Manipulate Boolean values with conditions

In this article I explore a common code smell related to conditionals and Boolean values and show how to fix it.

Manipulate Boolean values with conditions

The code smell with return

Consider the function is_even shown below.

def is_even(number):
if number % 2 == 0:
return True
else:
return False

print(is_even(2))  # True
print(is_even(3))  # False

What's the code smell it shows?

The code smell is that it uses an if statement to manipulate the Boolean value that we will return from the function. I'll explain in more detail what I mean, but it'll be easier if I show you the improved version:

def is_even(number):
return number % 2 == 0

print(is_even(2))  # True
print(is_even(3))  # False

Notice how we took the condition from the if statement and just plugged it into the return, there is no need for an actual if statement.

What people sometimes forget is that the condition of the if statement is actually an expression: it's a piece of code that produces an actual value that you can use. So, whenever you have an if statement to pick a Boolean value to return, you can rewrite the return to include the condition(s).

I hope this is making some sense. I have another example and an exercise for you, to make sure it is.

The code smell with assignments

The code smell we're talking about isn't restricted to return statements. Sometimes, the code smell shows itself in assignments like the one below:

def morning_routine(rodrigo, day):
if day in {"mon", "tue", "wed", "thu", "fri"}:
weekday = True
weekend = False
else:
weekday = False
weekend = True

rodrigo.shower()
if weekday:
rodrigo.shave()
rodrigo.walk_pet()
# ...

The point here is that we're still using an if statement to manipulate Boolean values instead of using the Boolean values directly. For instance, we could assign to weekday directly and then set weekend to the negation of the value in weekday:

def morning_routine(rodrigo, day):
weekday = day in {"mon", "tue", "wed", "thu", "fri"}
weekend = not weekday

rodrigo.shower()
if weekday:
rodrigo.shave()
rodrigo.walk_pet()
# ...

Maybe we don't even need the variable weekend! (Or maybe we do, it would depend on the remainder of the body of the function morning_routine.)

Exercise

Function to rewrite

Consider the function is_ordered_triple, shown below:

def is_ordered_triple(tup):
if len(tup) == 3:
if tup[0] <= tup[1] and tup[1] <= tup[2]:
return True
else:
return False
else:
return False

print(is_ordered_triple((1, 2)))  # False
print(is_ordered_triple((1, 3, 2)))  # False
print(is_ordered_triple((1, 2, 3)))  # True

Can you rewrite it in the best way possible, and specifically addressing the code smell I discussed above? After you're done, add your solution to the comments below!

Solution

The first thing you can do is rewrite the inner if statement so that we return the Boolean value directly:

def is_ordered_triple(tup):
if len(tup) == 3:
return tup[0] <= tup[1] and tup[1] <= tup[2]
else:
return False

print(is_ordered_triple((1, 2)))  # False
print(is_ordered_triple((1, 3, 2)))  # False
print(is_ordered_triple((1, 2, 3)))  # True

Then, we can chain comparison operators to make the condition slightly shorter:

def is_ordered_triple(tup):
if len(tup) == 3:
return tup[0] <= tup[1] <= tup[2]
else:
return False

print(is_ordered_triple((1, 2)))  # False
print(is_ordered_triple((1, 3, 2)))  # False
print(is_ordered_triple((1, 2, 3)))  # True

Finally, we can recognise that we're looking at the same code smell again!

Notice how we're using an if statement to pick a Boolean value to return! Instead, we can use the condition of the if statement (the code len(tup) == 3) and add it to the return statement:

def is_ordered_triple(tup):
return len(tup) == 3 and tup[0] <= tup[1] <= tup[2]

print(is_ordered_triple((1, 2)))  # False
print(is_ordered_triple((1, 3, 2)))  # False
print(is_ordered_triple((1, 2, 3)))  # True

If you're not sure how/why this works, you may want to read up on Boolean short-circuiting, a very common technique in many programming languages.

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