What is the best way to remove a set from a collection, but still keep the items that were removed in a separate collection?
I have written an extension method that does that, but I think there must be a better way. Here is my function:
public static List<T> FindAndRemove<T>(this List<T> lst, Predicate<T> match)
{
List<T> ret = lst.FindAll(match);
lst.RemoveAll(match);
return ret;
}
And you would use it like this:
List<String> myList = new List<String>();
myList.Add("ABC");
myList.Add("DEF");
myList.Add("ABC");
List<String> removed = myList.FindAndRemove(x => x == "ABC");
// myList now contains 1 item (DEF)
// removed now contains 2 items (ABC, ABC)
I'm not 100% sure what goes on behind the scenes in the FindAll
and RemoveAll
methods, but I imagine a better way would be to somehow "transfer" items from one list to the other.
remove() method is used to remove elements from a collection. It removes the element at the specified position in this list. Shifts any subsequent elements to the left by subtracts one from their indices.
The Java ArrayList removeAll() method removes all the elements from the arraylist that are also present in the specified collection. The syntax of the removeAll() method is: arraylist. removeAll(Collection c);
Op's answer is the best out of the proposed and suggested solutions so far. Here are timings on my machine:
public static class Class1
{
// 21ms on my machine
public static List<T> FindAndRemove<T>(this List<T> lst, Predicate<T> match)
{
List<T> ret = lst.FindAll(match);
lst.RemoveAll(match);
return ret;
}
// 538ms on my machine
public static List<T> MimoAnswer<T>(this List<T> lst, Predicate<T> match)
{
var ret = new List<T>();
int i = 0;
while (i < lst.Count)
{
T t = lst[i];
if (!match(t))
{
i++;
}
else
{
lst.RemoveAt(i);
ret.Add(t);
}
}
return ret;
}
// 40ms on my machine
public static IEnumerable<T> GuvanteSuggestion<T>(this IList<T> list, Func<T, bool> predicate)
{
var removals = new List<Action>();
foreach (T item in list.Where(predicate))
{
T copy = item;
yield return copy;
removals.Add(() => list.Remove(copy));
}
// this hides the cost of processing though the work is still expensive
Task.Factory.StartNew(() => Parallel.ForEach(removals, remove => remove()));
}
}
[TestFixture]
public class Tester : PerformanceTester
{
[Test]
public void Test()
{
List<int> ints = Enumerable.Range(1, 100000).ToList();
IEnumerable<int> enumerable = ints.GuvanteSuggestion(i => i % 2 == 0);
Assert.That(enumerable.Count(), Is.EqualTo(50000));
}
}
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