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How to exit loop when input is nothing

I'm trying to work out the average of numbers that the user will input. If the user inputs nothing (as in, no value at all) I want to then calculate the average of all numbers that have been input by the user upto that point. Summing those inputs and finding the average is working well, but I'm getting value errors when trying to break the loop when the user inputs nothing. For the if statement I've tried
if number == ''
First attempt that didn't work, also tried if number == int("")
if len(number) == 0
This only works for strings
if Value Error throws up same error
Full code below
sum = 0
while True :
number = int(input('Please enter the number: '))
sum += number
if number == '' :
break
print(sum//number)
Error I'm getting is
number = int(input('Please enter the number: '))
ValueError: invalid literal for int() with base 10:>
Any help much appreciated!
EDIT: Now getting closer thanks to the suggestions in that I can get past the problems of no value input but my calculation of average isn't working out.
Trying this code calculates fine but I'm adding the first input twice before I move to the next input
total = 0
amount = 0
while True :
user_input = input('Please enter the number: ')
try:
number = int(user_input)
total = number + number
amount += 1
except:
break
total += number
print(total/amount)
Now I just want to figure out how I can start the addition from the second input instead of the first.

sum = 0
while True :
number = input('Please enter the number: '))
if number == '' :
break
sum += int(number)
print(sum//number)
try like this
the issue is using int() python try to convert input value to int. So, when its not integer value, python cant convert it. so it raise error. Also you can use Try catch with error and do the break.

You will always get input as a string, and if the input is not a int then you cant convert it to an int. Try:
sum = 0
while True :
number = input('Please enter the number: ')
if number == '' :
break
sum += int(number)
print(sum//number)

All of the answers dont work since the print statement referse to a string.
sum = 0
while True :
user_input = input('Please enter the number: ')
try:
number = int(user_input)
except:
break
sum += number
print(sum//number)
including a user_input will use the last int as devisor.
My answer also makes sure the script does not crash when a string is entered.

The user has to always input something (enter is a character too) for it to end or you will have to give him a time limit.
You can convert character into int after you see it isn't a character or
use try & except.
sum = 0
i = 0
while True :
try:
number = int(input('Please enter the number: '))
except ValueError:
break
i += 1
sum += number
try:
print(sum/number)
except NameError:
print("User didn't input any number")
If you try to convert a character into int it will show ValueError.
So if this Error occurs you can break from the loop.
Also, you are trying to get the average value.
So if a user inputs nothing you get NameError so you can print an Error message.

How to Input numbers in python until certain string is entered

I am completing questions from a python book when I came across this question.
Write a program which repeatedly reads numbers until the user enters "done". Once done is entered, print out total, count, and average of the numbers.
My issue here is that I do not know how to check if a user specifically entered the string 'done' while the computer is explicitly checking for numbers. Here is how I approached the problem instead.
#Avg, Sum, and count program
total = 0
count = 0
avg = 0
num = None
# Ask user to input number, if number is 0 print calculations
while (num != 0):
try:
num = float(input('(Enter \'0\' when complete.) Enter num: '))
except:
print('Error, invalid input.')
continue
count = count + 1
total = total + num
avg = total / count
print('Average: ' + str(avg) + '\nCount: ' + str(count) + '\nTotal: ' + str(total))
Instead of doing what it asked for, let the user enter 'done' to complete the program, I used an integer (0) to see if the user was done inputting numbers.

Keeping your Try-Except approach, you can simply check if the string that user inputs is done without converting to float, and break the while loop. Also, it's always better to specify the error you want to catch. ValueError in this case.
while True:
num = input('(Enter \'done\' when complete.) Enter num: ')
if num == 'done':
break
try:
num = float(num)
except ValueError:
print('Error, invalid input.')
continue

I think a better approach that would solve your problem would be as following :
input_str = input('(Enter \'0\' when complete.) Enter num: ')
if (input_str.isdigit()):
num = float(input_str)
else:
if (input_str == "done"):
done()
else:
error()
This way you control cases in which a digit was entered and the cases in which a string was entered (Not via a try/except scheme).

Does While loop ends when the program return value?

I am a new learner for Python. I have a question about while loop.
I wrote a program below to look for square roots.
When I input anything but integers, the message "is not an integer" shows up and it repeats itself until I input correct data(integers).
My question is, why does it end loop when it return value on line 5, return(int(val))?
Thank you for your attention.
def readInt():
while True:
val = input('Enter an integer: ')
try:
return(int(val))
except ValueError:
print(val, "is not an integer")
g = readInt()
print("The square of the number you entered is", g**2)

To answer your original question, 'return' effectively exits the loop and provide the result that follows the 'return' statement, but you have to explicity print it like so:
def read_int(num1, num2):
while True:
return num1 + num2
print(read_int(12, 15))
If you simply put 'read_int(12, 14)' instead of 'print(read_int(12, 15))' in this scenario, you won't print anything but you will exit the loop.
If you allow me, here are some modifications to your original code:
def read_int(): # functions must be lowercase (Python convention)
while True:
val = input('Enter an integer: ')
try:
val = int(val) # converts the value entered to an integer
minimum_value = 0 # There is no need to evaluate a negative number as it will be positive anyway
maximum_value = 1000000 # Also, a number above 1 million would be pretty big
if minimum_value <= val <= maximum_value:
result = val ** 2
print(f'The square of the number you entered is {result}.')
# This print statement is equivalent to the following:
# print('The square of the number you entered is {}.'.format(result))
break # exits the loop: else you input an integer forever.
else:
print(f'Value must be between {minimum_value} and {maximum_value}.')
except ValueError: # If input is not an integer, print this message and go back to the beginning of the loop.
print(val, 'is not an integer.')
# There must be 2 blank lines before and after a function block
read_int()
With the final 'print' that you actually have at the end of your code, entering a string of text in the program generates an error. Now it doesn't ;). Hope this is useful in some way. Have a great day!

Checking if input value is integer, command, or just bad data (Python)

I'm trying to do something (admittedly) simple. I want my function to take an input, run a few lines if it's an integer or print an error message if it's a string or exit the loop if it's a specific command (done). My problem is that it never detects an integer, instead always displaying the error message unless I'm exiting the loop.
#The starting values of count (total number of numbers)
#and total (total value of all the numbers)
count = 0
total = 0
while True:
number = input("Please give me a number: ")
#the first check, to see if the loop should be exited
if number == ("done"):
print("We are now exiting the loop")
break
#the idea is that if the value is an integer, they are to be counted, whereas
#anything else would display in the error message
try:
int(number)
count = count + 1
total = total + number
continue
except:
print("That is not a number!")
continue
#when exiting the code the program prints all the values it has accumulated thus far
avarage = total / count
print("Count: ", count)
print("Total: ", total)
print("Avarage: ", avarage)
From poking around a bit with the code, it seems like the problem lies with the (count = count + 1) and (total = total + 1), but I'm unable to see why. Any help greatly appreciated.

You aren't assigning your int(number) to anything, so it remains a string.
Two things you need to do. Change your exception handling to print the actual error, so that you have some clue as to what's going on. This code does the following.
except Exception as e:
print("That is not a number!", e)
continue
Output:
That is not a number! unsupported operand type(s) for +: 'int' and 'str'
Which means you are adding a string and an integer together, which you can't do. Looking at your code, you do that at:
try:
int(number) <------- This is not doing anything for your program
count = count + 1
total = total + number
You think it's permanently changing number to an int, so you can use it later, but it's not. It's just for that one line, so you'll need to move it two lines down like this:
try:
count = count + 1
total = total + int(number)

How to convert a string to a float?

I'm currently working on a program in Python and I need to figure out how to convert a string value to a float value.
The program will ask the user to enter a number, and uses a loop to continue asking for more numbers. The user must enter 0 to stop the loop (at which point, the program will give the user the average of all the numbers they entered).
What I want to do is allow the user to enter the word 'stop' instead of 0 to stop the loop. I've tried making a variable for stop = 0, but this causes the program to give me the following error message:
ValueError: could not convert string to float: 'stop'
So how do I make it so that 'stop' can be something the user can enter to stop the loop? Please let me know what I can do to convert the string to float. Thank you so much for your help! :)
Here is some of my code:
count = 0
total = 0
number = float(input("Enter a number (0, or the word 'stop', to stop): "))
while (number != 0):
total += number
count += 1
print("Your average so far is: " , total / count)
number = float(input("Enter a number (0, or the word 'stop', to stop): "))
if (number == 0):
if (count == 0):
print("")
print("Total: 0")
print("Count: 0")
print("Average: 0")
print("")
print("Your average is equal to 0. Cool! ")
else:
print("")
print("Total: " , "%.0f" % total)
print("Count: " , count)
print("Average: " , total / count)
Please let me know what I should do. Thanks.

I'd check the input to see if it equals stop first and if it doesn't I'd try to convert it to float.
if input == "stop":
stop()
else:
value = float(input)
Looking at your code sample I'd do something like this:
userinput = input("Enter a number (0, or the word 'stop', to stop): ")
while (userinput != "stop"):
total += float(userinput) #This is not very faulttolerant.
...

You could tell the user to enter an illegal value - like maybe your program has no use for negative numbers.
Better, would be to test if the string you've just read from sys.stdin.readline() is "stop" before converting to float.

You don't need to convert the string to a float. From what you've said it appears that entering 0 already stops the loop, so all you need to do is edit you're currently existing condition check, replacing 0 with "stop".

Note a few things: if the input is stop it will stop the loop, if it's not a valid number, it will just inform the user that the input were invalid.
while (number != 0):
total += number
count += 1
print("Your average so far is: " , total / count)
user_input = input("Enter a number (0, or the word 'stop', to stop): ")
try:
if str(user_input) == "stop":
number = 0
break
else:
number = float(user_input)
except ValueError:
print("Oops! That was no valid number. Try again...")
PS: note that keeped your code "as is" mostly, but you should be aware to not use explicit counters in python search for enumerate...