How to throw exception when comparing with None on Python 2.7? -

in python 2.7:

expression 1 (fine):

>>> 2 * none traceback (most recent call last):   file "<stdin>", line 1, in <module> typeerror: unsupported operand type(s) *: 'int' , 'nonetype' 

expression 2 (fine):

>>> none none true 

expression 3 (bad):

>>> 2 < none false 

expression 4 (bad):

>>> none<none false 

expression 5 (bad):

>>> none==none true 

i want somehow force expressions 3, 4 , 5 throw typeerror (same in expression 1).

python 3.4 there - expression 5 returns true.

i need in 2.7.

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my use case (if interested):

i'v made application evaluates expressions (example):

d = a+b e = int(a>b) f = if a<b else b if b<c else c 

expressions based on values (a, b , c) comes deserialized json. most of time values (a, b , c) integers, (when value missing) none. when value missing , used in expressions expression should return none (i thinking catching typeerror exception).

in case when a=none, b=1, c=2, exptected result is: d=none, e=none, f=none.

in case when a=1, b=2, c=none, expected result is: d=3, e=0, f=1

you can't change behavior of none. however, can implement own type has behavior want. starting point:

class calcnum(object):     def __init__(self, value):         self.value = value      def __add__(self, o):         return calcnum(self.value + o.value)      def __lt__(self, o):         if self.value none or o.value none:             # pick one:             # return calcnum(none)             raise typeerror("can't compare nones")          return calcnum(self.value < o.value)