python 运算符优先级 in 和比较 [英] python operator precedence of in and comparison

问题描述

以下比较产生True:

<预><代码>>>>11"中的1"真的>>>(11"中的1")==真真的

用括号反之，我得到一个类型错误:

<预><代码>>>>'1' in ('11' == True)回溯(最近一次调用最后一次):文件<stdin>"，第 1 行，在 <module> 中类型错误:bool"类型的参数不可迭代

那么如何获得没有括号的 False?

<预><代码>>>>11"中的1"==真错误的

解决方案

Python 手册说 in 和 == 是相同的 优先级.因此，默认情况下从左到右评估它们，但也需要考虑链接.您上面的表达式('1' in '11' == True)实际上被评估为...

('11' 中的'1') and ('11' == True)

当然是False.如果您不知道什么是链接，那么它可以让您执行诸如...

如果 0 <＜1:

在 Python 中，这意味着您所期望的(a 大于 0 但小于 1").

The following comparisons produce True:

>>> '1' in '11'
True
>>> ('1' in '11') == True
True

And with the parentheses the other way, I get a TypeError:

>>> '1' in ('11' == True)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: argument of type 'bool' is not iterable

So how do I get False with no parentheses?

>>> '1' in '11' == True
False

解决方案

The Python manual says in and == are of equal precedence. Thus, they're evaluated from left to right by default, but there's also chaining to consider. The expression you put above ('1' in '11' == True) is actually being evaluated as...

('1' in '11') and ('11' == True)

which of course is False. If you don't know what chaining is, it's what allows you to do something like...

if 0 < a < 1:

in Python, and have that mean what you expect ("a is greater than 0 but less than 1").

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