Exercise: Any Number

Exercise 8 Very easy

Prerequisites for the exercise

  1. Python Addition Calculator Exercise
  2. Python Number Basics
  3. All previous chapters


Extend the Addition Calculator exercise to add any two numbers; not just two integers.


Back in the exercise Addition Calculator, we created a program that took as input two integers and then printed out their sum.

Recall that the entered values were converted into integers using int(). This meant that floating-point numbers couldn't be entered — they would've otherwise lead to an error.

Now, in this exercise, your task is to rewrite that simple program such that it accepts any two numbers; not just integers.

Furthermore, after performing the addition, if the result is an integer (whose fractional part is zero), output the result as an integer.

Shown below are two examples:

x: 5 y: 2.5 The sum is: 7.5

Here, since the sum 7.5 is not an integer, it is output as it is.

x: 1.5 y: 2.5 The sum is: 4

However, here the sum 4.0 is indeed an integer, and is likewise output as an integer — 4.

View Solution

New file

Inside the directory you created for this course on Python, create a new folder called Exercise-8-Any-Number and put the .py solution files for this exercise within it.


The description clearly says that we can't use int() to convert the input strings into a number, since it throws an error when given a string containing a floating-point number.

This can be seen as follows:

Traceback (most recent call last): File "stdin", line 1, in <module> int('1.5') ValueError: invalid literal for int() with base 10: '1.5'

So with int() gone, do we have any other options to convert the input strings into numbers?

float()? Well, yes, it's the only option we've got right now. And guess what, it's just what's required.

float() can convert a stringified integer as well as a stringified float into an actual floating-point number. And then we could perform arithmetic on this float easily.

Therefore, to start with, we replace the calls to int() with float() in the program we created in the Addition Calculator exercise:

x = input('x: ')
y = input('y: ')

x = float(x)
y = float(y) print('The sum is:', x + y)

But the story doesn't end here! The exercise requires one additional thing.

If the result of the addition is an integer, it should be output as an integer. For instance, if the result is 5.0, then it should be output as 5.

How to determine whether a float is an integer or not? Recall anything?

We'll need the is_integer() method of floats.

The idea is to call is_integer() on the expression x + y. If the method returns True, we convert x + y into an integer and then print it. However, if this is not the case, then we continue on with our normal float output.

x = input('x: ')
y = input('y: ')

x = float(x)
y = float(y)

if (x + y).is_integer():
    print('The sum is:', int(x + y))
    print('The sum is:', x + y)
In line 7, the expression x + y in enclosed in parentheses to call the is_integer() method on the result of x + y. Without the parentheses, we'd have the expression x + y.is_integer(), which is adding x and y.is_integer(). Doesn't seem right, does it?

This solves our exercise.

Improving the code

Despite the fact that the code above gives the correct output, there are a couple of problems in it.

The expression x + y is written thrice. Secondly, the print() statement, with exactly the same format, is written twice. This goes against the DRY principle, which we talked about earlier in the Rudimentary Arithmetic Exercise.

DRY (Don't Repeat Yourself) states not to repeat code unnecessarily — which is undoubtedly happening in the code above.

How can we prevent this repetition?

For x + y, we could just compute it once and then save it in a variable. That's it.

We'll call this variable result. First, let's get done with this thing in our code:

x = input('x: ')
y = input('y: ')

x = float(x)
y = float(y)

result = x + y if result.is_integer(): print('The sum is:', int(result)) else: print('The sum is:', result)

Next, let's analyse the print() statements.

How could we improve on them? What do you think is changing in the print() statements?

The difference between both print() calls above is that the first one has int(result) as the second arg, while the second one has result as the second arg. The rest is the same.

Whether the condition for if is true or false, a print() is regardless called, and something is, likewise, output. So, why not take the print() statement out of if and else.

What we could do is to write one print() statement after the whole block of if..else conditionals, and use those conditionals to determine the second arg to print(), instead:

x = input('x: ')
y = input('y: ')

x = float(x)
y = float(y)

result = x + y

if result.is_integer():
result = int(result) else:
result = result
print('The sum is:', result)

In words, the conditionals here say that 'if result is an integer, assign to it the integer value of result, or else just keep it as it is.'

Beyond these conditionals, result is in the desired format, and is directly output using print().

Amazing! We've improved our code a lot.

But there's one thing still left. Let's see whether you could figure it out..

Notice the else block above. If result is not an integer, result is assigned back to result. In other words, result remains whatever it is. This piece of code is redundant. Even if we were to remove it, result would remain whatever it is — it won't just change on its own!

And this is just what we'll do — remove the redundant else block:

x = input('x: ')
y = input('y: ')

x = float(x)
y = float(y)

result = x + y if result.is_integer(): result = int(result) print('The sum is:', result)

Now our code is simpler and much more flexible than the one we created previously.

One thing to keep in mind is that these small improvements at this novice stage of learning Python won't do wonders in the speed of execution of the program.

However, they'll teach you how to write clean code and apply coding principles to it. This is an essential skill to have when you go on to write highly complex applications in Python, or any other programming language.

"I created Codeguage to save you from falling into the same learning conundrums that I fell into."

— Bilal Adnan, Founder of Codeguage