说,我有一个通用函数f(input, kind)
和一长串kind
。我想为列表中的每个f_kind(input)
创建函数kind
。可以不用通过f_kind1 = partial(f, kind=kind1)
手动进行操作,而是可以通过遍历列表的循环(类似于类中的setattr(class_name, method_name, method)
)来更动态地创建它们?
答案 0 :(得分:0)
您的问题让我回想起了Guido van Rossum很久以前读过的一篇有关在Python中实现multimethods的文章-因为它的核心是您正在做的事情。
需要修改它以适合在课堂上工作,但我认为这可能比使用setattr()
来实现目标更好。
mm.py:
''' MultiMethod module. '''
_registry = {}
class MultiMethod(object):
def __init__(self, name):
self.name = name
self.typemap = {}
def __call__(self, *args):
types = tuple(arg.__class__ for arg in args)
function = self.typemap.get(types)
if function is None:
raise TypeError("no match")
return function(*args)
def register(self, types, function):
if types in self.typemap:
raise TypeError("duplicate registration")
print('registering: {!r} for args: {}'.format(function.__name__, types))
self.typemap[types] = function
def multimethod(*types):
def register(function):
name = function.__name__
mm = _registry.get(name)
if mm is None:
mm = _registry[name] = MultiMethod(name)
mm.register(types, function)
return mm
return register
样品用量。
mm_test.py:
from mm import multimethod
@multimethod(int)
def f(input):
print('f_{}({!r}) called'.format(type(input).__name__, input))
@multimethod(str)
def f(input):
print('f_{}({!r}) called'.format(type(input).__name__, input))
f('answer')
f(42)
@multimethod(float)
def f(input):
print('f_{}({!r}) called'.format(type(input).__name__, input))
f(3.141529)
输出:
registering: 'f' for args: (<class 'int'>,)
registering: 'f' for args: (<class 'str'>,)
f_str('answer') called
f_int(42) called
registering: 'f' for args: (<class 'float'>,)
f_float(3.141529) called