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# How to compute two aggregate functions with Java streams?

## How to compute two aggregate functions with Java streams? Tag : java , By : Paul McKee Date : November 23 2020, 01:01 AM

I wish this help you I have a list of "Request" objects (id, amount, price) like this , Sure. You'll need the custom aggregation, but not the Pair:
`````` static class Aggregate {
private long count = 0L;
private double sum = 0L;
double average() { return sum / count; }
void merge(Aggregate other) {
count += other.count;
sum += other.sum;
}
void add(int count, double value) {
this.count += count;
this.sum += count * value;
}
}
}

requests.stream().collect(
Aggregate::new,
Aggregate::merge)
.average();
``````
``````requests.stream()
.mapToDouble(request -> request.getCount() * request.getPrice())
.sum()
/ requests.stream().mapToLong(Request::getCount).sum();
``````

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## Using two streams in Java lambda to compute covariance

Tag : java , By : Waynew
Date : March 29 2020, 07:55 AM
this one helps. In other programming languages, there is some kind of zip function, that takes several iterables, and returns an iterator that aggregates elements from each of the iterables. See for example the function zip in the Python Library.
Although it would be possible to make a similar function in Java, it's hard to implement it in such a way, that it supports efficient parallel execution. However, there is a commonly used pattern in Java, that is a bit different. In your case, it might look as follows:
``````public static double covariance(double[] xs, double[] ys) {
double xmean = mean(xs);
double ymean = mean(ys);
return IntStream.range(0, Math.min(xs.length, ys.length))
.parallel()
.mapToDouble(i -> {
double numerator = (xs[i] - xmean) * (ys[i] - ymean);
double denominator = ...;
return numerator / denominator;
})
.sum();
}
``````

## Aggregate information using Java 8 streams

Tag : java , By : ezzze
Date : March 29 2020, 07:55 AM
I wish this help you You're on the right track. The uses of Collectors.summingInt are examples of downstream collectors of the outer groupingBy collector. This operation extracts one of the integer values from each VisitSummaryBySource instance in the same group, and sums them. This is essentially a reduction over integers.
The problem, as you note, is that you can extract/reduce only one of the integer values, so you have to perform a second pass to extract/reduce the other integer values.
``````static VisitSummaryBySource merge(VisitSummaryBySource a,
VisitSummaryBySource b) {
assert a.getSource().equals(b.getSource());
return new VisitSummaryBySource(a.getSource(),
a.getRecordCount() + b.getRecordCount(),
a.getErrorCount() + b.getErrorCount());
}
``````
``````    Map<String, Optional<VisitSummaryBySource>> map =
data.stream()
.collect(groupingBy(VisitSummaryBySource::getSource,
reducing(VisitSummaryBySource::merge)));
``````
``````    List<VisitSummaryBySource> output =
data.stream()
.collect(groupingBy(VisitSummaryBySource::getSource,
reducing(VisitSummaryBySource::merge)))
.values().stream()
.map(Optional::get)
.collect(toList());
``````

## What is the best way to aggregate Streams into one DISTINCT with Java 8

Tag : java , By : Allen
Date : March 29 2020, 07:55 AM
will be helpful for those in need I think the parallel overhead is much greater than the actual work as you stated in the comments. So let your Streams do the job in sequential manner.
FYI: You should prefer using Stream::concat because slicing operations like Stream::limit can be bypassed by Stream::flatMap.
``````Stream<AppStory> appStr1 =StreamSupport.stream(splititerato1, false)
.map(this::vertexToStory1);
Stream<AppStory> appStr2 =StreamSupport.stream(splititerato2, false)
.map(this::vertexToStory2);
Stream<AppStory> appStr3 =StreamSupport.stream(splititerato3, false)
.map(this::vertexToStory3);

List<AppStory> stories = Stream.concat(Stream.concat(appStr1, appStr2), appStr3)
.distinct().collect(Collectors.toList());
// assuming AppStory::getLastUpdateTime is of type `long`
stories.sort(Comparator.comparingLong(AppStory::getLastUpdateTime));
``````

## How do aggregate operations work on Java streams?

Tag : java , By : Ravenal
Date : March 29 2020, 07:55 AM
it helps some times The Collectors do not “act”, hence do not act in an order. It’s not like one Collector processes all data before passing it to another using an intermittent type.
You are composing a single Collector with these factories, which will do the entire work at once when being passed to Stream.collect.
``````Map<Person.Sex, List<String>> namesByGender = new HashMap<>();
for(Person p: roster)