例子:
List二,多个属性在某个List集合中进行过滤orgs = sysOrgMapper.queryOrgsPaging(map); List orgIds=orgs.stream().map(QueryOrgsResp::getOrgId) .collect(Collectors.toList()); 还可以用Set集合,例如: Set orgIds = orgs.stream().map(QueryOrgsResp::getOrgId) .collect(Collectors.toSet());
List三,把一个List通过一个属性分组形成MapsysOrgs = new LambdaQueryChainWrapper<>(sysOrgMapper) .list(); List checkStudents = students.stream() .filter(SchoolStudent -> SchoolStudent.getStudentName().equals(schoolStudent.getStudentName())) .filter(SchoolStudent -> !SchoolStudent.getStudentId().equals(schoolStudent.getStudentId())) .filter(SchoolStudent -> SchoolStudent.getSchoolId().equals(schoolStudent.getSchoolId())) .collect(Collectors.toList());
List四,Guava用法总结subjectDatas = new ArrayList<>(); Map > gradeMap = subjectDatas.stream().collect(Collectors.groupingBy(SubjectRes -> SubjectRes.getGradeId()));//年级分组 gradeMap.forEach((gradeId, gradeList) -> { });
https://blog.csdn.net/u013278314/article/details/85100115
常用的几种:
1,集合的创建
List list =Lists.newArrayList();
List list2 = Lists.newArrayList(“a”,“b”,“c”);
Set set = Sets.newHashSet();
2,Lists.partition可以将一个大的集合分割成小集合,适用于分批查询、插入等场景。
List listStr = Lists.newArrayList(“1”, “2”, “3”,“4”,“5”,“6”,“7”);
List batchList = Lists.partition(listStr,3);
//被分割成了: [[1, 2, 3], [4, 5, 6], [7]]
3,字符串处理
3.1拼接
Joiner 可以快速地把多个字符串或字符串数组连接成为用特殊符号连接的字符串。
List list = Lists.newArrayList(“a”,“b”,“c”);
String value =Joiner.on("-").skipNulls().join(list);
System.out.println(value);
//输出为: a-b-c
3.2切割 Splitter用来分割字符串
String testString = “Monday,Tuesday,Thursday,Friday,”;
//英文分号分割;忽略空字符串
Splitter splitter = Splitter.on(",").omitEmptyStrings().trimResults();
System.out.println(splitter.split(testString).toString());
//转换为了:[Monday, Tuesday, Thursday, Friday]
1,String转list
//字符串转list
String str = “asdfghjkl”;
List lis = Arrays.asList(str.split(""));
2,list转字符串
String.join("", lis);
https://blog.csdn.net/weixin_33941350/article/details/91477207
public static void main(String[] args) { List七,一个List中某个属性相同进行和并,形成一个新的Listlist1 = new ArrayList (); list1.add("1"); list1.add("2"); list1.add("3"); list1.add("5"); list1.add("6"); List list2 = new ArrayList (); list2.add("2"); list2.add("3"); list2.add("7"); list2.add("8"); // 交集 List intersection = list1.stream().filter(item -> list2.contains(item)).collect(toList()); System.out.println("---交集 intersection---"); intersection.parallelStream().forEach(System.out :: println); // 差集 (list1 - list2) List reduce1 = list1.stream().filter(item -> !list2.contains(item)).collect(toList()); System.out.println("---差集 reduce1 (list1 - list2)---"); reduce1.parallelStream().forEach(System.out :: println); // 差集 (list2 - list1) List reduce2 = list2.stream().filter(item -> !list1.contains(item)).collect(toList()); System.out.println("---差集 reduce2 (list2 - list1)---"); reduce2.parallelStream().forEach(System.out :: println); // 并集 List listAll = list1.parallelStream().collect(toList()); List listAll2 = list2.parallelStream().collect(toList()); listAll.addAll(listAll2); System.out.println("---并集 listAll---"); listAll.parallelStream().forEachOrdered(System.out :: println); // 去重并集 List listAllDistinct = listAll.stream().distinct().collect(toList()); System.out.println("---得到去重并集 listAllDistinct---"); listAllDistinct.parallelStream().forEachOrdered(System.out :: println); System.out.println("---原来的List1---"); list1.parallelStream().forEachOrdered(System.out :: println); System.out.println("---原来的List2---"); list2.parallelStream().forEachOrdered(System.out :: println); }
List八,判断一个List中的元素值重复newTeacherSimpleDtoList = new ArrayList<>();//定义新的集合 //TODO list转换 for (TeacherSimpleDto oldTeacherSimpleDto : teachers) { // 遍历新的List,看是否存在,存在则拼接,不存在则放入新的List boolean flag = false; for (TeacherSimpleDto newTeacherSimpleDto : newTeacherSimpleDtoList) { // 新的List中存在名字相同的,存在则拼接 if (newTeacherSimpleDto.getTeacherMobile().equals(oldTeacherSimpleDto.getTeacherMobile())) { newTeacherSimpleDto.setTeacherMobile(oldTeacherSimpleDto.getTeacherMobile()); newTeacherSimpleDto.setTeacherName(oldTeacherSimpleDto.getTeacherName()); } flag = true; } } // 新的List中不存在 if (!flag) { newTeacherSimpleDtoList.add(oldTeacherSimpleDto); } }
// 根据指定属性分组,并统计数量(key:指定属性,value:数量) Map九,解决n+1问题
Map十,集合中多条件排序问题curStudentIdMap = db.select(qSchoolClassStudent.studentId).from(qSchoolClassStudent) .where(qSchoolClassStudent.classId.eq(oldbaseClassId)) .fetch(String.class).stream().collect(Collectors.toMap(s -> s, s -> s)); curStudentIdMap.get(studentInfo.getStudentId()))
map分组的有序排序 用linkedHashMap实现 另外List多条件排序用stream实现
Map> classMap = list.stream().sorted(Comparator.comparing(ExportStudentResp::getGradeId) .thenComparing(ExportStudentResp::getClassType) .thenComparing(ExportStudentResp::getClassCode)) .collect(Collectors.groupingBy(ExportStudentResp -> ExportStudentResp.getClassId(), linkedHashMap::new, Collectors.toList()));//班级分组 classMap.forEach((subClassId, studentList) -> { });
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