From OCP : Oracle Certified Professional Java SE 8 Programmer || Study Guide : Exam 1z0-809 […] Depending on the number of CPUs available in your environment the following is a possible output of the code using a parallel stream […] Even better, the results scale with the number of processors. Scaling is the property … Read more
There are two possbile solutions. The first one is to override the equals method. Simply add: public class DataRecord { […..] private String TUN; @Override public boolean equals(Object o) { if (o instanceof DataRecord) { DataRecord that = (DataRecord) o; return Objects.equals(this.TUN, that.TUN); } return false; } @Override public int hashCode() { return Objects.hashCode(TUN); } … Read more
You can use the .indexOf(Object o) method which returns, as the java 8 documentation specifies: the index of the first occurrence of the specified element in this list, or -1 if this list does not contain the element You can also use .lastIndexOf(Object o) which, as you would guess, will return the last occurrence of … Read more
map.keySet().stream().map(Object::toString).collect(toCollection(LinkedHashSet::new)); solved Keyset to Set of string [closed]
I think the reason is pretty clear. ints.forEach((i) -> { System.out.print(ints.get(i-1) + ” “); }); Translates approximately to: for (Integer i : ints) { System.out.println(ints.get(i – 1) + ” “); } Which will cause IndexOutOfBoundsExceptions because i refers to the elements of each list, and each of those elements – 1 will give an index … Read more
Something like this… List<String> result = Stream.of(“A”, “A”, “BB”, “C”, “BB”, “D”) .collect(Collectors.groupingBy( Function.identity(), Collectors.counting())) .entrySet() .stream() .filter(x -> x.getValue() == 1L) .map(Entry::getKey).collect(Collectors.toList()); System.out.println(result); // [C, D] 2 solved return the list with strings with single occurrence [closed]
Use .orElse(null) if you want null. The idea of Optional is to force you to explicitly handle the absence of the value. 0 solved Java 8 Optional get
Something like this should do it for you : serviceConfigList.stream() .filter(ser -> ser.getValid().equals(“true”)) .forEach(ser -> { Connection.SERVICE_PORT = ser.getPort(); Connection.SERVICE_ADDRESS = ser.getIp(); // other code.. }); PS : I don’t have your complete code. So can’t help more. 4 solved Using Java 8 lambda and Stream for information extraction
There are many ways to compute managerNotFoundInManagerStoreIdList. Here’s the simplest one: Create a set of manager and look it up: Set<String> managers = managerStoreIdList.stream().map(ManagerStoreId::getManager) .collect(Collectors.toSet()); List<String> managerNotFoundInManagerStoreIdList = Manager.stream() .filter(m -> !managers.contains(m)) .collect(Collectors.toList()); solved Create a list comprised of items missing in a list [closed]
You can just combine all four conditions into one using logical operators. For example by using logical and && and logical or ||. It could then look like: if ((first && second) || (third && fourth)) { return true; } Or with all conditions substituted: if ((Obj.getSource().equals(“abc”) && Obj.getDest().equals(“bcd”)) || (Obj.getSource().equals(“abd”) && Obj.getDest().equals(“gdc”))) { return … Read more
As Benjamin M correctly stated in a comment, 01 would be a string (not an int). The correct way to convert a date to a string goes through a formatter. This one gives you a two-digit month — so 01 for January through 12 for December: private static final DateTimeFormatter monthFormatter = DateTimeFormatter.ofPattern(“MM”); To use … Read more
As already mentioned in a comment by the user “soon”, this problem can be easily solved with flatMap and map: List<TestSession> list = mapList.entrySet().stream() .flatMap(e1 -> e1.getValue().stream() .map(e2 -> new TestSession(e1.getKey(), e2.getKey(), e2.getValue()))) .collect(Collectors.toList()); solved Convert Map to List with new java 8 streams
tl;dr LocalDate.parse( “2022-03-14” ).datesUntil( LocalDate.parse( “2022-04-15” ) ).filter( localDate -> localDate.getDayOfMonth() == 1 ).count() See this code run live at IdeOne.com. 1 Details First, parse the inputs as LocalDate objects. LocalDate start = LocalDate.parse( “2022-03-14” ); LocalDate end = LocalDate.parse( “2022-04-15” ); The LocalDate#datesUntil method creates a stream of LocalDate objects between the two values. … Read more
This is one way (it does create two streams and does merge the two lines of code): Map<String, Long> map = Arrays.stream(“some text some spaces”.split(” “)) .collect(Collectors.groupingBy(Function.identity(), Collectors.counting())) .entrySet() .stream() .sorted(Map.Entry.<String, Long>comparingByValue().reversed()) .collect(Collectors.toMap(e -> e.getKey(), e -> e.getValue(), (v1, v2) -> v2, LinkedHashMap::new)); System.out.println(map); // This prints: {some=2, spaces=1, text=1} solved Word count in descending … Read more
It’s because you’re splitting by a single character, you would have to do this eagerly: String[] Operators = stringNumbers.split(“[0-9.]*”); Or you can filter the results: String[] Operators = Arrays.stream(stringNumbers.split(“[0-9.]”)) .filter(str -> !str.equals(“”)) .toArray(String[]::new); 5 solved Why there are empty string while trying to split a string in Java? And how to fix it?