python 在python中传递命名变量参数
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Passing on named variable arguments in python
提问by Staale
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 methodAas follows, the second argument will be passed on as positional rather than named variable arguments.
在方法A中，我希望调用方法B，传递kwargs。但是，似乎如果我methodA如下定义，第二个参数将作为位置参数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？
回答by Cristian
Put the asterisks before the kwargs variable. This makes Python pass the variable (which is assumed to be a dictionary) as keyword arguments.
将星号放在 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)
回答by Sebastian Rittau
As an aside: When using functions instead of methods, you could also use functools.partial:
顺便说一句：当使用函数而不是方法时，你也可以使用 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”，当调用该函数时，将调用 foo 并将第一个参数设置为“argvalue”，并且所有其他函数刚刚传递：
bar(5, myarg="value")
will call
将会通知
foo("argvalue", 5, myarg="value")
Unfortunately that will not work with methods.
不幸的是，这不适用于方法。
回答by Staale
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...
现在看起来很明显...
