Python3有条件地装饰？

Question

是否可以根据条件修饰函数？

翼:

if she.weight() == duck.weight(): 
    @burn
def witch():
    pass

我只是想知道,如果逻辑可以使用(当witch被调用？)要弄清楚是否要装饰witch用@burn？

如果没有,是否可以在装饰器中创建条件以达到相同的效果？(witch被称为未修饰的.)

Answer 1

你可以创建一个'有条件'装饰器:

>>> def conditionally(dec, cond):
    def resdec(f):
        if not cond:
            return f
        return dec(f)
    return resdec

用法示例如下:

>>> def burn(f):
    def blah(*args, **kwargs):
        print 'hah'
        return f(*args, **kwargs)
    return blah

>>> @conditionally(burn, True)
def witch(): pass
>>> witch()
hah

>>> @conditionally(burn, False)
def witch(): pass
>>> witch()

Answer 2

可以通过重新分配来启用/禁用装饰器。

def unchanged(func):
    "This decorator doesn't add any behavior"
    return func

def disabled(func):
    "This decorator disables the provided function, and does nothing"
    def empty_func(*args,**kargs):
        pass
    return empty_func

# define this as equivalent to unchanged, for nice symmetry with disabled
enabled = unchanged

#
# Sample use
#

GLOBAL_ENABLE_FLAG = True

state = enabled if GLOBAL_ENABLE_FLAG else disabled
@state
def special_function_foo():
    print "function was enabled"

Answer 3

装饰器只是用于重新定义函数的语法糖,例如:

def wrapper(f):
    def inner(f, *args):
        return f(*args)
    return lambda *args: inner(f, *args)

def foo():
    return 4
foo = wrapper(foo)

这意味着你可以在语法糖存在之前以旧方式完成:

def foo():
    return 4
if [some_condition]:
    foo = wrapper(foo)