使用stream判断两个list元素的属性并输出
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/*** 使用stream判断两个list中元素不同的item*/@Testpublic 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@AllArgsConstructorpublic static class Param{private Integer id;private String name;} |
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/** * 使用stream判断两个list中元素不同的item */@Testpublic 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@EqualsAndHashCodepublic 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
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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)
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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
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//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服务器之家。
原文链接：https://enjoy-day.blog.csdn.net/article/details/116456140
