Develop Reference

Having trouble with while loops, input and strings in python

I am learning python and practicing my skills my making a simple text based adventure game.
In the game, I want to ask the player if they are ready to begin. I did this by creating a begin() function:
def begin():
print(raw_input("Are you ready to begin? > "))
while raw_input() != "yes":
if raw_input() == "yes":
break
print(start_adventure())
else:
print("Are you ready to begin? > ")
print(begin())
below this in my code is the function start_adventure()
def start_adventure():
print("Test, Test, Test")
When I run the program it starts up and I get to the point where it asks if I am ready to begin. Then it just loops infinitely and I can only exit the program if I completely close Powershell and restart Powershell. What am I doing wrong? How can I get the loop to stop once the player inputs "yes"?

What do you expect this to do? The solution to your problem is to try to understand what the code does, instead of just throwing stuff together. (Don't worry; at least 80% of us were at that stage at one point!)
As an aside, I strongly recommend using Python 3 instead of Python 2; they made a new version of Python because Python 2 was full of really strange, confusing stuff like input causing security vulnerabilities and 10 / 4 equalling 2.
What do you want this to do?
Repeatedly ask the user whether they are ready to begin until they answer "yes".
Call start_adventure().
Ok. Let's put what we've got so far into a function:
def begin():
while something:
raw_input("Are you ready to begin? > ")
start_adventure()
There are a lot of gaps in here, but it's a start. Currently, we're getting the user's input and throwing it away, because we're not storing it anywhere. Let's fix that.
def begin():
while something:
answer = raw_input("Are you ready to begin? > ")
start_adventure()
This is starting to take shape. We only want to keep looping while answer != "yes"...
def begin():
while answer != "yes":
answer = raw_input("Are you ready to begin? > ")
start_adventure()
Hooray! Let's see if this works!
Traceback (most recent call last):
File "example", line 2, in <module>
while answer != "yes":
NameError: name 'answer' is not defined
Hmm... We haven't set a value for answer yet. In order to make the loop run, it has to be something that isn't equal to "yes". Let's go with "no":
def begin():
answer = "no"
while answer != "yes":
answer = raw_input("Are you ready to begin? > ")
start_adventure()
This will work!

Python 3 Solution
You should not be calling raw_input() multiple times. Simply instantiate x and then wait until the user inputs Y to call your start_adventure function. This should get you started:
def start_adventure():
print('We have started!')
#do something here
def begin():
x = None
while x!='Y':
x = input('Are you ready to begin (Y/N)?')
if x=='Y':
start_adventure()
begin()

Your Raw input function (I'm assuming it works correctly) is never assigned to a variable. Instead you call it in your print statement, print the result of it and then you call it again in your while loop condition.
You never actually satisfy the while loop condition because your input isn't assigned to a variable. Assign Raw_input("Are you ready to begin? >") to a variable to store the input. Then while loop with the variable. Make sure in your while loop when the condition is met you reset the variable to something else.
Your program flow is wrong too, you need to call your raw input function inside the while loop. This will change the while loop condition so that when the condition is met (user types "yes") it won't loop infinitely. Hope this helps!
Example of what you need in code form:
//initialize the condition to no value
condition = None;
#check the condition
while condition != "yes"
#change the condition here based on user input **inside the loop**
condition = raw_input("are you ready to begin? >")
if condition == "yes":
#condition is met do what you need
else:
#condition not met loop again
#nothing needs to go here to print the message again

How do I use an elif statement properly?

Please don't judge me.. I've only been working with Python for a month now.
While laying in my bed I thought of making this and created it in a few minutes but I made to many else and if statements and my code just looks so messy, I kept adding things that weren't needed..(For fun :D)
Anyways, here is my code.. If you could tell me how to use the "elif" statements properly that'd be awesome.(I'm still learning python)
Question: I've tried using an elif statement multiple times and I keep getting an error. How do I fix this?
key = True # Game key, if this is false the program won't work.
print("Please type a password: ") # Asking for users password
Pass = input()
print("Thank you for typing your password, please make sure it's secure by trying again..") # Ask them to confirm their password by re-typing it
Again = input()
if Pass == Again:
print("Thank you for choosing a working password, please create your character")
print("Please type your username without numbers")
else:
print("Something's wrong with your password or username!")
# Has user confirm if his information is correct
User = input()
print("checking..")
if User.isalpha() and key == True:
print("So your Username is " + User + " and your chosen password is: " + str(Pass))
else:
print("Either your key is broken or something is wrong..")
if len(User) >= 4: # Checking if the chosen username has 4 or more characters in it
print("The length of your Username is: ")
print(str(len(User)))
print("If this information is correct please type 'true' or 'false'")
else:
print("Please type a username longer than 4 characters!")
answer = input() # I kinda fucked up because my coding is dirty and unorganized lol..
if answer == str(True):
print("Thank you, we're setting up your account! :D")
else:
print("Please re-run the program and fix your information!")

We can't debug code you haven't posted, and (as is to be expected - you are on a learning exercise here, and there's a lot to think about) your program structure isn't very helpful. For example, when the user enters non-matching passwords you tell them about it, but nevertheless continue to ask them for their username. Don't worry about this, you will soon learn how to fix it.
Since you ask about the elif, it is basically a syntax abbreviation for else if that avoids going to multiple indentation levels. Suppose you wanted a value of '1' or '2' to take different actions, and to declare other values invalid. You could write
if value == '1':
#take action appropriate to 1
else:
if value == '2':
# take action appropriate to 2
else:
raise ValueError("Allowed inputs are '1' or '2'")
Note that the different actions are at different indentation levels. The more cases you have to consider, the more levels of indentation you have to introduce. So it's generally felt to be more readable to write
if value == '1':
# Take action appropriate to 1
elif value == '2':
# Take action appropriate to 2
else:
raise ValueError("Allowed inputs are '1' or '2'")
Now all the actions and decisions are at the same indentation levels. That's pretty much all there is to it. If you leave the else case off then you won't take any actions at all, so it's normally used to specify the default action, in this case raising an exception.
PS: If you want to be sure the user has entered two matching passwords before you proceed, look at the while loop, which allows you to repeat a set of actions until some condition (in this case the passwords being equal) is true.

Here is an example if if/elif/else statement in python3:
test = 'mytest'
if test == 'not_my_test':
print('nope')
elif test == 'mytest':
print('yay')
else:
print('something else')
You can find more information here : https://docs.python.org/3/tutorial/controlflow.html
EDIT:
As a general remark, you should not define variable using capital letters (PEP convention: https://www.python.org/dev/peps/pep-0008/?)

So using Elif
if Pass == Again:
print("Thank you for choosing a working password, please create your character")
print("Please type your username without numbers")
elif Pass != Again:
print("Something's wrong with your password or username!")
Though what is this error you're getting, we can't really help you without it.

converting logic tests into try statements (python3)

I have been trying to convert some code into a try statement but I can't seem to get anything working.
Here is my code in pseudo code:
start
run function
check for user input ('Would you like to test another variable? (y/n) ')
if: yes ('y') restart from top
elif: no ('n') exit program (loop is at end of program)
else: return an error saying that the input is invalid.
And here is my code (which works) in python 3.4
run = True
while run == True:
spuriousCorrelate(directory)
cont = True
while cont == True:
choice = input('Would you like to test another variable? (y/n) ')
if choice == 'y':
cont = False
elif choice == 'n':
run = False
cont = False
else:
print('This is not a valid answer please try again.')
run = True
cont = True
Now what is the proper way for me to convert this into a try statement or to neaten my code somewhat?
This isn't a copy of the mentioned referenced post as I am trying to manage two nested statements rather than only get the correct answer.

If you want to make your code neater, you should consider having
while run:
instead of
while run == True:
and also remove the last two lines, because setting run and cont to True again isn't necessary (their value didn't change).
Furthermore, I think that a try - except block would be useful in the case of an integer input, for example:
num = input("Please enter an integer: ")
try:
num = int(num)
except ValueError:
print("Error,", num, "is not a number.")
In your case though I think it's better to stick with if - elif - else blocks.

Ok so as a general case I will try to avoid try...except blocks
Don't do this. Use the right tool for the job.
Use raise to signal that your code can't (or shouldn't) deal with the scenario.
Use try-except to process that signal.
Now what is the proper way for me to convert this into a try statement?
Don't convert.
You don't have anything that raises in your code, so there is no point of try-except.
What is the proper way to neaten my code somewhat?
Get rid of your flag variables (run, cont). You have break, use it!
This is prefered way of imlementing do-while, as Python docs says; unfortunately, I cannot find it to link it right now.
If someone finds it, feel free to edit my answer to include it.
def main()
while True: # while user wants to test variables
spuriousCorrelate(directory) # or whatever your program is doing
while True: # while not received valid answer
choice = input('Would you like to test another variable? (y/n) ')
if choice == 'y':
break # let's test next variable
elif choice == 'n':
return # no more testing, exit whole program
else:
print('This is not a valid answer please try again.')

How to loop back to the beginning of a programme - Python [duplicate]

This question already has answers here:
How to make program go back to the top of the code instead of closing [duplicate]
(7 answers)
Closed 8 years ago.
I've written a BMI calculator in python 3.4 and at the end I'd like to ask whether the user would like to use the calculator again and if yes go back to the beginning of the code. I've got this so far. Any help is very welcome :-)
#Asks if the user would like to use the calculator again
again =input("Thank you again for using this calculator, would you like to try again? Please type y for yes or n for no-")
while(again != "n") and (again != "y"):
again =input("Please type a valid response. Would you like to try again? Please type y for yes or n for no-")
if again == "n" :
print("Thank you, bye!")
elif again == "y" :
....

Wrap the whole code into a loop:
while True:
indenting every other line by 4 characters.
Whenever you want to "restart from the beginning", use statement
continue
Whenever you want to terminate the loop and proceed after it, use
break
If you want to terminate the whole program, import sys at the start of your code (before the while True: -- no use repeating the import!-) and whenever you want to terminate the program, use
sys.exit()

You just need to call the function if the user wants to start again:
def calculate():
# do work
return play_again()
def play_again():
while True:
again = input("Thank you again for using this calculator, would you like to try again? Please type y for yes or n for no-")
if again not in {"y","n"}:
print("please enter valid input")
elif again == "n":
return "Thank you, bye!"
elif again == "y":
# call function to start the calc again
return calculate()
calculate()

I can't seem to get user input to work in my if, elif, else loop in Python [duplicate]

This question already has answers here:
How to test multiple variables for equality against a single value?
(31 answers)
Closed 8 years ago.
So I am working on a program that essentially asks a person if they want to hear a joke; if they say yes, then it will proceed to tell a joke; if they say no, or stop/quit the program will stop; and if they don't say yes or no, it will give them an error message. However, the program seems to just want to proceed on with the joke no matter what they input. I will produce an example of the code.
import sys
print("Want to hear a joke?")
answer = input()
if answer == "Yes" or "yes" or "y":
print("Here is a joke")
elif answer == "No" or "no" or "n":
sys.exit("Bye!")
else:
print("Please input a different response.")

This is NOT how you do a conditional
if answer == "Yes" or "yes" or "y": # this will always evaluate to true
do this instead
if answer in ["Yes" ,"yes" , "y"]:
print "Here is a joke"