为什么空字典大于1？

Question

Why is the following code true?

>>> foo = {}
>>> foo > 1
True
>>> foo < 1
False
>>> foo == 0
False
>>> foo == -1
False
>>> foo == 1
False

I understand what I wanted was len(foo) > 1, but as a beginner this surprised me.

Answer 1

From the docs :

The operators <, >, ==, >=, <=, and != compare the values of two objects. The objects need not have the same type. If both are numbers, they are converted to a common type. Otherwise, objects of different types always compare unequal, and are ordered consistently but arbitrarily. You can control comparison behavior of objects of non-builtin types by defining a __cmp__ method or rich comparison methods like __gt__ , described in section 3.4.

(This unusual definition of comparison was used to simplify the definition of operations like sorting and the in and not in operators. In the future, the comparison rules for objects of different types are likely to change.)

Answer 2

rich comparison between incompatible types is based on the name(?) of the type in python2.x and has been disallowed in python3.x.

In any event, in python2.x, the ordering is guaranteed to give the same results for a particular python implementation and version, but the ordering itself is not defined.

Answer 3

I think this may have arisen from the fact that comparison operators only need to be partially defined to be derived, ie if you can test for == and < then you can derive the rest of the operators, <= is (< or ==), > is not <=, etc. so in the case of foo = {} you can get:

Python 2:

>>> foo == 0
False
>>> foo < 0
False
>>> not (foo <= 0)
True
so:
>>> foo > 0
True

python 3:

>>> foo = {}
>>> foo < 0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: dict() < int()
>>> foo > 0
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: dict() > int()
>>> foo == 0
False
>>>