You want something like this: List<Map<String, String>> result = four.stream() // iterate the Map with values .map(map -> map.entrySet() // .. for each one .stream() // .. stream and collect .collect(Collectors.toMap( // .. into a new Map e -> one.get(e.getKey()), // .. with a key from the `one` Map Map.Entry::getValue))) // .. and the same … Read more
Based on your exception, you aren’t passing any arguments to your class’ main method – System.out.println(Arrays.toString(args)); // <– to display your arguments. // Class<?> c = Class.forName(args[0]); // <– you should have a default Class<?> c = Class.forName(args.length > 0 ? args[0] : “java.lang.Object”); 3 solved Reflection – Exception in thread “main” java.lang.ArrayIndexOutOfBoundsException: 0
This should be working. List<Counter> toRemove = new ArrayList<Counter>(); Collections.sort(interval, (first, second) -> { int c = 0; if (first.compareWith(second)) { if (first.getCount() <= second.getCount()()) toRemove.add(first); else if (first.getCount() >= second.getCount()) toRemove.add(second); } else c = first.getCount().compareTo(second.getCount()); return c; } ); interval.removeAll(toRemove); This is the compareWith function in the Counter class. public boolean compareWith(Counter second) … Read more
To be brief, a hash table (dictionary) does not maintain a total order of insertions because it doesn’t need to. The abstract data type supports ammortized O(1) insertions, deletions, and searches, but does not support enumeration, and does not impose any order on the elements in the key set. HashTable implements a dictionary, and total … Read more
Assuming you want a Map<User.userId, List<UserData.customerId>> you can use this: Map<Long, List<Long>> result = users.stream() .collect(Collectors.toMap( User::getUserId, u -> u.getUserData().stream() .map(UserData::getCustomerId) .collect(Collectors.toList()) )); 3 solved Java 8 – collecting into a map from a list of objects with nested list
Read a line: Read String line by line in Java Split each line using a space character: Java split string to array The result of #2 will give you an array of String Objects. Use the HashMap data structure (stores a key/value pair) to store each element: HashMap hashMap = new HashMap(); hashMap.put(key, value); ie. … Read more
I’m guessing you mean by not writing your own bean for GSON to deserialize to. GSON provides support by having certain JSON types that you can make use of: Gson gson = new Gson(); final JsonArray jsonElements = gson.fromJson(JSON, JsonArray.class); 0 solved Could anyone help me to convert the below JSON to java object (List), … Read more
The main problem is the presence of the category name before the headers and data. This creates a condition to consider while looping over the lines because when we read a category name we need to consider 4 rows while when reading the transaction data we have to loop over 3 rows. However there is … Read more
Using explicit implementation of interfaces, you can do something like: public interface IMyClass { object this[int index] { get; set; } } public class MyClass : IMyClass { public string this[int index] { get { return “”; } set { } } object IMyClass.this[int index] { get { return “”; } set { } } … Read more