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I have an array list which contains objects of this class:
public class SearchCriteria {
private String key;
private String operation;
private Object value;
}
How to count length of all Strings in all objects in this ArrayList?
It can be done in foreach, but I suppose it also can be done in lambda, but I dont know how do it sneaky and modern.
now my solution is:
Integer sum=0;
for (SearchCriteria s: builder.getParams()
) {
sum+=s.getKey().length();
sum+=s.getOperation().length();
sum+=s.getValue().toString().length();
}
You can use flatMapToInt to convert a Stream<SearchCriteria > to an IntStream containing the lengths of all the properties of the elements of the original Stream:
int sum = builder.getParams()
.stream()
.flatMapToInt(sc -> IntStream.of(sc.getKey().length(),sc.getOperation().length(),sc.getValue().toString().length()))
.sum();
I don't know why this is useful to you, but you can try this:
List<SearchCriteria> criterion = builder.getParams();
int sum = criterion.stream().map(s -> s.getKey() + s.getOperation() + s.getValue())
.mapToInt(String::length).sum();
You can use:
List<SearchCriteria> list = ...
int sum = list.stream()
.flatMapToInt(crit -> Arrays.stream(new int[] {
crit.getKey().length(),
crit.getOperation().length(),
crit.getValue().toString().length()}))
.sum();
This makes an int stream of all length values and simply sums them.
You cannot sum in a straight way different things of the streamed element.
So as alternative, I would extract the computation logic in a SearchCriteria method such as :
public int computeAllLength(){
return key.length() + operation.length() + value.toString().length();
}
And I would use it in this way :
int sum = builder.getParams().stream()
.mapToInt(SearchCriteria::computeAllLength)
.sum();
Select sum(paidAmount), count(paidAmount), classificationName,
From tableA
Group by classificationName;
How can i do this in Java 8 using streams and collectors?
Java8:
lineItemList.stream()
.collect(Collectors.groupingBy(Bucket::getBucketName,
Collectors.reducing(BigDecimal.ZERO,
Bucket::getPaidAmount,
BigDecimal::add)))
This gives me sum and group by. But how can I also get count on the group name ?
Expectation is :
100, 2, classname1
50, 1, classname2
150, 3, classname3
Using an extended version of the Statistics class of this answer,
class Statistics {
int count;
BigDecimal sum;
Statistics(Bucket bucket) {
count = 1;
sum = bucket.getPaidAmount();
}
Statistics() {
count = 0;
sum = BigDecimal.ZERO;
}
void add(Bucket b) {
count++;
sum = sum.add(b.getPaidAmount());
}
Statistics merge(Statistics another) {
count += another.count;
sum = sum.add(another.sum);
return this;
}
}
you can use it in a Stream operation like
Map<String, Statistics> map = lineItemList.stream()
.collect(Collectors.groupingBy(Bucket::getBucketName,
Collector.of(Statistics::new, Statistics::add, Statistics::merge)));
this may have a small performance advantage, as it only creates one Statistics instance per group for a sequential evaluation. It even supports parallel evaluation, but you’d need a very large list with sufficiently large groups to get a benefit from parallel evaluation.
For a sequential evaluation, the operation is equivalent to
lineItemList.forEach(b ->
map.computeIfAbsent(b.getBucketName(), x -> new Statistics()).add(b));
whereas merging partial results after a parallel evaluation works closer to the example already given in the linked answer, i.e.
secondMap.forEach((key, value) -> firstMap.merge(key, value, Statistics::merge));
As you're using BigDecimal for the amounts (which is the correct approach, IMO), you can't make use of Collectors.summarizingDouble, which summarizes count, sum, average, min and max in one pass.
Alexis C. has already shown in his answer one way to do it with streams. Another way would be to write your own collector, as shown in Holger's answer.
Here I'll show another way. First let's create a container class with a helper method. Then, instead of using streams, I'll use common Map operations.
class Statistics {
int count;
BigDecimal sum;
Statistics(Bucket bucket) {
count = 1;
sum = bucket.getPaidAmount();
}
Statistics merge(Statistics another) {
count += another.count;
sum = sum.add(another.sum);
return this;
}
}
Now, you can make the grouping as follows:
Map<String, Statistics> result = new HashMap<>();
lineItemList.forEach(b ->
result.merge(b.getBucketName(), new Statistics(b), Statistics::merge));
This works by using the Map.merge method, whose docs say:
If the specified key is not already associated with a value or is associated with null, associates it with the given non-null value. Otherwise, replaces the associated value with the results of the given remapping function
You could reduce pairs where the keys would hold the sum and the values would hold the count:
Map<String, SimpleEntry<BigDecimal, Long>> map =
lineItemList.stream()
.collect(groupingBy(Bucket::getBucketName,
reducing(new SimpleEntry<>(BigDecimal.ZERO, 0L),
b -> new SimpleEntry<>(b.getPaidAmount(), 1L),
(v1, v2) -> new SimpleEntry<>(v1.getKey().add(v2.getKey()), v1.getValue() + v2.getValue()))));
although Collectors.toMap looks cleaner:
Map<String, SimpleEntry<BigDecimal, Long>> map =
lineItemList.stream()
.collect(toMap(Bucket::getBucketName,
b -> new SimpleEntry<>(b.getPaidAmount(), 1L),
(v1, v2) -> new SimpleEntry<>(v1.getKey().add(v2.getKey()), v1.getValue() + v2.getValue())));
I want to use Java 8 Lambda expression in following scenario but I am getting Local variable fooCount defined in an enclosing scope must be final or effectively final. I understand what the error message says, but I need to calculate percentage here so need to increment fooCount and barCount then calculate percentage. So what's the way to achieve it:
// key is a String with values like "FOO;SomethinElse" and value is Long
final Map<String, Long> map = null;
....
private int calculateFooPercentage() {
long fooCount = 0L;
long barCount = 0L;
map.forEach((k, v) -> {
if (k.contains("FOO")) {
fooCount++;
} else {
barCount++;
}
});
final int fooPercentage = 0;
//Rest of the logic to calculate percentage
....
return fooPercentage;
}
One option I have is to use AtomicLong here instead of long but I would like to avoid it, so later if possible I want to use parallel stream here.
There is a count method in stream to do counts for you.
long fooCount = map.keySet().stream().filter(k -> k.contains("FOO")).count();
long barCount = map.size() - fooCount;
If you want parallelisation, change .stream() to .parallelStream().
Alternatively, if you were trying to increment a variable manually, and use stream parallelisation, then you would want to use something like AtomicLong for thread safety. A simple variable, even if the compiler allowed it, would not be thread-safe.
To get both numbers, matching and non-matching elements, you can use
Map<Boolean, Long> result = map.keySet().stream()
.collect(Collectors.partitioningBy(k -> k.contains("FOO"), Collectors.counting()));
long fooCount = result.get(true);
long barCount = result.get(false);
But since your source is a Map, which knows its total size, and want to calculate a percentage, for which barCount is not needed, this specific task can be solved as
private int calculateFooPercentage() {
return (int)(map.keySet().stream().filter(k -> k.contains("FOO")).count()
*100/map.size());
}
Both variants are thread safe, i.e. changing stream() to parallelStream() will perform the operation in parallel, however, it’s unlikely that this operation will benefit from parallel processing. You would need humongous key strings or maps to get a benefit…
I agree with the other answers indicating you should use countor partitioningBy.
Just to explain the atomicity problem with an example, consider the following code:
private static AtomicInteger i1 = new AtomicInteger(0);
private static int i2 = 0;
public static void main(String[] args) {
IntStream.range(0, 100000).parallel().forEach(n -> i1.incrementAndGet());
System.out.println(i1);
IntStream.range(0, 100000).parallel().forEach(n -> i2++);
System.out.println(i2);
}
This returns the expected result of 100000 for i1 but an indeterminate number less than that (between 50000 and 80000 in my test runs) for i2. The reason should be pretty obvious.
I have a class Something which contains an instance variable Anything.
class Anything {
private final int id;
private final int noThings;
public Anything(int id, int noThings) {
this.id = id;
this.noThings = noThings;
}
}
class Something {
private final int parentId;
private final List<Anything> anythings;
private int getParentId() {
return parentId;
}
private List<Anything> getAnythings() {
return anythings;
}
public Something(int parentId, List<Anything> anythings) {
this.parentId = parentId;
this.anythings = anythings;
}
}
Given a list of Somethings
List<Something> mySomethings = Arrays.asList(
new Something(123, Arrays.asList(new Anything(45, 65),
new Anything(568, 15),
new Anything(145, 27))),
new Something(547, Arrays.asList(new Anything(12, 123),
new Anything(678, 76),
new Anything(98, 81))),
new Something(685, Arrays.asList(new Anything(23, 57),
new Anything(324, 67),
new Anything(457, 87))));
I want to sort them such that the Something objects are sorted depending on the total descending sum of the (Anything object) noThings, and then by the descending value of the (Anything object) noThings
123 = 65+15+27 = 107(3rd)
547 = 123+76+81 = 280 (1st)
685 = 57+67+87 = 211 (2nd)
So that I end up with
List<Something> orderedSomethings = Arrays.asList(
new Something(547, Arrays.asList(new Anything(12, 123),
new Anything(98, 81),
new Anything(678, 76))),
new Something(685, Arrays.asList(new Anything(457, 87),
new Anything(324, 67),
new Anything(23, 57))),
new Something(123, Arrays.asList(new Anything(45, 65),
new Anything(145, 27),
new Anything(568, 15))));
I know that I can get the list of Anythings per parent Id
Map<Integer, List<Anythings>> anythings
= mySomethings.stream()
.collect(Collectors.toMap(p->p.getParentId(),
p->p.getAnythings()))
;
But after that I'm a bit stuck.
Unless I'm mistaken, you can not do both sorts in one go. But since they are independent of each other (the sum of the nothings in the Anythings in a Something is independent of their order), this does not matter much. Just sort one after the other.
To sort the Anytings inside the Somethings by their noThings:
mySomethings.stream().map(Something::getAnythings)
.forEach(as -> as.sort(Comparator.comparing(Anything::getNoThings)
.reversed()));
To sort the Somethings by the sum of the noThings of their Anythings:
mySomethings.sort(Comparator.comparing((Something s) -> s.getAnythings().stream()
.mapToInt(Anything::getNoThings).sum())
.reversed());
Note that both those sorts will modify the respective lists in-place.
As pointed out by #Tagir, the second sort will calculate the sum of the Anythings again for each pair of Somethings that are compared in the sort. If the lists are long, this can be very wasteful. Instead, you could first calculate the sums in a map and then just look up the value.
Map<Something, Integer> sumsOfThings = mySomethings.stream()
.collect(Collectors.toMap(s -> s, s -> s.getAnythings().stream()
.mapToInt(Anything::getNoThings).sum()));
mySomethings.sort(Comparator.comparing(sumsOfThings::get).reversed());
The problem of other solutions is that sums are not stored anywhere during sorting, thus when sorting large input, sums will be calculated for every row several times reducing the performance. An alternative solution is to create intermediate pairs of (something, sum), sort by sum, then extract something and forget about sum. Here's how it can be done with Stream API and SimpleImmutableEntry as pair class:
List<Something> orderedSomethings = mySomethings.stream()
.map(smth -> new AbstractMap.SimpleImmutableEntry<>(smth, smth
.getAnythings().stream()
.mapToInt(Anything::getNoThings).sum()))
.sorted(Entry.<Something, Integer>comparingByValue().reversed())
.map(Entry::getKey)
.collect(Collectors.toList());
There's some syntactic sugar available in my free StreamEx library which makes the code a little bit cleaner:
List<Something> orderedSomethings = StreamEx.of(mySomethings)
.mapToEntry(smth -> smth
.getAnythings().stream()
.mapToInt(Anything::getNoThings).sum())
.reverseSorted(Entry.comparingByValue())
.keys().toList();
As for sorting the Anything inside something: other solutions are ok.
In the end I added an extra method to the Something class.
public int getTotalNoThings() {
return anythings.stream().collect(Collectors.summingInt(Anything::getNoThings));
}
then I used this method to sort by total noThings (desc)
somethings = somethings.stream()
.sorted(Comparator.comparing(Something::getTotalNoThings).reversed())
.collect(Collectors.toList());
and then I used the code suggested above (thanks!) to sort by the Anything instance noThings
somethings .stream().map(Something::getAnythings)
.forEach(as -> as.sort(Comparator.comparing(Anything::getNoThings).reversed()));
Thanks again for help.
Java 8 introduced a Stream class that resembles Scala's Stream, a powerful lazy construct using which it is possible to do something like this very concisely:
def from(n: Int): Stream[Int] = n #:: from(n+1)
def sieve(s: Stream[Int]): Stream[Int] = {
s.head #:: sieve(s.tail filter (_ % s.head != 0))
}
val primes = sieve(from(2))
primes takeWhile(_ < 1000) print // prints all primes less than 1000
I wondered if it is possible to do this in Java 8, so I wrote something like this:
IntStream from(int n) {
return IntStream.iterate(n, m -> m + 1);
}
IntStream sieve(IntStream s) {
int head = s.findFirst().getAsInt();
return IntStream.concat(IntStream.of(head), sieve(s.skip(1).filter(n -> n % head != 0)));
}
IntStream primes = sieve(from(2));
Fairly simple, but it produces java.lang.IllegalStateException: stream has already been operated upon or closed because both findFirst() and skip() are terminal operations on Stream which can be done only once.
I don't really have to use up the stream twice since all I need is the first number in the stream and the rest as another stream, i.e. equivalent of Scala's Stream.head and Stream.tail. Is there a method in Java 8 Stream that I can use to achieve this?
Thanks.
Even if you hadn’t the problem that you can’t split an IntStream, you code didn’t work because you are invoking your sieve method recursively instead of lazily. So you had an infinity recursion before you could query your resulting stream for the first value.
Splitting an IntStream s into a head and a tail IntStream (which has not yet consumed) is possible:
PrimitiveIterator.OfInt it = s.iterator();
int head = it.nextInt();
IntStream tail = IntStream.generate(it::next).filter(i -> i % head != 0);
At this place you need a construct of invoking sieve on the tail lazily. Stream does not provide that; concat expects existing stream instances as arguments and you can’t construct a stream invoking sieve lazily with a lambda expression as lazy creation works with mutable state only which lambda expressions do not support. If you don’t have a library implementation hiding the mutable state you have to use a mutable object. But once you accept the requirement of mutable state, the solution can be even easier than your first approach:
IntStream primes = from(2).filter(i -> p.test(i)).peek(i -> p = p.and(v -> v % i != 0));
IntPredicate p = x -> true;
IntStream from(int n)
{
return IntStream.iterate(n, m -> m + 1);
}
This will recursively create a filter but in the end it doesn’t matter whether you create a tree of IntPredicates or a tree of IntStreams (like with your IntStream.concat approach if it did work). If you don’t like the mutable instance field for the filter you can hide it in an inner class (but not in a lambda expression…).
My StreamEx library has now headTail() operation which solves the problem:
public static StreamEx<Integer> sieve(StreamEx<Integer> input) {
return input.headTail((head, tail) ->
sieve(tail.filter(n -> n % head != 0)).prepend(head));
}
The headTail method takes a BiFunction which will be executed at most once during the stream terminal operation execution. So this implementation is lazy: it does not compute anything until traversal starts and computes only as much prime numbers as requested. The BiFunction receives a first stream element head and the stream of the rest elements tail and can modify the tail in any way it wants. You may use it with predefined input:
sieve(IntStreamEx.range(2, 1000).boxed()).forEach(System.out::println);
But infinite stream work as well
sieve(StreamEx.iterate(2, x -> x+1)).takeWhile(x -> x < 1000)
.forEach(System.out::println);
// Not the primes till 1000, but 1000 first primes
sieve(StreamEx.iterate(2, x -> x+1)).limit(1000).forEach(System.out::println);
There's also alternative solution using headTail and predicate concatenation:
public static StreamEx<Integer> sieve(StreamEx<Integer> input, IntPredicate isPrime) {
return input.headTail((head, tail) -> isPrime.test(head)
? sieve(tail, isPrime.and(n -> n % head != 0)).prepend(head)
: sieve(tail, isPrime));
}
sieve(StreamEx.iterate(2, x -> x+1), i -> true).limit(1000).forEach(System.out::println);
It interesting to compare recursive solutions: how many primes they capable to generate.
#John McClean solution (StreamUtils)
John McClean solutions are not lazy: you cannot feed them with infinite stream. So I just found by trial-and-error the maximal allowed upper bound (17793) (after that StackOverflowError occurs):
public void sieveTest(){
sieve(IntStream.range(2, 17793).boxed()).forEach(System.out::println);
}
#John McClean solution (Streamable)
public void sieveTest2(){
sieve(Streamable.range(2, 39990)).forEach(System.out::println);
}
Increasing upper limit above 39990 results in StackOverflowError.
#frhack solution (LazySeq)
LazySeq<Integer> ints = integers(2);
LazySeq primes = sieve(ints); // sieve method from #frhack answer
primes.forEach(p -> System.out.println(p));
Result: stuck after prime number = 53327 with enormous heap allocation and garbage collection taking more than 90%. It took several minutes to advance from 53323 to 53327, so waiting more seems impractical.
#vidi solution
Prime.stream().forEach(System.out::println);
Result: StackOverflowError after prime number = 134417.
My solution (StreamEx)
sieve(StreamEx.iterate(2, x -> x+1)).forEach(System.out::println);
Result: StackOverflowError after prime number = 236167.
#frhack solution (rxjava)
Observable<Integer> primes = Observable.from(()->primesStream.iterator());
primes.forEach((x) -> System.out.println(x.toString()));
Result: StackOverflowError after prime number = 367663.
#Holger solution
IntStream primes=from(2).filter(i->p.test(i)).peek(i->p=p.and(v->v%i!=0));
primes.forEach(System.out::println);
Result: StackOverflowError after prime number = 368089.
My solution (StreamEx with predicate concatenation)
sieve(StreamEx.iterate(2, x -> x+1), i -> true).forEach(System.out::println);
Result: StackOverflowError after prime number = 368287.
So three solutions involving predicate concatenation win, because each new condition adds only 2 more stack frames. I think, the difference between them is marginal and should not be considered to define a winner. However I like my first StreamEx solution more as it more similar to Scala code.
The solution below does not do state mutations, except for the head/tail deconstruction of the stream.
The lazyness is obtained using IntStream.iterate. The class Prime is used to keep the generator state
import java.util.PrimitiveIterator;
import java.util.stream.IntStream;
import java.util.stream.Stream;
public class Prime {
private final IntStream candidates;
private final int current;
private Prime(int current, IntStream candidates)
{
this.current = current;
this.candidates = candidates;
}
private Prime next()
{
PrimitiveIterator.OfInt it = candidates.filter(n -> n % current != 0).iterator();
int head = it.next();
IntStream tail = IntStream.generate(it::next);
return new Prime(head, tail);
}
public static Stream<Integer> stream() {
IntStream possiblePrimes = IntStream.iterate(3, i -> i + 1);
return Stream.iterate(new Prime(2, possiblePrimes), Prime::next)
.map(p -> p.current);
}
}
The usage would be this:
Stream<Integer> first10Primes = Prime.stream().limit(10)
You can essentially implement it like this:
static <T> Tuple2<Optional<T>, Seq<T>> splitAtHead(Stream<T> stream) {
Iterator<T> it = stream.iterator();
return tuple(it.hasNext() ? Optional.of(it.next()) : Optional.empty(), seq(it));
}
In the above example, Tuple2 and Seq are types borrowed from jOOλ, a library that we developed for jOOQ integration tests. If you don't want any additional dependencies, you might as well implement them yourself:
class Tuple2<T1, T2> {
final T1 v1;
final T2 v2;
Tuple2(T1 v1, T2 v2) {
this.v1 = v1;
this.v2 = v2;
}
static <T1, T2> Tuple2<T1, T2> tuple(T1 v1, T2 v2) {
return new Tuple<>(v1, v2);
}
}
static <T> Tuple2<Optional<T>, Stream<T>> splitAtHead(Stream<T> stream) {
Iterator<T> it = stream.iterator();
return tuple(
it.hasNext() ? Optional.of(it.next()) : Optional.empty,
StreamSupport.stream(Spliterators.spliteratorUnknownSize(
it, Spliterator.ORDERED
), false)
);
}
If you don't mind using 3rd party libraries cyclops-streams, I library I wrote has a number of potential solutions.
The StreamUtils class has large number of static methods for working directly with java.util.stream.Streams including headAndTail.
HeadAndTail<Integer> headAndTail = StreamUtils.headAndTail(Stream.of(1,2,3,4));
int head = headAndTail.head(); //1
Stream<Integer> tail = headAndTail.tail(); //Stream[2,3,4]
The Streamable class represents a replayable Stream and works by building a lazy, caching intermediate data-structure. Because it is caching and repayable - head and tail can be implemented directly and separately.
Streamable<Integer> replayable= Streamable.fromStream(Stream.of(1,2,3,4));
int head = repayable.head(); //1
Stream<Integer> tail = replayable.tail(); //Stream[2,3,4]
cyclops-streams also provides a sequential Stream extension that in turn extends jOOλ and has both Tuple based (from jOOλ) and domain object (HeadAndTail) solutions for head and tail extraction.
SequenceM.of(1,2,3,4)
.splitAtHead(); //Tuple[1,SequenceM[2,3,4]
SequenceM.of(1,2,3,4)
.headAndTail();
Update per Tagir's request -> A Java version of the Scala sieve using SequenceM
public void sieveTest(){
sieve(SequenceM.range(2, 1_000)).forEach(System.out::println);
}
SequenceM<Integer> sieve(SequenceM<Integer> s){
return s.headAndTailOptional().map(ht ->SequenceM.of(ht.head())
.appendStream(sieve(ht.tail().filter(n -> n % ht.head() != 0))))
.orElse(SequenceM.of());
}
And another version via Streamable
public void sieveTest2(){
sieve(Streamable.range(2, 1_000)).forEach(System.out::println);
}
Streamable<Integer> sieve(Streamable<Integer> s){
return s.size()==0? Streamable.of() : Streamable.of(s.head())
.appendStreamable(sieve(s.tail()
.filter(n -> n % s.head() != 0)));
}
Note - neither Streamable of SequenceM have an Empty implementation - hence the size check for Streamable and the use of headAndTailOptional.
Finally a version using plain java.util.stream.Stream
import static com.aol.cyclops.streams.StreamUtils.headAndTailOptional;
public void sieveTest(){
sieve(IntStream.range(2, 1_000).boxed()).forEach(System.out::println);
}
Stream<Integer> sieve(Stream<Integer> s){
return headAndTailOptional(s).map(ht ->Stream.concat(Stream.of(ht.head())
,sieve(ht.tail().filter(n -> n % ht.head() != 0))))
.orElse(Stream.of());
}
Another update - a lazy iterative based on #Holger's version using objects rather than primitives (note a primitive version is also possible)
final Mutable<Predicate<Integer>> predicate = Mutable.of(x->true);
SequenceM.iterate(2, n->n+1)
.filter(i->predicate.get().test(i))
.peek(i->predicate.mutate(p-> p.and(v -> v%i!=0)))
.limit(100000)
.forEach(System.out::println);
There are many interesting suggestions provided here, but if someone needs a solution without dependencies to third party libraries I came up with this:
import java.util.AbstractMap;
import java.util.Optional;
import java.util.Spliterators;
import java.util.stream.StreamSupport;
/**
* Splits a stream in the head element and a tail stream.
* Parallel streams are not supported.
*
* #param stream Stream to split.
* #param <T> Type of the input stream.
* #return A map entry where {#link Map.Entry#getKey()} contains an
* optional with the first element (head) of the original stream
* and {#link Map.Entry#getValue()} the tail of the original stream.
* #throws IllegalArgumentException for parallel streams.
*/
public static <T> Map.Entry<Optional<T>, Stream<T>> headAndTail(final Stream<T> stream) {
if (stream.isParallel()) {
throw new IllegalArgumentException("parallel streams are not supported");
}
final Iterator<T> iterator = stream.iterator();
return new AbstractMap.SimpleImmutableEntry<>(
iterator.hasNext() ? Optional.of(iterator.next()) : Optional.empty(),
StreamSupport.stream(Spliterators.spliteratorUnknownSize(iterator, 0), false)
);
}
To get head and tail you need a Lazy Stream implementation. Java 8 stream or RxJava are not suitable.
You can use for example LazySeq as follows.
Lazy sequence is always traversed from the beginning using very cheap
first/rest decomposition (head() and tail())
LazySeq implements java.util.List interface, thus can be used in
variety of places. Moreover it also implements Java 8 enhancements to
collections, namely streams and collectors
package com.company;
import com.nurkiewicz.lazyseq.LazySeq;
public class Main {
public static void main(String[] args) {
LazySeq<Integer> ints = integers(2);
LazySeq primes = sieve(ints);
primes.take(10).forEach(p -> System.out.println(p));
}
private static LazySeq<Integer> sieve(LazySeq<Integer> s) {
return LazySeq.cons(s.head(), () -> sieve(s.filter(x -> x % s.head() != 0)));
}
private static LazySeq<Integer> integers(int from) {
return LazySeq.cons(from, () -> integers(from + 1));
}
}
Here is another recipe using the way suggested by Holger.
It use RxJava just to add the possibility to use the take(int) method and many others.
package com.company;
import rx.Observable;
import java.util.function.IntPredicate;
import java.util.stream.IntStream;
public class Main {
public static void main(String[] args) {
final IntPredicate[] p={(x)->true};
IntStream primesStream=IntStream.iterate(2,n->n+1).filter(i -> p[0].test(i)).peek(i->p[0]=p[0].and(v->v%i!=0) );
Observable primes = Observable.from(()->primesStream.iterator());
primes.take(10).forEach((x) -> System.out.println(x.toString()));
}
}
This should work with parallel streams as well:
public static <T> Map.Entry<Optional<T>, Stream<T>> headAndTail(final Stream<T> stream) {
final AtomicReference<Optional<T>> head = new AtomicReference<>(Optional.empty());
final var spliterator = stream.spliterator();
spliterator.tryAdvance(x -> head.set(Optional.of(x)));
return Map.entry(head.get(), StreamSupport.stream(spliterator, stream.isParallel()));
}
If you want to get head of a stream, just:
IntStream.range(1, 5).first();
If you want to get tail of a stream, just:
IntStream.range(1, 5).skip(1);
If you want to get both head and tail of a stream, just:
IntStream s = IntStream.range(1, 5);
int head = s.head();
IntStream tail = s.tail();
If you want to find the prime, just:
LongStream.range(2, n)
.filter(i -> LongStream.range(2, (long) Math.sqrt(i) + 1).noneMatch(j -> i % j == 0))
.forEach(N::println);
If you want to know more, go to get abacus-common
Declaration： I'm the developer of abacus-common.