Java Streams for looping twice over same list
Multi tool use
Java Streams for looping twice over same list
I am new to java 8, and i have a problem understanding streams for some reason. Let's say we have a list of objects List< MyObject >, where MyObject has 2 fields : Long Id, Date insertTime, and i would like to remove elements with same ID and earlier time.
With 2 for loops it is something like this :
for(MyObject object : myObjects) {
for(MyObject tmpObject : myObjects) {
if(object.getId() == tmpObject.getId()) {
if(object.getInsertDate().after(tmpObject.getInsertDate()))
myObjects.remove(tmpObject);
else
myObjects.remove(object);
}
}
}
How would this look when using streams?
@Michael my bad. Than i would need another list or array to store the response data.
Let's say i have
So i need as a result
Thank you all.
On a side note, you should always use
{ and } on your statements as it will make your life easier in the long run.– notyou
Jul 2 at 7:31
{
}
@AxelH I suppose the
equals method is the wrong place for that. The equals method should return true for the same id (and optionally the same date)– Glains
Jul 2 at 7:39
equals
Yep @Glains, I didn't read that the date should be earlier ... so it is incorrect.
– AxelH
Jul 2 at 7:42
you can do somthing like below : List<Optional<MyObj>> collect = list.stream().collect(groupingBy(MyObj::getId, maxBy(Comparator.comparing(MyObj::getDate)))).values().stream().collect(toList());
– Laxmikant
Jul 2 at 8:13
2 Answers
2
group by id and then for each id pick the one with the largest insertion date. You can replace Sample by MyObject.
Map<Long, List<Sample>> map = list.stream().collect(Collectors.groupingBy(Sample::getId));
map.values().stream()
.map(samples -> Collections.max(
samples, Comparator.comparing(Sample::getInsertDate)))
.collect(Collectors.toList());
Instead of collecting each group into a
List, to select the maximum element in a second step, you can let the groupingBy operation collect the maximum element in the first place, like in this answer.– Holger
Jul 2 at 11:30
List
groupingBy
You can do it like so,
List<MyObject> latestObjects = myObjects.stream()
.collect(Collectors.collectingAndThen(
Collectors.groupingBy(MyObject::getId,
Collectors.maxBy(Comparator.comparing(MyObject::getInsertTime))),
map -> map.values().stream().map(op -> op.orElse(null)).collect(Collectors.toList())));
First group the objects by their Id value, then reduce it to keep only the latest object for each Id value, finally collect them into a List.
List
Update
This can further be improved as per the comment below. Here's the augmented version of it.
List<MyObject> latestObjects = myObjects.stream()
.collect(Collectors.collectingAndThen(
Collectors.toMap(MyObject::getId, Function.identity(),
BinaryOperator.maxBy(Comparator.comparing(MyObject::getInsertTime))),
map -> new ArrayList<MyObject>(map.values())));
Whenever the downstream collector is a reduction (i.e. all these cases where you have to deal with an
Optional afterwards), you may use toMap instead of groupingBy: List<MyObject> latestObjects = myObjects.stream() .collect(Collectors.collectingAndThen( Collectors.toMap(MyObject::getId, Function.identity(), BinaryOperator.maxBy(Comparator.comparing(MyObject::getInsertTime))), map -> new ArrayList<>(map.values())));– Holger
Jul 2 at 11:36
Optional
toMap
groupingBy
List<MyObject> latestObjects = myObjects.stream() .collect(Collectors.collectingAndThen( Collectors.toMap(MyObject::getId, Function.identity(), BinaryOperator.maxBy(Comparator.comparing(MyObject::getInsertTime))), map -> new ArrayList<>(map.values())));
Totally agree with you. Thanks for the invaluable feedback !
– Ravindra Ranwala
Jul 2 at 14:37
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"With 2 for loops it is something like this" I'm afraid it's not, because you can't remove from a list while you're iterating over it.
– Michael
Jul 2 at 7:24