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How to validate field inside list before doing a stream()?

On the lines where I create the cardType and groupId variables I get a "Call Optional#isPresent() before accessing the value" message from my Sonar. I tried to add ofNullable().isPresent to these fields in several ways but none of them made the message go away. Any ideas on how to make this code cleaner?
var customerResponse = customerPromise.get().getContent().stream().findFirst().orElse(null);
if (ofNullable(customerResponse).isPresent()) {
var cardValuesResponse = customerResponse.getCardValuesResponses();
if (ofNullable(cardValuesResponse).isPresent()) {
var cardType = cardValuesResponse.stream().map(CardValuesResponse::getCardType)
.findFirst().get();
var groupId = cardValuesResponse.stream().map(CardValuesResponse::getGroupId)
.findFirst().get();
var discountResponse = discountPromise.get().getContent()
.stream().filter(d -> d.getDescription().equals(cardType))
.filter(d -> d.getCustomerGroup().equals(groupId))
.map(d -> d.getPercent())
.findFirst();
}
}

This can be simplified a lot by using .ifPresent instead of .isPresent and .get, and you get the interim null checks for free through the .map steps. Start with this:
customerPromise.get()
.getContent()
.stream()
.findFirst()
.map(CustomerResponse::getCardValuesResponse)
.ifPresent(
cardValuesResponse -> {
// handle card data here
}
);

RxJava2 combine multiple observables to make them return single result

How to combine multiple results emmited by observables into one result and emit it once?
I have a Retrofit service:
public interface MyService {
#GET("url")
Observable<UserPostsResult> getUserPosts(#Query("userId") int id);
#GET("url")
Observable<UserPostsResult> getUserPosts(#Query("userId") int id, #Query("page") int pageId);
}
And I have a model:
public class UserPostsResult {
#SerializedName("posts")
List<UserPost> mPosts;
#SerializedName("nextPage")
int mPageId;
}
Also I have ids List<Integer> friendsIds;
My goal is to have a method like this one:
public Observable<Feed> /*or Single<Feed>*/ getFeed(List<Integer> ids) {
...
}
It returns one Observable or Single that does the following:
Combines all getUserPosts(idFromList) to one observable
For each UserPostsResult must do:
if (userPostResult.mPageId > -1)
getUserPosts(currentUserId, userPostResult.mPageId);
And merge this result to the previous userPostResult
Return one single model as result of all operations.
Result class:
public class Feed {
List<UserPost> mAllPostsForEachUser;
}
EDIT (More details):
My client specifications was that I must take from social network user posts with no logging in, no token requesting. So I must parse HTML pages. That's why I have this complex structure.
EDIT (Partial solution)
public Single<List<Post>> getFeed(List<User> users) {
return Observable.fromIterable(users)
.flatMap(user-> mService.getUserPosts(user.getId())
.flatMap(Observable::fromIterable))
.toList()
.doOnSuccess(list -> Collections.sort(list, (o1, o2) ->
Long.compare(o1.getTimestamp(), o2.getTimestamp())
));
}
This solution doesn't include pages problem. Thats why it is only partial solution

There are a number of operators which transform things into other things. fromIterable() will emit each item in the iterable, and flatMap() will convert one type of observable into another type of observable and emit those results.
Observable.fromIterable( friendsIds )
.flatMap( id -> getUserPosts( id ) )
.flatMap( userPostResult -> userPostResult.mPageId
? getUserPosts(currentUserId, userPostResult.mPageId)
: Observable.empty() )
.toList()
.subscribe( posts -> mAllPostsForEachUser = posts);

If you need join two response in one you should use Single.zip
Single.zip(firsSingle.execute(inputParams), secondSingle.execute(inputPrams),
BiFunction<FirstResponse, SecondResponse, ResponseEmitted> { firstResponse, secondResponse ->
//here you put your code
return responseEmmitted
}
}).subscribe({ response -> },{ })

Java 8 Streams - Filter More Than One Condition

Based on some sports results data, I have a Fixture object which has getHome() and getAway() method. I'd like to shorten this method which I've written to only use a single lambda function (instead of creating a new list and two lambdas), is this possible?
private Collection<FixtureResult> finalResults(Team team) {
List<FixtureResult>finalResults = new ArrayList<>();
List<FixtureResult> homeResults = resultList.stream().filter(fixture ->
fixture.getHome().equals(team))
.collect(toList());
List<FixtureResult> awayResults = resultList.stream().filter(fixture ->
fixture.getAway().equals(team))
.collect(toList());
finalResults.addAll(homeResults);
finalResults.addAll(awayResults);
return finalResults;
}

Simple enough
resultList.stream()
.filter(fixture -> fixture.getHome().equals(team) || fixture.getAway().equals(team)))
.collect(toList());
EDIT: This is on the assumption that order does not matter to you. If your final list needs to have home result and then away, have a look at Elliott Frisch's answer.

If you wan to get fancy with lambdas:
Predicate<FixtureResult> isHome = fr -> fr.getHome().equals(team)
Predicate<FixtureResult> isAway = fr -> fr.getAway().equals(team)
resultList.stream()
.filter(isHome.or(isAway))
.collect(toList()));
You could even extract the compose predicate to test it in isolation, with no streams involved, which is good for more complex predicates:
Predicate<FixtureResult> isHomeOrAway = isHome.or(isAway)
assertTrue(isHomeOrAway(homeFixture));
...

Assuming the order doesn't matter, you can do it on one line. Like,
private Collection<FixtureResult> finalResults(Team team) {
return resultList.stream()
.filter(fixture -> fixture.getHome().equals(team)
|| fixture.getAway().equals(team))
.collect(toList());
}
If the order matters (home results and then away), you can do it with a single List like
private Collection<FixtureResult> finalResults(Team team) {
List<FixtureResult> al = new ArrayList<>(resultList.stream()
.filter(fixture -> fixture.getHome().equals(team)).collect(toList()));
al.addAll(resultList.stream()
.filter(fixture -> fixture.getAway().equals(team)).collect(toList()));
return al;
}

You can simply create a conditions concatenations or can concatenate multiple filter call
Conditions concatenations
myList.stream()
.filter(element -> (condition1 && condition2 && condition3))
Multiple filter call
myList.stream()
.filter(element -> condition1)
.filter(element -> condition2)
.filter(element -> condition3)

You can do the following
someStream.filter(((Predicate<SomeClass>) someObject-> someCondition).or(someObject-> someOtherCondition))
Or you can define your own "or" function that won't cause such a deep hierarchy
#SuppressWarnings("unchecked")
<R> Predicate<R> or(Predicate<R> ...predicates) {
return r -> Arrays.stream(predicates).anyMatch(p -> p.test(r));
}
That gives you a cleaner interface without casting and the nesting
.filter(or(
yourObject -> {
return false;
},
yourObject -> {
return false;
},
yourObject -> {
return false;
},
yourObject -> {
return false;
}
))

RxJava: dynamically create Observables and send the final resut as Observable

I am using RxJava in which I want to dynamically create a number of Observables based on some condition. Once I'm done with creating, I want to do some processing on the different values returned by the observables and then send as a single Observable to which I can subscribe on. Here is how my code is :
List<String> valueList = ....
List<Observable<String>> listOfObservables = new ArrayList<Observable<String>>();
for(int i =; i <valueList.size(); i++){
listOfObservables.add(new SomeClass.doOperation(valueList(i)));
// SomeClass.doOperation will return an Observable<String>
}
return Observable.merge(listOfObservables);
But here , I want to do some operation on the values emitted by different Observables in the listOfObservable and finally return it as a single Observable<String>
Like in Observable.zip() , I can do this like
return Observable.zip(observable1, observable2, (string1, string2) -> {
// joining final string here
return string1 + string2;
But I know the number of arguments here. Please let me know how I can achieve this.

Use the zip overload that takes a variable number of arguments, it has a signature of
<R> Observable<R> zip(Iterable<? extends Observable<?>> ws,
FuncN<? extends R> zipFunction)
Example usage:
List<String> valueList = ....
return Observable.from(valueList)
.map(string -> SomeClass.doOperationThatReturnsObservable(string))
.toList()
.flatMap(listOfObs -> Observable.zip(listOfObs, (Object[] results) -> {
// do something with the strings in the array.
return Arrays.stream(results)
.map(Object::toString)
.collect(Collectors.joining(","));
}));

RxJava: How to .zip two Observable, then .merge them and eventually .reduce to aggregate all results

I have the following code:
public void foo() {
Long[] gData = new Long[] { 1L, 2L };
rx.Observable.from(gData)
.concatMap(data -> {
rx.Observable<GmObject> depositObs1 = depositToUserBalance(data, 1);
rx.Observable<GmObject> depositObs2 = depositToUserBalance(data, 2);
return rx.Observable.zip(depositObs1, depositObs2, (depositObj1, depositObj2) -> {
depositObj1.putNumber("seat_index", data);
depositObj2.putNumber("seat_index", data);
return rx.Observable.merge(
rx.Observable.just(depositObj1),
rx.Observable.just(depositObj2));
})
})
.reduce(new ArrayList<Long>(), (payoutArr, payoutObj) -> {
int seatIndex = ((GmObject) payoutObj).getNumber("seat_index").intValue();
long payout = ((GmObject) payoutObj).getNumber("payout").longValue();
payoutArr.add(seatIndex, payout);
return payoutArr;
})
.subscribe(results -> {
System.out.println(results);
});
}
This code uses .zip to emits to observables, and then it adds a 'seat_index' property and calls .merge in order to use .reduce so eventually all results would be aggregated into an ArrayList.
There is a problem with this code: When .reduce processes its input it gets it as Observable and not as GmObject ...What function can 'extract' the GmObject from its Observable wrap?
Does it make sense to use rxJava this way? or there is a better technique?
Thanks!

the zip operator take as third argument a lambda. this lambda is a 2 args function which return an object that result of the composition of args. And not a Observable of the result of the composition (but, of course, the object can be an Observable, but it's not what you want in your case).
So after your zip call, you'll have an Observable<Observable<GmObject>> but you expect an Observable<GmObject>.
I don't think that the zip operator is the operator you're looking for.
public void foo() {
Long[] gData = new Long[] { 1L, 2L };
rx.Observable.from(gData)
.concatMap(data -> {
rx.Observable<GmObject> depositObs1 = depositToUserBalance(data, 1).doOnNext(obj -> obj.putNumber("seat_index", data));
rx.Observable<GmObject> depositObs2 = depositToUserBalance(data, 2).doOnNext(obj -> obj.putNumber("seat_index", data));
return rx.Observable.merge(depositObs1, depositObs2);
})
.reduce(new ArrayList<Long>(), (payoutArr, payoutObj) -> {
int seatIndex = ((GmObject) payoutObj).getNumber("seat_index").intValue();
long payout = ((GmObject) payoutObj).getNumber("payout").longValue();
payoutArr.add(seatIndex, payout);
return payoutArr;
})
.subscribe(results -> System.out.println(results));
}