python - How to check if a number is in a interval -

suppose got:

first_var = 1 second_var = 5 interval = 2 

i want interval second_var second_var ± interval (from 3 7). wank check if first_var in interval. in specific case want false if first_var = 4, want true

i can this:

if (first_var > second_var-interval) , (first_var < second_var+interval):   #true 

is there more pythonic way this?

i use class __contains__ represent interval:

class interval(object):     def __init__(self, middle, deviation):         self.lower = middle - abs(deviation)         self.upper = middle + abs(deviation)      def __contains__(self, item):         return self.lower <= item <= self.upper 

then define function interval simplify syntax:

def interval(middle, deviation):     return interval(middle, deviation) 

then can call follows:

>>> 8 in interval(middle=6, deviation=2) true  >>> 8 in interval(middle=6, deviation=1) false 

with python 2 solution more efficient using range or xrange don't implement __contains__ , have search matching value.

python 3 smarter , range generating object efficient xrange, implements __contains__ doesn't have search valid value. xrange doesn't exist in python 3.

this solution works floats.

also, note, if use range need careful of off-by-1 errors. better encapsulate it, if you're going doing more once or twice.