Python optional function argument to default to another argument's value

问题

I want to define a function with some optional arguments, let's say A (mandatory) and B (optional). When B is not given, I want it to take the same value as A. How could I do that?

I have tried this, but it doesn't work (name 'B' is not defined):

def foo(A, B=A):
    do_something()

I understand that the values of the arguments are not assigned before the body of the function.

回答1:

You shall do this inside of your function.

Taking your original function:

def foo(A, B=A):
    do_something()

try something like:

def foo(A, B=None):
    if B is None:
        B = A
    do_something()

Important thing is, that function default values for function arguments are given at the time, the function is defined.

When you call the function with some value for A, it is too late as the B default value was already assigned and lives in function definition.

回答2:

You could do it like this. If B has a value of None then assign it from A

def foo(A, B=None):
    if B is None:
        B = A

    print 'A = %r' % A
    print 'B = %r' % B

来源：https://stackoverflow.com/questions/35026716/python-optional-function-argument-to-default-to-another-arguments-value