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Asking the user for input until they give a valid response
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Closed 2 years ago.
My goal is creating a factorial program in python that asks infinite user input to find factorial of the number user type, until I would like to quit the program. But there is probably discrepancy between the lines of the code to works for exit the program and integer numbers below it.
1) I tried to solve this to not write int(input) I wrote just
input('Enter a number. Type exit to stop:> ')
both above or below the while True statement but it didn't work.
2) I also want to use lower() function to quit the program but when I use it, the discrepancy happens again because I ask user input for an integer but when I turn it to a normal input and type it the above while True statement, problem occurs.
3) Also I want to user input as a number with using that isdigit() function tried to use like this but it didn't work well:
factorial = 1
user_input = input('Enter a number: ')
while user_input.lower() != 'exit':
while not user_input.isdigit():
print('This is not a number. Please try again:> ')
user_input = int(input('Try again:> '))
user_input = int(input('Enter a new number: '))
.
.
.
and this, too didn't work
My actual code is this:
while True:
factorial = 1
user_input = int(input('Enter a number. Type exit to stop:> '))
if user_input == 'exit':
print('You are quitting the program')
break
elif user_input < 0:
print("Sorry, factorial does not exist for negative numbers")
elif user_input == 0:
print("The factorial of 0 is 1")
else:
for i in range(1, user_input + 1):
factorial = factorial * i
print("The factorial of", user_input, "is", factorial)
The program works like this:
Enter a number. Type exit to stop:> 4
The factorial of 4 is 24
Enter a number. Type exit to stop:> 5
The factorial of 5 is 120
Enter a number. Type exit to stop:> 6
The factorial of 6 is 720
and when I type 'exit' to quit from program I am receiving this kind of error:
Traceback (most recent call last):
File "E:/Kodlar/Python/Taslak projeler/taslak177.py", line 5, in <module>
user_input = int(input('Enter a number. Type exit to stop:> '))
ValueError: invalid literal for int() with base 10: 'exit'
As you can see, code blocks work instead of quitting the program with user input. How can I fix this?
Can anyone help? Thanks already!
Edit: I reorganized the code and it works perfectly fine. Thanks for your all responses!
while True:
user_input = input("Enter a number:> ")
if user_input == "exit":
print('You are quitting the program...')
break
else:
try:
factorial = 1
user_input = int(user_input)
if user_input < 0:
print("Sorry, factorial does not exist for negative numbers")
elif user_input == 0:
print("The factorial of 0 is 1")
else:
for i in range(1, user_input + 1):
factorial = factorial * i
print(f'The factorial of {user_input} is {factorial}')
except ValueError:
print("Please provide a valid number")
You should check if the input is exit before converting it to int and if so, break the loop.
Try this instead:
while True:
user_input = input("Enter a number:")
if user_input == "exit":
print('You are quitting the program')
break
else:
try:
user_number = int(user_input)
if user_number < 0:
print("Sorry, factorial does not exist for negative numbers")
elif user_number == 0:
print("The factorial of 0 is 1")
else:
# calc factorial here
except ValueError:
print("Please provide a valid number")
Your program get int inputs,
user_input = int(input('Enter a new number: '))
try this instead, get string input
user_input = input('Enter a new number: ')
and convert it into int later
user_input = int(user_input)
Because you are casting the user's response (a string) into an int in both cases.
user_input = int(input('Enter a new number: '))
and later
user_input = int(input('Enter a number. Type exit to stop:> '))
Perhaps try a little tweak:
while True:
factorial = 1
user_input = input('Enter a number. Type exit to stop:> ')
if user_input.lower().strip() == 'exit':
print('You are quitting the program')
break
elif user_input.isnumeric() and user_input < 0:
print("Sorry, factorial does not exist for negative numbers")
elif user_input.isnumeric() and user_input == 0:
print("The factorial of 0 is 1")
elif user_input.isnumeric():
for i in range(1, user_input + 1):
factorial = factorial * i
print("The factorial of", user_input, "is", factorial)
else:
print("Please enter an integer or 'exit'")
You could also wrap another if so you don't duplicate the isnumeric() tests
Related
I'm trying to make a program in python so that when I input a number from 1 to 10, a specific set of program goes on and asks for another number from 1 to 10 and runs another program, until I enter 0(zero) and the program stops.
So my guess was using a while loop but it didn't quite work out.
user_input = input()
user_input = int(user_input)
while user_input != 0:
(program)
else:
quit()
Try this:
user_input = int(input())
while user_input != 0:
(program)
user_input = int(input())
quit()
With your current code you only ask for input once so the loop won't end. This way you can input a new number after every iteration.
Your current program only asks once and then the loop keeps repeating. You need to keep asking for input inside the loop.
def program():
print("Executing Task....")
user_input = int(input())
while user_input != 0:
program()
user_input = int(input())
printf("Program Terminated")
Here it is:
def program():
pass
user_input = int(input())
while user_input:
program()
user_input = int(input())
quit(0)
A different way using iter with a sentinel:
def program(number):
if number < 0 or number > 10:
print('Invalid number:', number)
else:
print('Valid number:', number)
def quit():
print('quitting')
def get_number():
return int(input('Enter a number from 1 to 10: '))
for number in iter(get_number, 0):
program(number)
else:
quit()
not_zero = True
while not_zero:
num = int(input("Enter a number: "))
if num == 0:
not_zero = False
you can stop your loop using a boolean value.
I am working on a small coding challenge which takes user input. This input should be checked to be a digit. I created a "try: ... except ValueError: ..." block which checks once whether the input is a digit but not multiple times. I would like it to basically checking it continuously.
Can one create a while-exception loop?
My code is the following:
try:
uinput = int(input("Please enter a number: "))
while uinput <= 0:
uinput = int(input("Number is negative. Please try again: "))
else:
for i in range(2, uinput):
if (uinput % i == 0):
print("Your number is a composite number with more than
one divisors other than itself and one.")
break
else:
print(uinput, "is a prime number!")
break
except ValueError:
uinput = int(input("You entered not a digit. Please try again: "))
flag = True
while flag:
try:
uinput = int(input("Please enter a number: "))
while uinput <= 0:
uinput = int(input("Number is negative. Please try again: "))
else:
flag=False
for i in range(2, uinput):
if (uinput % i == 0):
print("Your number is a composite number with more than one divisors other than itself and one.")
break
else:
print(uinput, "is a prime number!")
break
except ValueError:
print('Wrong input')
Output :
(python37) C:\Users\Documents>py test.py
Please enter a number: qwqe
Wrong input
Please enter a number: -123
Number is negative. Please try again: 123
123 is a prime number!
I add flag boolean to not make it repeat even when the input is correct and deleted input in except because it would ask 2 times.
If you press Enter only, the loop terminated:
while True:
uinput = input("Please enter a number: ")
if uinput.strip()=="":
break
try:
uinput=int(uinput)
except:
print("You entered not a digit. Please try again")
continue
if uinput<=0:
print("Not a positive number. Please try again")
continue
for i in range(2, uinput):
pass; # put your code here
I'm trying to make a simple number guesser program, it works pretty well however if I enter 'a' twice instead of a valid int it crashes out. Can someone explain what I'm doing wrong here.
import random
def input_sanitiser():
guess = input("Please enter a number between 1 and 10: ")
while True:
if type(guess) != int:
guess = int(input("That isn't a number, try again: "))
elif guess not in range (1,11):
guess = int(input("This is not a valid number, try again: "))
else:
break
def main():
number = random.randrange(1,10)
guess = 0
input_sanitiser()
while guess != number:
if guess < number:
print("This number is too low!")
input_sanitiser()
if guess > number:
print("This number is too high!")
input_sanitiser()
else:
break
print ("Congratulations, you've guessed correctly")
if __name__ == "__main__":
main()
You want to check the input before trying to convert it to int:
int(input("This is not a valid number, try again: "))
I would write:
while True:
try:
guess = int(input("This is not a valid number, try again: "))
except ValueError:
pass
else:
break
Side note: the code isn't working as expected:
def main():
number = random.randrange(1,10)
guess = 0
input_sanitiser() # <<<<<<<<<<
while guess != number:
Note that input_sanitiser does not modify the variable guess in main, you need some other way round, like processing the input then returning the result from input_sanitiser, like this:
def input_sanitiser():
guess = input("Please enter a number between 1 and 10: ")
while True:
try:
guess = int(input("This is not a valid number, try again: "))
except ValueError:
continue # keep asking for a valid number
if guess not in range(1, 11):
print("number out of range")
continue
break
return guess
def main():
number = random.randrange(1,10)
guess = input_sanitiser()
while guess != number:
if guess < number:
print("This number is too low!")
guess = input_sanitiser()
if guess > number:
print("This number is too high!")
guess = input_sanitiser()
else:
break
print ("Congratulations, you've guessed correctly")
I'm a newbie in Python3 coding and I have a problem here.
In line 14, I intended to end this program by printing "Thank you! Goodbye" at the part where you answer "n" to "try again?". However, it turned out that I would start all over again even if I've inserted "break" under it. Now, the only solution I can come up is to end the whole program with sys.exit(0), but I don't consider it an ideal solution since it just closes the whole program down.
import sys
while True:
x=int(input("Enter the coins you expected="))
f=int(input("Enter first coin="))
while f!=1 and f!=5 and f!=10 and f!=25:
print("invalid number")
f=int(input("Enter first coin="))
if x>f:
while x>f:
n=input("Enter next coin=")
if not n:
print("Sorry-you only entered",f,"cents")
again=input("Try again (y/n)?=")
if again=="y":
True
elif again=="n":
print("Thank you, goodbye!")
sys.exit(0)
break
while int(n)!=1 and int(n)!=5 and int(n)!=10 and int(n)!=25:
print("invalid number")
n=input("Enter next coin=")
f=f+int(n)
Replace your whole code with this:
import sys
Stay = True
while Stay:
x = int(input("Enter the coins you expected = "))
f = int(input("Enter first coin = "))
while f != 1 and f != 5 and f != 10 and f != 25:
f = int(input("Invalid number entered./nEnter first coin = "))
while x > f and Stay:
n = input("Enter next coin = ")
if not n:
print("Sorry, you only entered " + str(f) + " cents")
again = input("Try again (y/n)?=")
if again == "n":
print("Thank you, goodbye!")
Stay = False
if Stay:
n = int(n)
while n != 1 and n != 5 and n != 10 and n != 25:
print("Invalid number entered.")
n = int(input("Enter next coin = "))
f += n
I made your code more readable and fixed your problem by using a Boolean flag (Stay). This basically means that the program runs while Stay is True, and Stay becomes False when the user enters 'n'.
I need to be able to prompt the user to enter an empty string so it can check if the answer is correct. but every time I do that I can error saying invalid literal for int()
so I need to change my user_input so it can accept int() and strings(). how do I make that possible ?
# program greeting
print("The purpose of this exercise is to enter a number of coin values")
print("that add up to a displayed target value.\n")
print("Enter coins values as 1-penny, 5-nickel, 10-dime,and 25-quarter.")
print("Hit return after the last entered coin value.")
print("--------------------")
#print("Enter coins that add up to 81 cents, one per line.")
import sgenrand
#prompt the user to start entering coin values that add up to 81
while True:
total = 0
final_coin= sgenrand.randint(1,99)
print ("Enter coins that add up to", final_coin, "cents, on per line")
user_input = int(input("Enter first coin: "))
if user_input != 1 and user_input!=5 and user_input!=10 and user_input!=25:
print("invalid input")
else:
total = total + user_input
while total <= final_coin:
user_input = int(input("Enter next coin:"))
if user_input != 1 and user_input!=5 and user_input!=10 and user_input!=25:
print("invalid input")
else:
total = total + user_input
if total > final_coin :
print("Sorry - total amount exceeds", (final_coin))
elif total < final_coin:
print("Sorry - you only entered",(total))
else:
print("correct")
goagain= input("Try again (y/n)?:")
if goagain == "y":
continue
elif goagain == "n":
print("Thanks for playing ... goodbye!" )
break
Store the value returned by input() in a variable.
Check that the string is not empty before calling int().
if it's zero, that's the empty string.
otherwise, try int()ing it.