Efficient way to round to arbitrary precision in Python [closed]

Question

What is a Pythonic solution to the following?

I'm reading a temperature sensor that has .5 resolution. I need to write to it (it has a programmable thermostat output), also with .5 resolution.

So I wrote this function (Python 2.7) to round off a float as input to the to the nearest .5:

def point5res(number):
    decimals = number - int(number)
    roundnum = round(number, 0)
    return roundnum + .5 if .25 <= decimals < .75 else roundnum

print point5res (6.123)

print point5res(6.25)

print point5res(6.8)

Which works fine, outputs 6.0, 6.5 and 7.0, respectively. That's just what I want.

I'm relatively new to Python. The line

return roundnum + .5 if .25 <= decimals < .75 else roundnum

has me drooling with admiration for it implementors. But is it Pythonic?

Edit: since posting, I have learned a little more about what is and isn't 'Pythonic'. My code isn't. Cmd's anwwer, below, is. Thank you!

Answer 1

They are considered pythonic if you keep the expressions simple otherwise it becomes difficult to read.

I would round to the nearest 0.5 like this:

round(number*2) / 2.0

or more generically:

def roundres(num, res):
    return round(num / res) * res