python中是否有内置的标识函数? [英] Is there a builtin identity function in python?

问题描述

我想指出一个什么都不做的函数:

def identity(*args)返回参数

我的用例是这样的

尝试:gettext.find(...)..._ = gettext.gettext别的:_ = 身份

当然，我可以使用上面定义的 identity，但是内置的肯定会运行得更快(并避免我自己引入的错误)​​.

显然，map 和 filter 使用 None 作为标识，但这特定于它们的实现.

<预><代码>>>>_=无>>>_(你好")回溯(最近一次调用最后一次):文件<stdin>"，第 1 行，在 <module> 中TypeError: 'NoneType' 对象不可调用

解决方案

做了更多的研究，没有，在 issue 1673203 来自 Raymond Hettinger 说不会有:

<块引用>

最好让人们编写自己的琐碎传递并考虑签名和时间成本.

所以实际上更好的方法是(lambda 避免命名函数):

_ = lambda *args: args

• 优点:接受任意数量的参数
• 缺点:结果是参数的盒装版本

或

_ = lambda x: x

• 优点:不改变参数的类型
• 缺点:只需要 1 个位置参数

I'd like to point to a function that does nothing:

def identity(*args)
    return args

my use case is something like this

try:
    gettext.find(...)
    ...
    _ = gettext.gettext
else:
    _ = identity

Of course, I could use the identity defined above, but a built-in would certainly run faster (and avoid bugs introduced by my own).

Apparently, map and filter use None for the identity, but this is specific to their implementations.

>>> _=None
>>> _("hello")
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'NoneType' object is not callable

解决方案

Doing some more research, there is none, a feature was asked in issue 1673203 And from Raymond Hettinger said there won't be:

Better to let people write their own trivial pass-throughs and think about the signature and time costs.

So a better way to do it is actually (a lambda avoids naming the function):

_ = lambda *args: args

• advantage: takes any number of parameters
• disadvantage: the result is a boxed version of the parameters

OR

_ = lambda x: x

• advantage: doesn't change the type of the parameter
• disadvantage: takes exactly 1 positional parameter

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