python:匿名函数的替代品

Question

Python不支持复杂的匿名函数.什么是好的选择？例如:

class Calculation:
    def __init__(self, func):
        self.func = func

    def __call__(self, data):
        try:
        # check if the value has already been calculated
        # if it has, it would be cached under key = self.func
            return data[self.func]
        except KeyError:
            pass # first-time call; calculate and cache the values
        data[self.func] = self.func(data)
        return data[self.func]

# with a simple function, which can be represented using lambda, this works great
f1 = Calculation(lambda data : data['a'] * data['b'])

# with a complicated function, I can do this:
def f2_aux:
   # some complicated calculation, which isn't suitable for a lambda one-liner
f2 = Calculation(f2_aux) 

这是一个合理的设计吗？

如果是这样,有没有办法避免我在模块中定义的每个f*的f*_aux的丑陋？

更新:

使用示例:

d = {'a' : 3, 'b' : 6}

# computes 3 * 6
# stores 18 in d under a key <function <lambda> at ...>
# returns 18
f1(d)

# retrieves 18 from d[<function <lambda> at ...>]
# returns 18, without having to recalculate it
f1(d)

更新:

仅仅为了我的理解,我添加了一个使用内部函数的版本.

def memoize(func):
    def new_func(data):
        try:
        # check if the value has already been calculated
        # if it has, it would be cached under key = self.func
            return data[func]
        except KeyError:
            pass # first-time call; calculate and cache the values
        data[func] = func(data)
        return data[func]
    return new_func

@memoize
def f1(data):
  return data['a'] * data['b']

Answer 1

您不需要匿名功能.此外,memoization已经做得比这更好,你可能没有理由自己推出.

但要回答这个问题:您可以将您的班级用作装饰者.

@Calculation
def f2():
    ...

这简单地定义了函数,将其包装Calculation并存储为其结果f2.装饰器语法定义为等效于:

_decorator = Calculation # a fresh identifier
# not needed here, but in other cases (think properties) it's useful
def f2():
    ...
f2 = _decorator(f2)