我有一个并行流,因为任务真的很慢,我将在下面粘贴代码。情况就是这样。
我有一个arrayList,我需要对该列表中的每个对象做一些事情(这很慢),然后将该对象添加到临时列表中,我认为流中的过程结束了,因为我可以看到每个对象用日志处理。
流结束时,有时临时列表中有n-1个对象或一个为null。
有什么主意吗?
使用此示例代码不会发生错误,但是逻辑是相同的,但是没有业务逻辑。
public class SampleCode {
public List<SomeObject> example(List<SomeObject> someObjectList) {
List<SomeObject> someObjectListTemp = new ArrayList<>();
someObjectList.parallelStream().forEach(someObject -> {
List<ExtraData> extraDataList = getExtraData(someObject.getId());
if (extraDataList.isEmpty()) {
someObjectListTemp.add(someObject);
} else {
for (ExtraData extraData : extraDataList) {
SomeObject someObjectTemp = null;
someObjectTemp = (SomeObject) cloneObject(someObject);
if (extraData != null) {
someObjectTemp.setDate(extraData.getDate());
someObjectTemp.setData2(extraData.getData2());
}
if (someObjectTemp == null) {
System.out.println("Warning null object"); //I NEVER see this
}
someObjectListTemp.add(someObjectTemp);
System.out.println("Added object to list"); //I Always see this the same times as elements in original list
}
}
});
if (someObjectListTemp.size() < 3) {
System.out.println("Error: There should be at least 3 elements"); //Some times one object is missing in the list
}
for (SomeObject someObject : someObjectListTemp) {
if (someObject == null) {
System.out.println("Error: null element in list"); //Some times one object is null in the list
}
}
return someObjectListTemp;
}
答案 0 :(得分:1)
您可以尝试使用flatMap
方法而不是foreach
吗?
flatMap
获取一个列表列表,并将其所有元素放在一个列表中。
这样,您就不会使用另一个ArrayList
来存储临时对象。
我认为这可能是个问题,因为parallelStream
是多线程的,而ArrayList
没有同步
List<SomeObject> someObjectListTemp = someObjectList.parallelStream()
.map(so -> processSomeObject(so)) // makes a stream of lists (Stream<List<SomeObject>>)
.flatMap(Collection::stream) // groups all the elements of all the lists in one stream (Stream<Someobject>)
.collect(Collectors.toList()); // transforms the stream into a list (List<SomeObject>)
并将代码粘贴在单独的方法processSomeObject
中,该方法将返回SomeObject
的列表:
static List<SomeObject> processSomeObject(SomeObject someObject) {
List<ExtraData> extraDataList = getExtraData(someObject.getId());
List<SomeObject> someObjectListTemp = new ArrayList<>();
if (extraDataList.isEmpty()) {
someObjectListTemp.add(someObject);
} else {
for (ExtraData extraData : extraDataList) {
SomeObject someObjectTemp = (SomeObject) cloneObject(someObject);
if (extraData != null) {
someObjectTemp.setDate(extraData.getDate());
someObjectTemp.setData2(extraData.getData2());
}
someObjectListTemp.add(someObjectTemp);
System.out.println("Added object to list");
}
}
return someObjectListTemp;
}
答案 1 :(得分:1)
一个小例子是
public static void main(String[] args) {
List<Object> test = new ArrayList<>();
IntStream.range(0, 100000).parallel().forEach(i -> test.add(new Object()));
for(Object o : test) {
System.out.println(o.getClass());
}
}
是因为ArrayList不是线程安全的并且内部数组被拧紧了