如何在Python 2.7中将用户输入限制为特定的整数，并跟踪异常？

Question

EDIT: The suggested duplicate is incredibly helpful in regards to basic input validation. While it does cover a lot, my specific problem (failing to assign int(evaluation) to a variable) is only explicitly addressed here. I'm marking this separately in case anyone else has made a similarly silly mistake :)

I've spent the last few weeks playing with Python 2.7 and having a lot of fun. To learn more about while loops, I've created a small script which asks the user for an integer between 1 and 10.

My goal is to then be able to respond to cases in which the user responds with unexpected input, like a non-integer, or an integer outside the specified range. I've been able to fix a lot of my issues with help from other StackOverflow threads, but now I'm stumped.

First, I created a variable, idiocy , to keep track of exceptions. (The script is supposed to be sassy, but until I get it working, I'm the one it's making fun of.)

idiocy = 0

while 1:
    evaluation = raw_input("> ")
    try:
        int(evaluation)
        if evaluation < 1 or evaluation > 10:
            raise AssertionError
    except ValueError:
        idiocy += 1
        print "\nEnter an INTEGER, dirtbag.\n"
    except AssertionError:
        idiocy += 1
        print "\nI said between 1 and 10, moron.\n"
    else:
        if idiocy == 0:
            print "\nOkay, processing..."
        else:
            print "\nDid we finally figure out how to follow instructions?"
            print "Okay, processing..."
        break

As you can see, I'm trying to handle two different errors -- a ValueError for the input type, and an AssertionError for the integer range -- and keep track of how many times they're raised. (Really, I only care about knowing whether or not they've been raised at least once; that's all I need to insult the user.)

Anyways, when I run the script in its current form, the error response works just fine ('dirtbag' for non-integers, 'moron' for out-of-range). The problem is that even when I input a valid integer, I still get an out-of-range AssertionError .

I suspect that my issue has to do with my while logic, but I'm not sure what to do. I've added a break here or there but that doesn't seem to help. Any suggestions or blatant errors? Again, total Python beginner here, so I'm half winging it.

//If anyone has simpler, cleaner, or prettier ways to do this, feel free to let me know too. I'm here to learn!

Answer 1

Your problem is you're not saving the int version of evaluation to evaluation like this:

idiocy = 0

while 1:
    evaluation = raw_input("> ")
    try:
        evaluation = int(evaluation) <--- here
        if evaluation < 1 or evaluation > 10:
            raise AssertionError
    except ValueError:
        idiocy += 1
        print "\nEnter an INTEGER, dirtbag.\n"
    except AssertionError:
        idiocy += 1
        print "\nI said between 1 and 10, moron.\n"
    else:
        if idiocy == 0:
            print "\nDid we finally figure out how to follow instructions?"
            print "Okay, processing..."
        else:
            print "\nOkay, processing..."

If you wanted to track the types of exceptions raised, you could use collections.Counter for idiocy and change the code like this:

from collections import Counter

idiocy = Counter()

while 1:
    evaluation = raw_input("> ")
    try:
        evaluation = int(evaluation)
        if evaluation < 1 or evaluation > 10:
            raise AssertionError
    except ValueError as e:
        idiocy[e.__class__] += 1 
        print "\nEnter an INTEGER, dirtbag.\n"
    except AssertionError as e:
        idiocy[e.__class__] += 1
        print "\nI said between 1 and 10, moron.\n"
    else:
        if idiocy == 0:
            print "\nDid we finally figure out how to follow instructions?"
            print "Okay, processing..."
        else:
            print "\nOkay, processing..."

>>> idiocy
Counter({AssertionError: 2, ValueError: 3})

And you can access the error counts by key like idiocy[AssertionError]

Answer 2

You have int(evalutation) , but you're not assigning it to anything.

Try

try:
    evaluation = int(evaluation)
    assert 0 < evaluation < 10
except ValueError:
    ...
except AssertionError:

Answer 3

Your range test can be refactored as

assert 1 <= evaluation <= 10

You could keep your insults in a dictionary

insults = {AssertionError : "\nI said between 1 and 10, moron.\n",
           ValueError : "\nEnter an INTEGER, dirtbag.\n"
           }

And write the try/except like this

try:
    ...
except (AssertionError, ValueError) as e:
    print(insults[type(e)])

When you change the user input to an int, you need to assign it to something

evaluation = int(evaluation)

---

assert was meant for debugging - you are using it incorrectly.

• You should use TypeError for non-integer repsonses - the type is wrong.
• You use ValueError for responses outside of a range - the value is wrong

Answer 4

In your code: int(evaluation) is not typecasting evaluation variable to int type. The output is:

> 2
<type 'str'>
I said between 1 and 10, moron.

Try this:

idiocy = 0

while 1:
    try:
        evaluation = int(raw_input("> "))
        if evaluation < 1 or evaluation > 10:
            raise AssertionError
    except ValueError:
        idiocy += 1
        print "\nEnter an INTEGER, dirtbag.\n"
    except AssertionError:
        idiocy += 1
        print "\nI said between 1 and 10, moron.\n"
    else:
        if idiocy == 0:
            print "\nOkay, processing..."
        else:
            print "\nDid we finally figure out how to follow instructions?"
            print "Okay, processing..."
        break

By the way you can use tuple to store all your exceptions. Example:

idiocy = 0

all_exceptions = (ValueError, AssertionError)
while 1:
    try:
        evaluation = int(raw_input("> "))
        if evaluation < 1 or evaluation > 10:
            raise AssertionError("\nI said between 1 and 10, moron.\n")
    except all_exceptions as e:
        idiocy += 1
        print str(e)
    else:
        if idiocy == 0:
            print "\nOkay, processing..."
        else:
            print "\nDid we finally figure out how to follow instructions?"
            print "Okay, processing..."
        break

Hope it helps.