How can I get the last element of a stream or list in the following code?
Where data.careas is a List<CArea>:
CArea first = data.careas.stream()
.filter(c -> c.bbox.orientationHorizontal).findFirst().get();
CArea last = data.careas.stream()
.filter(c -> c.bbox.orientationHorizontal)
.collect(Collectors.toList()).; //how to?
As you can see getting the first element, with a certain filter, is not hard.
However getting the last element in a one-liner is a real pain:
It seems I cannot obtain it directly from a Stream. (It would only make sense for finite streams)
It also seems that you cannot get things like first() and last() from the List interface, which is really a pain.
I do not see any argument for not providing a first() and last() method in the List interface, as the elements in there, are ordered, and moreover the size is known.
But as per the original answer: How to get the last element of a finite Stream?
Personally, this is the closest I could get:
int lastIndex = data.careas.stream()
.filter(c -> c.bbox.orientationHorizontal)
.mapToInt(c -> data.careas.indexOf(c)).max().getAsInt();
CArea last = data.careas.get(lastIndex);
However it does involve, using an indexOf on every element, which is most likely not you generally want as it can impair performance.
It is possible to get the last element with the method Stream::reduce. The following listing contains a minimal example for the general case:
Stream<T> stream = ...; // sequential or parallel stream
Optional<T> last = stream.reduce((first, second) -> second);
This implementations works for all ordered streams (including streams created from Lists). For unordered streams it is for obvious reasons unspecified which element will be returned.
The implementation works for both sequential and parallel streams. That might be surprising at first glance, and unfortunately the documentation doesn't state it explicitly. However, it is an important feature of streams, and I try to clarify it:
The Javadoc for the method Stream::reduce states, that it "is not constrained to execute sequentially".
The Javadoc also requires that the "accumulator function must be an associative, non-interfering, stateless function for combining two values", which is obviously the case for the lambda expression (first, second) -> second.
The Javadoc for reduction operations states: "The streams classes have multiple forms of general reduction operations, called reduce() and collect() [..]" and "a properly constructed reduce operation is inherently parallelizable, so long as the function(s) used to process the elements are associative and stateless."
The documentation for the closely related Collectors is even more explicit: "To ensure that sequential and parallel executions produce equivalent results, the collector functions must satisfy an identity and an associativity constraints."
Back to the original question: The following code stores a reference to the last element in the variable last and throws an exception if the stream is empty. The complexity is linear in the length of the stream.
CArea last = data.careas
.stream()
.filter(c -> c.bbox.orientationHorizontal)
.reduce((first, second) -> second).get();
If you have a Collection (or more general an Iterable) you can use Google Guava's
Iterables.getLast(myIterable)
as handy oneliner.
One liner (no need for stream;):
Object lastElement = list.isEmpty() ? null : list.get(list.size()-1);
Guava has dedicated method for this case:
Stream<T> stream = ...;
Optional<T> lastItem = Streams.findLast(stream);
It's equivalent to stream.reduce((a, b) -> b) but creators claim it has much better performance.
From documentation:
This method's runtime will be between O(log n) and O(n), performing
better on efficiently splittable streams.
It's worth to mention that if stream is unordered this method behaves like findAny().
list.stream().sorted(Comparator.comparing(obj::getSequence).reversed()).findFirst().get();
reverse the order and get the first element from the list. here object has sequence number, Comparator provides multiple functionalities can be used as per logic.
Another way to get the last element is by using sort.
Optional<CArea> num=data.careas.stream().sorted((a,b)->-1).findFirst();
You can also use skip() function as below...
long count = data.careas.count();
CArea last = data.careas.stream().skip(count - 1).findFirst().get();
it's super simple to use.
One more approach. Pair will have first and last elements:
List<Object> pair = new ArrayList<>();
dataStream.ForEach(o -> {
if (pair.size() == 0) {
pair.add(o);
pair.add(o);
}
pair.set(1, o);
});
If you need to get the last N number of elements. Closure can be used.
The below code maintains an external queue of fixed size until, the stream reaches the end.
final Queue<Integer> queue = new LinkedList<>();
final int N=5;
list.stream().peek((z) -> {
queue.offer(z);
if (queue.size() > N)
queue.poll();
}).count();
Another option could be to use reduce operation using identity as a Queue.
final int lastN=3;
Queue<Integer> reduce1 = list.stream()
.reduce(
(Queue<Integer>)new LinkedList<Integer>(),
(m, n) -> {
m.offer(n);
if (m.size() > lastN)
m.poll();
return m;
}, (m, n) -> m);
System.out.println("reduce1 = " + reduce1);
Related
Given:
List<String> str = Arrays.asList ("my", "pen", "is", "your", "pen");
Predicate<String> test = s -> {
int i = 0;
boolean result = s.contains ("pen");
System.out.print((i++) + ":");
return result;
};
This prints: 0:0:
str.stream().filter(test).findFirst();
And this prints 0:0:0:0:0:
str.stream().filter(test).collect(Collectors.toList());
What confuses me is, no matter it is findFirst or collect, no matter the terminal operation is short-circuiting or not, they both should iterate through each item in the list, right?
So why "0:" is printed twice, not just once, not 5 times, in the first example findFirst?
What confuses me is, no matter it is findFirst or collect, no matter the terminal operation is short-circuiting or not, they both should iterate through each item in the list, right?
Ability to return without examining the entire stream is exactly what makes an operation (like findFirst) short-circuiting. See Streams javadoc.
Streams documentation actually makes no promises about how or whether your predicate will be evaluated. It just guarantees that it will return the first element of the stream that matches the predicate.
The idea of the first Stream is iterate to find the first occurrence of "pen", so the Predicate iterates the array until finds the first element which matches the condition of the filter. For that reason, you see only two "0:". Official documentation
I am trying to think of an alternative to the List method for remove(int index) and remove(T element). Where I take a list and do some filtering and return a new list without the element requested to be removed. I want to do it functionally, as I don't want to mutate the original list.
Here is my attempt.
List<Integer> integers = Arrays.asList(2, 4, 1, 2, 5, 1);
Integer elementToRemove = 1;
List<Integer> collect =
integers.stream()
.filter(elements -> !elements.equals(elementToRemove))
.collect(Collectors.toList());
This will remove all the 1's.
I wont to remove just the first 1, so I will be left with a list like [2,4,2,5,1]
I know how to do it using indexOf() and sublist() and addAll(). But I feel this is not as good as using streams.
Looking for functional solutions using streams for implementing remove(int index) and remove(T element).
I want to do it functionally, as I dont want to mutate the original
list.
You can still perform the removal operation and not mutate the source without going functional.
But I feel this is not as good as using streams.
Quite the opposite as this is done better without streams:
List<Integer> result = new ArrayList<>(integers); // copy the integers list
result.remove(Integer.valueOf(elementToRemove)); // remove from the new list leaving the old list unmodified
I agree with #Aominè but this can be a alternative in stream API
IntStream.range(0,integers.size())
.filter(i->i != integers.indexOf(elementToRemove))
.mapToObj(i->integers.get(i))
.collect(Collectors.toList());
As #Aominè commented for optimize, find index of elementToRemove firstly then use it in the filter.
While my other answer is definitely the way I recommend to proceed with and #Hadi has also provided the "stream" alternative which is also valid. I decided to play about with different ways to achieve the same result using features as of JDK-8.
In JDK-9 there is a takeWhile and dropWhile methods where the former returns a stream consisting of the longest prefix of elements taken from a stream that match a given predicate.
The latter returns a stream consisting of the remaining elements of a given stream after dropping the longest prefix of elements that match a given predicate.
The idea here is to consume the elements while it's not equal to the elementToRemove:
integers.stream()
.takeWhile(e -> !Objects.equals(e, elementToRemove))
and drop the elements while it's not equal to the elementToRemove and skip(1) to exclude the elementToRemove:
integers.stream()
.dropWhile(e -> !Objects.equals(e, elementToRemove))
.skip(1)
hence yielding two streams where the first stream is all the preceding numbers to elementToRemove and the second stream plus the skip(1) is all the elements after the elementToRemove then we simply concatenate them and collect to a list implementation.
List<Integer> result = Stream.concat(integers.stream()
.takeWhile(e -> !Objects.equals(e, elementToRemove)),
integers.stream()
.dropWhile(e -> !Objects.equals(e, elementToRemove))
.skip(1))
.collect(Collectors.toList());
Assuming the element to remove does not exist in the list the takeWhile will consume all the elements and the dropWhile will drop all the elements and when we merge these two streams we get back the initial elements.
Overall this will accomplish the same result as the other answers.
However, do not use this solution in production code as it's suboptimal and not obvious to the eye what the code does. it's only here to show different ways to accomplish the said requirement.
I have the following code:
ArrayList <String> entries = new ArrayList <String>();
entries.add("0");
entries.add("1");
entries.add("2");
entries.add("3");
String firstNotHiddenItem = entries.stream()
.filter(e -> e.equals("2"))
.findFirst()
.get();
I need to know what is the index of that first returned element, since I need to edit it inside of entries ArrayList. As far as I know get() returns the value of the element, not a reference. Should I just use
int indexOf(Object o)
instead?
You can get the index of an element using an IntStream like:
int index = IntStream.range(0, entries.size())
.filter(i -> "2".equals(entries.get(i)))
.findFirst().orElse(-1);
But you should use the List::indexOf method which is the preferred way, because it's more concise, more expressive and computes the same results.
You can't in a straightforward way - streams process elements without context of where they are in the stream.
However, if you're prepared to take the gloves off...
int[] position = {-1};
String firstNotHiddenItem = entries.stream()
.peek(x -> position[0]++) // increment every element encounter
.filter("2"::equals)
.findFirst()
.get();
System.out.println(position[0]); // 2
The use of an int[], instead of a simple int, is to circumvent the "effectively final" requirement; the reference to the array is constant, only its contents change.
Note also the use of a method reference "2"::equals instead of a lambda e -> e.equals("2"), which not only avoids a possible NPE (if a stream element is null) and more importantly looks way cooler.
A more palatable (less hackalicious) version:
AtomicInteger position = new AtomicInteger(-1);
String firstNotHiddenItem = entries.stream()
.peek(x -> position.incrementAndGet()) // increment every element encounter
.filter("2"::equals)
.findFirst()
.get();
position.get(); // 2
This will work using Eclipse Collections with Java 8
int firstIndex = ListIterate.detectIndex(entries, "2"::equals);
If you use a MutableList, you can simplify the code as follows:
MutableList<String> entries = Lists.mutable.with("0", "1", "2", "3");
int firstIndex = entries.detectIndex("2"::equals);
There is also a method to find the last index.
int lastIndex = entries.detectLastIndex("2"::equals);
Note: I am a committer for Eclipse Collections
Yes, you should use indexOf("2") instead. As you might have noticed, any stream based solution has a higher complexity, without providing any benefit.
In this specific situation, there is no significant difference in performance, but overusing streams can cause dramatic performance degradation, e.g. when using map.entrySet().stream().filter(e -> e.getKey().equals(object)).map(e -> e.getValue()) instead of a simple map.get(object).
The collection operations may utilize their known structure while most stream operation imply a linear search. So genuine collection operations are preferable.
Of course, if there is no collection operation, like when your predicate is not a simple equality test, the Stream API may be the right tool. As shown in “Is there a concise way to iterate over a stream with indices in Java 8?”, the solution for any task involving the indices works by using the indices as starting point, e.g. via IntStream.range, and accessing the list via List.get(int). If the source in not an array or a random access List, there is no equally clean and efficient solution. Sometimes, a loop might turn out to be the simplest and most efficient solution.
I'm wondering if I can add an operation to a stream, based off of some sort of condition set outside of the stream. For example, I want to add a limit operation to the stream if my limit variable is not equal to -1.
My code currently looks like this, but I have yet to see other examples of streams being used this way, where a Stream object is reassigned to the result of an intermediate operation applied on itself:
// Do some stream stuff
stream = stream.filter(e -> e.getTimestamp() < max);
// Limit the stream
if (limit != -1) {
stream = stream.limit(limit);
}
// Collect stream to list
stream.collect(Collectors.toList());
As stated in this stackoverflow post, the filter isn't actually applied until a terminal operation is called. Since I'm reassigning the value of stream before a terminal operation is called, is the above code still a proper way to use Java 8 streams?
There is no semantic difference between a chained series of invocations and a series of invocations storing the intermediate return values. Thus, the following code fragments are equivalent:
a = object.foo();
b = a.bar();
c = b.baz();
and
c = object.foo().bar().baz();
In either case, each method is invoked on the result of the previous invocation. But in the latter case, the intermediate results are not stored but lost on the next invocation. In the case of the stream API, the intermediate results must not be used after you have called the next method on it, thus chaining is the natural way of using stream as it intrinsically ensures that you don’t invoke more than one method on a returned reference.
Still, it is not wrong to store the reference to a stream as long as you obey the contract of not using a returned reference more than once. By using it they way as in your question, i.e. overwriting the variable with the result of the next invocation, you also ensure that you don’t invoke more than one method on a returned reference, thus, it’s a correct usage. Of course, this only works with intermediate results of the same type, so when you are using map or flatMap, getting a stream of a different reference type, you can’t overwrite the local variable. Then you have to be careful to not use the old local variable again, but, as said, as long as you are not using it after the next invocation, there is nothing wrong with the intermediate storage.
Sometimes, you have to store it, e.g.
try(Stream<String> stream = Files.lines(Paths.get("myFile.txt"))) {
stream.filter(s -> !s.isEmpty()).forEach(System.out::println);
}
Note that the code is equivalent to the following alternatives:
try(Stream<String> stream = Files.lines(Paths.get("myFile.txt")).filter(s->!s.isEmpty())) {
stream.forEach(System.out::println);
}
and
try(Stream<String> srcStream = Files.lines(Paths.get("myFile.txt"))) {
Stream<String> tmp = srcStream.filter(s -> !s.isEmpty());
// must not be use variable srcStream here:
tmp.forEach(System.out::println);
}
They are equivalent because forEach is always invoked on the result of filter which is always invoked on the result of Files.lines and it doesn’t matter on which result the final close() operation is invoked as closing affects the entire stream pipeline.
To put it in one sentence, the way you use it, is correct.
I even prefer to do it that way, as not chaining a limit operation when you don’t want to apply a limit is the cleanest way of expression your intent. It’s also worth noting that the suggested alternatives may work in a lot of cases, but they are not semantically equivalent:
.limit(condition? aLimit: Long.MAX_VALUE)
assumes that the maximum number of elements, you can ever encounter, is Long.MAX_VALUE but streams can have more elements than that, they even might be infinite.
.limit(condition? aLimit: list.size())
when the stream source is list, is breaking the lazy evaluation of a stream. In principle, a mutable stream source might legally get arbitrarily changed up to the point when the terminal action is commenced. The result will reflect all modifications made up to this point. When you add an intermediate operation incorporating list.size(), i.e. the actual size of the list at this point, subsequent modifications applied to the collection between this point and the terminal operation may turn this value to have a different meaning than the intended “actually no limit” semantic.
Compare with “Non Interference” section of the API documentation:
For well-behaved stream sources, the source can be modified before the terminal operation commences and those modifications will be reflected in the covered elements. For example, consider the following code:
List<String> l = new ArrayList(Arrays.asList("one", "two"));
Stream<String> sl = l.stream();
l.add("three");
String s = sl.collect(joining(" "));
First a list is created consisting of two strings: "one"; and "two". Then a stream is created from that list. Next the list is modified by adding a third string: "three". Finally the elements of the stream are collected and joined together. Since the list was modified before the terminal collect operation commenced the result will be a string of "one two three".
Of course, this is a rare corner case as normally, a programmer will formulate an entire stream pipeline without modifying the source collection in between. Still, the different semantic remains and it might turn into a very hard to find bug when you once enter such a corner case.
Further, since they are not equivalent, the stream API will never recognize these values as “actually no limit”. Even specifying Long.MAX_VALUE implies that the stream implementation has to track the number of processed elements to ensure that the limit has been obeyed. Thus, not adding a limit operation can have a significant performance advantage over adding a limit with a number that the programmer expects to never be exceeded.
There is two ways you can do this
// Do some stream stuff
List<E> results = list.stream()
.filter(e -> e.getTimestamp() < max);
.limit(limit > 0 ? limit : list.size())
.collect(Collectors.toList());
OR
// Do some stream stuff
stream = stream.filter(e -> e.getTimestamp() < max);
// Limit the stream
if (limit != -1) {
stream = stream.limit(limit);
}
// Collect stream to list
List<E> results = stream.collect(Collectors.toList());
As this is functional programming you should always work on the result of each function. You should specifically avoid modifying anything in this style of programming and treat everything as if it was immutable if possible.
Since I'm reassigning the value of stream before a terminal operation is called, is the above code still a proper way to use Java 8 streams?
It should work, however it reads as a mix of imperative and functional coding. I suggest writing it as a fixed stream as per my first answer.
I think your first line needs to be:
stream = stream.filter(e -> e.getTimestamp() < max);
so that your using the stream returned by filter in subsequent operations rather than the original stream.
I known it is a bit too late, but I had the same question myself and didn't find the satisfying answer, however, inspired by this question and answers I came to the following solution:
return Stream.of( ///< wrap target stream in other stream ;)
/*do regular stream stuff*/
stream.filter(e -> e.getTimestamp() < max)
).flatMap(s -> limit != -1 ? s.limit(limit) : s) ///< apply limit only if necessary and unwrap stream of stream to "normal" stream
.collect(Collectors.toList()) ///< do final stuff
I am looking for a memory-efficient way in Java to find top n elements from a huge collection. For instance, I have a word, a distance() method, and a collection of "all" words.
I have implemented a class Pair that implements compareTo() so that pairs are sorted by their values.
Using streams, my naive solution looks like this:
double distance(String word1, String word2){
...
}
Collection<String> words = ...;
String word = "...";
words.stream()
.map(w -> new Pair<String, Double>(w, distance(word, w)))
.sorted()
.limit(n);
To my understanding, this will process and intermediately store each element in words so that it can be sorted before applying limit(). However, it is more memory-efficient to have a collection that stores n elements and whenever a new element is added, it removes the smallest element (according to the comparable object's natural order) and thus never grows larger than n (or n+1).
This is exactly what the Guava MinMaxPriorityQueue does. Thus, my current best solution to the above problem is this:
Queue<Pair<String, Double>> neighbours = MinMaxPriorityQueue.maximumSize(n).create();
words.stream()
.forEach(w -> neighbours.add(new Pair<String, Double>(w, distance(word, w)));
The sorting of the top n elements remains to be done after converting the queue to a stream or list, but this is not an issue since n is relatively small.
My question is: is there a way to do the same using streams?
A heap-based structure will of course be more efficient than sorting the entire huge list. Luckily, streams library is perfectly happy to let you use specialized collections when necessary:
MinMaxPriorityQueue<Pair<String, Double>> topN = words.stream()
.map(w -> new Pair<String, Double>(w, distance(word, w)))
.collect(toCollection(
() -> MinMaxPriorityQueue.maximumSize(n).create()
));
This is better than the .forEach solution because it's easy to parallelize and is more idiomatic java8.
Note that () -> MinMaxPriorityQueue.maximumSize(n).create() should be possible to be replaced with MinMaxPriorityQueue.maximumSize(n)::create but, for some reason, that won't compile under some conditions (see comments below).