How to find type of user input and print different values depending on the type of input in Python 3.x

Question

Develop a Python function which either returns the float square of its parameter x if the parameter is a number, or prints the string "Sorry Dave, I'm afraid I can't do that" if the parameter is a string, and then returns 0.0.

What am I doing wrong? I'm a first year CS student and I have no previous programming background.

I created a function that takes user input, evaluates what type of input it is and print different out puts for number and strings.

For that I used eval(var) func. I also the type(var) == type to verify the type and a if-else loop.

def findt():
    userin = input("Input: ")       # Takes user input
    inpeval = eval(userin)          # Evaluates input type    

    if type(inpeval) == int:        # If input is an int
        userfloat = float(inpeval)  # Modifies into a float
        print(userfloat ** 2)       # Prints the square of the value
    elif type(inpeval) == float:    # If input is a float
        print(inpreval ** 2)        # Prints the square of the value
    elif type(userin) == str:       # If input is a string
        print("Sorry Dave, I'm afraid I can't do that")   # Print a string
        return 0.0                  # Return 0.0
    else:
        print("Invalid Input")    
findt()

When I run my code it works well when input is an int, a float or a char. But if I write more than one char it returns me an error:

NameError: name 'whateverinput' is not defined.

Possible duplicate of [Asking the user for input until they give a valid response](https://stackoverflow.com/questions/23294658/asking-the-user-for-input-until-they-give-a-valid-response) — Ofer Sadan, Oct 23 '19 at 20:38
Be _very_ careful of using [`eval`](https://stackoverflow.com/a/1832957/4739755) - there's almost always a better way to do whatever you're using `eval` to do. You should also look into the python string methods, specifically [`string.isnumeric`](https://docs.python.org/3/library/stdtypes.html#str.isnumeric). The return of `input` will always be a string initially. — b_c, Oct 23 '19 at 20:46
To be honest I don't really understand the use of `eval`. It was a suggestion of a class mate. My understanding of python is still very limited. I learned only a bit about basic operations, functions, data types, while and for loops. I never heard about `string.isnumeric` which looks like can solve my problem. I just want to solve the problem with the my limited knowledge on the subject — RicardoGoncalves, Oct 23 '19 at 20:55
Use the **isinstance()** built-in function to test data types. It takes two parameters, your variable and the expected type. Very simple. — S3DEV, Oct 23 '19 at 21:08

score 0 · Answer 1 · answered Oct 23 '19 at 21:04

0

You're trying to eval input before you know it's needed. Get rid of it entirely:

def findt():
    userin = input("Input: ")       # Takes user input

    if type(userin) == int:        # If input is an int
        userfloat = float(userin)  # Modifies into a float
...

The root problem is that you can't evaluate an undefined name. If your input is a string that is not the name of an object in your program, it will fail. eval requires everything you feed it to be defined.

answered Oct 23 '19 at 21:04

Prune

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The problem then is that every input will be identified as a string as by default `input()` stores the value as a string. So how can I test if the user input was a number or a string? – RicardoGoncalves Oct 23 '19 at 21:11

codrelphi · Answer 2 · 2019-10-23T22:18:10.967

0

Here is a better way to achieve your goal by using the string method isnumeric() to test if the input is numeric or not.

def findt():
    userin = input("Input: ")

    if userin.isnumeric():
        # userin is numeric
        result = float(userin) ** 2
        print(result)

    else:
        try:
            # userin is a float
            result = float(userin) ** 2
            print(result)
        except ValueError:
            # userin is a string
            print("Sorry Dave, I'm afraid I can't do that")
            return 0.0

findt()

Update: a concise version:

def findt():
    userin = input("Input: ")

    try:
        # userin is an int or float
        result = float(userin) ** 2
        print(result)
    except ValueError:
        # userin is a string
        print("Sorry Dave, I'm afraid I can't do that")
        return 0.0

findt()

edited Oct 23 '19 at 22:18

answered Oct 23 '19 at 21:10

codrelphi

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it still doesn't work. If the input is a float it treats it as a string – RicardoGoncalves Oct 23 '19 at 21:14
Thank you very much for your help. I followed your suggestion and found my own solution that I posted. – RicardoGoncalves Oct 23 '19 at 22:10
Ok. So, you can upvote my post or accept it. It will be helpful for others... – codrelphi Oct 23 '19 at 22:11

score 0 · Accepted Answer · answered Oct 23 '19 at 22:08

I found the solution for my problem. The way I did it I take the input from the user and i try to convert it to a float. If it is a number it will convert and print a float that is the square of the input. If the input is a string it cannot be converted to a float and will give me an error so I use an except ValueError: to print the string I want and return 0.0.

def whattype():
    user_input = input(">>> ")
    try:                                                
            user_input = float(user_input)                      
            print(user_input ** 2)
    except ValueError:
            print("Sorry Dave, I'm afraid I can't do that")
            return 0.0

whattype()

Thank you all for the suggestions and help

How to find type of user input and print different values depending on the type of input in Python 3.x

3 Answers3