如何使用Java 8流通过使用带有过滤器的其他列表对象值来创建新列表？

这是代码，在注释中有解释。

import java.util.List;
import java.util.stream.Collectors;
import java.util.stream.Stream;

//Note - Object.toString() was defined for AAA & BBB classes. Their member variables are public just to reduce code.
public class Test {
    public static void main(String [] args){
        //Generate some test data.
        List<AAA> testData = Stream
                .of(1,2,3,4,5)
                .map(i -> new AAA(i+"", (i%3)+"", i*3+"", i*4+""))
                .collect(Collectors.toList());//% is modulo operator.

        System.out.println("Test data:\n");
        testData.forEach(System.out::println);

        List<BBB> processedData = testData
                .stream()
                //If a.A1 = a.A2, then allow it.
                .filter(a -> a.A1.equals(a.A2))
                //Take an AAA and convert it into a BBB.
                .map(a -> new BBB(a.A3, a.A4))
                //Collect all the BBBs and put them in a list.
                .collect(Collectors.toList());

        System.out.println("Processed data:\n");
        processedData.forEach(System.out::println);
    }
}

输出：

Test data:

AAA{A1='1', A2='1', A3='3', A4='4'}
AAA{A1='2', A2='2', A3='6', A4='8'}
AAA{A1='3', A2='0', A3='9', A4='12'}
AAA{A1='4', A2='1', A3='12', A4='16'}
AAA{A1='5', A2='2', A3='15', A4='20'}
Processed data:

BBB{A3='3', A4='4'}
BBB{A3='6', A4='8'}

回答来源：Stack Overflow