5

How do I get python to accept both integer and float?

This is how I did it:

def aud_brl(amount,From,to):
    ER = 0.42108
    if amount == int:
        if From.strip() == 'aud' and to.strip() == 'brl':
            ab = int(amount)/ER
         print(ab)
        elif From.strip() == 'brl' and to.strip() == 'aud':
            ba = int(amount)*ER
         print(ba)
    if amount == float:
        if From.strip() == 'aud' and to.strip() == 'brl':
            ab = float(amount)/ER
         print(ab)
        elif From.strip() == 'brl' and to.strip() == 'aud':
            ba = float(amount)*ER
         print(ba)

def question():
    amount = input("Amount: ")
    From = input("From: ")
    to = input("To: ")

    if From == 'aud' or 'brl' and to == 'aud' or 'brl':
        aud_brl(amount,From,to)

question()

This is another example of how I did it:

number = input("Enter a number: ")

if number == int:
    print("integer")
if number == float:
    print("float")

But those two ways don't work.

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4
  • 1
    I changed your title and headings to lowercase. Please refrain from shouting at us :) Commented Apr 22, 2017 at 20:14
  • if type(number) is int But that will always be false, since number will always be a string. Commented Apr 22, 2017 at 20:21
  • @juanpa.arrivillaga no it is not. He's using input to read from user, type(numer) is str. Commented Apr 22, 2017 at 20:23
  • 2
    Just so you know, the line if From == 'aud' or 'brl' and to == 'aud' or 'brl' will always evaluate to True, since 'brl' is truthy in both conditions. If you're looking to see whether From is 'aud' or 'brl', you need something like this: if From == 'aud' or From == 'brl' ... Commented Apr 22, 2017 at 20:49

7 Answers 7

Reset to default
6

It looks like you just want to make sure the value passed in can be operated upon like a float, regardless of whether the input is 3 or 4.79 for example, correct? If that's the case, then just cast the input as a float before operating on it.

Here's your modified code:

def aud_brl(amount, From, to):
    ER = 0.42108 
    if From.strip() == 'aud' and to.strip() == 'brl': 
        result = amount/ER 
    elif From.strip() == 'brl' and to.strip() == 'aud': 
        result = amount*ER 

    print(result)

def question(): 
    amount = float(input("Amount: "))
    From = input("From: ") 
    to = input("To: ")

    if (From == 'aud' or From == 'brl') and (to == 'aud' or to == 'brl'): 
        aud_brl(amount, From, to)

question()
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Comments

4

Use the isinstance function, which is built in

if isinstance(num, (int, float)):
    #do stuff

Also, you should refrain from using reserved keywords for variable names. The keyword from is a reserved keyword in Python

Finally, there is one other error I noticed:

if From == 'aud' or 'brl'

Should be

if From == 'aud' or From == 'brl'

Lastly, to clean up the if statements you could theoretically use the list (if you have more currencies in the future, this might be better.

currencies = ['aud', 'brl']     #other currencies possible
if From in currencies and to in currencies:
    #do conversion
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4 Comments

isinstance(num, (int, float)) can be done directly... and it looks as if OPs input starts with strings.
@hiroprotagonist The true Pythonic solution here, of course, would be duck typing and catching an error if a non int/float is passed!
How would I be able to convert it, if the currencies are not specified in the if statement?
@KGarcia you would have to restructure your code significantly for that, which is beyond the scope of this comment, but if you are planning on having multiple currencies being converted, the way you have laid out your function is quite inefficient
4

this is how you could check the given string and accept int or float (and also cast to it; nb will be an int or a float):

number = input("Enter a number: ")

nb = None
for cast in (int, float):
    try:
        nb = cast(number)
        print(cast)
        break
    except ValueError:
        pass

but in your case just using float might do the trick (as also string representations of integers can be converted to floats: float('3') -> 3.0):

number = input("Enter a number: ")

nb = None
try:
    nb = float(number)
except ValueError:
    pass

if nb is None you got something that could not be converted to a float.

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2 Comments

Why did the other guy said, "The true Pythonic solution here, of course, would be duck typing and catching an error if a non int/float is passed!"? Can you please explain? I'm new in programming.
this has nothing to do with duck-typing. i just try to convert a string to a float in a manner that will never crash. one of the python philosophies is EAFP as opposed to LBYL. so python coders wil often try something instead of checking things first. (e.g. if you wanted to convert a str to an int you could check first if the string consists of digits only; that is not a pythonic thing to do).
1

amount==int doesn't make sense. input gives us a string. int (and float) is a function. A string never equals a function.

In [42]: x=input('test')
test12.23
In [43]: x
Out[43]: '12.23'
In [44]: int(x)
....
ValueError: invalid literal for int() with base 10: '12.23'
In [45]: float(x)
Out[45]: 12.23

float('12.23') returns a float object. int('12.23') produces an error, because it isn't a valid integer string format.

If the user might give either '12' or '12.23', it is safer to use float(x) to convert it to a number. The result will be a float. For many calculations you don't need to worry whether it is a float or integer. The math is the same.

You can convert between int and floats if needed:

In [45]: float(x)
Out[45]: 12.23
In [46]: float(12)
Out[46]: 12.0
In [47]: int(12.23)
Out[47]: 12
In [48]: round(12.23)
Out[48]: 12

You can also do instance tests

In [51]: isinstance(12,float)
Out[51]: False
In [52]: isinstance(12.23,float)
Out[52]: True
In [53]: isinstance(12.23,int)
Out[53]: False
In [54]: isinstance(12,int)
Out[54]: True

But you probably don't need to do any those.

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Comments

0

These seem to work well.

def getInt():
        """
                input returns a str,
                coerce return to required type 
        """
    x = str()
    while type(x) != int:
        try:
            return int(input('enter an integer: '))
        except ValueError: continue


def getFloat():
        """
                input returns a str,
                coerce return to required type 
        """
    x = str()
    while type(x) != float:
        try:
            return float(input('enter a float: '))
        except ValueError: continue
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2 Comments

Please use the code blocks when writing/sharing code.
As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center.
0 
# let's choose a random number
# from input, its type is string by default
number = input() 

try:
    
    x==int(number)      # check whether it is an int or float
    
except:
    
    x=float(number)
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1 Comment

And what happens if input is non-numeric? Ex. input is "abc"
-1

I think you should just use float(input() function. And if a float or an int is entered, the interpreter will print the right values, unless a non float or non int is entered. E.g look at this simple function and you will see that both float and integers are taken:

num1 = float(input("Enter first number: "))
num2 = float(input("Enter second number: "))
num3 = float(input("Enter third number: "))
print(f"The sum is: {num1 + num2 + num3}")

or this example:

while True: try: num = float(input("Enter number one: ")) break # Exit the loop if input is valid except ValueError: print("Please enter a valid number (integer or float).")

Now you can use the num variable which will be a float

print("You entered:", num)

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2 Comments

The solution to use float(...) is just duplicating the other earlier answers.
That answer has already been given years ago. Please, avoid code only answer, and add some explanation. Especially when answering to old questions, it is important to explain why your answer is different, and even better than existing answers.

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