AnyLogic findAll facility within certain distance from people and return best facility - anylogic

Question

I am trying to use findAll function that takes two arguments: collection, condition. My collection will be facilities (Pharma) but condition is based on population distance. After finding all, I want to choose best (service) one. So far, I fried following approach. But an error and probably on condition in findAll; saying method is not applicable.... Could anybody help with this? or any other ideas?
Pharma nearest = this.getNearestAgent(main.pharmas);
for (int i=0; i<main.people.size(); i++){
Person p = main.people.get(i);
if (p.getConnectionsNumber()>0){
List <Pharma> x = findAll(main.pharmas, p.agentsInRange(100));//[Problem is here]
Pharma y = top(x, s->s.serviceQuality);
return y;
}else{
return nearest;
}
}

Answer 1

The find all function in AnyLogic requires you to use a predicate as the second argument for find all.
As per the comments you need to use
x = findAll(main.pharmas,p -> p.agentsInRange(100))
You can read more about using predicates here - https://www.theanylogicmodeler.com/post/using-predicates-in-anylogic

Answer 2

Below is the code that solved my purpose. I wanted that there is a certain percentage of possibility that some people may go beyond the nearest facility for some reasons like better service quality.
Pharma nearest = this.getNearestAgent(main.pharmas);
if (this.getConnectionsNumber()>0){
List <Pharma> x = findAll(main.pharmas, l->l.distanceTo(this)<400);
Pharma y = top(x, s->s.serviceQuality);
if (randomTrue(0.4)){
return y;
}else{
return nearest;
}
}else{
return nearest;
}