random boolean by percentage

Question

I'm trying to get a get a random boolean but with a weighted percentage. For instance, I want the user to pass in a percentage (i.e. 60) and the generator will randomly select true 60% of the time.

What I have is this:

def reset(percent=50):
    prob = random.randrange(0,100)
    if prob > percent:
        return True
    else:
        return False

Is there a better way to do this? This seems inefficient and cumbersome. I don't need it to be perfect since it is just used to simulate data, but I need this to be as fast as possible.

I've searched (Google/SO) and have not found any other questions for this.

Answer 1

Just return the test:

def reset(percent=50):
    return random.randrange(100) < percent

because the result of a < lower than operator is already a boolean. You do not need to give a starting value either.

Note that you need to use lower than if you want True to be returned for a given percentage; if percent = 100 then you want True all of the time, e.g. when all values that randrange() produces are below the percent value.

Answer 2

How about:

def reset(percent=50):
    return random.randrange(0, 100) > percent