如何设置星形参数的默认值？

Question

我有一个接受的函数*args,但我想设置一个默认元组,以防没有提供.(这是不可能的def f(*args=(1, 3, 5)),这会引发一个SyntaxError.)最好的方法是什么？预期的功能如下所示.

f()
# I received 1, 2, 3!

f(1)
# I received 1!

f(9, 3, 72)
# I received 9, 3, 72!

以下功能g将提供正确的功能,但我更喜欢*args.

def g(args=(1, 2, 3)):
    return "I received {}!".format(', '.join(str(arg) for arg in args))

g()
# I received 1, 2, 3!

g((1,))
# I received 1!

g((9, 3, 72))
# I received 9, 3, 72!

Answer 1

你可以检查args你的函数是否真实:

def g(*args):
    if not args:
        args = (1, 2, 3)
    return "I received {}!".format(', '.join(str(arg) for arg in args))

如果没有args传递给函数,它将导致一个空元组,其计算结果为False.

Answer 2

如果没有收到任何参数,args将是一个空元组.您不能在方法签名本身中添加默认值,但可以检查是否args为空并将其替换为函数内的回退值.

def g(*args):
    if not args: 
        args = (1, 2, 3)
    return 'I received {}!'.format(', '.join(str(arg) for arg in args))