java使用stream判断两个list元素的属性并输出方式
作者：enjoy嚣士
这篇文章主要介绍了java使用stream判断两个list元素的属性并输出方式，具有很好的参考价值，希望对大家有所帮助。如有错误或未考虑完全的地方，望不吝赐教
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<Param> stringList1 = new LinkedList<Param>(){{
add(new Param(1,"1111"));
add(new Param(2, "2222"));
add(new Param(3, "3333"));
}};
List<Param> stringList2 = new LinkedList<Param>(){{
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<Integer, String> 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<Param> stringList1 = new LinkedList<Param>(){{
add(new Param(1,"1111", "b"));
add(new Param(2, "2222", "c"));
add(new Param(3, "3333", "a"));
}};
List<Param> stringList2 = new LinkedList<Param>(){{
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<Integer, String> 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);
// 如果需要判断多个值，直接将对象加入进去
Map<Integer, Param> 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<ProfileServiceDto> profileServiceDtoList= new ArrayList<>(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<String> 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.");
}以上为个人经验，希望能给大家一个参考，也希望大家多多支持脚本之家。