I want to pass all the arguments passed to a function(func1) as arguments to another function(func2) inside func1 This can be done with *args, *kwargs in the called func1 and passing them down to func2, but is there another way?
我希望将传递给函数（func1）的所有参数作为func1中另一个函数（func2）的参数传递。这可以通过调用func1中的* args，* kwargs并将它们传递给func2来完成，但还有另一种方法？
Originally
本来
def func1(*args, **kwargs):
func2(*args, **kwargs)
but if my func1 signature is
但如果我的func1签名是
def func1(a=1, b=2, c=3):
how do I send them all to func2, without using
如何在不使用的情况下将它们全部发送到func2
def func1(a=1, b=2, c=3):
func2(a, b, c)
Is there a way as in javascript callee.arguments?
有没有像javascript callee.arguments那样的方法？
2 个解决方案
#1
38
Explicit is better than implicit but if you really don't want to type a few characters:
显式优于隐式，但如果你真的不想输入几个字符：
def func1(a=1, b=2, c=3):
func2(**locals())
locals() are all local variables, so you can't set any extra vars before calling func2 or they will get passed too.
locals（）都是局部变量，因此在调用func2之前不能设置任何额外的变量，否则它们也会被传递。
#2
9
Provided that the arguments to func1 are only keyword arguments, you could do this:
如果func1的参数只是关键字参数，您可以这样做：
def func1(a=1, b=2, c=3):
func2(**locals())
#1
38
Explicit is better than implicit but if you really don't want to type a few characters:
显式优于隐式，但如果你真的不想输入几个字符：
def func1(a=1, b=2, c=3):
func2(**locals())
locals() are all local variables, so you can't set any extra vars before calling func2 or they will get passed too.
locals（）都是局部变量，因此在调用func2之前不能设置任何额外的变量，否则它们也会被传递。
#2
9
Provided that the arguments to func1 are only keyword arguments, you could do this:
如果func1的参数只是关键字参数，您可以这样做：
def func1(a=1, b=2, c=3):
func2(**locals())