Java8 Stream流操作在用户系统中的妙用

在做目前这个项目的时候，发现以前有一个筛选的需求，老程序员是这么做的，先请求Http服务器得到一长串json数据，大概用A4纸打了40多页那么多，然后将这些对象写入到sqlite数据库中，再用数据库查询语句根据筛选条件查出来。最后将数据库丢弃。把我们这些新程序员看的目瞪口呆。自从接触了Java8之后，发现可以像操作数据库一样操作内存，而且在Stream操作中对内存的开销十分友善，操作方式十分灵活，减少了IO的支出，简直爽歪歪。

传统的数据处理都是用循环来解决，而不是像搜索数据库那样有具体的搜索语句，而Java8的Stream提供了很好的方案，往往一行就搞定了，而且Stream还可以链式操作，一行代码实现多个循环的功能，代码风格十分像nosql数据库，但是在实际应用中发现一个巨大的问题，就是执行耗时特别长，时间开销是传统方法的几百倍，这是一个巨大的问题。

本文主要来讨论一下如何发挥Stream的优势展示对用户管理操作

首先我们制造一个User类用来代表用户，里面有姓名年龄密码等常用字段，顺道再写个构造函数和toString()，如下

public class User{public int age;//年龄public String name;//姓名private String password;//密码public short gendar;//性别，0未知，1男，2女public boolean hasMarried;//是否已婚public String getPassword() {return password;}public User(int age, String name, String password, short gendar,boolean hasMarried) {super();this.age = age;this.name = name;this.password = password;this.gendar = gendar;this.hasMarried = hasMarried;}@Overridepublic String toString() {return "{\"age\":\"" + age + "\", \"name\":\"" + name+ "\", \"password\":\"" + password + "\", \"gendar\":\""+ gendar + "\", \"hasMarried\":\"" + hasMarried + "\"} \n";}}

现在我们伪造一点数据，暂时就用我大学同学的名字吧：

ArrayList<User> users = new ArrayList<User>();users.add(new User(22, "王旭", "123456", (short)1, true));users.add(new User(22, "王旭", "123456", (short)1, true));users.add(new User(22, "王旭", "123456", (short)1, true));users.add(new User(21, "孙萍", "a123456", (short)2, false));users.add(new User(23, "步传宇", "b123456", (short)1, false));users.add(new User(18, "蔡明浩", "c123456", (short)1, true));users.add(new User(17, "郭林杰", "d123456", (short)1, false));users.add(new User(5, "韩凯", "e123456", (short)1, true));users.add(new User(22, "韩天琪", "f123456", (short)2, false));users.add(new User(21, "郝玮", "g123456", (short)2, false));users.add(new User(19, "胡亚强", "h123456", (short)1, false));users.add(new User(14, "季恺", "i123456", (short)1, false));users.add(new User(17, "荆帅", "j123456", (short)1, true));users.add(new User(16, "姜有琪", "k123456", (short)1, false));

场景一、对用户进行排序

首先我们制定一个排序规则：按照年龄大小进行排序，设计一个Comparator

Comparator<User> ageComparator = new Comparator<User>() {@Overridepublic int compare(User o1, User o2) {// TODO Auto-generated method stubif(o1.age>o2.age)return 1;if(o1.age<o2.age)return -1;return 0;}};

传统方式排序：

time = System.currentTimeMillis();Collections.sort(users, ageComparator);System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(users);

输出结果：

耗时0
[{"age":"5", "name":"韩凯", "password":"e123456", "gendar":"1", "hasMarried":"true"}
, {"age":"14", "name":"季恺", "password":"i123456", "gendar":"1", "hasMarried":"false"}
, {"age":"16", "name":"姜有琪", "password":"k123456", "gendar":"1", "hasMarried":"false"}
, {"age":"17", "name":"郭林杰", "password":"d123456", "gendar":"1", "hasMarried":"false"}
, {"age":"17", "name":"荆帅", "password":"j123456", "gendar":"1", "hasMarried":"true"}
, {"age":"18", "name":"蔡明浩", "password":"c123456", "gendar":"1", "hasMarried":"true"}
, {"age":"19", "name":"胡亚强", "password":"h123456", "gendar":"1", "hasMarried":"false"}
, {"age":"21", "name":"孙萍", "password":"a123456", "gendar":"2", "hasMarried":"false"}
, {"age":"21", "name":"郝玮", "password":"g123456", "gendar":"2", "hasMarried":"false"}
, {"age":"22", "name":"王旭", "password":"123456", "gendar":"1", "hasMarried":"true"}
, {"age":"22", "name":"王旭", "password":"123456", "gendar":"1", "hasMarried":"true"}
, {"age":"22", "name":"王旭", "password":"123456", "gendar":"1", "hasMarried":"true"}
, {"age":"22", "name":"韩天琪", "password":"f123456", "gendar":"2", "hasMarried":"false"}
, {"age":"23", "name":"步传宇", "password":"b123456", "gendar":"1", "hasMarried":"false"}
]

Java8的方式排序：

long time = System.currentTimeMillis();List<User> sortedUsers = users.stream().sorted(ageComparator).collect(Collectors.toList());System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(sortedUsers);

输出结果相同，耗时234ms

从结果来看，java8 Stream操作的耗时至少是传统方法的200多倍，时间成本较大。

场景一（2）选出年龄最小的三个人

有时候我们也许并不需要获得排序的所有结果，只需要获得前几名就可以了，比如我想获得年龄最小的三个人

传统方法排序限制：

首先进行上面的排序，然后取出数组的前三个元素

Collections.sort(users, ageComparator);users.subList(0, 2);

耗时0

Java8方式排序限制：

long time = System.currentTimeMillis();List<User> resultArr = users.stream().sorted(ageComparator).limit(3).collect(Collectors.toList());System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(resultArr);

结果：耗时375
[{"age":"5", "name":"韩凯", "password":"e123456", "gendar":"1", "hasMarried":"true"}
, {"age":"14", "name":"季恺", "password":"i123456", "gendar":"1", "hasMarried":"false"}
, {"age":"16", "name":"姜有琪", "password":"k123456", "gendar":"1", "hasMarried":"false"}
]

场景二：去除重复数据

为了识别两个对象是否重复，需要复写User的equals方法

@Overridepublic boolean equals(Object obj) {// TODO Auto-generated method stubif(!(obj instanceof User))return false;User u = (User)obj;if(age != u.age|| gendar!=u.gendar|| hasMarried!=u.hasMarried|| !name.equals(u.name)|| !password.equals(u.getPassword()))return false;return true;}

要实现去重，必须首先将数组的顺序严格排好，也就是相似的数据要放在一起，便于排序，所以我们要重写一下比较器

Comparator<User> equalComparator = new Comparator<User>() {@Overridepublic int compare(User o1, User o2) {// TODO Auto-generated method stub//首先比较年龄大小，因为年龄的区分度比较高if(o1.age>o2.age)return 1;if(o1.age<o2.age)return -1;//如果年龄相同就比较性别，女的排在前面if(o1.gendar>o2.gendar)return 1;if(o1.gendar<o2.gendar)return -1;//如果性别也一样就比较是否已婚if(o1.hasMarried == true && o2.hasMarried==false)return 1;//结婚的排在前面if(o1.hasMarried == false && o2.hasMarried==true)return 1;//结婚的排在前面//最后比较姓名，因为字符串比较耗时较长if(o1.name.hashCode()>o2.name.hashCode())return 1;if(o1.name.hashCode()<o2.name.hashCode())return -1;return 0;}};

传统方法去重：

Collections.sort(users, ageComparator);time = System.currentTimeMillis();int length = users.size();for(int i=1;i<length;i++){if(users.get(i).equals(users.get(i-1))){users.remove(i);i--;length--;}}System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(users);

输出结果：

耗时0
[{"age":"5", "name":"韩凯", "password":"e123456", "gendar":"1", "hasMarried":"true"}
, {"age":"14", "name":"季恺", "password":"i123456", "gendar":"1", "hasMarried":"false"}
, {"age":"16", "name":"姜有琪", "password":"k123456", "gendar":"1", "hasMarried":"false"}
, {"age":"17", "name":"郭林杰", "password":"d123456", "gendar":"1", "hasMarried":"false"}
, {"age":"17", "name":"荆帅", "password":"j123456", "gendar":"1", "hasMarried":"true"}
, {"age":"18", "name":"蔡明浩", "password":"c123456", "gendar":"1", "hasMarried":"true"}
, {"age":"19", "name":"胡亚强", "password":"h123456", "gendar":"1", "hasMarried":"false"}
, {"age":"21", "name":"孙萍", "password":"a123456", "gendar":"2", "hasMarried":"false"}
, {"age":"21", "name":"郝玮", "password":"g123456", "gendar":"2", "hasMarried":"false"}
, {"age":"22", "name":"王旭", "password":"123456", "gendar":"1", "hasMarried":"true"}
, {"age":"22", "name":"韩天琪", "password":"f123456", "gendar":"2", "hasMarried":"false"}
, {"age":"23", "name":"步传宇", "password":"b123456", "gendar":"1", "hasMarried":"false"}
]

Java8 去重：

Stream去重有一个先决条件，就是要去重的对象必须实现comparable接口，不能使用比较器，于是让user类implement comparable，并复写其方法compareTo

public int compareTo(User o2) {// TODO Auto-generated method stub//首先比较年龄大小，因为年龄的区分度比较高User o1 = this;if(o1.age>o2.age)return 1;if(o1.age<o2.age)return -1;//如果年龄相同就比较性别，女的排在前面if(o1.gendar>o2.gendar)return 1;if(o1.gendar<o2.gendar)return -1;//如果性别也一样就比较是否已婚if(o1.hasMarried == true && o2.hasMarried==false)return 1;//结婚的排在前面if(o1.hasMarried == false && o2.hasMarried==true)return 1;//结婚的排在前面//最后比较姓名，因为字符串比较耗时较长if(o1.name.hashCode()>o2.name.hashCode())return 1;if(o1.name.hashCode()<o2.name.hashCode())return -1;return 0;}

然后先调用stream的sort方法进行排序，再调用disdinct方法进行去重，结果同上面一样，耗时249ms

场景三：按条件筛选

这种场景也许是最常见的一种应用场景了，在许多元素构成的数组中筛选出我们需要的满足特定条件的元素，在这里我们把所有姓韩的筛选出来

传统方法：

long time = System.currentTimeMillis();ArrayList<User> resultArr = new ArrayList<User>();//用于存放结果for(User u:users){if(u.name.startsWith("韩"))resultArr.add(u);}System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(resultArr);

结果：耗时0
[{"age":"5", "name":"韩凯", "password":"e123456", "gendar":"1", "hasMarried":"true"}
, {"age":"22", "name":"韩天琪", "password":"f123456", "gendar":"2", "hasMarried":"false"}
]

Java8方法：

long time = System.currentTimeMillis();List<User> resultArr = users.stream().filter(t->t.name.startsWith("韩")).collect(Collectors.toList());System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(resultArr);

结果相同，耗时266ms

其中，java8采用的泛型进行处理，上面的t->t.name中的t是Stream<T>的泛型，而这个T又是List<T>中的泛型，t可以换成其他任何字母，并且也可以点出User类的相关方法，并且还可以支持复合筛选，比如我们要筛选姓韩的女生，可以这样写

List<User> resultArr = users.stream().filter(t->t.name.startsWith("韩") && t.gendar==2).collect(Collectors.toList());

结果：耗时281
[{"age":"22", "name":"韩天琪", "password":"f123456", "gendar":"2", "hasMarried":"false"} ]

也可以这样写

List<User> resultArr = users.stream().filter(t->t.name.startsWith("韩")).filter(t->t.gendar==2).collect(Collectors.toList());

耗时296ms

还可以改变条件的顺序：

List<User> resultArr = users.stream().filter(t->t.gendar==2).filter(t->t.name.startsWith("韩")).collect(Collectors.toList());

耗时265ms

场景四：只列出所有人的名字和婚姻状况

这次要用的.map()函数，map()就是为了只显示对象的一部分信息而准备的。

传统方式：

long time = System.currentTimeMillis();ArrayList<String> marryStatus = new ArrayList<String>();for(User u:users){marryStatus.add(u.name+":".concat(u.hasMarried?"已婚":"未婚")+"\n");}System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(marryStatus);

耗时：0

[王旭:已婚
, 王旭:已婚
, 王旭:已婚
, 孙萍:未婚
, 步传宇:未婚
, 蔡明浩:已婚
, 郭林杰:未婚
, 韩凯:已婚
, 韩天琪:未婚
, 郝玮:未婚
, 胡亚强:未婚
, 季恺:未婚
, 荆帅:已婚
, 姜有琪:未婚
]

java8方式：

long time = System.currentTimeMillis();List<String> marryStatus = users.stream().map(t->t.name+":".concat(t.hasMarried?"已婚":"未婚")+"\n").collect(Collectors.toList());System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(marryStatus);

结果相同：耗时234

场景五：判断当前数组是否包含某些特定元素

如果我要看看现在的用户中是否有未成年人怎么办呢

传统方法：

long time = System.currentTimeMillis();boolean isChild = false;for(User u:users){if(u.age<18){isChild = true;break;}}System.out.println("耗时"+(System.currentTimeMillis()-time)+hasChild);

结果：耗时0 true

Java8方法：

long time = System.currentTimeMillis();boolean isChild = users.stream().anyMatch(t->t.age<18);System.out.println("耗时"+(System.currentTimeMillis()-time)+isChild);

耗时78ms 结果相同

场景六：确认所有元素均满足某一条件

这里以查看所有人是否都已婚为例

传统方法：

long time = System.currentTimeMillis();boolean allMarried = true;for(User u:users){if(!u.hasMarried){allMarried = false;break;}}System.out.println("耗时"+(System.currentTimeMillis()-time)+allMarried);

结果：耗时0 false

java8方法：

long time = System.currentTimeMillis();boolean allMarried = users.stream().allMatch(t->t.hasMarried);System.out.println("耗时"+(System.currentTimeMillis()-time)+allMarried);

结果相同，耗时46ms

场景七：求和求平均值

求和这种操作在用户管理上十分频繁，java8的流操作省去了循环，节省了大量代码，比如我们要求所有用户的平均年龄

传统方法:

long time = System.currentTimeMillis();int sum = 0;for(User u:users){sum+=u.age;}System.out.println("耗时"+(System.currentTimeMillis()-time)+sum/users.size());

结果：耗时0 平均年龄18

java8方法：

这里先用map方法把所有元素的age取出来，然后调用Integer.sum方法进行聚合（reduce函数），得到年龄和，返回是一个OptionalInt对象，这里面包含一个int，但也有可能为null，注意这里reduce()函数的参数是一个方法，注意Java8支持将函数作为参数传入了，有点像c++写法，规则是完整类名::方法名(方法参数...)

long time = System.currentTimeMillis();OptionalInt sum = users.stream().mapToInt(t->t.age).reduce(Integer::sum);System.out.println("耗时"+(System.currentTimeMillis()-time)+sum.getAsInt()/users.size());

结果相同，耗时63ms

场景八：分组

比如我们要按用户的年龄进行分组，相同年龄的人分在同一组，用一个Map<Integer,List<User>>存放，key是年龄，value是该年龄的所有用户

传统方法：

long time = System.currentTimeMillis();Map<Integer,List<User>> group = new HashMap<Integer,List<User>>();for(User u:users){List<User> list = group.get(u.age);if(list==null){list = new ArrayList<User>();group.put(u.age,list);}list.add(u);}System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(group);

结果：

耗时0
{16=[{"age":"16", "name":"姜有琪", "password":"k123456", "gendar":"1", "hasMarried":"false"} 
], 17=[{"age":"17", "name":"郭林杰", "password":"d123456", "gendar":"1", "hasMarried":"false"} 
, {"age":"17", "name":"荆帅", "password":"j123456", "gendar":"1", "hasMarried":"true"} 
], 18=[{"age":"18", "name":"蔡明浩", "password":"c123456", "gendar":"1", "hasMarried":"true"} 
], 19=[{"age":"19", "name":"胡亚强", "password":"h123456", "gendar":"1", "hasMarried":"false"} 
], 21=[{"age":"21", "name":"孙萍", "password":"a123456", "gendar":"2", "hasMarried":"false"} 
, {"age":"21", "name":"郝玮", "password":"g123456", "gendar":"2", "hasMarried":"false"} 
], 5=[{"age":"5", "name":"韩凯", "password":"e123456", "gendar":"1", "hasMarried":"true"} 
], 22=[{"age":"22", "name":"王旭", "password":"123456", "gendar":"1", "hasMarried":"true"} 
, {"age":"22", "name":"王旭", "password":"123456", "gendar":"1", "hasMarried":"true"} 
, {"age":"22", "name":"王旭", "password":"123456", "gendar":"1", "hasMarried":"true"} 
, {"age":"22", "name":"韩天琪", "password":"f123456", "gendar":"2", "hasMarried":"false"} 
], 23=[{"age":"23", "name":"步传宇", "password":"b123456", "gendar":"1", "hasMarried":"false"} 
], 14=[{"age":"14", "name":"季恺", "password":"i123456", "gendar":"1", "hasMarried":"false"} 
]}

Java8方法：

long time = System.currentTimeMillis();Map<Integer,List<User>> group = users.stream().collect(Collectors.groupingBy(t->t.age));System.out.println("耗时"+(System.currentTimeMillis()-time));System.out.println(group);

结果相同，耗时62ms

如果想按是否结婚分组，也就是key变成bool，那就应该这么写

Map<Boolean,List<User>> group = users.stream().collect(Collectors.partitioningBy(t->t.hasMarried));

耗时也是62ms

场景九：链式操作

如果我们需要打印所有女生的名字，那么同样可以一行代码搞定，思路是先通过源Stream通过筛选得到一个新Stream，再对这个新的Stream进行操作，如此循环,注意这里使用的forEach()函数是遍历Stream中的每一个元素，参数是方法，注意Java8支持将函数作为参数传入了，有点像c++写法，规则是完整类名::方法名(方法参数...)

long time = System.currentTimeMillis();users.stream().filter(t->t.gendar==2).map(t->t.name).forEach(System.out::println);System.out.println("耗时"+(System.currentTimeMillis()-time));

结果：孙萍
韩天琪
郝玮
耗时47