在python中传递命名变量参数

Say I have the following methods:

说我有以下方法:

def methodA(arg, **kwargs):
    pass

def methodB(arg, *args, **kwargs):
    pass

In methodA I wish to call methodB, passing on the kwargs. However, it seems that if I define methodA as follows, the second argument will be passed on as positional rather than named variable arguments.

在methodA中,我希望调用methodB,传递kwargs。但是,似乎如果我按如下方式定义methodA,则第二个参数将作为位置而不是命名变量参数传递。

def methodA(arg, **kwargs):
    methodB("argvalue", kwargs)

How do I make sure that the **kwargs in methodA gets passed as **kwargs to methodB?

如何确保methodA中的** kwargs作为** kwargs传递给methodB?

3 个解决方案

#1

Put the asterisks before the kwargs variable. This makes Python pass the variable (which is assumed to be a dictionary) as keyword arguments.

把星号放在kwargs变量之前。这使得Python将变量(假定为字典)作为关键字参数传递。

methodB("argvalue", **kwargs)

#2

As an aside: When using functions instead of methods, you could also use functools.partial:

暂且不说:当使用函数而不是方法时,您也可以使用functools.partial:

import functools

def foo(arg, **kwargs):
    ...

bar = functools.partial(foo, "argvalue")

The last line will define a function "bar" that, when called, will call foo with the first argument set to "argvalue" and all other functions just passed on:

最后一行将定义一个函数“bar”,当调用它时,它将调用foo,第一个参数设置为“argvalue”,所有其他函数都被传递:

bar(5, myarg="value")

will call

foo("argvalue", 5, myarg="value")

Unfortunately that will not work with methods.

不幸的是,这不适用于方法。

#3

Some experimentation and I figured this one out:

一些实验,我想出了这个:

def methodA(arg, **kwargs): methodB("argvalue", **kwargs)

def methodA(arg,** kwargs):methodB(“argvalue”,** kwargs)

Seems obvious now...

现在似乎显而易见......

#1