This code takes the users input and changes it to an integer and then checks if the int is between 0 and 10. I would also like this code to validate the users input against floats and non-numerical strings and loop back if the user enters a bad input. EX: user inputs 3.5 or "ten" and gets an Error and loops again.
pyramid = int(input("Please enter an integer between 0 and 10 to begin the sequence: "))
while pyramid < 0 or pyramid > 10:
print("That value is not in the correct range. Please try again.")
pyramid = int(input("Please enter an integer between 0 and 10 to begin the sequence: "))
I'd suggest to try to:
loop indefinitely (while 1)
Cast the input to a float, if this succeeds you can check if it has any decimals (pyramid % 1 != 0) and print the appropriate error in this case.
Cast the input to a integer and break the loop, if it is.
print an error that the input is not an integer
while 1:
str_in = input("Please enter an integer between 0 and 10 to begin the sequence: ")
try:
pyramid = float(str_in)
if(pyramid % 1 != 0):
print("That value is a float not an integer. Please try again.")
continue
except:
pass
try:
pyramid = int(str_in)
if pyramid >= 0 and pyramid <= 10:
break
except:
pass
print("That value is a string not an integer. Please try again.")
print("Your value is {}".format(pyramid))
Related
We want to create a program that prompts the user to enter a number between 1 and 10. As long as the number is out of range the program reprompts the user for a valid number. Complete the following steps to write this code.
a.Write a line of code the prompts the user for number between 1 and 10.
number = float(input("Enter a number between 1 and 10: "))
b. Write a Boolean expression that tests the number the user entered by the code in step "a." to determine if it is not in range.
x = (number > 10 or number < 1)
c.Use the Boolean expression created in step b to write a while loopthat executes when the user input is out of range. The body of the loop should tell the user that they enteredan invalid number and prompt them for a valid number again.
while x == True:
print("you printed an invalid number")
number = float(input("please enter the number again, this time between 1 and 10"))
d.Write the code that prints a message telling the user that they entered a valid number.
if x == False:
print("wow, you printed a number between 1 and 10!")
I answered the stuff for the question, but my problem is that whenever the user enters a wrong number on their first try and a correct number on their second try, the program still considers it as an invalid input. How do I fix this???
Rewrite this line in the while loop:
x = (number > 10 or number < 1)
so it becomes
while x == True:
print("you printed an invalid number")
number = float(input("please enter the number again, this time between 1 and 10"))
x = (number > 10 or number < 1)
This changes the value of x so it doesn't stay at True
If you use a while True construct, you won't need to repeat any code. Something like this:
LO, HI = 1, 10
while True:
input_ = input(f'Enter a number between {LO} and {HI}: ')
try:
x = float(input_)
if LO <= x <= HI:
print(f'Wow! You entered a number between {LO} and {HI}')
break
print(f'{input_} is not in range. Try again')
except ValueError:
print(f'{input_} is not a valid number. Try again')
Note:
When asking for numeric input from the user, don't assume that their input can always be converted properly. Always check
The following code snippet should do all you need:
number = float(input("Please input a number: "))
while (number > 10 or number < 0):
number = float(input("Error. Please input a new number: "))
Use an infinite loop, so that you can prompt for the input only once.
Use break to terminate the loop after the number in the correct range is entered.
Use f-strings or formatted string literals to print the message.
while True:
num = float(input('Enter a number between 1 and 10: '))
if 1 <= num <= 10:
print('Wow, you printed a number between 1 and 10!')
break
else:
print(f'You printed an invalid number: {num}!')
I'm processing integer inputs from a user and would like for the user to signal that they are done with inputs by typing in 'q' to show they are completed with their inputs.
Here's my code so far:
(Still very much a beginner so don't flame me too much)
def main():
print("Please enter some numbers. Type 'q' to quit.")
count = 0
total = 0
num=[]
num = int(input("Enter a number: "))
while num != "q":
num = int(input("Enter a number: "))
count += 1
total += num
del record[-1]
print (num)
print("The average of the numbers is", total / count)
main()
Any feedback is helpful!
You probably get a ValueError when you run that code. That's python's way of telling you that you've fed a value into a function that can't handle that type of value.
Check the Exceptions docs for more details.
In this case, you're trying to feed the letter "q" into a function that is expecting int() on line 6. Think of int() as a machine that only handles integers. You just tried to stick a letter into a machine that's not equipped to handle letters and instead of bursting into flames, it's rejecting your input and putting on the breaks.
You'll probably want to wrap your conversion from str to int in a try: statement to handle the exception.
def main():
num = None
while num != "q":
num = input("Enter number: ")
# try handles exceptions and does something with them
try:
num = int(num)
# if an exception of "ValueError" happens, print out a warning and keep going
except ValueError as e:
print(f'that was not a number: {e}')
pass
# if num looks like an integer
if isinstance (num, (int, float)):
print('got a number')
Test:
Enter number: 1
got a number
Enter number: 2
got a number
Enter number: alligator
that was not a number: invalid literal for int() with base 10: 'alligator'
Enter number: -10
got a number
Enter number: 45
got a number
Enter number: 6.0222
that was not a number: invalid literal for int() with base 10: '6.0222'
I'll leave it to you to figure out why 6.02222 "was not a number."
changing your code as little as possible, you should have...
def main():
print("Please enter some numbers. Type 'q' to quit.")
count = 0
total = 0
num=[]
num.append(input("Enter a number: "))
while num[-1] != "q":
num.append(input("Enter a number: "))
count += 1
try:
total += int(num[-1])
except ValueError as e:
print('input not an integer')
break
print (num)
print("The average of the numbers is", total / count)
main()
You may attempt it the way:
def main():
num_list = [] #list for holding in user inputs
while True:
my_num = input("Please enter some numbers. Type 'q' to quit.")
if my_num != 'q':
num_list.append(int(my_num)) #add user input as integer to holding list as long as it is not 'q'
else: #if user enters 'q'
print(f'The Average of {num_list} is {sum(num_list)/len(num_list)}') #calculate the average of entered numbers by divide the sum of all list elements by the length of list and display the result
break #terminate loop
main()
As part of a larger menu-driven program, I'd like to test user input to see if that input:
is an integer AND
if it is an integer, if it is within the range 1 to 12, inclusive.
number = 0
while True:
try:
number = int(input("Enter a whole number between 1 and 12 >>> "))
except ValueError:
print("Invlaid input, please try again >>> ")
continue
else:
if not (1<= number <=12):
print("Need a whole number in range 1-12 >>> ")
continue
else:
print("You selected:",number)
break
I'm using Python 3.4.3, and wanted to know if there's a more succinct (fewer lines, better performance, more "Pythonic", e.g.) way to achieve this? Thanks in advance.
You don't need anything bar one if in the try:
while True:
try:
number = int(input("Enter a whole number between 1 and 12 >>> "))
if 1 <= number <= 12:
print("You selected:", number)
break
print("Need a whole number in range 1-12 >>> ")
except ValueError:
print("Invlaid input, please try again >>> ")
Bad input will mean you go straight to the except, if the input is good and is in your accepted range, the print("You selected:", number) and will be executed then we break or else print("Need a whole number in range 1-12 >>> ") will be executed if is outside the range.
Your code looks pretty good to me. Minor fix-ups (spelling, indentation, unnecessary continues):
while True:
try:
number = int(input("Enter a whole number between 1 and 12 >>> "))
except ValueError:
print("Invalid input, please try again >>> ")
else:
if 1 <= number <= 12:
print("You selected: {}".format(number))
break
else:
print("Need a whole number in range 1-12 >>> ")
Use isdigit() to check for non-digit characters. Then you shouldn't need to catch the exception. There's only one if and it uses operator short-circuiting to avoid doing int(blah) if blah contains non-digits.
while True:
num_str = raw_input("Enter a whole number between 1 and 12 >>> ")
if num_str.isdigit() and int(num_str) in range(1,13):
print("You selected:",int(num_str))
break
else:
print("Need a whole number in range 1-12 >>> ")
I don't think you need a whole try/except block. Everything can be fit into a single condition:
number = raw_input("Enter a whole number between 1 and 12 >>> ")
while not (number.isdigit() and type(eval(number)) == int and 1<= eval(number) <=12):
number = raw_input("Enter a whole number between 1 and 12 >>> ")
print("You selected:",number)
So I need to gather input from a user but I have to ensure that they can only enter a number between 1 and 0 before the program accepts their input.
Here is what I've got so far:
def user_input():
try:
initial_input = float(input("Please enter a number between 1 and 0"))
except ValueError:
print("Please try again, it must be a number between 0 and 1")
user_input()
Could someone edit this or explain to me how I can add another rule as well as the ValueError so that it only accepts numbers between 1 and 0?
You can't check the value in the same line as you catch the exception.
Try this:
def user_input():
while True:
initial_input = input("Please enter a number between 1 and 0")
if initial_input.isnumeric() and (0.0 <= float(initial_input) <= 1.0):
return float(initial_input)
print("Please try again, it must be a number between 0 and 1")
EDIT
Removed the try/except and used isnumeric()instead.
If I'm asking for a user input of numbers which continues as long as an empty string is not entered, if an empty string is entered then the program ends.
My current code is:
n=0
while n != "":
n = int(input("Enter a number: "))
But obviously this isn't exactly what I want. I could remove the int input and leave it as a regular input, but this will allow all types of inputs and i just want numbers.
Do i ago about this a different way?
calling int() on an empty string will cause a ValueError so you can encapsulate everything in a try block:
>>> while True:
try:
n = int(input('NUMBER: '))
except ValueError:
print('Not an integer.')
break
NUMBER: 5
NUMBER: 12
NUMBER: 64
NUMBER:
not a number.
this also has the added benefit of catching anything ELSE that isn't an int.
I would suggest using a try/except here instead.
Also, with using a try/except, you can instead change your loop to using a while True. Then you can use break once an invalid input is found.
Also, your solution is not outputting anything either, so you might want to set up a print statement after you get the input.
Here is an example of how you can put all that together and test that an integer is entered only:
while True:
try:
n = int(input("Enter a number: "))
print(n)
except ValueError:
print("You did not enter a number")
break
If you want to go a step further, and handle numbers with decimals as well, you can try to cast to float instead:
while True:
try:
n = float(input("Enter a number: "))
print(n)
except ValueError:
print("You did not enter a number")
break