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After further thought: This section is about the syntax operators, not the special methods. The syntax operators never evaluate to NotImplemented as they (apparently) interpret its return from a special method the same as a raising of TypeError, and always raise TypeError when neither the op or its reflection is supported. So there should be no mention of NotImplemented here. Just a reference to 3.3. #15997 is related to my 'wonder' but not directly relevant to a patch for this. Please submit a draft patch when you have one.
I determined that 'raise TypeError' and 'return NotImplemented' both result in the call of the reflected method, at least for a couple of cases. (And same seems true for arithmetic ops too.)
class C():
 def __ge__(self, other):
 # print("in C.__ge__", end='')
 return True
 def __add__(self, other):
 return 44
 __radd__ = __add__
class O():
 def __le__(self, other):
 # print ("in O.__le__")
 return NotImplemented
 def __add__(self, other):
 return NotImplemented
 
c = C()
o = O()
ob = object() 
print(c >= o, o <= c, ob <= c)
# True True True
# print(ob <= ob) # raises TypeError
print(c + o, o + c, ob + c)
# 44 44 44
# print(ob + ob) # raises TypeError
# print(ob >= o) # with O.__le__ print uncommented
# in O.__le__ # so non-implemented reflected o <= ob *is* called
# TypeError: unorderable types: object() >= O()