Failing While loop in Python

Question

I'm fairly new to python (coding in general really), but I mostly grasp what's put in front of me. I'm trying to create an input that will only accept integers, so I put together the following function:

def askFor(request):
 """Program asks for input, continues to ask until an integer is given"""
 num = ' '
 while (isinstance(num, int) == False):
 num = input(request)
 else:
 return num

it works fine when i enter a number, but anything else breaks it instead of looping it back. Am I missing an element, or am I just best off to go with

str(input('wargharble: '))

• Pieter Witvoet: Just a stylistic note: while not isinstance(num, int): is a little more Pythonic than while (isinstance(num, int) == False):. :)
• martineau: The problem is that in Python 2 input() runs eval() on whatever the user types in (see the docs), so if it's not valid Python syntax for a constant or literal expression, like 42, 1+2, or "Bob", an exception will be raised which your program will need to catch. I suggest you instead use num = int(raw_input(request)) within a try/except clause because it doesn't use eval() and is therefore safer.

Answer 1

You are on the safer side with raw_input in Python 2, because it will always return a string, whereas input will try to evaluate the given string, which is dangerous.

def askFor(prompt):
 """ ask for input, continues to ask until an integer is given
 """
 number = None
 while True:
 try:
 number = int(raw_input(prompt))
 break
 except ValueError:
 pass # or display some hint ...
 print(number) # or return or what you need ...

• Volatility: I was just in the middle of writing the exact same thing! But anyway, you might consider mentioning that in python 2 input evaluates the input as python code, while raw_input returns a string.
• miku: @Volatility, you might consider mentioning ...: my thought exactly :)
• DJ80: this didn't seem to work with the rest of the code that I had. Whenever I entered a non-integer, it would accept it, and not ask for the same data again, then break when it's line came up.
• miku: @DJ80, the snippet works for me as it is posted here - it asks for input as long as the user does not input an int.
• DJ80: @Miku, I've copy/pasted your coding to a new file with the rest of my program, and it now functions as you said. I likely just missed something simple like changing input to raw_input. Now I just need to wrap my brain around what makes it work.

Answer 2

The problem is that input() returns the raw user input evaluated as a python expression. So when you input something like "someinput" on the console here is what happens:

1. The raw input is captured from the console as a string
2. That string is then parsed as a python expression (so imagine you simply type this string into your program, in this case, it becomes a variable with the name "someinput"
3. When isinstance() is called on this python expression, the python interpreter can't find a variable named "someinput" which is defined - so it throws an exception.

What you really want to do is use raw_input(), this will return a string representing the user's input. Since you are receiving a string from the console now though, you need some way of checking if a string is an int (because isinstance will always return a string)

Now to determine if the string is a number, you could use something like isdigit() - but be careful - this function will validate any string that contains only digits (so 0393 would be valid).

Here are the corrections that need to be made:

def askFor(request):
 """Program asks for input, continues to ask until an integer is given"""
 num = ' '
 while (num.isdigit() == False):
 num = raw_input(request)
 else:
 return num

• Will C.: Agreed, I defer to miku's solution for parsing the string as an integer

Answer 3

This is because input() is failing on non-integer input (this has to do with the fact that input() is calling eval() on the string your inputting (see python builtins).

raw_input() does not. (in fact, input() calls raw_input())

Try this instead:

while True:
 num = raw_input(request)
 if num.isdigit():
 break

The isdigit() function checks to see if each character in a string is a digit ('0' - '9')

Answer 4

input evaluates whatever's typed in as Python code, so if the user types something non-sensical in, it'll produce an error. Or worse, it can run arbitrary commands on your computer. Usually, you want to use raw_input and convert the string value it returns safely yourself. Here, you just need to call int on the string, which can still produce an error if the user enters a non-int, but we can handle that by looping.

And voila:

def read_int(request):
 while True:
 try:
 return int(raw_input(request))
 except ValueError:
 pass

Note, there's no need for break or else; just return the value once you have parsed it successfully!

Answer 5

Well first off, your indentation if a bit off (the while loop's indentation level does not match the previous level).

def askFor(request):
 """Program asks for input, continues to ask until an integer is given"""
 num, isNum = '', False # multiple assignment
 while not isNum:
 num = input(request)
 if type(num) in (int, long): # if it is any of these
 isNum = True
 break 
 else: isNum = False
 return num # at the end of the loop, it will return the number, no need for an else

Another way to do it:

def askFor(request):
 """Program asks for input, continues to ask until an integer is given"""
 num, isNum = '', False
 while not isNum:
 num = raw_input(request)
 try: # try to convert it to an integer, unless there is an error
 num = int(num)
 isNum = True
 break
 except ValueError:
 isNum = False
 return num

• miku: -1: In you first snippet you use type, which is a kind of last resort, if nothing else works (see also: stackoverflow.com/a/1549854/89391); in your second example you introduce too much clutter.
• DJ80: The first way didn't work for me, but the second way seems to work perfectly. I see I need to read up more on try statements. Many thanks.
• Rushy Panchal: Yeah the problem with the first one is that say the user enters 'two' for the input statement, two is used as a variable. As it is not defined, you will get a NameError. You could use try and except with that as well, or use raw_input; raw_input returns the user's input as a string, so then you can use a try to convert it to an integer, unless there is an error (most likely TypeError); then prompt them again.