.
Now based on your goal as you mentioned. You can do it this way where you can add the break.
if first_number == 'q':
keep_going = False
break
# Prompt for user to input second number and store it in a variable.
second_number = input("Second number: ")
# Create a break when entering the second number.
if second_number == 'q':
keep_going = False
break
I'd also like to suggest have it this way, it's just more specific in terms of what is to happen on the code, but of course it is up to you.
first_number = input("First number: ")
# Create a break when entering the first number.
if first_number == 'q':
keep_going = False
break
# Prompt for user to input second number and store it in a variable.
# Create a break when entering the second number.
else:
second_number = input("Second number: ")
if second_number =='q':
keep_going = False
break

While loops execute until their given condition is false. A loop will only check its condition when required (program execution is moved to the top of the loop). In almost every case, this occurs when the full body of the loop has run. See here:
keep_going = True
while keep_going:
keep_going = False
# keep_going is False, but this will still print once
# because the loop has not checked its condition again.
print("Will execute once")
"Will execute once" prints a single time even after keep_going is set to False. This happens because the while loop does not re-check its condition until its entire body has run.
However, break statements are different. A break statement will cause a loop to exit immediately no matter what.
keep_going = True
while keep_going:
break # Exits the while loop immediately. The condition is not checked.
print("Will never print")
Here, nothing is printed even though keep_going is True the whole time. break made the loop exit regardless of the condition.
A continue statement will move program execution back to the start of the loop, and cause your condition to be checked again.
In this example, continue sends program execution back to the start of the loop. Since keep_going was set to False, nothing will print because the while loop will exit after realizing its condition evaluates to false.
keep_going = True
while keep_going:
keep_going = False
continue
print("Will never print")

First off, hope you have a great time learning Python!
Both ways will work and stop the loop, but there is a difference:
In the first method, you are changing the keep_going variable to false, therefore, the loop will stop when the the while loops finds out that keep_going had become False. However, the checking only happens at the end of the loop (In your case, it is after you have done your except or else part), the loop will not stop right away even when you entered q for your variable first_number.
In the second solution, you are using the break keyword in Python, to break away from the loop right away after you entered q for first_number.
Technically speaking, you will want to break if you want to break off from the loop right away when q is detected, otherwise, setting keep_going to False if you want the whole loop to be completed, but not run again for the next round.
In scenario 1 the result, even when you entered q,
Please enter a number!
Will always show, but not for scenario 2.

this is a little different approach to your script:
def main():
print("Enter two numbers and I will tell you the sum of the numbers.")
print("Press 'q' at anytime to exit.")
val = []
while True:
check_value = lambda x: 'quit' if x.lower() == 'q' or x.lower() == 'quit' else int(x)
if not val:
value = input("First number: ")
elif len(val) == 2:
answer = sum(val)
print(f"\nThe answer is: {answer}")
print('==='*15 + ' < ' + f'PROGRAM RESTARTING' + ' > ' + '==='*15)
val[:] = []
continue
else:
value = input("Second number: ")
try:
check_ = check_value(value)
val.append(check_)
except ValueError:
print("Please enter a number!")
continue
finally:
if check_ == 'quit':
print('Program is stopping....')
break
else:
pass
if __name__ == '__main__':
main()
It check at anytime the user's input, whether is a 'q' 'Q' or ('quit' or 'QUIT') or any combination or capital letter because run the check x.lower()
I suggest you to have a look at realpython.com especially the paragraph "he Python break and continue Statements."
Long story short:
Use Break to terminate the loop at any given time.
Use continue to roll back where you left and repeat the loop again (I used it in my code if the value is not a Int)
User pass to keep the loop running with no stop.

Using nested if statements within a while loop (python)

Here's my code:
#Greeting
print('\n---Sales Discount Calculator---\n')
#User Input
packageCount = input('Packages Purchased: ')
#To check whether packageCount is an integer or not
#loop continues until a positive integer is inputted
validInput = False
while (validInput == False):
if (packageCount.isdigit()):
packageCount = int(packageCount)
if (packageCount != 0):
validInput = True
print(packageCount)
else:
print('Invalid Input. Try again...')
packageCount = input('Packages Purchased: ')
I'm trying to see if the input from the user is a positive integer and also not a zero. I need to check if it's an integer first(the first if-statement) because you can't compare type string to numbers. If it is an integer, then I need to check if it is a zero or not(second if-statement). When the second if-statement is not present, the code checks for both a positive integer and if it's a string or not, but somehow when I include the second if statement and type in a string, it continues without printing the 'Invalid Input'(the else). *This causes issues later on.

This is because you have two seperate if-else statements. So the first checks for a digit, and sets packageCount to an int explicitly, however if this isn't hit then packageCount stays as a string and continues into the next if statement either way
if (packageCount != 0):
I believe you probably want to indent this in so your code becomes:
#Greeting
print('\n---Sales Discount Calculator---\n')
#User Input
packageCount = input('Packages Purchased: ')
#To check whether packageCount is an integer or not
#loop continues until a positive integer is inputted
validInput = False
while (validInput == False):
if (packageCount.isdigit()):
packageCount = int(packageCount)
if (packageCount != 0):
validInput = True
print(packageCount)
else:
print('Invalid Input. Try again...')
packageCount = input('Packages Purchased: ')
else:
print('Invalid Input. Try again...')
packageCount = input('Packages Purchased: ')
You will notice you now have repeated code, but I will leave that to you to remove :)
---edit---
The extra indents we have added ensures the second if-else is only called if the first is successful. The reason I've added a new else in is so that the error message and input request is maintained.
So the first if checks for an integer, if this fails (ie. the input was a string), then we hit the second else.
If this succeeds we then check if packageCount != 0 and either succeed or hit the first else in the code.

Function won't be called upon

When I run the program everything is fine until the password validity function is ran and all the conditions of length and uppercase and lowercase are equal to true. What the program is supposed to do here is run the creditcard function but it doesn't as seen here:
if length and upper and lower == True:
creditcard()
It doesn't give me any errors or syntax messages about it when the conditions of length, upper, and lower case are all true but still will not run the creditcard function, the program simply ends. however when any of the conditions of length uppercase or lowercase are not met using or logic then "The password's invalid" is still successfully printed.
def creditcard():
balence = input("What is your credit card balence this month")
choice = input("Would you like to make a payment or make the standard deduction? y or n")
intrest = .1596/12
amountdue = balence*interest
enter code here
if choice == "y":
payment = input("How much would you like to pay?")
if payment == "n":
if 0.3 * amountdue>25:
payment = 0.3 * amountdue
elif 0.3 * amountdue<25:
payment = 25
balence-=payment
print("Your balence is now", balence)
def validity(password):
length=()
upper=()
lower=()
if len(password)<5:
length = True
else: lenghth = False
for char in password:
if char.isupper():
upper = True
else: upper = False
for char in password:
if char.islower():
lower = True
else: lower = False
if length and upper and lower == True:
creditcard()
if length or upper or lower == False:
print("The password's invalid")
password=""
def main():
password=""
while password=="":
password= input("Please enter a valid password")
validity(password)`enter code here`
main()

The problem is that your if condition is never being met. You can check this quickly by putting something like print("Going to call creditcard") between the if and creditcard() lines. You'll see the message is never printed. To find out why, you could print the values of length, upper and lower before the ifs to see if the values are what you expect.
That's how I'd recommend trying to diagnose your problem with quick and dirty debugging. As for explaining what your problem actually is:
When you do for char in password this is going through each character in the password and then running the test for is upper/lower for that character. This means if you have the password aB then on the first run it will check a and so will set lower to True, but then it will test the next character, B, and set lower to False. So basically, it's only checking the case of the last character, and as a single character can't be both upper and lower your condition for upper and lower being true will never be met.
A better way to do it is to initialise all the conditions as being false and then set them to true when met, e.g.
upper = False
for char in password:
if char.isupper():
upper = True
­
Edit: As deceze pointed out, you also need to be careful with the line if length or upper or lower == False. Because that is not the same as if (length == False) or (upper == False) or (lower == False). Instead, it evaluates it like: if length or upper or (lower == False), and an or operator returns true if either condition is true.
As an example, if the length = False, upper = False and lower = True, the condition would then be: if False or False or (True == False), which gets evaluated to False or False or False which gets evaluated to False. This may be a bit confusing, check the link deceze commented for a fuller description, but in your case a simpler way would be to just use an else.
There are a few more issues I can see, I realise you may not have got to this stage of checking your code yet, but just a heads up:
There are a few typos which can trip you up if you spell a variable differently throughout your code.
I assume that you want to length to be greater than 5? Currently you're using less than.
When you use input() the value is always a string (text) this will cause problems when you try to do your calculations with balance and payment. It's easy to overcome by telling Python to cast the data to a number by using int() for whole numbers, or float() for decimals, like so: balance = int(input("What is your credit card balance this month?"))
And finally (of the things that I can see), it looks like you're trying to make a loop so the user can keep re-entering the password if their previous attempt was invalid? But the condition on the while loop is just checking if the password is empty, so will terminate as soon as the user types in any password. You could create a new boolean called created_password initialised to False and use that as the condition? If I were you I'd do this by making validity just return True or False (depending on if the password is or isn't valid), then call creditcard() from main once out of the while loop.
I'd recommend you try to work through the errors yourself, this way you'll be able to actually see the error messages and learn how to fix them. If you get stuck I've made a version which works here: https://trinket.io/python3/b6b8c0ca31 which you can compare with your code.
Or a slightly more compact version: https://trinket.io/python3/d934a3a48e

Loop - Input numbers to list, don't allow alphabetic input, break out by pressing enter ..?

been at this all day and made absolutely no progress, it's surely supposed to be pretty simple too but I'm new to Python. Google hasn't brought up much so this is my last resort!
It's supposed to be some basic code for manually putting numeric values into a list. If I add a 'print' line to the script I can see the values are being successfully entered as I go along, but I can't seem to add the right script for breaking out of the loop by leaving the input blank. Currently if I set anything to make it break, the script seems to freeze while it's running and I have to reset the console completely.
Also I'm wondering is there a way of ensuring the input is always an integer? It would be good to make things a bit cleaner and bring up an error message or something if the user enters anything non-numeric.
Here's the code.
values = []
while True:
a = raw_input('Enter numeric values. Leave blank to stop: ')
if a == (??????) :
break
values.append(float(a))
Thanks!

You can restrict to only numbers with
if a.isdigit():
So your function could look like
def accept_inputs():
values = []
while True:
a = raw_input('Enter numeric values. Leave blank to stop: ')
if a.isdigit():
values.append(float(a))
if not a:
return values
Testing
>>> accept_inputs()
Enter numeric values. Leave blank to stop: 5
Enter numeric values. Leave blank to stop: 42
Enter numeric values. Leave blank to stop: 73
Enter numeric values. Leave blank to stop: ab
Enter numeric values. Leave blank to stop: abc
Enter numeric values. Leave blank to stop:
[5, 42, 73]

My approach is similar to #CoryKramer, with small changes
>>> values = []
>>> while True:
val = raw_input('Enter Number : ')
if not val:
print values
elif val.isdigit():
values.append(int(val))
else:
print 'you have entered non - digit value'
Enter Number : 2
Enter Number : a
you have entered non - digit value
Enter Number : 3
Enter Number : 5
Enter Number : 6
Enter Number : 22
Enter Number : 546
Enter Number : 31s
you have entered non - digit value
Enter Number : 345678
Enter Number :
>>> values
[2, 3, 5, 6, 22, 546, 345678]

Strings have a built in function isdigit() that returns true if all the characters are numbers.
To break out if nothing is entered, use the len() function to detect if the string is empty.
More info here
You're code would look like this:
if a.isdigit():
#keep going, and add it to the list
elif len(a) == 0:
break #stop the loop

Getting raw input from user

I am getting input from users via raw_input (using a for loop).
What code can I use to ensure that if a user presses only Enter it keeps the default values (i.e. raw_input returns an empty string '' when no value has been entered)?
So for the default values are in the form of variables:
age_years = 2
cash_earned_last_year = 1000
growth = 0.02

If i understand you correctly you need to iterate over values inserted by user and replace with an empty string if the user type only Enter key.
def values_input():
return raw_input("Please enter value or 'quit' to finish: ")
values = []
for code in iter(values_input, "quit"):
if len(code) == 0:
values +=['']
else:
values += [code]
print values

You can use an if to check if the user pressed only enter. As jonrsharpe
said, only Enter would set your input equal to empty string, that is, ''.
if user_answer_1 == '':
age_years = 2
cash_earned_last_year = 1000
growth = 0.02
If user is supposed to either press enter, or give numerous inputs, then you can use break to skip the rest of the questions if his first answer was Enter.
while True:
user_age_answer = input('\nAge?')
if user_age_answer == '':
age_years = 2
cash_earned_last_year = 1000
growth = 0.02
# Skips the rest of the questions.
break
# Here go the rest of the questions.
user_cash_answer = input('Cash?')
# And then you do what you want with the answers.
age_years = user_age_answer
# This break is used to exit loop.
break