java使用stream判断两个list元素的属性并输出方式

目录使用stream判断两个list元素的属性并输出stream判断列表是否包含某几个元素/重复元素代码SHOWjava stream判断列表是否包含重复元素

使用stream判断两个list元素的属性并输出

/**

* 使用stream判断两个list中元素不同的item

*/

@Test

public void test1(){

List stringList1 = new LinkedList(){{

add(new Param(1,"1111"));

add(new Param(2, "2222"));

add(new Param(3, "3333"));

}};

List stringList2 = new LinkedList(){{

add(new Param(1,"1111"));

add(new Param(2, "4444"));

add(new Param(3, "5555"));

}};

// 判断key相同，value相同的元素

Map tmpList2 = stringList2.stream().collect(Collectors.toMap(Param::getId, Param::getName));

var tmplist = stringList1.stream().filter(item -> (tmpList2.get(item.getId()) != null && tmpList2.get(item.getId()).equals(item.getName()))).collect(Collectors.toList());

System.out.println(tmplist);

}

@Setter

@Getter

@ToString

@AllArgsConstructor

public static class Param{

private Integer id;

private String name;

}

/**

* 使用stream判断两个list中元素不同的item

*/

@Test

public void test1(){

List stringList1 = new LinkedList(){{

add(new Param(1,"1111", "b"));

add(new Param(2, "2222", "c"));

add(new Param(3, "3333", "a"));

}};

List stringList2 = new LinkedList(){{

add(new Param(1,"1111", "c"));

add(new Param(2, "4444", "b"));

add(new Param(3, "5555", "a"));

}};

// 判断key相同，value相同的元素

Map tmpList2 = stringList2.stream().collect(Collectors.toMap(Param::getId, http://Param::getName));

var tmplist = stringList1.stream().filter(item -> (tmpList2.get(item.getId()) != null && tmpList2.get(item.getId()).equals(item.getName()))).collect(Collectors.toList());

System.out.println(tmplist);

// 如果需要判断多个值，直接将对象加入进去

Map tmpList3 = stringList2.stream().collect(Collectors.toMap(Param::getId, Function.identity()));

var tmplist2 = stringList1.stream().filter(item -> (tmpList3.get(item.getId()) != null && tmpList3.get(item.getId()).getType().equals(item.getType()))).collect(Collectors.toList());

System.out.println(tmplist2);

}

@Setter

@Getter

@ToString

@AllArgsConstructor

@EqualsAndHashCode

public static class Param{

private Integer id;

private String name;

private String type;

}

stream判断列表是否包含某几个元素/重复元素

(需求经过修改过)判断一个profile是否包含PROFILE-IN-A和PROFILE-IN-B且都是Enable=1打勾的.

既然已经JDK8了,那就用lambda吧,如果是foreach可能比较难处理,用stream的filter则可以这样做.

核心代码可以这么写

int intCheck = profileServiceDtoList.stream().filter(e ->

"1".equals(e.getEnable())

&&(("PROFILE-IN-MOSHOW".equals(e.getServiceIdentifier()))||("PROFILE-IN-ADC".equals(e.getServiceIdentifier())))

).collect(Collectors.toList()).size();

代码SHOW

新建三个不同类型的profile,其中两个是要判断的,一个是干扰的.通过steam进行filter,找出是否包含这两个元素(相当于把要判断的元素过滤进去)判断list的size大小(intCheck>1找到两个则代表同时出现)

public static void main(String[] args) {

List profileServiceDtoList= new ArrayList&UMnBAxSUlt;>(3);

ProfileServiceDto profileService1 = new ProfileServiceDto();

profileService1.setServiceId(1001L);

profileService1.setServiceIdentifier("PROFILE-IN-MOSHOW");

profileService1.setEnable("1");

profileServiceDtoList.add(profileService1);

ProfileServiceDto profileService2 = new ProfileServiceDto();

profileService2.setServiceId(1002L);

profileService2.setServiceIdentifier("PROFILE-IN-ADC");

profileService2.setEnable("1");

profileServiceDtoList.add(profileService2);

ProfileServiceDto profileService3 = new ProfileServiceDto();

profileService3.setServiceId(1003L);

profileService3.setServiceIdentifier("PROFILE-XXX-ABC");

profileService3.setEnable("1");

profileServiceDtoList.add(profileService3);

int intCheck = profileServiceDtoList.stream().filter(e ->

"1".equals(e.getEnable())&&(("PROFILE-IN-MOSHOW".equals(e.getServiceIdentifier()))||("PROFILE-IN-ADC".equals(e.getServiceIdentifier())))

).collect(Collectors.toList()).size();

System.out.println("intCheck->"+intCheck);

if(intCheck>1){

System.error.println("In one profile, cannot contain two more PROFILE-IN profile.");

}

}

Java stream判断列表是否包含重复元素

思路是通过一个distinct的list,然后跟原先的list来判断大小,如果不一致(原先list.size>distinctList.size)则表示有重复元素

//profileServiceDtoList路上,不累赘

//多了一个profileService1.setGroupId("A");profileService1.setGroupId("B");profileService3.setGroupId("A");

List groupList = new ArrayList<>(4);

profileServiceDtoList.stream().forEach(e -> {

if ("Y".equals(e.getEnable()) && StringUtils.isNotEmpty(e.getGroupId())) {

groupList.add(e.getGroupId());

}

});

int distinctGroupSize = groupList.stream().distinct().collect(Collectors.toList()).size();

if (groupList.size() > distinctGroupSize) {

throw new ValidationException("100001","In one profile, the services with the same groupId cannot co-exist.");

}

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