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: http://stackoverflow.com/questions/5424716/python-how-to-check-if-input-is-a-number — Knells, Feb 02 '16 at 04:01
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, Feb 04 '16 at 22:51

score 1 · Answer 1 · answered Feb 02 '16 at 04:14

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 02 '16 at 04:14

pzp

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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") – Dylan Feb 04 '16 at 22:50

GIRISH RAMNANI · Answer 2 · 2016-02-02T04:01:36.263

0

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

something = input()

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

edited Feb 02 '16 at 04:01

answered Feb 02 '16 at 03:58

GIRISH RAMNANI

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Won't your `if` always return `true`? Because you check if `number` is an `int` right after casting it as an `int`. – Knells Feb 02 '16 at 04:00
thats what the user is doing – GIRISH RAMNANI Feb 02 '16 at 04:02

score 0 · Answer 3 · answered Feb 02 '16 at 04:04

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 02 '16 at 04:04

Fernando Vaz

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3

wouldn't this throw a `ValueError` if user passes a string eg. "hello" – GIRISH RAMNANI Feb 02 '16 at 04:09
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") – Fernando Vaz Feb 02 '16 at 04:15
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. – GIRISH RAMNANI Feb 02 '16 at 04:18
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. – Fernando Vaz Feb 02 '16 at 04:24

Python input processing when input is a float or string

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