If you know n in advance, I think it's more idiomatic to use one of the stream sources that creates a stream that is known to have exactly n elements. Despite the appearances, using limit(10) doesn't necessarily result in a SIZED stream with exactly 10 elements -- it might have fewer. Having a SIZED stream improves splitting in the case of parallel execution.

As Sotirios Delimanolis noted in a comment, you could do something like this:

List<MyClass> list = IntStream.range(0, n)
    .mapToObj(i -> new MyClass())
    .collect(toList());

An alternative is this, though it's not clear to me it's any better:

List<MyClass> list2 = Collections.nCopies(10, null).stream()
    .map(o -> new MyClass())
    .collect(toList());

You could also do this:

List<MyClass> list = Arrays.asList(new MyClass[10]);
list.replaceAll(o -> new MyClass());

But this results in a fixed-size, though mutable, list.

Something like:

Map<String,Integer> m =
    list.stream()
        .collect(Collectors.toMap(e -> e.getUser, // key extractor
                                  e -> getDetail(e.getId()).getBody().getData() // value extractor
                                 )
                );    

----EDIT----

Ho ho... This is exactly Holger's comment. Give credit where credit is due.

You can use IntStream as shown below and explained in the comments:

(1) Iterate IntStream range from 1 to 1000

(2) Convert to parallel stream

(3) Apply Predicate condition to allow integers with (i+1)%100 == 0

(4) Now convert the integer to a string "Test number "+i+" started."

(5) Output to console

IntStream.range(1, 1000). //iterates 1 to 1000
   parallel().//converts to parallel stream
   filter( i -> ((i+1)%100 == 0)). //filters numbers & allows like 99, 199, etc..)
   mapToObj((int i) -> "Test number "+i+" started.").//maps integer to String
   forEach(System.out::println);//prints to the console

Here's another technique I ran across the other day:

Collections.nCopies(8, 1)
           .stream()
           .forEach(i -> System.out.println(i));

The Collections.nCopies call creates a List containing n copies of whatever value you provide. In this case it's the boxed Integer value 1. Of course it doesn't actually create a list with n elements; it creates a "virtualized" list that contains only the value and the length, and any call to get within range just returns the value. The nCopies method has been around since the Collections Framework was introduced way back in JDK 1.2. Of course, the ability to create a stream from its result was added in Java SE 8.

Big deal, another way to do the same thing in about the same number of lines.

However, this technique is faster than the IntStream.generate and IntStream.iterate approaches, and surprisingly, it's also faster than the IntStream.range approach.

For iterate and generate the result is perhaps not too surprising. The streams framework (really, the Spliterators for these streams) is built on the assumption that the lambdas will potentially generate different values each time, and that they will generate an unbounded number of results. This makes parallel splitting particularly difficult. The iterate method is also problematic for this case because each call requires the result of the previous one. So the streams using generate and iterate don't do very well for generating repeated constants.

The relatively poor performance of range is surprising. This too is virtualized, so the elements don't actually all exist in memory, and the size is known up front. This should make for a fast and easily parallelizable spliterator. But it surprisingly didn't do very well. Perhaps the reason is that range has to compute a value for each element of the range and then call a function on it. But this function just ignores its input and returns a constant, so I'm surprised this isn't inlined and killed.

The Collections.nCopies technique has to do boxing/unboxing in order to handle the values, since there are no primitive specializations of List. Since the value is the same every time, it's basically boxed once and that box is shared by all n copies. I suspect boxing/unboxing is highly optimized, even intrinsified, and it can be inlined well.

Here's the code:

    public static final int LIMIT = 500_000_000;
    public static final long VALUE = 3L;

    public long range() {
        return
            LongStream.range(0, LIMIT)
                .parallel()
                .map(i -> VALUE)
                .map(i -> i % 73 % 13)
                .sum();
}

    public long ncopies() {
        return
            Collections.nCopies(LIMIT, VALUE)
                .parallelStream()
                .mapToLong(i -> i)
                .map(i -> i % 73 % 13)
                .sum();
}

And here are the JMH results: (2.8GHz Core2Duo)

Benchmark                    Mode   Samples         Mean   Mean error    Units
c.s.q.SO18532488.ncopies    thrpt         5        7.547        2.904    ops/s
c.s.q.SO18532488.range      thrpt         5        0.317        0.064    ops/s

There is a fair amount of variance in the ncopies version, but overall it seems comfortably 20x faster than the range version. (I'd be quite willing to believe that I've done something wrong, though.)

I'm surprised at how well the nCopies technique works. Internally it doesn't do very much special, with the stream of the virtualized list simply being implemented using IntStream.range! I had expected that it would be necessary to create a specialized spliterator to get this to go fast, but it already seems to be pretty good.

You can apply the doubling and tripling within the stream:

public static void main(String[] args) {
    int[] numbers = {1, 2, 3, 4, 5, 6, 7, 8};

    System.out.println("Double Numbers");
    Arrays.stream(numbers).map(x -> x * 2).forEach(System.out::println);

    System.out.println("Triple Numbers");
    Arrays.stream(numbers).map(x -> x * 3).forEach(System.out::println);
}

though technically that's still iterating over the array twice.

As a trick you could instead collect the results in a Map:

Map<Integer, Integer> m = Arrays.stream(numbers)
        .boxed()
        .collect(Collectors.toMap(
                x -> x * 2,
                x -> x * 3
        ));

The problem is, you're creating the List(s) inside the forEach and forEach returns void. This would be the equivalent of the following for loop:

for (Car car : cars) {
    List<Car> filteredCars = new ArrayList<>();
    for (Wheel wheel : wheels) {

        if (car.getColor() == wheel.getColor() &&
                wheel.isWorking() == true ) {

            filteredCars.add(car);
            break;
        }
    }
}

return filteredCars; // whoops cannot be accessed (scope) !!!

You could use filter on the cars stream and collect the use collect on the filtered stream to achieve the desired results:

Predicate<Car> carCheck = car -> wheels.stream().anyMatch(wheel -> car.getColor() == wheel.getColor() && wheel.isWorking());

List<Car> filteredCars = cars.stream().filter(carCheck).collect(Collectors.toList());

The following code should be equivalent, except that the value of the boolean is reversed so it's false if there are missing nodes.

First all the user node Ids are collected to a TreeSet (if Id implements hashCode() and equals() you should use a HashSet). Then we stream itemSetNodeList to see if all those nodeIds are contained in the set.

TreeSet<Id> all = allUserNodes
                        .stream()
                        .map(n -> n.getNodeId())
                        .collect(Collectors.toCollection(TreeSet::new));

boolean isAllNodes = itemSetNodeList
                        .stream()
                        .allMatch(n -> all.contains(n.getNodeId()));

There are many ways to write equivalent (at least to outside eyes) code, this uses a Set to improve the lookup so we don't need to keep iterating the allUserNodes collection constantly.

You want to avoid using a stream in a loop, because that will turn your algorithm into O(n²) when you're doing a linear loop and a linear stream operation inside it. This approach is O(n log n), for the linear stream operation and O(log n) TreeSet lookup. With a HashSet this goes down to just O(n), not that it matters much unless you're dealing with large amount of elements.

One of the reasons to use IntStream is to add parallel-ism, assuming you understand the impact of this.

IntStream.range(0, LIMIT).parallel().forEach($ -> {
    // some thing thread safe.
});