Python input processing when input is a float or string

Question

I was hoping someone could help or hint to where I'm going wrong with this Python homework assignment:

number = int(input("Enter a number"))
if number == int or float:
    print(number * number, number * number * number, number ** 4)
elif number != int or float:
    print("This is not a valid number")

It runs fine with a whole number, but not with a float or a string. I think it's because number is set to look for an integer, but I'm not sure what to substitute that with in order to make it work.

Take a look at this question for a start: stackoverflow.com/questions/5424716/… — Knells
– Knells, Commented Feb 2, 2016 at 4:01
FInal answer: number = input("Enter a number") try: number = int(number) print(number * number, number * number * number, number ** 4) except ValueError: try: number = float(number) print(number * number, number * number * number, number ** 4) except ValueError: print("This isn't an integer or float") — Dylan
– Dylan, Commented Feb 4, 2016 at 22:51

pzp · Accepted Answer · 2016-02-02 04:14:20Z

1

You want to use a try... except... else block:

try:
    number = float(input("Enter a number"))
except ValueError:
    print("This is not a valid number")
else:
    print(number * number, number * number * number, number ** 4)

answered Feb 2, 2016 at 4:14

pzp

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1 Comment

Dylan Over a year ago

I wound up extrapolating on this to get my final answer, so thanks pzp. This is what my final code looks like: number = input("Enter a number") try: number = int(number) print(number * number, number * number * number, number ** 4) except ValueError: try: number = float(number) print(number * number, number * number * number, number ** 4) except ValueError: print("This isn't an integer or float")

GIRISH RAMNANI · Accepted Answer · 2016-02-02 04:01:36Z

0

you can wrap around input around a try ... except.

something = input()

try:
    something = int(something)
except:
    print("not an int")

edited Feb 2, 2016 at 4:01

answered Feb 2, 2016 at 3:58

GIRISH RAMNANI

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2 Comments

Knells Over a year ago

Won't your if always return true? Because you check if number is an int right after casting it as an int.

GIRISH RAMNANI Over a year ago

thats what the user is doing

Fernando Vaz · Accepted Answer · 2016-02-02 04:04:54Z

0

Can you just replace the int with a float? like this: "number = float(input("Enter a number"))"

I think it solves your problem. Anyway we could use a little more description.

Good luck!

answered Feb 2, 2016 at 4:04

Fernando Vaz

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4 Comments

GIRISH RAMNANI Over a year ago

wouldn't this throw a ValueError if user passes a string eg. "hello"

Fernando Vaz Over a year ago

Yeah i didn't get it. He wants to do what python already does? Because python throws a "not a number" error when a string is passed.. If so i think try... except would be the solution Anyway, here's a solution with try... except try: number = float(input("Enter a number")) print(number * number, number * number * number, number ** 4) except: print("This is not a valid number")

GIRISH RAMNANI Over a year ago

maybe he wants to like sanitize the user's input. like if the user sends something other than a int or a float then the program shouldnt break and rather show an error and ask again.

Fernando Vaz Over a year ago

yes, i'm struggling with the markdown editing here I made a code that catches the error: try: number = float(input("Enter a number")) print(number * number, number * number * number, number ** 4) except: print("This is not a valid number") I hope you can read this.

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