在Drools中，如何找到匹配次数最少的关键字？

Question

我有一些这样的标记化字符串

declare UserAddress
    tokens : List
end

我有一个像这样的可识别标记列表

rule "Count Frequency of RealTokens occurrence in RealAddresses"
salience -10
when
    $token : RealToken();
    $count : List() from collect(RealAddress(tokens contains $token.getToken()));   
then
    modify($token) { setCount($count) };
end

使用 drools，我如何确定给定 UserAddress 中的哪个 tokens 与 $count 值可能最低的 RealToken 相匹配？

我已经试过了：

rule "Find the Most Statistically Significant Token for each User Address"
salience -20
    $ua : UserAddress();
    $ut : String( length() > 0) from $ua.getTokens().subList(1, $ua.getTokens().size());
    $rt : RealToken(token == $ut);
    not(RealToken(token == $ut && this.count < $rt.count));
then
    System.out.println("MSST: " + $ua.toString() + " = " + $rt.toString());
end

但无法解决 DRL 语法问题：

java.lang.IllegalArgumentException: Found errors in package builder
[85,1]: [ERR 102] Line 85:1 mismatched input '$ua' in rule "Find the Most Statistically Significant Token for each User Address"
[0,0]: Parser returned a null Package

Answer 1

错误信息很清楚，如果你不厌其烦地看第 85 行，你会发现

rule "Find the Most Statistically Significant Token for each User Address"
salience -20
    $ua : UserAddress();  ############################### line 85
    $ut : String( length() > 0) from $ua.getTokens().subList(1, $ua.getTokens().size());
    $rt : RealToken(token == $ut);

第一个模式之前没有关键字 when。

在Drools中，如何找到匹配次数最少的关键字？

In Drools, how to find the keyword with the least number of matches?

rule-engine

drools