在 Python 中比较字符串和数字

发布于 2024-12-24 20:56:49 字数 133 浏览 0 评论 0原文

为什么下面的片段会表现得像它的行为？

>>> '10' > 100
True
>>> 100 < '10'
True

难道不应该引发异常吗？

原文

Why does the following piece behave like it behaves?

>>> '10' > 100
True
>>> 100 < '10'
True

Shouldn't it raise an exception?

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飘然心甜 2024-12-31 20:56:49

来自文档：

CPython实现细节：除数字之外的不同类型的对象按其类型名称排序；不支持正确比较的相同类型的对象按其地址排序。

所以这只是 CPython 中发生的事情 ('int' < 'str')，但不保证在其他实现中发生。

事实上，这个行为在python3中已经被删除了：

>>> '10' > 100
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: str() > int()
>>> 100 < '10'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: int() < str()

From the documentation:

CPython implementation detail: Objects of different types except numbers are ordered by their type names; objects of the same types that don’t support proper comparison are ordered by their address.

So it's just something that happens in CPython ('int' < 'str'), but that isn't guaranteed to happen in other implementations.

In fact, this behaviour has been removed in python3:

>>> '10' > 100
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: str() > int()
>>> 100 < '10'
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unorderable types: int() < str()

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怕倦 2024-12-31 20:56:49

来自 manual：

CPython implementation detail: Objects of different types except numbers are 
ordered  by their type names; objects of the same types that don’t 
support proper comparison are ordered by their address.

因此，如果您比较这两种类型： int / string 你有一个按元素类型的字典顺序

From manual:

CPython implementation detail: Objects of different types except numbers are 
ordered  by their type names; objects of the same types that don’t 
support proper comparison are ordered by their address.

So if you compare this two types: int / string you have a lexicographic order bye the type of elements

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握住我的手 2024-12-31 20:56:49

不同的运算符被调用，在某个时刻 int 的 __gt__，在另一个时刻，str 的 __lt__

检查这一点：

class t(int):
  def __gt__(self, other):
    print 'here', v
    return v

class t2(str):
  def __lt__(self, other):
    v = super(t2, self).__lt__(other)
    print 'ohere', v
    return v

if __name__ == '__main__':
  a = t('10')
  b = t2(100)

  a > b
  b < a

different operators are getting called, at one point int's __gt__, at another, str's __lt__

check this:

class t(int):
  def __gt__(self, other):
    print 'here', v
    return v

class t2(str):
  def __lt__(self, other):
    v = super(t2, self).__lt__(other)
    print 'ohere', v
    return v

if __name__ == '__main__':
  a = t('10')
  b = t2(100)

  a > b
  b < a

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