将包装函数定义中收到的** kwargs传递给封闭的(即包装的)函数调用的参数 [英] Passing **kwargs received in a wrapper-function definition, to arguments of an enclosed (i.e. wrapped) function call

问题描述

哦，亲爱的，我希望我能拿到冠军头衔.:)

Oh dear, I hope I got the title right. :)

如何将提供给包装器功能 定义的** kwargs传递给它包装的另一个(封闭的)函数 call .例如:

How can one pass the **kwargs supplied to a wrapper-function definition, to another (enclosed) function call that it wraps. For example:

def wrapped_func(**kwargs):
   # Do some preparation stuff here.
   func('/path/to/file.csv', comma_separated_key=value_injected_here)
   # Do some other stuff.

例如，此调用:

wrapped_func(error_bad_lines=True, sep=':', skip_footer=0, ...)

应导致以下结果:

func('/path/to/file.csv', error_bad_lines=True, sep=':', skip_footer=0, ...)

在过去的几个小时中，我尝试了多种方法，但每个方法都暴露了 type-preservation 漏洞(针对这些值).我以前没有使用过这种特殊的包装模式，并且想知道社区是否可以提供帮助.预先谢谢你.

I've tinkered with a variety of approaches over the past couple of hours, but each exposed type-preservation vulnerabilities (for the values). I've not used this particular wrapper pattern before, and was wondering if the community could give some help. Thank you in advance.

推荐答案

** kwargs 是dict，这意味着您可以使用double splat( ** )来将其解压缩为关键字参数列表.因此，您的包装函数可能是这样的:

**kwargs is a dict, meaning you can use the double splat (**) to unpack it as a list of keyword arguments. So your wrapper function could be like this:

def wrapped_func(**kwargs):
   # Do some preparation stuff here.
   func('/path/to/file.csv', **kwargs)
   # Do some other stuff.

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