Python:如何编写所有需要相同条件的方法

Question

我有class很多方法依赖于另一个特定实例的可能但不确定的存在class.这些方法将按以下方式进行调整:if other_instance return X else return None.实际上,我写了几个相同的if语句作为每个方法的代码的一部分.(另请注意,__init__由于继承问题,我不能使用自定义.)除了None如下所示的依赖方法(在极大简化的模型上)之外,是否有一种实用或适当的方法来限制这种冗余编码？

class QuarterResult(ParentClass):
    result = None
    prior = None

    def get_prior_quarter(self):
        # code to find prior_quarter
        if prior_quarter:
            self.prior = prior_quarter
            return prior_quarter

        # commence None-ing out unusable methods
        # Note: the actual model has 10+ methods which need prior

        unusable_methods = ('change_in_results', 'quarterly_return')
        for mthd in unusable_methods:
            setattr(self, mthd, None)

    # assume this method is ALWAYS run first
    def calculate_result(self, *args):
        # result = some calculation based on the given args
        _ = self.get_prior_quarter()
        self.result = result

    @property
    def change_in_results(self):
        return self.prior.result - self.result

    @property
    def quarterly_return(self):
        return self.change_in_results / self.prior.results

    @property
    def real_result(self):
        if self.result == 42:
            return "You are right: it's 42"
        else:
            return 'What do I know?'

Answer 1

您可以使用自定义装饰器

def dependent(meth):
    def depmeth(self, *args, **kwargs):
        if self.other_instance:
            return meth(self, *args, **kwargs)
        else:
            return None
    return depmeth

然后你的班级就会变成

class ...:
    ...
    def meth1(self, x):
        return "always working"

    @dependent
    def meth2(self, y):
        return "code executed only if other_instance is present"