I have two conditions in my story using the question that I made to ask the users.
The first condition is true and second condition is false. In the first condition, if it's true, I want the program to finish. In the second condition, if it's false, I want to loop back to the question that I asked.
I have the following code, but so far it will loop back if the first is true and the second is false.
Any ideas?
invalid = ValueError
def age():
user_input = input("Enter your Age")
try:
val = int(user_input)
print("Input is an integer number. Number = ", val)
except ValueError:
print("No.. input is not a number. It's a string")
while invalid:
age()
I'd make things easier and just use a boolean as your continuation flag, rather than the ValueError:
ask_again = True
def age():
user_input = input("Enter your Age")
try:
val = int(user_input)
print("Input is an integer number. Number = ", val)
ask_again = False
except ValueError:
print("No.. input is not a number. It's a string")
while ask_again:
age()
Does that look like what you want? Happy Coding!
The simplest way to execute a while loop is to just use while True and then break from the loop once the condition is satisfied
def age():
while True:
try:
val = int(input("Enter your Age"))
print("Input is an integer number. Number = ", val)
break
except ValueError:
print("No.. input is not a number. It's a string")
age()
Related
I am trying to create a function that would take a user inputted number and determine if the number is an integer or a floating-point depending on what the mode is set to. I am very new to python and learning the language and I am getting an invalid syntax error and I don't know what to do. So far I am making the integer tester first. Here is the code:
def getNumber(IntF, FloatA, Msg, rsp):
print("What would you like to do?")
print("Option A = Interger")
print("Option B = Floating Point")
Rsp = int(input("What number would like to test as an interger?"))
A = rsp
if rsp == "A":
while True:
try:
userInput = int(input("What number would like to test as an interger"))
except ValueError as ve:
print("Not an integer! Try again.")
continue
else:
return userInput
break
The problem with the code you shared is :
The syntax error you mentioned is probably because the except clause has to be at the same indentation level as try, and same for if and else of the same if/else clause. All the code in the function should be indented 1 level too. Python requires all this to identify blocks of code.
You don't need to give 4 arguments to the getNumber() function if you're not using them. This isn't really a problem, but you'll have to pass it some 4 values each time you call it (for example getNumber(1,2,3,4) etc...) to avoid missing argument errors; and these won't matter because you're not doing anything with the given values inside the function - so it's a little wasted effort. I rewrote it in the example below so that you aren't dealing with more variables than you need - it makes the code clearer/simpler.
You also don't need break after a return statement because the return will exit the enitre function block, including the loop.
Try this and see if it makes sense - I've changed a lot of the code :
def getNumber():
while True:
try:
userInput = int(input("What number would like to test as an integer ? "))
return userInput
except ValueError as ve:
print("Not an integer! Try again.")
continue
print("What would you like to do?")
print("Option A = Interger")
print("Option B = Floating Point")
chosen_option = input()
if chosen_option == 'A':
integer_received = getNumber()
print(integer_received, "was an int !")
else:
print("You did not choose 'A' or 'B'")
To determine whether a number is a float or integer, you can use this approach
float is nothing but the integer with floating-point(.).
to determine this we first need to convert it to string and find does it contain a point or not.
number = input("Enter a numbeer\n")
if number.find(".") == -1:
# find will return -1 when the value is not in string
print("it is integer")
else:
print("it is float")
I have looked on here for an idea in order to loop the user's invalid input and I haven't found one.
Ok, I know how to loop the first_number and second_number together but I am a bit confused on how to loop the separately if need be. So if the user inputs a bad second_number and loops that instead of the whole thing again.
I've attached the part I need help with (yes this is a school assignment):
def get_numbers():
first_number = 0.0
second_number = 0.0
while True:
try:
first_number = float(input("Please enter the first number: "))
second_number = float(input("Please enter the second number: "))
except ValueError:
print("Not a valid input, please try again.")
continue
return first_number, second_number
To use 1 loop, you may need to recognize the differences:
You use 2 simple variables for the results though you could use 1 list.
The input string is almost the same except for "first" and "second" words.
Concept:
First you want a list.
Then use a for loop to use "first", then "second" words.
Then use a while loop which processes the inputs and uses the list to extend with the replies. Use break to get out of the while loop after each good reply.
def get_numbers():
result = []
for item in ("first", "second"):
while True:
try:
number = float(input("Please enter the " + item + " number: "))
except ValueError:
print("Not a valid input, please try again.")
else:
result += [number]
break
return tuple(result)
Returning as a tuple as you have done.
First, you want to use two loops, one for each number, and break if the input is valid instead of continuing if it's invalid. Then you need to move the return statement to immediately following the second loop.
By the way, you can leave out the initial assignments to first_number and second_number (commented out below). Assigning them to float(input(...)) is enough.
def get_numbers():
#first_number = 0.0
#second_number = 0.0
while True:
try:
first_number = float(input("Please enter the first number: "))
break
except ValueError:
print("Not a valid input, please try again.")
while True:
try:
second_number = float(input("Please enter the second number: "))
break
except ValueError:
print("Not a valid input, please try again.")
return first_number, second_number
I am very new to programming, please advise me if my code is correct.
I am trying to 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.
largest = None
smallest = None
while True:
num = input("Enter a number: ")
if num == 'done':
break
try:
fnum = float(num)
except:
print("Invalid input")
continue
lst = []
numbers = int(input('How many numbers: '))
for n in range(numbers):
lst.append(num)
print("Maximum element in the list is :", max(lst), "\nMinimum element in the list is :", min(lst))
Your code is almost correct, there are just a couple things you need to change:
lst = []
while True:
user_input = input('Enter a number: ')
if user_input == 'done':
break
try:
lst.append(int(user_input))
except ValueError:
print('Invalid input')
if lst:
print('max: %d\nmin: %d' % (max(lst), min(lst)))
Also, since you said you're new to programming, I'll explain what I did, and why.
First, there's no need to set largest and smallest to None at the beginning. I actually never even put those values in variables because we only need them to print them out.
All your code is then identical up to the try/except block. Here, I try to convert the user input into an integer and append it to the list all at once. If any of this fails, print Invalid input. My except section is a little different: it says except ValueError. This means "only run the following code if a ValueError occurs". It is always a good idea to be specific when catching errors because except by itself will catch all errors, including ones we don't expect and will want to see if something goes wrong.
We do not want to use a continue here because continue means "skip the rest of the code and continue to the next loop iteration". We don't want to skip anything here.
Now let's talk about this block of code:
numbers = int(input('How many numbers: '))
for n in range(numbers):
lst.append(num)
From your explanation, there is no need to get more input from the user, so none of this code is needed. It is also always a good idea to put int(input()) in a try/except block because if the user inputs something other than a number, int(input()) will error out.
And lastly, the print statement:
print('max: %d\nmin: %d' % (max(lst), min(lst)))
In python, you can use the "string formatting operator", the percent (%) sign to put data into strings. You can use %d to fill in numbers, %s to fill in strings. Here is the full list of characters to put after the percent if you scroll down a bit. It also does a good job of explaining it, but here are some examples:
print('number %d' % 11)
x = 'world'
print('Hello, %s!' % x)
user_list = []
while True:
user_input = int(input())
if user_input < 0:
break
user_list.append(user_input)
print(min(user_list), max(user_list))
I'm writing a basic program in IDLE with a menu choice with options from 1 to 4.
If a user input anything else then a number, it gives a ValueError: invalid literal for int() with base 10: 'a'
How can I check if the input is not a letter, and if it is, print a error message of my own?
def isNumber (value):
try:
floatval = float(value)
if floatval in (1,2,3,4):
return True
else:
return False
except:
return False
number_choice = input('Please choose a number: 1, 2, 3 or 4.\n')
while isNumber(number_choice) == False:
number_choice = input('Please choose a number: 1, 2, 3 or 4.\n')
else:
print('You have chosen ' + number_choice + '.\n')
This will check if the number is 1,2,3 or 4 and if not will ask the user to input the number again until it meets the criteria.
I am slightly unclear on whether you wish to test whether something is an integer or whether it is a letter, but I am responding to the former possibility.
user_response = input("Enter an integer: ")
try:
int(user_response)
is_int = True
except ValueError:
is_int = False
if is_int:
print("This is an integer! Yay!")
else:
print("Error. The value you entered is not an integer.")
I am fairly new to python, so there might very well be a better way of doing this, but that is how I have tested whether or not input values are integers in the past.
isalpha() - it is a string method which checks that whether a string entered is alphabet or words(only alphabets, no spaces or numeric) or not
while True:
user_response = input("Enter an integer : ")
if user_response.isalpha():
print("Error! The value entered is not an integer")
continue
else:
print("This is an integer! Yay!")
break
This program is having infinite loop i.e. until you enter an integer this program will not stop. I have used break and continue keyword for this.
I need a function to check that different user input variables are integers.
The results should be confirmed to the user at the end.
The check works in that it keeps looping until integer is typed in,
but cannot get the results to display...
def chkint(msg):
while True:
try:
n = input(msg)
return(int(n))
except ValueError:
print("Please enter an actual integer.")
number1 = input (chkint("Please enter first value:"))
number2 = input (chkint("Please enter second value:"))
results = (number1, number2)
print ("I have accepted: " + str (results))
No answer, so I just played about with this and hey presto, it works...
def chkint(msg):
while 1:
try:
n = input(msg)
return(int(n))
except ValueError:
print("Please enter an integer.")
number1 = chkint("Please enter first value:")
number2 = chkint("Please enter second value:")
results = [number1, number2]
print ("I have accepted: " + str (results))
Casting it to int() in a try: block is a good way to check a number. In your original attempt you were asking for an input whose message relied on further input.
Simplified version of the mistake:
def getMessage():
return input() # this asks the user what to ask the user for
input(getMessage()) # this waits for getmessage to finish before asking the user
Removing the input() statements was the easiest fix, as you did.
But a more readable fix would be to make chkint(msg) do nothing but return true or false based on whether or not the string was a number, like this
def chkint(msg): # returns true if the string can be converted, false otherwise
try:
int(msg)
except ValueError:
return False
return True