python 非标准可选参数默认值
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Non-Standard Optional Argument Defaults
提问by ooboo
I have two functions:
我有两个功能：
def f(a,b,c=g(b)):
blabla
def g(n):
blabla
cis an optional argument in function f. If the user does not specify its value, the program should compute g(b) and that would be the value of c. But the code does not compile - it says name 'b' is not defined. How to fix that?
c是 function 中的一个可选参数f。如果用户没有指定它的值，程序应该计算 g(b)，这就是 的值c。但是代码没有编译——它说名称“b”没有定义。如何解决？
Someone suggested:
有人建议：
def g(b):
blabla
def f(a,b,c=None):
if c is None:
c = g(b)
blabla
But this doesn't work. Maybe the user intended cto be None and then cwill have another value.
但这不起作用。也许用户打算c成为 None ，然后c会有另一个值。
回答by Paolo Bergantino
def f(a,b,c=None):
if c is None:
c = g(b)
If Nonecan be a valid value for cthen you do this:
如果None可以是一个有效值，c那么你可以这样做：
sentinel = object()
def f(a,b,c=sentinel):
if c is sentinel:
c = g(b)
回答by codeape
You cannot do it that way.
你不能那样做。
Inside the function, check if c is specified. If not, do the calculation.
在函数内部，检查是否指定了 c。如果不是，请进行计算。
def f(a,b,c=None):
if c == None:
c = g(b)
blabla
回答by SilentGhost
value of cwill be evaluated (g(b)) at compilation time. You need gdefined before ftherefore. And of course you need a global bvariable to be defined at that stage too.
的值c将g(b)在编译时评估 ( )。因此，您需要g事先定义f。当然，您还需要b在该阶段定义一个全局变量。
b = 4
def g(a):
return a+1
def test(a, c=g(b)):
print(c)
test(b)
prints 5.
打印 5.
回答by Thomas 'PointedEars' Lahn
The problem with
问题与
sentinel = object()
def f(a, b, c=sentinel):
if c is sentinel:
c = g(b)
is that sentinelis global/public unless this code is part of a function/method. So someone might still be able to call f(23, 42, sentinel). However, if fis global/public, you can use a closure to make sentinellocal/private so that the caller cannot use it:
是sentinel全局/公共的，除非此代码是函数/方法的一部分。所以有人可能仍然可以打电话f(23, 42, sentinel)。但是，如果f是 global/public，则可以使用闭包使sentinellocal/private 使调用者无法使用它：
def f():
sentinel = object()
def tmp(a, b, c=sentinel):
if c is sentinel:
c = g(b)
return tmp
f = f()
If you are concerned that static code analyzers could get the wrong idea about fthen, you can use the same parameters for the factory:
如果您担心静态代码分析器可能会对此产生错误的想法f，您可以对工厂使用相同的参数：
def f(a, b, c=object()): #@UnusedVariable
sentinel = object()
def tmp(a, b, c=sentinel):
if c is sentinel:
c = g(b)
return tmp
f = f(23, 42)
回答by Maiku Mori
def f(a,b,*args):
if len(args) == 1:
c = args[0]
elif len(args) == 0:
c = g(b)
else:
raise Exception('Function takes 2 or 3 parameters only.')
blabla
def g(n):
blabla
You can probably structure it better, but that's the main idea. Alternatively you can use **kwargsand use the function like f(a,b,c=Something), you just have to modify faccordingly.
您可能可以更好地构建它，但这是主要思想。或者，您可以使用**kwargs和使用类似的功能f(a,b,c=Something)，您只需要相应地进行修改f。
