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

我有一个像这样的公认令牌列表

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，如何确定给定tokens中哪个UserAddress与RealToken的{​​{1}}值匹配？{/ p>

我已经尝试过了：

$count

但无法解决DRL语法问题：

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 

Answer 1

错误信息非常清楚，当你不愿意看第85行时，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。