More pythonic way of checking if user input exists and consists of integers

Question

I have a GUI where I ask the user to enter some values. When the user submits the data I do some validation:

first I check if the user has entered a value for each input
then I check if each of the inputted values are integers

Trying not to repeat myself to much I came up with this, but the second part of the validation looks more like a hack. Is there a more pythonic way of rewriting this, short of spelling everything out like in the first part of the validation?

    errors = []
    # 1) check if values exist
    if not self.startInput.GetValue():
        errors.append("Please provide a start")
    if not self.stopInput.GetValue():
        errors.append("Please provide a stop")
    if not self.valueInput.GetValue():
        errors.append("Please provide a value")
    # 2) check if values are integers
    try:
        self.start = int(self.startInput.GetValue())
        self.stop = int(self.stopInput.GetValue())
        self.value = int(self.valueInput.GetValue())
    except ValueError as err:
        tb = traceback.format_exc()
        func = re.search('self\.(.*) =', tb).groups()[0]
        errors.append("Value for {0} needs to be an integer.".format(func)) 
    if errors:
        raise RuntimeError('\n'.join(errors))

Joel Cornett · Accepted Answer · 2013-07-17 08:55:37Z

4

Since you're checking for integers, and not floats, you can simply do:

if self.start.GetValue().strip().isdigit():
    pass

isdigit() returns False for both cases where the input is an empty string, and when the input contains non-digits.

If you want to send specific errors for incorrect assignment:

startValue = self.start.GetValue().strip()

if not startValue: 
    errors.append("Please provide a start.")

if not startValue.isdigit():
    errors.append("Value of start must be an integer.")

edited Jul 17, 2013 at 8:55

answered Jul 17, 2013 at 8:29

Joel Cornett

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2 Comments

BioGeek Over a year ago

This seems like a good option, but then I don't have a way of giving different error messages based on whether the data is missing ("Please provide a start value") versus the data exists but is not an integer ("Value for start needs to be an integer"). Correct?

Joel Cornett Over a year ago

@BioGeek: True, if you want that functionality, I would go with either two if statements, or a try...except block. Either one is perfectly pythonic.

perreal · Accepted Answer · 2013-07-17 08:32:36Z

1

If you have these inputs in a dictionary called inputs you can do:

errors = []
for inputname, inputval in self.inputs.items():
    if not inputval:
        errors.append("Please provide a {}".format(inputname))
    try:
        setattr(self, inputname, int(inputval.GetValue()))
    except:
        errors.append("Value for {0} needs to be an integer.".format(inputname)) 
if errors:
    raise RuntimeError('\n'.join(errors))

answered Jul 17, 2013 at 8:32

perreal

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1 Comment

Henrik Over a year ago

+1 for including all incorrectly formatted values in the error list, not just the first one

hivert · Accepted Answer · 2013-07-17 08:58:43Z

1

I think the try: ... except is perfectly Pythonic. I would have uses an helper function instead of searching through an error message get_int_of_name(name, value, error) which return an int and update error if needed:

def get_int_of_name(name, value, error):
    try:
        res = int(value)
    except ValueError:
        error.append("...")
        return 0
    else:
        return res

edited Jul 17, 2013 at 8:58

answered Jul 17, 2013 at 8:30

hivert

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3 Comments

Joel Cornett Over a year ago

I agree that try...except is beautifully explicit and therefore pythonic. You should, however, avoid catching naked exceptions. I would do except ValueError:, instead of except:.

BioGeek Over a year ago

@Joel: It is more the parsing with a regex of the traceback when there is a ValueError that seemed unpythonic to me.

hivert Over a year ago

@Joel Cornett: Sure ! I was checking which exception I wanted to put here and forgot to put the right one. I fixed it now.

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