案例:
某公司的开发部门,分为开发一部和开发二部,现在需要进行年中数据结算。
分析:
- 员工信息至少包含了(名称,性别,工资,奖金,处罚记录)。
- 开发一部个员工,开发二部有5名员工。
- 分别筛选出2个部门的最高工资的员工信息,封装成优秀员工对象Topperformer。
- 分别统计出2个部门的平均月收入,要求去掉最高和最低工资。
- 统计2个开发部门整体的平均工资,去掉最低和最高工资的平均值。
员工信息类:
public class Employee {
private String name;//名称
private char sex;//性别
private double salary;//工资
private double bonus;//奖金
private String punish;//处罚记录
public Employee() {
}
public Employee(String name, char sex, double salary, double bonus, String punish) {
this.name = name;
this.sex = sex;
this.salary = salary;
this.bonus = bonus;
this.punish = punish;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public char getSex() {
return sex;
}
public void setSex(char sex) {
this.sex = sex;
}
public double getSalary() {
return salary;
}
public void setSalary(double salary) {
this.salary = salary;
}
public double getBonus() {
return bonus;
}
public void setBonus(double bonus) {
this.bonus = bonus;
}
public String getPunish() {
return punish;
}
public void setPunish(String punish) {
this.punish = punish;
}
@Override
public String toString() {
return "Employee{" +
"name='" + name + '\'' +
", sex=" + sex +
", salary=" + salary +
", bonus=" + bonus +
", punish='" + punish + '\'' +
'}';
}
}
封装成优秀员工对象Topperformer:
public class Topperformer {
private String name;
private double money;//月薪
public Topperformer() {
}
public Topperformer(String name, double money) {
this.name = name;
this.money = money;
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public double getMoney() {
return money;
}
public void setMoney(double money) {
this.money = money;
}
@Override
public String toString() {
return "Topperformer{" +
"name='" + name + '\'' +
", money=" + money +
'}';
}
}
案例实现:
import java.math.BigDecimal;
import java.math.RoundingMode;
import java.util.ArrayList;
//import java.util.Comparator;
import java.util.List;
import java.util.stream.Stream;
/**
* Stream的综合案例:某公司的开发部门,分为开发一部和开发二部,现在需要进行年中数据结算。
* 1.员工信息至少包含了(名称,性别,工资,奖金,处罚记录)。
* 2.开发一部个员工,开发二部有5名员工。
* 3.分别筛选出2个部门的最高工资的员工信息,封装成优秀员工对象Topperformer。
* 4.分别统计出2个部门的平均月收入,要求去掉最高和最低工资。
* 5.统计2个开发部门整体的平均工资,去掉最低和最高工资的平均值。
*/
public class StreamDemo {
//4.创建共享变量,用于统计出2个部门的平均月收入中的求和
public static double allMoney;
public static double allMoney2;//两个部门综合
public static void main(String[] args) {
//2.定义两个部门
//2.1开发一部
List<Employee> one = new ArrayList<>();
one.add(new Employee("小小1",'男',30000,25000,null));
one.add(new Employee("小小2",'男',25000,1000,"顶撞上司"));
one.add(new Employee("小小3",'男',20000,20000,null));
one.add(new Employee("小小4",'男',20000,25000,null));
//2.2开发二部
List<Employee> two = new ArrayList<>();
two.add(new Employee("大大1",'男',15000,9000,null));
two.add(new Employee("大大2",'男',25000,10000,null));
two.add(new Employee("大大3",'男',50000,100000,"顶撞上司"));
two.add(new Employee("大大4",'女',3500,1000,"顶撞上司"));
two.add(new Employee("大大4",'女',20000,0,"顶撞上司"));
//3.筛选最高工资员工 ==》制定大小规则
// System.out.println(one.stream().max(new Comparator<Employee>() {
// @Override
// public int compare(Employee e1, Employee e2) {
// return Double.compare(e1.getSalary() + e1.getBonus(), e2.getSalary() + e2.getBonus());
// }
// }).get());
//Employee{name='小小1', sex=男, salary=30000.0, bonus=25000.0, punish='null'}
//简化:
// System.out.println(one.stream().max(( e1, e2) -> Double.compare(e1.getSalary() + e1.getBonus(), e2.getSalary() + e2.getBonus())).get());
Employee e = one.stream().max((e1,e2)->Double.compare(e1.getSalary()+ e1.getBonus(), e2.getSalary()+ e2.getBonus())).get();
System.out.println(e);//Employee{name='小小1', sex=男, salary=30000.0, bonus=25000.0, punish='null'}
Employee ee = two.stream().max((e1,e2)->Double.compare(e1.getSalary()+ e1.getBonus(), e2.getSalary()+ e2.getBonus())).get();
System.out.println(ee);//Employee{name='大大3', sex=男, salary=50000.0, bonus=100000.0, punish='顶撞上司'}
//封装成优秀员工对象Topperformer==>使用map加工
Topperformer t = one.stream().max((e1,e2)->Double.compare(e1.getSalary()+ e1.getBonus(), e2.getSalary()+ e2.getBonus()))
.map(e1 -> new Topperformer(e1.getName(),e1.getSalary()+e1.getBonus())).get();
System.out.println(t);//Topperformer{name='小小1', money=55000.0}
Topperformer tt = two.stream().max((e1,e2)->Double.compare(e1.getSalary()+ e1.getBonus(), e2.getSalary()+ e2.getBonus()))
.map(e1 -> new Topperformer(e1.getName(),e1.getSalary()+e1.getBonus())).get();
System.out.println(tt);//Topperformer{name='大大3', money=150000.0}
//4.分别统计出2个部门的平均月收入,要求去掉最高和最低工资。
one.stream().sorted((e1,e2) -> Double.compare(e1.getSalary()+ e1.getBonus(), e2.getSalary()+ e2.getBonus()))
.skip(1).limit(one.size() - 2).forEach(e1 ->{
//求出总和,剩余工资的总和
allMoney += (e1.getSalary()+e1.getBonus());
});
System.out.println("开发一部的平均工资是:" + allMoney/(one.size() - 2));
two.stream().sorted((e1,e2) -> Double.compare(e1.getSalary()+ e1.getBonus(), e2.getSalary()+ e2.getBonus()))
.skip(1).limit(two.size() - 2).forEach(e1 ->{
//求出总和,剩余工资的总和
allMoney += (e1.getSalary()+e1.getBonus());
});
//System.out.println("开发二部的平均工资是:" + allMoney/(two.size() - 2));
//开发二部的平均工资是:54666.666666666664 ==》精度失真 ==》解决办法如下
BigDecimal a = BigDecimal.valueOf(allMoney);
BigDecimal b = BigDecimal.valueOf(two.size() - 2);
System.out.println("开发二部的平均工资是:" + a.divide(b,2, RoundingMode.HALF_UP));//结果四舍五入保留两位小数
//开发二部的平均工资是:54666.67
//5.统计2个开发部门整体的平均工资,去掉最低和最高工资的平均值。
//5.1合并两个集合流,再统计
Stream<Employee> s1 = one.stream();
Stream<Employee> s2 = two.stream();
Stream<Employee> s3 = Stream.concat(s1,s2);
s3.sorted((e1,e2) -> Double.compare(e1.getSalary()+ e1.getBonus(), e2.getSalary()+ e2.getBonus()))
.skip(1).limit((one.size()+two.size()) - 2).forEach(e1 ->{
//求出总和,剩余工资的总和
allMoney2 += (e1.getSalary()+e1.getBonus());
});
System.out.println("2个开发部门的平均工资是:" + allMoney2/((one.size()+two.size()) - 2));
}
}