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())