Second highest voted answer doesn’t do proper merge. If second array contains an unique property, it is not taken into account. This approach does a proper merge. Lodash var a = [ { userId:”p1″, item:1}, { userId:”p2″, item:2}, { userId:”p3″, item:4} ]; var b = [ { userId:”p1″, profile:1}, { userId:”p2″, profile:2}, { userId:”p4″, profile:4} … Read more
You need to use the iterator directly, and remove the item via that iterator. for (Iterator<String> iterator = list.iterator(); iterator.hasNext(); ) { String fruit = iterator.next(); if (“banane”.equals(fruit)) { iterator.remove(); } System.out.println(fruit); }
You can always iterate without an enumerator. Which means a regular for loop, and when you remove an object:- decrement the index variable and continue;. if you are caching the array’s count before entering the for-loop, then make sure you decrement that too when removing an object. Anyway, I do not see why an array … Read more
While the purity argument does make the method get(Object) suspect, the underlying intent is not moot. There are various class and interface families that slightly redefine equals(Object). One need look no further than the collections interfaces. For example, an ArrayList and a LinkedList can be equal; their respective contents merely need to be the same … Read more
Collection is also the supertype of List, Queue, Deque, and others, so it gives you more options. For example, I try to use Collection as a parameter to library methods that shouldn’t explicitly depend on a certain type of collection. Generally, you should use the right tool for the job. If you don’t want duplicates, … Read more
Total ordering means all values can be compared to all other values. For example, if you have a collection of BigDecimal and String there is no natural total order (but you could invent one) In Java, the Natural order is defined as the ordering provided by the JVM. This might not match what a people … Read more
You asked: Also what do you gain with the java.util.Arrays.ArrayList implementation ? It is because the Arrays$ArrayList returned by Arrays.asList is just a view on the original array. So when the original array is changed then the view is changed too. If one would use an real ArrayList then the elements will be copied, and … Read more
I’m not certain from your wording whether you want the keys or the values. Either way, it’s pretty straightforward. Use either the Keys or Values property of the dictionary and the ToArray extension method. var arrayOfAllKeys = yourDictionary.Keys.ToArray(); var arrayOfAllValues = yourDictionary.Values.ToArray();
If you really wanted to create your own IGrouping<TKey, TElement>, it is a simple interface to implement: public class Grouping<TKey, TElement> : List<TElement>, IGrouping<TKey, TElement> { public Grouping(TKey key) : base() => Key = key; public Grouping(TKey key, int capacity) : base(capacity) => Key = key; public Grouping(TKey key, IEnumerable<TElement> collection) : base(collection) => Key … Read more
If you use immutable maps, you can use the – method to create a new map without the given key: val mx = Map(“data” -> “sumi”, “rel” -> 2, “privacy” -> 0) val m = mx(“privacy”) match { case 0 => mx – “data” case _ => mx } => m: scala.collection.immutable.Map[String,Any] = Map(rel -> … Read more