Question

比方说，我有红色和绿色的篮子，里面装有盒草莓和葡萄：

public class Basket {
    public enum Color { RED, GREEN };

    private Color color;
    private List<FruitBox> fruitBoxes;

    //Assume appropriate constructors, getters exist
}

public class FruitBox {
    public enum Fruit { STRAWBERRY, GRAPE };

    private Fruit fruit;

    //Assume appropriate constructors, getters exist
}

我需要使用Drools来打印以下计算：

如果篮子中的GRAPE或STRAWBERRY FruitBoxs数量超过两个，则
按“水果盒”水果和“篮子颜色”分组的“水果盒”总数

我该如何按照简单高效的Drools规则编码（1.）和（2.）？阅读文档后，我提出了以下解决方案：

rule "Two or more of same fruit in any Basket"
    when
        //Bind the contents of each basket
        $Basket : Basket($FruitBoxes : fruitBoxes)

        //Compute A list of the unique fruits in current basket contents
        $fruits : Object() from accumulate( FruitBox($f : fruit) from $FruitBoxes, collectSet($f) )

        //Match and bind each unique fruit one by one
        $fruit : FruitBox.Fruit() from $fruits

        //Find the number of times the unique fruit occurs in this Basket
        $count : Number(intValue >= 2) from accumulate(
            $f : FruitBox(fruit == $fruit) from $FruitBoxes,
            count($f) )
    then
        System.out.println($Basket + " has " + $count + " of the fruit " + $fruit);
end

对于一个测试数据集，成功打印出的输出如下：

com.sample.DroolsTest$Basket@10cd6753 has 4 of the fruit STRAWBERRY
com.sample.DroolsTest$Basket@10cd6753 has 3 of the fruit GRAPE
com.sample.DroolsTest$Basket@2e060819 has 2 of the fruit GRAPE

我担心的是，我的规则漫长而复杂，并且似乎对每个篮子进行了多次迭代（一旦在其中找到独特的水果，然后再对每个水果进行计数）。我怀疑这是我缺乏Drools知识的产物，并且存在更好的解决方案。在纯Java中，我将使用HashMap一次存储每个水果及其数量，这既简单又快速。有什么方法可以重写我的Drools代码以提高性能和可读性？同样的问题也适用于我的第二条规则：

rule "Totals by fruit and by basket color"
    when
        //Calculate the cross product of all fruits and colors
        $fruit : FruitBox.Fruit()
        $color : Basket.Color()

        //Count the number of FruitBoxes of the chosen fruit in Baskets of the chosen color
        $count : Number() from accumulate(
            $fb : FruitBox(fruit == $fruit) 
            and exists Basket(color == $color, fruitBoxes contains $fb),
            count($fb) )
    then
        System.out.println($count + " " + $fruit + " in " + $color + " baskets");
end

哪个会产生如下输出：

5 STRAWBERRY in GREEN baskets
6 STRAWBERRY in RED baskets
6 GRAPE in GREEN baskets
4 GRAPE in RED baskets

但是，不幸的是，看起来它在每个FruitBox的所有篮子中搜索，而在一次遍历篮子的同时计算一个（或全部）种类的FruitBox会更有意义。就性能而言，此规则是否有更好的语法？

感谢您的帮助！

Answer 1

嗨，第一个问题，我将使用两个单独的规则：

rule "strawberriesInBasket"
dialect "mvel"
when
    $basket : Basket($fruitBoxes : fruitBoxes)
    Number($count : intValue() > 2) from accumulate ($fruitBox : FruitBox(fruit.name == "strawberry") from $fruitBoxes, count($fruitBox)) 
then
    System.out.println("Count of strawberry boxes in basket is: " + $count )
end

和

rule "grapesInBasket"
dialect "mvel"
when
    $basket : Basket($fruitBoxes : fruitBoxes)
    Number($count : intValue() > 2) from accumulate ($fruitBox : FruitBox(fruit.name == "grape") from $fruitBoxes, count($fruitBox)) 
then
    System.out.println("Count of grape boxes in basket is: " + $count )
end

请注意，我已将枚举水果更改为具有字段名称的数据对象。

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