python - 基于多个数据点的评分

Question

需要有关最简单方法的指导。我有多个 Assets ，每个 Assets 都有多个数据点。我希望根据每个数据点的决策为每个 Assets 分配一个值。举个例子，每个 Assets 是一栋房子，数据点包括 window 数量、车库门、破损的栅栏等，每个房子都会有一个分数。

除了数百个 if 语句和从分数中添加/减去之外，是否有推荐的编码方式？

我计划如何执行此操作的示例

def score_house(house):
    # score house
    score = 0
    if (house.windows > 2): score += 10
    if (house.garage): score += 10
    if (house.fence == 'broken'): score += 80
    return score

Answer 1

我认为您也可以在这里使用“责任链”模式:

The pattern allows multiple objects to handle the request without coupling sender class to the concrete classes of the receivers. The chain can be composed dynamically at runtime with any handler that follows a standard handler interface.

使用此模式的好处在于，您可以在单独的模块中定义和扩展不同的评分器，并在运行时根据问题条件动态组合它们。以下是您可以如何做到这一点。首先，定义一个父记分器类:

from functools import reduce


class BaseScorer(object):

    def __init__(self):
        self._next_scorer = None

    def set_next(self, scorer):
        self._next_scorer = scorer

        return scorer

    def _call_next(self, house, score):

        if self._next_scorer is None:
            return score

        return self._next_scorer.score(house, score)

    def score(self, house, score=0):
        raise NotImplementedError

    @staticmethod
    def chain(scorers):
        reduce(lambda x, y: x.set_next(y), scorers)

        return scorers[0]

然后，定义各种记分器类，例如:

class WindowScorer(BaseScorer):

    def score(self, house, score=0):

        if house.windows > 2:
            score = score + 10

        return self._call_next(house, score)


class GarageScorer(BaseScorer):

    def score(self, house, score=0):

        if house.garage:
            score = score + 10

        return self._call_next(house, score)


class FenceScorer(BaseScorer):

    def score(self, house, score=0):

        if house.fence == 'broken':
            score = score - 5

        return self._call_next(house, score)

这就是它的使用方式:

scorer = BaseScorer.chain([
    WindowScorer(),
    GarageScorer(),
    FenceScorer()
])

house = House(windows=4, garage=True, fence='nice')
score = scorer.score(house)