[英]How to compare Two List of Map to identify the matching and non matching records with multiple filter predicates in Java8 Streams
要求是使用流使用多个匹配条件从 Map 列表中获取所有匹配和不匹配的记录。 即,不是使用单个过滤器来仅比较“电子邮件”,而是需要比较两个列表以匹配记录,并使用多个过滤器谓词来比较 Email 和 Id。
清单 1:
[{"Email","naveen@domain.com", "Id": "A1"},
{"Email":"test@domain.com","id":"A2"}]
列表2:
[{"Email","naveen@domain.com", "Id": "A1"},
{"Email":"test@domain.com","id":"A2"},
{"Email":"test1@domain.com","id":"B1"}]
使用流,我可以在 Email: Matching Records 上使用 Single filter predicate 找到匹配和不匹配的记录:
[{"Email","naveen@domain.com", "Id": "A1"},
{"Email":"test@domain.com","id":"A2"}]
非匹配记录:
[{"Email":"test1@domain.com","id":"B1"}]]
有没有办法比较电子邮件和 ID 比较而不仅仅是电子邮件
dbRecords.parallelStream().filter(searchData ->
inputRecords.parallelStream().anyMatch(inputMap ->
searchData.get("Email").equals(inputMap.get("Email")))).
collect(Collectors.toList());
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.stream.Collectors;
public class ListFiltersToGetMatchingRecords {
public static void main(String[] args) {
long startTime = System.currentTimeMillis();
List<Map<String, Object>> dbRecords = createDbRecords();
List<Map<String, Object>> inputRecords = createInputRecords();
List<Map<String,Object>> matchinRecords = dbRecords.parallelStream().filter(searchData ->
inputRecords.parallelStream().anyMatch(inputMap ->
searchData.get("Email").equals(inputMap.get("Email")))).
collect(Collectors.toList());
List<Map<String,Object>> notMatchinRecords = inputRecords.parallelStream().filter(searchData ->
dbRecords.parallelStream().noneMatch( inputMap ->
searchData.get("Email").equals(inputMap.get("Email"))
)).collect(Collectors.toList());
long endTime = System.currentTimeMillis();
System.out.println("Matching Records: " + matchinRecords.size());
matchinRecords.forEach(record -> {
System.out.println(record.get("Email"));
});
System.out.println("Non Matching Records" + notMatchinRecords.size());
notMatchinRecords.forEach(record -> {
System.out.println(record.get("Email"));
});
System.out.println("Non Matching Records" + notMatchinRecords.size());
System.out.println("Matching Records: " + matchinRecords.size());
System.out.println("TotalTImeTaken =" + ((endTime-startTime) /1000) + "sec");
}
private static List<Map<String, Object>> createDbRecords() {
List<Map<String, Object>> dbRecords = new ArrayList<>();
for(int i =0; i< 100; i+=2) {
Map<String, Object> dbRecord = new HashMap<>();
dbRecord.put("Email","naveen" + i +"@gmail.com");
dbRecord.put("Id", "ID" + i);
dbRecords.add(dbRecord);
}
return dbRecords;
}
private static List<Map<String, Object>> createInputRecords() {
List<Map<String, Object>> dbRecords = new ArrayList<>();
for(int i =0; i< 100; i++) {
Map<String, Object> dbRecord = new HashMap<>();
dbRecord.put("Email", "naveen" + i +"@gmail.com");
dbRecord.put("ID", "ID" + i);
dbRecords.add(dbRecord);
}
return dbRecords;
}
}
如果您关心性能,则不应将线性搜索与另一个线性搜索结合起来; 当列表变大时,无法通过并行处理修复由此产生的时间复杂度。
您应该首先构建一个允许高效查找的数据结构:
Map<List<?>,Map<String, Object>> inputKeys = inputRecords.stream()
.collect(Collectors.toMap(
m -> Arrays.asList(m.get("ID"),m.get("Email")),
m -> m,
(a,b) -> { throw new IllegalStateException("duplicate "+a+" and "+b); },
LinkedHashMap::new));
List<Map<String,Object>> matchinRecords = dbRecords.stream()
.filter(m -> inputKeys.containsKey(Arrays.asList(m.get("ID"),m.get("Email"))))
.collect(Collectors.toList());
matchinRecords.forEach(m -> inputKeys.remove(Arrays.asList(m.get("ID"),m.get("Email"))));
List<Map<String,Object>> notMatchinRecords = new ArrayList<>(inputKeys.values());
此解决方案将保留Map
的身份。
如果您只对与"Email"
键关联的值感兴趣,那就简单多了:
Map<Object,Object> notMatchinRecords = inputRecords.stream()
.collect(Collectors.toMap(
m -> m.get("ID"),
m -> m.get("Email"),
(a,b) -> { throw new IllegalStateException("duplicate"); },
LinkedHashMap::new));
Object notPresent = new Object();
Map<Object,Object> matchinRecords = dbRecords.stream()
.filter(m -> notMatchinRecords.getOrDefault(m.get("ID"), notPresent)
.equals(m.get("Email")))
.collect(Collectors.toMap(
m -> m.get("ID"),
m -> m.get("Email"),
(a,b) -> { throw new IllegalStateException("duplicate"); },
LinkedHashMap::new));
notMatchinRecords.keySet().removeAll(matchinRecords.keySet());
System.out.println("Matching Records: " + matchinRecords.size());
matchinRecords.forEach((id,email) -> System.out.println(email));
System.out.println("Non Matching Records" + notMatchinRecords.size());
notMatchinRecords.forEach((id,email) -> System.out.println(email));
第一个变体可以扩展以轻松支持更多/其他地图条目:
List<String> keys = Arrays.asList("ID", "Email");
Function<Map<String,Object>,List<?>> getKey
= m -> keys.stream().map(m::get).collect(Collectors.toList());
Map<List<?>,Map<String, Object>> inputKeys = inputRecords.stream()
.collect(Collectors.toMap(
getKey,
m -> m,
(a,b) -> { throw new IllegalStateException("duplicate "+a+" and "+b); },
LinkedHashMap::new));
List<Map<String,Object>> matchinRecords = dbRecords.stream()
.filter(m -> inputKeys.containsKey(getKey.apply(m)))
.collect(Collectors.toList());
matchinRecords.forEach(m -> inputKeys.remove(getKey.apply(m)));
List<Map<String,Object>> notMatchinRecords = new ArrayList<>(inputKeys.values());
您只需要在比较中添加一个条件
dbRecords.parallelStream().filter(searchData ->
inputRecords.parallelStream().anyMatch(inputMap ->
searchData.get("Email").equals(inputMap.get("Email"))
&& searchData.get("id").equals(inputMap.get("id"))))
.collect(Collectors.toList());
noneMatch()
添加相同的noneMatch()
。System.nanoTime()
计算平均时间,更准确.parallelStream()
(只是.stream()
)因为不确定它对你有帮助)List<Map<String,String>> unMatchedRecords = dbRecords.parallelStream().filter(searchData ->
inputRecords.parallelStream().noneMatch( inputMap ->
searchData.entrySet().stream().noneMatch(value ->
inputMap.entrySet().stream().noneMatch(value1 ->
(value1.getKey().equals(value.getKey()) &&
value1.getValue().equals(value.getValue()))))
)).collect(Collectors.toList());
笔记:
如果上面使用的 <Map<String,String> 是 <Map<Object,Object>,不要忘记为 .getKey() 和 value.getKey() 应用 .toString()。
如此获得的不匹配记录可以很容易地从列表(即,dbRecords 或 inputRecords)中减去以检索匹配结果,并且操作很快。
干杯,
舒巴姆·乔汉
为什么不在anyMatch
使用&&
:
anyMatch(inputMap -> searchData.get("Email").equals(inputMap.get("Email"))
&& searchData.get("Id").equals(inputMap.get("Id")))
而且我怀疑您是否真的需要parallelStream
,另一方面您确实需要System.nanoTime
而不是currentTimeMillis
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.