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Multiplication guessing game in Python | if wrong guess until it is correct

I am trying to write a program as follows:
Python generates random multiplications (factors are random numbers from 1 to 9) and asks to the users to provide the result
The user can quit the program if they input "q" (stats will be calculated and printed)
If the user provides the wrong answer, they should be able to try again until they give the correct answer
if the user responds with a string (e.g. "dog"), Python should return an error and ask for an integer instead
It seems I was able to perform 1) and 2).
However I am not able to do 3) and 4).
When a user gives the wrong answer, a new random multiplication is generated.
Can please somebody help me out?
Thanks!
import random
counter_attempt = -1
counter_win = 0
counter_loss = 0
while True:
counter_attempt += 1
num_1 = random.randint(1, 9)
num_2 = random.randint(1, 9)
result = str(num_1 * num_2)
guess = input(f"How much is {num_1} * {num_2}?: ")
if guess == "q":
print(f"Thank you for playing, you guessed {counter_win} times, you gave the wrong answer {counter_loss} times, on a total of {counter_attempt} guesses!!!")
break
elif guess == result:
print("Congratulations, you got it!")
counter_win += 1
elif guess != result:
print("Wrong! Please try again...")
counter_loss += 1

Hi my Idea is to put the solving part in a function:
import random
counter_attempt = -1
counter_win = 0
counter_loss = 0
def ask(num1, num2, attempt, loss, win):
result = str(num1 * num2)
guess = input(f"How much is {num1} * {num2}?: ")
if guess == "q":
print(
f"Thank you for playing, you guessed {win} times, you gave the wrong answer {loss} times, on a total of {attempt} guesses!!!")
return attempt, loss, win, True
try:
int(guess)
except ValueError:
print("Please insert int.")
return ask(num1, num2, attempt, loss, win)
if guess == result:
print("Congratulations, you got it!")
win += 1
return attempt, loss, win, False
elif guess != result:
print("Wrong! Please try again...")
loss += 1
attempt += 1
return ask(num1, num2, attempt, loss, win)
while True:
num_1 = random.randint(1, 9)
num_2 = random.randint(1, 9)
counter_attempt, counter_loss, counter_win, escape = ask(num_1, num_2, counter_attempt, counter_loss, counter_win)
if escape:
break
Is that what you asked for?

Note that everything withing your while loop happens every single iteration. Specifically, that includes:
num_1 = random.randint(1, 9)
num_2 = random.randint(1, 9)
So you are, indeed, generating new random numbers every time (and then announcing their generation to the user with guess = input(f"How much is {num_1} * {num_2}?: "), which is also within the loop).
Assuming you only intend to generate one pair of random numbers, and only print the "how much is...?" message once, you should avoid placing those within the loop (barring the actual input call, of course: you do wish to repeat that, presumably, otherwise you would only take input from the user once).
I strongly recommend "mentally running the code": just go line-by-line with your finger and a pen and paper at hand to write down the values of variables, and make sure that you understand what happens to each variable & after every instruction at any given moment; you'll see for yourself why this happens and get a feel for it soon enough.
Once that is done, you can run it with a debugger attached to see that it goes as you had imagined.
(I personally think there's merit in doing it "manually" as I've described in the first few times, just to make sure that you do follow the logic.)
EDIT:
As for point #4:
The usual way to achieve this in Python would be the isdigit method of str:
if not guess.isdigit():
print('Invalid input. Please enter an integer value.')
continue # Skip to next iteration
An alternative method, just to expose you to it, would be with try/except:
try:
int(guess) # Attempt to convert it to an integer.
except ValueError: # If the attempt was unsuccessful...
print('Invalid input. Please enter an integer value.')
continue # Skip to next iteration.
And, of course, you could simply iterate through the string and manually ensure each of its characters is a digit. (This over-complicates this significantly, but I think it is helpful to realise that even if Python didn't support neater methods to achieve this result, you could achieve it "manually".)
The preferred way is isdigit, though, as I've said. An important recommendation would be to get yourself comfortable with employing Google-fu when unsure how to do something in a given language: a search like "Python validate str is integer" is sure to have relevant results.
EDIT 2:
Make sure to check if guess == 'q' first, of course, since that is the one case in which a non-integer is acceptable.
For instance:
if guess == "q":
print(f"Thank you for playing, you guessed {counter_win} times, you gave the wrong answer {counter_loss} times, on a total of {counter_attempt} guesses!!!")
break
elif not guess.isdigit():
print('Invalid input. Please enter an integer value.')
continue # Skip to next iteration
elif guess == result:
...
EDIT 3:
If you wish to use try/except, what you could do is something like this:
if guess == "q":
print(f"Thank you for playing, you guessed {counter_win} times, you gave the wrong answer {counter_loss} times, on a total of {counter_attempt} guesses!!!")
break
try:
int(guess)
except ValueError:
print('Invalid input. Please enter an integer value.')
continue # Skip to next iteration
if guess == result:
...

You are generating a new random number every time the user is wrong, because the
num_1 = random.randint(1, 9)
num_2 = random.randint(1, 9)
result = str(num_1 * num_2)
Is in the while True loop.
Here is the fixed Code:
import random
counter_attempt = 0
counter_win = 0
counter_loss = 0
while True:
num_1 = random.randint(1, 9)
num_2 = random.randint(1, 9)
result = str(num_1 * num_2)
while True:
guess = input(f"How much is {num_1} * {num_2}?: ")
if guess == "q":
print(f"Thank you for playing, you guessed {counter_win} times, you gave the wrong answer {counter_loss} times, on a total of {counter_attempt} guesses!!!")
input()
quit()
elif guess == result:
print("Congratulations, you got it!")
counter_win += 1
break
elif guess != result:
print("Wrong! Please try again...")
counter_loss += 1

How do I stop a while loop if something happens inside a function that should stop the itterations?

Hello fellow programmers! I am a beginner to python and a couple months ago, I decided to start my own little project to help my understanding of the whole development process in Python. I briefly know all the basic syntax but I was wondering how I could make something inside a function call the end of the while loop.
I am creating a simple terminal number guessing game, and it works by the player having several tries of guessing a number between 1 and 10 (I currently made it to be just 1 to test some things in the code).
If a player gets the number correct, the level should end and the player will then progress to the next level of the game. I tried to make a variable and make a true false statement but I can't manipulate variables in function inside of a while loop.
I am wondering how I can make it so that the game just ends when the player gets the correct number, I will include my code down here so you guys will have more context:
import random
import numpy
import time
def get_name(time):
name = input("Before we start, what is your name? ")
time.sleep(2)
print("You said your name was: " + name)
# The Variable 'tries' is the indication of how many tries you have left
tries = 1
while tries < 6:
def try_again(get_number, random, time):
# This is to ask the player to try again
answer = (input(" Do you want to try again?"))
time.sleep(2)
if answer == "yes":
print("Alright!, well I am going to guess that you want to play again")
time.sleep(1)
print("You have used up: " + str(tries) + " Of your tries. Remember, when you use 5 tries without getting the correct number, the game ends")
else:
print("Thank you for playing the game, I hope you have better luck next time")
def find_rand_num(get_number, random, time):
num_list = [1,1]
number = random.choice(num_list)
# Asks the player for the number
ques = (input("guess your number, since this is the first level you need to choose a number between 1 and 10 "))
print(ques)
if ques == str(number):
time.sleep(2)
print("Congratulations! You got the number correct!")
try_again(get_number, random, time)
elif input != number:
time.sleep(2)
print("Oops, you got the number wrong")
try_again(get_number, random, time)
def get_number(random, try_again, find_rand_num, time):
# This chooses the number that the player will have to guess
time.sleep(3)
print("The computer is choosing a random number between 1 and 10... beep beep boop")
time.sleep(2)
find_rand_num(get_number, random, time)
if tries < 2:
get_name(time)
tries += 1
get_number(random, try_again, find_rand_num, time)
else:
tries += 1
get_number(random, try_again, find_rand_num, time)
if tries > 5:
break
I apologize for some of the formatting in the code, I tried my best to look as accurate as it is in my IDE. My dad would usually help me with those types of questions but it appears I know more python than my dad at this point since he works with front end web development. So, back to my original question, how do I make so that if this statement:
if ques == str(number):
time.sleep(2)
print("Congratulations! You got the number correct!")
try_again(get_number, random, time)
is true, the while loop ends? Also, how does my code look? I put some time into making it look neat and I am interested from an expert's point of view. I once read that in programming, less is more, so I am trying to accomplish more with less with my code.
Thank you for taking the time to read this, and I would be very grateful if some of you have any solutions to my problem. Have a great day!

There were too many bugs in your code. First of all, you never used the parameters you passed in your functions, so I don't see a reason for them to stay there. Then you need to return something out of your functions to use them for breaking conditions (for example True/False). Lastly, I guess calling functions separately is much more convenient in your case since you need to do some checking before proceeding (Not inside each other). So, this is the code I ended up with:
import random
import time
def get_name():
name = input("Before we start, what is your name? ")
time.sleep(2)
print("You said your name was: " + name)
def try_again():
answer = (input("Do you want to try again? "))
time.sleep(2)
# Added return True/False to check whether user wants to play again or not
if answer == "yes":
print("Alright!, well I am going to guess that you want to play again")
time.sleep(1)
print("You have used up: " + str(tries) + " Of your tries. Remember, when you use 5 tries without getting the correct number, the game ends")
return True
else:
print("Thank you for playing the game, I hope you have better luck next time")
return False
# Joined get_number and find_random_number since get_number was doing nothing than calling find_rand_num
def find_rand_num():
time.sleep(3)
print("The computer is choosing a random number between 1 and 10... beep beep boop")
time.sleep(2)
num_list = [1,1]
number = random.choice(num_list)
ques = (input("guess your number, since this is the first level you need to choose a number between 1 and 10 "))
print(ques)
if ques == str(number):
time.sleep(2)
print("Congratulations! You got the number correct!")
# Added return to check if correct answer is found or not
return "Found"
elif input != number:
time.sleep(2)
print("Oops, you got the number wrong")
tries = 1
while tries < 6:
if tries < 2:
get_name()
res = find_rand_num()
if res == "Found":
break
checker = try_again()
if checker is False:
break
# Removed redundant if/break since while will do it itself
tries += 1

how can I reduce my code? A lot of it is repetitive

The code creates a random addition problem and spits out "Congratulations" if correct and "sorry...." if the inputted value is wrong. The while loop repeats this process until the user inserts "N" for the question "continue (Y/N):, at the same time it keeps track of how many questions have been answered, and which ones are correct. The code works fine, my problem is it has repetitive code. I was wondering if there is a way to shrink it.
**I appreciate everyone one's help and advice. I"m a noob that's just learning python **
import random
correct=0
count=1
num1=random.randint(0,100)
num2=random.randint(0,100)
print(format(num1,'4d'))
print('+',num2)
answer=int(input('='))
sum=num1+num2
if answer==sum:
print('Congraulations!')
correct+=1
else:
print('Sorry the correct answer is',sum)
c=input('Continue (Y/N):')
while c == "Y":
count+=1
num1=random.randint(0,100)
num2=random.randint(0,100)
print(format(num1,'4d'))
print('+',num2)
answer=int(input('='))
sum=num1+num2
if answer==sum:
print('Congraulations!')
correct+=1
else:
print('Sorry the correct answer is',sum)
c=input('Continue (Y/N):')
else:
print('Your final score is',correct,'/',count)

A first start, would be eliminating the code before the while, by initializing the count variable (which keeps track of the turns), in zero, and allowing the while loop to run the first turn, we just need to have a variable like want_to_play and by default it's True, so the first time we'll be playing, and at the end of the game If I don't input Y or y it will asume I don't want to play any more and set the variable to false, that way I can have all the turns ran by the while loop.
and you'll be getting something like this.:
from random import sample
correct = 0
count = 0 # STartint in turn zero
want_to_play = True # Control Variable
while want_to_play:
count += 1
# First turn this is zero, and adds one.
[num1, num2] = sample(range(0, 101), 2)
# Just another way of getting two random numbers from 1 up to (including) 100.
# Printing could be done in one line.
print(format(num1, '5d') + '\n+' + format(num2, '4d'))
answer = int(input('= '))
# The comparison really doesn't really hurt if you do it this way.
if answer == num1 + num2:
print('Congraulations!')
correct += 1
else:
print('Sorry the correct answer is', sum)
# HERE you ask if you want to play again or not, using a one line if
# you decide.
want_to_play = (True if 'y' == input('Continue (Y/N).lower():')
else False)
else:
print('Your final score is',correct,'/',count)

By initializing the variable c as "Y", the condition is met and the loop can be executed:
import random
correct=0
count=1
c = "Y"
while c == "Y":
count+=1
num1=random.randint(0,100)
num2=random.randint(0,100)
print(format(num1,'4d'))
print('+',num2)
answer=int(input('='))
sum=num1+num2
if answer==sum:
print('Congraulations!')
correct+=1
else:
print('Sorry the correct answer is',sum)
c=input('Continue (Y/N):')
c = c.upper()
else:
print('Your final score is',correct,'/',count)
I also added the method upper() to the Y/N input so the user can also type it in lowercase

Try to move as much of the processing as possible into the loop. The first "paragraph" of your code was basically a duplicate of the main-loop. By creating the continuation variable c so that it drops straight into the loop, most of that first block could be removed.
import random
correct=0
count=0
c = 'Y'
while c == "Y":
count+=1
num1=random.randint(0,100)
num2=random.randint(0,100)
print(format(num1,'4d'))
print('+',num2)
answer=int(input('='))
sum=num1+num2
if answer==sum:
print('Congratulations!')
correct+=1
else:
print('Sorry the correct answer is',sum)
c=input('Continue (Y/N):')
else:
print('Your final score is',correct,'/',count)
The two formula printing statements can also be reduced to a single one:
print(format(num1,'4d'))
print('+',num2)
could be
print( format(num1,'4d') + '+', num2 )
The variable sum could be removed, but it does make the code self-documenting, which is a good thing.

How to use a loop to rerun a basic calculator [closed]

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Edit the question to include desired behavior, a specific problem or error, and the shortest code necessary to reproduce the problem. This will help others answer the question.
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I'm very new to programming and as a student mathematics I'm going to follow Python programming. To be well prepared I thought I would already dive into the program, using some youtube videos and online material. Now the question.
I'm building a basic calculator. It works fine for the three functions I described in it. But if someone mistyped the function he or she wants to use (for example typed "multifly" i.s.o "multiply") it returns a sentence telling the user it has made a typo. I want to represent this line but also let it rerun from the start. That is, if you made a typo, to get back to line 1 asking the user what he wants to do.
I know I have to use a for or a while loop but I have no idea how actually to get it in, working. Please give me some advice :)
choice = input("Welcome to the basic calculator, please tell me if you want to add, substract or muliply: ")
if choice == "add":
print("You choose to add")
input_1 = float(input("Now give me your first number: "))
input_2 = float(input("And now the second: "))
result = (input_1 + input_2)
if (result).is_integer() == True:
print("Those two added makes " + str(int(result)))
else:
print("Those two added makes " + str(result))
elif choice == "substract":
print("You choose to substract")
input_1 = float(input("Now give me your first number: "))
input_2 = float(input("And now the second: "))
result = (input_1 - input_2)
if (result).is_integer() == True:
print("Those two substracted makes " + str(int(result)))
else:
print("Those two substracted makes " + str(result))
elif choice == "multiply":
print("You choose to multiply")
input_1 = float(input("Now give me your first number: "))
input_2 = float(input("And now the second: "))
result = (input_1 * input_2)
if (result).is_integer() == True:
print("Those two multiplied makes " + str(int(result)))
else:
print("Those two multiplied makes " + str(result))
else:
print("I think you made a typo, you'd have to try again.")

choices={"add","substract","multiply"}
while 1:
choice = input("Welcome to the basic calculator, please tell me if you want to add, substract or muliply: ")
if choice not in choices:
print("I think you made a typo, you'd have to try again.")
else: break
if choice == "add":
...
You do the input and verification part in the loop, then your code with the calculating. (Assuming the calculator only calculates once and then exits. If not, you could put the whole thing in another loop for more calculations, with maybe an exit command.)
In my example, verification is done with a set (choices) containing your possible commands, and checking for membership of the input.

OP = ("add", "subtract", "multiply")
while True:
choice = input("Pick an operation {}: ".format(OP))
if choice not in OP:
print("Invalid input")
else:
break
if choice == OP[0]:
#...

I need to figure out how to make my program repeat. (Python coding class)

I am a beginner student in a python coding class. I have the majority of the done and the program itself works, however I need to figure out a way to make the program ask if wants a subtraction or an adding problem, and if the user would like another question. I asked my teacher for assistance and he hasn't gotten back to me, so I'm simply trying to figure out and understand what exactly I need to do.
import random
x = int(input("Please enter an integer: "))
if x < 0:
x = 0
print('Negative changed to zero')
elif x == 0:
print('Zero')
elif x == 1:
print('Single')
else:
print('More')
maximum = 10 ** x;
maximum += 1
firstnum = random.randrange(1,maximum) # return an int from 1 to 100
secondnum = random.randrange(1, maximum)
compsum = firstnum + secondnum # adds the 2 random numbers together
# print (compsum) # print for troubleshooting
print("What is the sum of", firstnum, " +", secondnum, "?") # presents problem to user
added = int(input("Your answer is: ")) # gets user input
if added == compsum: # compares user input to real answer
print("You are correct!!!")
else:
print ("Sorry, you are incorrect")

You'll want to do something like this:
def foo():
print("Doing good work...")
while True:
foo()
if input("Want to do more good work? [y/n] ").strip().lower() == 'n':
break
I've seen this construct (i.e., using a break) used more often than using a sentinel in Python, but either will work. The sentinel version looks like this:
do_good_work = True
while do_good_work:
foo()
do_good_work = input("Want to do more good work? [y/n] ").strip().lower() != 'n'
You'll want to do more error checking than me in your code, too.

Asking users for input is straightforward, you just need to use the python built-in input() function. You then compare the stored answer to some possible outcomes. In your case this would work fine:
print('Would you like to test your adding or subtracting skills?')
user_choice = input('Answer A for adding or S for subtracting: ')
if user_choice.upper() == 'A':
# ask adding question
elif user_choice.upper() == 'S':
# ask substracting question
else:
print('Sorry I did not understand your choice')
For repeating the code While loops are your choice, they will repeatedly execute a statement in them while the starting condition is true.
while True: # Condition is always satisfied code will run forever
# put your program logic here
if input('Would you like another test? [Y/N]').upper() == 'N':
break # Break statement exits the loop
The result of using input() function is always a string. We use a .upper() method on it which converts it to UPPERCASE. If you write it like this, it doesn't matter whether someone will answer N or n the loop will still terminate.

If you want the possibility to have another question asked use a while loop and ask the user for an input. If you want the user to input whether (s)he want an addition or substraction you already used the tools to ask for such an input. Just ask the user for a string.