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[英]Filter elements in list1 that are not in list2 java without using streams
[英]How to compare two ArrayList and get list1 with filter using java streams
我有两个列表list1和list2类型列表
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(list1.psid!= list2.psid)。
我尝试了这个,但它不起作用
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());
}
我想得到list1中满足fallowing条件的所有记录if(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
您正在寻找过滤谓词中的内部anyMatch
:
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());
}
解决这个问题的另一种方法是使用groupingBy
和mapping
创建一个sid
Map
到任何列表中的一Set
psid
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());
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