如何获取与流中的条件匹配的第一个元素?我试过这个但是没有用
this.stops.stream().filter(Stop s-> s.getStation().getName().equals(name));
该条件不起作用,过滤器方法在Stop之外的其他类中调用。
public class Train {
private final String name;
private final SortedSet<Stop> stops;
public Train(String name) {
this.name = name;
this.stops = new TreeSet<Stop>();
}
public void addStop(Stop stop) {
this.stops.add(stop);
}
public Stop getFirstStation() {
return this.getStops().first();
}
public Stop getLastStation() {
return this.getStops().last();
}
public SortedSet<Stop> getStops() {
return stops;
}
public SortedSet<Stop> getStopsAfter(String name) {
// return this.stops.subSet(, toElement);
return null;
}
}
import java.util.ArrayList;
import java.util.List;
public class Station {
private final String name;
private final List<Stop> stops;
public Station(String name) {
this.name = name;
this.stops = new ArrayList<Stop>();
}
public String getName() {
return name;
}
}
答案 0 :(得分:159)
这可能就是你要找的东西:
yourStream
.filter(/* your criteria */)
.findFirst()
.get();
一个例子:
public static void main(String[] args) {
class Stop {
private final String stationName;
private final int passengerCount;
Stop(final String stationName, final int passengerCount) {
this.stationName = stationName;
this.passengerCount = passengerCount;
}
}
List<Stop> stops = new LinkedList<>();
stops.add(new Stop("Station1", 250));
stops.add(new Stop("Station2", 275));
stops.add(new Stop("Station3", 390));
stops.add(new Stop("Station2", 210));
stops.add(new Stop("Station1", 190));
Stop firstStopAtStation1 = stops.stream()
.filter(e -> e.stationName.equals("Station1"))
.findFirst()
.get();
System.out.printf("At the first stop at Station1 there were %d passengers in the train.", firstStopAtStation1.passengerCount);
}
输出是:
At the first stop at Station1 there were 250 passengers in the train.
答案 1 :(得分:4)
当您编写lambda表达式时,->
左侧的参数列表可以是带括号的参数列表(可能为空),也可以是没有任何括号的单个标识符。但在第二种形式中,标识符不能使用类型名称声明。因此:
this.stops.stream().filter(Stop s-> s.getStation().getName().equals(name));
语法不正确;但
this.stops.stream().filter((Stop s)-> s.getStation().getName().equals(name));
是对的。或者:
this.stops.stream().filter(s -> s.getStation().getName().equals(name));
如果编译器有足够的信息来确定类型,也是正确的。
答案 2 :(得分:2)
我认为这是最好的方法:
this.stops.stream().filter(s -> Objects.equals(s.getStation().getName(), this.name)).findFirst().orElse(null);