Limiting a number without a conditional

Question

Consider the following function:

def absolute_value(some_number):
    if some_number < 0:
        return -some_number

    return some_number

and this one:

def absolute_value(some_number):
     return (some_number**2)**0.5

We can say that (in a naive manner) these two more or less achieve the same result (ignoring the performance issues) but the latter one has less cyclomatic complexity due to lack of the conditional. What I want is, for the sake of reduced cyclomatic complexity, I want to implement the following function without the conditional:

def limit(some_number):
    if some_number > 1
        return 1

    return some_number

Is there an elegant way of doing this?

Thank you for your interest.

Asked By: Tuğberk Özdemir

||

Source

Answer 1

Aside from min(1, some_number) you can also do this:

def limit(x):
    return (x > 1) + (x <= 1) * x

But that is in my opinion way harder to understand than your code. That also holds for your absolute_value example. I think such tricks to reduce cyclomatic complexity are beyond the point of cyclomatic complexity. In the end it doesn’t make your code easier to understand, but cyclomatic complexity is just unable to capture this. But maybe that is what you want to demonstrate…

Answered By: cherrywoods

Answer 2

If you are trying to convert negative number into positive number then here is a simple way to do so.

abs(number)

It should give positive value for a negative number.
And if you want the number to be in a limit. For example if you want the number to be less than 4 , you can use modulus 4.

n%=4

no matter what is the value of n, it will always return 0-3. Hope it helps.

Answered By: Jeson Pun

Limiting a number without a conditional

Question:

Answers: