集合List和Set使用

問題

多線程 處理數據，每個子線程返回的結果都一樣。理論上返回結果集應該不一樣。

排查

檢查線程中處理邏輯是否有問題，InsuranceDetailCallable中call()方法主體邏輯正確，可能存在的傳到每個子線程中的List insuranceDetails數據是一樣的，導致每個線程的數據結果都一樣。

/*** 數據處理 operate_type link_id*/ private class InsuranceDetailCallable implements Callable<Set<Long>> {private CountDownLatch countDownLatch;private List<FinalStatementInsuranceDetail> insuranceDetails;public InsuranceDetailCallable(CountDownLatch countDownLatch,List<FinalStatementInsuranceDetail> insuranceDetails) {this.countDownLatch = countDownLatch;this.insuranceDetails = insuranceDetails;}@Overridepublic Set<Long> call() {Set<Long> result = new HashSet<>();try {//代碼省略} catch (Exception e) {System.out.println("子線程異常：" + e.getMessage());} finally {countDownLatch.countDown();}return result;} }

父線程中將獲取到的list數據根據statementId字段進行分組得到Map集合，接著將Map中的key進行分組得到List<List>組，子線程分別對這些group中集合進行處理數據。問題出現在代碼

List<List<Long>> groupList = Stream.iterate(0, n -> n + 1).limit(set.size()).parallel().map(a -> set.stream().skip((long) a * finalSize).limit(finalSize) .parallel().collect(Collectors.toList())).filter(CollUtil::isNotEmpty).collect(Collectors.toList());

處理的集合Set是無序的，并行處理分組數據，導致分組完的groupList中數據List部分數據一樣，子線程中處理的數據也一樣。List是有序的，將Set轉換成List，可以解決這個問題。

/*** 修改前父線程代碼*/ public void testMaintainNormalOperateType() {//獲取還未處理的數據List<FinalStatementInsuranceDetail> insuranceDetailList = insuranceDetailService.listOperateTypeIsNull();if (CollUtil.isEmpty(insuranceDetailList)) {return;}Map<Long, List<FinalStatementInsuranceDetail>> mapList = insuranceDetailList.stream().collect(Collectors.groupingBy(FinalStatementInsuranceDetail::getStatementId));Set<Long> set = mapList.keySet();int size = set.size() / 100;if (size == 0) {size = set.size();}int finalSize = size;List<List<Long>> groupList = Stream.iterate(0, n -> n + 1).limit(list.size()).parallel().map(a -> set.stream().skip((long) a * finalSize).limit(finalSize).parallel().collect(Collectors.toList())).filter(CollUtil::isNotEmpty).collect(Collectors.toList()); }

修改后的父線程代碼

public void testMaintainNormalOperateType() {long start = System.currentTimeMillis();List<FinalStatementInsuranceDetail> insuranceDetailList = insuranceDetailService.listOperateTypeIsNull(true);if (CollUtil.isEmpty(insuranceDetailList)) {return;}Map<Long, List<FinalStatementInsuranceDetail>> mapList = insuranceDetailList.stream().collect(Collectors.groupingBy(FinalStatementInsuranceDetail::getStatementId));List<Long> list = new ArrayList<>(mapList.keySet());try {int size = list.size() / 100;if (size == 0) {size = list.size();}int finalSize = size;List<List<Long>> groupList = Stream.iterate(0, n -> n + 1).limit(list.size()).parallel().map(a -> list.stream().skip((long) a * finalSize).limit(finalSize).parallel().collect(Collectors.toList())).filter(CollUtil::isNotEmpty).collect(Collectors.toList());ExecutorService executorService = Executors.newFixedThreadPool(groupList.size());CountDownLatch countDownLatch = new CountDownLatch(groupList.size());List<Future<Set<Long>>> futureList = new ArrayList<>();for (List<Long> statementIdList : groupList) {List<FinalStatementInsuranceDetail> insuranceDetails = new ArrayList<>();for (Long statementId : statementIdList) {insuranceDetails.addAll(mapList.getOrDefault(statementId, new ArrayList<>()));}InsuranceDetailCallable callable = new InsuranceDetailCallable(countDownLatch, insuranceDetails);Future<Set<Long>> future = executorService.submit(callable);futureList.add(future);}executorService.shutdown();for (Future<Set<Long>> future : futureList) {Set<Long> setLong = future.get();System.out.println("輸出：" + setLong.size() + " " + JSONObject.toJSONString(setLong));}} catch (Exception e) {System.out.println("父線程異常：" + e.getMessage());}long end = System.currentTimeMillis();System.out.println("耗時 = " + (end - start) / 1000); }

另外一種方式使用apache.commons夾包ListUtils.partition(List list, int size)方法

<!--對應maven依賴 --> <dependency><groupId>org.apache.commons</groupId><artifactId>commons-collections4</artifactId><version>4.1</version> </dependency>

總結

List是有序的，不唯一；Set無序，數據唯一。

總結

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