本文介绍了如何使用Java流比较两个ArrayList并使用过滤器获取list1的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有两个列表list1&列表类型的list2
I have two lists list1 & list2 of type List
Term{
long sId;
int rowNum;
long psid;
String name;
}
List<Term> list1 = new ArrayList<>();
List<Term> list2 = new ArrayList<>();
我想从list1返回所有项目,其中(list1.psid!= list2.psid).
I want to return all the items from list1 where (list1.psid != list2.psid).
我尝试了这个,但是没有用
I tried this but its not working
public List<Term> getFilteredRowNum(List<Term> list1, List<Term> list2) {
List<Long> psid = list2.stream().map(x -> x.getPsid()).collect(Collectors.toList());
return list1.stream().filter(x -> !psid.contains(x.getPsid())).map(x -> x.getRowNum()+1).collect(Collectors.toList());
}
我想获取满足休憩条件的list1中的所有记录if(list1.psid!= list2.psid)
I want to get all the records in list1 which satisfies fallowing conditionif(list1.psid != list2.psid)
Sample Date:
List1: rowNum psId name sid
1 1288 home 101
1 9012 home 101
2 1296 office 150
3 1290 park 161
List2: rowNum psId name sid
1 9012 home 101
2 1296 office 150
3 1290 park 161
List1 psId not in list2 so I am expecting fallowing list as result from list1
Expected: List1
rowNum psId name sid
1 1288 home 101
推荐答案
您正在过滤谓词中查找内部anyMatch
为:
You are looking for an inner anyMatch
in the filter predicate as:
public List<Term> getFilteredRowNum(List<Term> termList1, List<Term> termList2) {
return termList1.stream()
.filter(term1 -> termList2.stream()
.anyMatch(term2 -> term1.getSId() == term2.getSId()
&& term1.getPsid() != term2.getPsid()))
.collect(Collectors.toList());
}
另一种解决方法是使用groupingBy
和mapping
Another way to solve that would be to create a Map
of sid
to a Set
of psid
s present in any of the list using groupingBy
and mapping
Map<Long, Set<Long>> sIdToPsIdsMap = termList2.stream()
.collect(Collectors.groupingBy(Term::getSId,
Collectors.mapping(Term::getPsid, Collectors.toSet())));
并进一步将其用于filter
条件
return termList1.stream()
.filter(term1 -> sIdToPsIdsMap.containsKey(term1.getSId())
&& !sIdToPsIdsMap.get(term1.getSId()).contains(term1.getPsid()))
.collect(Collectors.toList());
这篇关于如何使用Java流比较两个ArrayList并使用过滤器获取list1的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!