为什么在python 2.x中int（50）< str（5）？ [英] Why is int(50)<str(5) in python 2.x?

问题描述

在python 3中， int（50）<'2'导致 TypeError ，它应该是。但是，在python 2.x中， int（50）<'2'返回 True （这也是其他数字格式的情况，但py2和py3中都存在 int 。那么，我的问题有几个部分：

In python 3, int(50)<'2' causes a TypeError, and well it should. In python 2.x, however, int(50)<'2' returns True (this is also the case for other number formats, but int exists in both py2 and py3). My question, then, has several parts:

1. 为什么Python 2.x（< 3？）允许这种行为？




• （谁认为允许这个开头是个好主意？）

• 是指 ord / chr ？


• 是否有一些不太明显的二进制格式？

• Is it referring to ord / chr?
• Is there some binary format which is less obvious?

推荐答案

它就像这样 ¹ 。

>>> float() == long() == int() < dict() < list() < str() < tuple()
True

数字比较小于容器。数字类型将转换为通用类型，并根据其数值进行比较。容器按其名称的字母值进行比较。 ²

Numbers compare as less than containers. Numeric types are converted to a common type and compared based on their numeric value. Containers are compared by the alphabetic value of their names.²

来自 docs ：

CPython实现细节：不同的对象除数字以外的类型按其类型名称排序;不支持正确比较的相同类型的对象按其地址排序。

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.

不同内置类型的对象按字母顺序按类型名称进行比较 int 以'i'开头， str 以<$ c $开头c> s 所以 int 小于任何 str 。。

Objects of different builtin types compare alphabetically by the name of their type int starts with an 'i' and str starts with an s so any int is less than any str..

1. 我不知道。
   
   
   
   • 醉酒大师。
   
   

• 这是指任意订单。


• 否。

回应关于 long<的评论int

>>> int < long
True

您可能意味着这些类型的值，在这种情况下数字比较适用。

You probably meant values of those types though, in which case the numeric comparison applies.

^{1这一切都在Python 2.6.5上}

^{2感谢kRON为我解决这个问题。我以前从未想过将数字与 dict 进行比较，数字的比较就是那些很容易被忽视的事情之一。}

^{2 Thank to kRON for clearing this up for me. I'd never thought to compare a number to a dict before and comparison of numbers is one of those things that's so obvious that it's easy to overlook.}

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