对于每对元素,我们将键和值放入结果 HashMap 中,就像前一个示例中一样。 7. 结论 在本文中,我们通过示例学习了三种将两个给定List合并为Map的方法。 首先,我们基于随机访问的列表使用了 for 循环和Stream解决了这个问题。然后,我们讨论了随机访问方法的性能问题,当我们的输入是LinkedList时。 最后,我们看到了基于I...
map1.put("num1",1); Map<String,Object> map2=new HashMap<>(); map2.put("date","2021-11-25"); map2.put("num1",2); Map<String,Object> map3=new HashMap<>(); map3.put("date","2021-11-26"); map3.put("num1",3); Map<String,Object> map4=new HashMap<>(); map4.pu...
val map1 = mutable.Map("a"->1, "b"->2, "c"->3) val map2 = mutable.Map("a"->4, "b"->5, "d"->6) val map3: mutable.Map[String, Int] = map2.foldLeft(map1) { (map, kv) => { val k = kv._1 val v = kv._2 map(k) = map.getOrElse(k, 0) + v map } ...
get("a_id"); })).entrySet().stream().map(o->{ //合并 Map<String, Object> map = o.getValue().stream().flatMap(m->{ return m.entrySet().stream(); }).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (a,b)->b)); //为没有的key赋值0 set.stream().forEac...
map<String,Object> map1 = new HashMap<>(); map1.put("a_id",1); map1.put("in_num",10); map<String,Object> map2 = new HashMap<>(); map1.put("a_id",3); map1.put("in_num",10); map<String,Object> map3 = new HashMap<>(); map1.put("a_id",4); map1.put("in...
JAVA合并两个具有相同key的map为list,不多说,直接上代码: 代码语言:javascript 复制 publicclassMapUtil{publicstaticvoidmain(String[]args){List<Map<String,String>>osvList=newArrayList<>();Map<String,String>map1=newHashMap<>();map1.put("osV","5.1");map1.put("gaidNum","100");Map<String,Str...
// 将list转换成Map类型 Map<String, String> map = list.stream().collect(Collectors.toMap(Person::getId, Person::getName)); // 如果报 map里的value空指针异常,则需要在value,也就是toMap()的第二个参数进行空(null)值的判断逻辑;例如:也就是 Person::getName 改成 p -> p.getName()==null?
{//合并Map<String,Object> map =o.getValue().stream().flatMap(m ->{returnm.entrySet().stream();}).collect(Collectors.toMap(Map.Entry::getKey,Map.Entry::getValue,(a,b)->b));//为没有key的赋值0set.stream().forEach(k->{if(!map.containsKey(k))map.put(k,0);});returnmap;})...
1 2 3 4 5 6 publicstaticvoidmain(String[] args) {<br> List<String> idList =newArrayList<>(Arrays.asList("姓名","性别")); List<String> nameList =newArrayList<>(Arrays.asList("杰克","男"));<br> Map<String,String> listMap = idList.stream().collect(Collectors.toMap(key->key, key...
Map<String,Stu> stus = new HashMap<String,Stu>();for (Stu stu : list1) { if (stus.containsKey(stu.getName())) { Stu temp = stus.get(stu.getName());temp.setDesc(temp.getDesc()+stu.getDesc());stus.put(stu.getName(),temp);} else { stus.put(stu.getName(), ...