print("Give me two numbers, I'll sum them")
print("Enter 'q' to quit")
while True:
num1 = input("Please, enter a number here: ")
if num1 == 'q':
break
num2 = input("Please, enter a number here: ")
if num2 == 'q':
break
try:
sum = int(num1) + int(num2)
except ValueError:
print("'q' entered, program exit 0")
else:
print(sum)
Hi the above program in Python3 works fine when numbers are inputted.
But when I input q, it just exits with no exception.
May you please assist me with this issue?
Thank you very much indeed.
the break statement will exit your loop without printing anything, because your try-except test is after breaks.
Here is what you can do:
print("Give me two numbers, I'll sum them")
print("Enter 'q' to quit")
while True:
num1 = input("Please, enter a number here: ")
if num1 == 'q':
print("'q' entered, program exit 0")
break
num2 = input("Please, enter a number here: ")
if num2 == 'q':
print("'q' entered, program exit 0")
break
if not num1.isdecimal() or not num2.isdecimal():
print('Wrong input, please enter decimal numbers !')
continue
sum = int(num1) + int(num2)
print(sum)
Related
My task is:
Write a program that repeatedly prompts a user for integer numbers until the user enters 'done'. Once 'done' is entered, print out the largest and smallest of the numbers. If the user enters anything other than a valid number catch it with a try/except and put out an appropriate message and ignore the number. Enter 7, 2, bob, 10, and 4 and match the output below.
I want to ignore any invalid integer and print 'Invalid Output' message after calculating the maximum and minimum number.But it always prints the invalid message right after user input. How am i supposed to solve this?
Thanks in advance.
Code:
largest = None
smallest = None
while True:
num = input("Enter a number: ")
if num == "done":
break
try:
n = int(num)
except:
print('Invalid input')
if largest is None:
largest=n
elif n>largest:
largest=n
elif smallest is None:
smallest=n
elif n<smallest:
smallest=n
print("Maximum", largest)
print('Minimum', smallest)
You could store that invalid output as a boolean variable
largest = None
smallest = None
is_invalid=False
while True:
num = input("Enter a number: ")
if num == "done":
break
try:
n = int(num)
except:
is_invalid=True
if largest is None:
largest=n
elif n>largest:
largest=n
elif smallest is None:
smallest=n
elif n<smallest:
smallest=n
print("Maximum", largest)
print('Minimum', smallest)
if is_invalid:
print('Invalid Input')
This can be one solution
largest = None
smallest = None
errors = False
while True:
num = input('Please type a number : ')
if num == 'done':
break
try:
number = int(num)
#your logical operations and assignments here
except ValueError:
errors = True
continue
if errors:
print('Invalid input')
else:
print('Your Results')
Hope this helps :)
If you are familiar with lists,you can use list to solve your problem effectively,
here is a modified version of your code which uses list,
num1=[]
while True:
num = input("Enter a number: ")
num1.append(num)
if num == "done":
break
for i in num1:
try:
i = int(i)
except:
print('Invalid input:',i)
num1.remove(i)
print("Maximum", max(num1))
print('Minimum', min(num1))
output:
Enter a number: 34
Enter a number: 57
Enter a number: 89
Enter a number: ds
Enter a number: 34
Enter a number: do
Enter a number: done
Invalid input: ds
Invalid input: do
Maximum 89
Minimum 34
hope this helps,let me know if anything is incorrect.
Try removing the (int) in try block. Because you want only integer input in try block so if does not satisfy the condition of integer input it will execute the except block.
Your code should look like this in try block:
try:
n = num
When the user enters nothing, it is supposed to loop back and ask the question again. It performs correctly with every other type of input.
Here is the code:
string = ""
def str_analysis(string):
while True:
if string.isdigit():
if int(string) > 99:
print(str(string)+ " is a pretty big number!")
break
else:
print(str(string)+ " is a smaller number than expected")
break
elif string.isalpha():
print(string + " is an alphabetical character")
break
elif string == "":
print("")
else:
print(string + " is a surprise! It's neither all alpha nor all digit characters!")
break
print(str_analysis(input("Enter word or integer: ")))
There are a few things in your code that make no sense.
1:
print(str_analysis(input("Enter word or integer: ")))
You are trying to print the output of a function that has no return value
2:
It cant loop back and ask the question again, because the input is not taken inside of the function.
3:
If the string is empty you dont break the code but constantly print newlines.
Here is some code wich I think should do what you wanted:
def str_analasys():
while True:
string = input("Enter word or integer: ")
if string == '':
continue
elif string.isdigit():
if int(string) > 99:
print(str(string)+ " is a pretty big number!")
else:
print(str(string)+ " is a smaller number than expected")
elif string.isalpha():
print(string + " is an alphabetical character")
else:
print(string + " is a surprise! It's neither all alpha nor all digit characters!")
break
str_analasys()
This because when string is empty you are just printing an empty ""
elif string == "":
print("")
# break it here or take input again
# break or string = input()
I am trying to make a simple guess the number program in python. When I run this code,an error generates saying that,"local variable 'chance' referenced before assignment". I looked up for a solution on internet but I could not rectify my error. Please help with this problem. How can I use the variable globally which is declared inside a function?
I am beginner in programming, so plese explain in simple words.
Here is the code..
Since I am a beginner,I will be pleased if my code can be rectified
import random
def Random():
chance = 3
number = random.randint(0,20)
return chance
return number
def main():
while chance > 0:
UserInput = int(input('Guess the number: '))
if UserInput == number:
print('You have guesses the secret number!')
elif UserInput > 20 and UserInput < 0:
print('Your guess is out of range!\n Try again!')
else:
chance -= 1
if chance == 1:
print('You are out of chances!')
print('Wrong Guess!\nTry again!')
print(f'You have {chance} chances left!')
Random()
main()
playAgain = input('Want to play again? ')
if playAgain == 'yes' or 'YES' or 'Yeah' or 'yeah':
Random()
main()
else:
print('Thanks for playing!')
You can return a list or a tuple to the outside word:
import random
def example():
chance = 3
number = random.randint(0,20)
return (chance, number) # return both numbers as a tuple
chance, randNr = example() # decomposes the returned tuple
print(chance, randNr)
prints:
3, 17
There are more bugs in your program, f.e.:
if playAgain == 'yes' or 'YES' or 'Yeah' or 'yeah':
is always True and you'll never be able to leave the game. Better would be
if playAgain.lower() in {'yes', 'yeah'}:
etc.
Here is a working example for your programs purpose:
import random
while True:
chances = 3
number = random.randint(0,20)
while chances > 0:
guess = int(input("Guess number: "))
if guess == number:
print("Correct")
break
else:
chances -= 1
print("Wrong, ", chances, " more tries to get it right.")
if chances == 0:
print ("You failed")
if not input("Play again? ")[:1].lower() == "y":
break
print("Bye.")
Read about tuples
Output:
Guess number: 1
Wrong, 2 more tries to get it right.
Guess number: 4
Correct
Play again? y
Guess number: 1
Wrong, 2 more tries to get it right.
Guess number: 2
Wrong, 1 more tries to get it right.
Guess number: 3
Wrong, 0 more tries to get it right.
You failed
Play again? n
Bye.
import random
def Random():
chance = 3
number = random.randint(0,20)
main(chance,number)
def main(chance,number):
while chance > 0:
UserInput = int(input('Guess the number: '))
if UserInput == number:
print('You have guesses the secret number!')
elif UserInput > 20 and UserInput < 0:
print('Your guess is out of range!\n Try again!')
else:
chance -= 1
if chance == 1:
print('You are out of chances!')
print('Wrong Guess!\nTry again!')
print('You have',chance,'chances left!')
Random()
playAgain = input('Want to play again? ')
if playAgain == 'yes' or 'YES' or 'Yeah' or 'yeah':
Random()
else:
print('Thanks for playing!')
Trying to exit the program by importing sys.exit(), break and ask == False, but nothing works. Full code here
#import sys
def body_cycle(*args):
if option == "/":
error_func_for_zero(first_number, second_number, option)
print(division_option(first_number, second_number))
print()
print(begin)
def error_func_for_zero(*args):
try:
first_number / 0 or second_number / 0
except ZeroDivisionError:
print("YOU CANNOT DIVIDE BY ZERO!")
print(begin)
def division_option(*args):
return first_number / second_number
begin = " "
while begin:
print("Hello, I am calculator. ")
print("Please, enter your numbers (just integers) ! ")
print()
first_number = int(input("First number: "))
print()
second_number = int(input("Second number: "))
print()
option = input("Remember: you can't divide by zero.\nChoose your option (+, -, *, /): ")
print(body_cycle(first_number, second_number, option))
ask = " "
while ask:
exit_or_continue = input("If you want continue press 'Y', 'y'. For break press 'N' or 'n'? \nChoice: ")
if exit_or_continue == "Y" or "y":
print("OK")
elif exit_or_continue == "N" or "n":
#break
ask == False
else:
print("Break program. ")
break
You just want to replace ask == False by ask = False.
In addition, you could really use a simpler code structure. The whole thing before begin can be compacted down to:
def operation(a, b, option):
if option == "+":
return a + b
elif option == "-":
return a - b
elif option == "*":
return a * b
elif option == "/":
try:
return a / b
except ZeroDivsionError
return "YOU CANNOT DIVIDE BY ZERO!"
The rest can be put in a single loop instead of two, like so:
print("Hello, I am calculator. ")
while True:
print("Please, enter your numbers (just integers) ! ")
print()
first_number = int(input("First number: "))
print()
second_number = int(input("Second number: "))
print()
option = input("Remember: you can't divide by zero.\n
Choose your option (+, -, *, /): ")
# Result.
print(operation(first_number, second_number, option))
exit_or_continue = input("If you want continue press 'Y', 'y'. For break press 'N' or 'n'.\n
Choice: ").lower()
if exit_or_continue == "y":
print("OK")
elif exit_or_continue == "n":
break
I'm trying to get a guessing game with the user input as an answer and if the user type exit the game will show the amount of time the player tried but the program won't run because it can only take either interger or string type.
import random
while True:
number = random.randint(1,9)
guess = int(input('guess the number: '))
time = 0
time += 1
if guess == number:
print('you guessed correct')
elif guess < number:
print('your guessed is lower than the actual number')
elif guess > number:
print('your guessed is higher than the actual number')
elif guess == 'exit':
print(time)
break
something like this
import random
time = 0
number = random.randint(1,9)
while True:
guess = input('guess the number: ')
time += 1
if guess == "exit":
print(time)
break
elif int(guess) < number:
print('your guessed is lower than the actual number')
elif int(guess) > number:
print('your guessed is higher than the actual number')
elif int(guess) == number:
print('you guessed correct')
print(time)
break
note that time and number have to be initializate outside the while loop, because if not, we would get different random numbers for each iteration, and also time would be initializate to 0 each time.
You can test the input as a string first before converting it to an integer:
while True:
response = input('guess the number: ')
if response == 'exit':
break
guess = int(response)
# the rest of your code testing guess as a number