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Asking the user for input until they give a valid response
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Closed 5 years ago.
I'm learning python, and am stuck on a project. The main part of the project was to code the collatz sequence, which wasn't a problem. The next part is to validate the user input using try and except, in order to make sure only an integer value is given in.
So far, I have the following code:
def collatz(number):
if number % 2 == 0:
return number // 2
elif number % 2 == 1:
return (3*number) + 1
print('Please enter a number:')
number = ''
try:
number = int(input())
except ValueError:
print('Incorrect type of data entered.\nPlease enter an integer!')
while number is not int:
try:
number = int(input())
except ValueError:
print('Incorrect type of data entered.\nPlease enter an integer!')
output = number
while output != 1:
output = collatz(output)
print(output)
My problem is I'm not sure how to repeat the try/except statement until I get an integer from the user. Right now, if I enter a string instead of an integer, the program goes into a loop and entering an integer subsequently does not help. I read quite a few threads on the topic, but they didn't shed light as to the aforementioned problem.
Would really like to understand where I'm going wrong.
You can use:
valid=False
while not valid:
try:
number=int(input())
valid=True
except ValueError:
print('Incorrect type of data entered.\nPlease enter an integer!')
Just use isinstance(x, int), where x is input.
It returns True if x is an int. Just another solution, in case you were looking for a different way!
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'm studying Zed shaw's learn python 3 the hard way book and there's a part in the code i'm trying to improve.
My aim was to have an if statement with the condition that the input the user gives is of int type.
def gold_room():
print("This room is full of gold. How much do you take?")
choice = eval(input("> "))
if type(choice) is int:
how_much = choice
else:
print("Man, learn to type a number.")
if how_much < 50:
print("Nice, you're not greedy, you win!")
exit(0)
else:
print("You greedy bastard!")
gold_room()
It works if the input is indeed an integer but if I type in a string i get the error:
NameError: name 'string' is not defined
I tried to use int() but then if the input is a string I get an error.
Is there a way to do this?
Use choice = int(input("> ")). The int function will convert the string that the input function gives you into an integer. But if it can't, it will raise an exception (ValueError), that you may want to try-except.
Using #Lenormju 's answer I wrote this and it worked like a charm.
def gold_room():
print("This room is full of gold. How much do you take?")
# used try to check whether the code below raises any errors
try:
choice = int(input("> "))
how_much = choice
# except has code which would be executed if there is an error
except:
dead("Man, learn to type a number.")
# else has code which would be executed if no errors are raised
else:
if how_much < 50:
print("Nice, you're not greedy, you win!")
exit(0)
else:
dead("You greedy bastard!")
Thank you to everyone who took the time out to respond
Apologies if similar questions have been asked but I wasn't able to find anything to fix my issue. I've written a simple piece of code for the Collatz Sequence in Python which seems to work fine for even numbers but gets stuck in an infinite loop when an odd number is enter.
I've not been able to figure out why this is or a way of breaking out of this loop so any help would be greatly appreciate.
print ('Enter a positive integer')
number = (int(input()))
def collatz(number):
while number !=1:
if number % 2 == 0:
number = number/2
print (number)
collatz(number)
elif number % 2 == 1:
number = 3*number+1
print (number)
collatz(number)
collatz(number)
Your function lacks any return statements, so by default it returns None. You might possibly wish to define the function so it returns how many steps away from 1 the input number is. You might even choose to cache such results.
You seem to want to make a recursive call, yet you also use a while loop. Pick one or the other.
When recursing, you don't have to reassign a variable, you could choose to put the expression into the call, like this:
if number % 2 == 0:
collatz(number / 2)
elif ...
This brings us the crux of the matter. In the course of recursing, you have created many stack frames, each having its own private variable named number and containing distinct values. You are confusing yourself by changing number in the current stack frame, and copying it to the next level frame when you make a recursive call. In the even case this works out for your termination clause, but not in the odd case. You would have been better off with just a while loop and no recursion at all.
You may find that http://pythontutor.com/ helps you understand what is happening.
A power-of-two input will terminate, but you'll see it takes pretty long to pop those extra frames from the stack.
I have simplified the code required to find how many steps it takes for a number to get to zero following the Collatz Conjecture Theory.
def collatz():
steps = 0
sample = int(input('Enter number: '))
y = sample
while sample != 1:
if sample % 2 == 0:
sample = sample // 2
steps += 1
else:
sample = (sample*3)+1
steps += 1
print('\n')
print('Took '+ str(steps)+' steps to get '+ str(y)+' down to 1.')
collatz()
Hope this helps!
Hereafter is my code snippet and it worked perfectly
#!/usr/bin/python
def collatz(i):
if i % 2 == 0:
n = i // 2
print n
if n != 1:
collatz(n)
elif i % 2 == 1:
n = 3 * i + 1
print n
if n != 1:
collatz(n)
try:
i = int(raw_input("Enter number:\n"))
collatz(i)
except ValueError:
print "Error: You Must enter integer"
Here is my interpretation of the assignment, this handles negative numbers and repeated non-integer inputs use cases as well. Without nesting your code in a while True loop, the code will fail on repeated non-integer use-cases.
def collatz(number):
if number % 2 == 0:
print(number // 2)
return(number // 2)
elif number % 2 == 1:
result = 3 * number + 1
print(result)
return(result)
# Program starts here.
while True:
try:
# Ask for input
n = input('Please enter a number: ')
# If number is negative or 0, asks for positive and starts over.
if int(n) < 1:
print('Please enter a positive INTEGER!')
continue
#If number is applicable, goes through collatz function.
while n != 1:
n = collatz(int(n))
# If input is a non-integer, asks for a valid integer and starts over.
except ValueError:
print('Please enter a valid INTEGER!')
# General catch all for any other error.
else:
continue
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 working my way through this set of tutorials. In section 3.2a - While Loops the following code is supposed to loop until the user enters the target number (7) then display a congratulations message however regardless of what number is entered Python either gives a right answer or a wrong answer, even 7 will sometimes flag a wrong answer. I know there are other ways to perform this sort of task but I would like to get the code from the tutorial working.
targetNumber = 7
guess = input("Guess a number between 1 and 10 ")
while guess != targetNumber:
print("Wrong, try again ")
guess = input("Guess a number between 1 and 10 ")
print("Congratulations - that's right!")
You should convert the target numger to an string before comparison. Also, you should exclude the congratulations message from the loop. I would suggest :
targetNumber = str(7)
guess = input("Guess a number between 1 and 10 ")
while guess != targetNumber:
print("Wrong, try again ")
guess = input("Guess a number between 1 and 10 ")
print("Congratulations - that's right!")
The detail is that input returns a string and if you compare a string to an integer, it will always return false.
Python's input function (or raw_input in Python 2.x) returns a string entered by the user. targetNumber, on the other hand, is an integer. In the Python Interpreter, try:
>>> 7 == "7"
False
You need to cast the user's input to an integer first.
try:
guess = int(input("Please enter a number: "))
except ValueError:
print("That is not a valid number!")