How to take all but the last element in a sequence using LINQ?
The Enumerable.SkipLast(IEnumerable<TSource>, Int32)
method was added in .NET Standard 2.1. It does exactly what you want.
IEnumerable<int> sequence = GetSequenceFromExpensiveSource();
var allExceptLast = sequence.SkipLast(1);
From https://docs.microsoft.com/en-us/dotnet/api/system.linq.enumerable.skiplast
Returns a new enumerable collection that contains the elements from source with the last count elements of the source collection omitted.
I don't know a Linq solution - But you can easily code the algorithm by yourself using generators (yield return).
public static IEnumerable<T> TakeAllButLast<T>(this IEnumerable<T> source) {
var it = source.GetEnumerator();
bool hasRemainingItems = false;
bool isFirst = true;
T item = default(T);
do {
hasRemainingItems = it.MoveNext();
if (hasRemainingItems) {
if (!isFirst) yield return item;
item = it.Current;
isFirst = false;
}
} while (hasRemainingItems);
}
static void Main(string[] args) {
var Seq = Enumerable.Range(1, 10);
Console.WriteLine(string.Join(", ", Seq.Select(x => x.ToString()).ToArray()));
Console.WriteLine(string.Join(", ", Seq.TakeAllButLast().Select(x => x.ToString()).ToArray()));
}
Or as a generalized solution discarding the last n items (using a queue like suggested in the comments):
public static IEnumerable<T> SkipLastN<T>(this IEnumerable<T> source, int n) {
var it = source.GetEnumerator();
bool hasRemainingItems = false;
var cache = new Queue<T>(n + 1);
do {
if (hasRemainingItems = it.MoveNext()) {
cache.Enqueue(it.Current);
if (cache.Count > n)
yield return cache.Dequeue();
}
} while (hasRemainingItems);
}
static void Main(string[] args) {
var Seq = Enumerable.Range(1, 4);
Console.WriteLine(string.Join(", ", Seq.Select(x => x.ToString()).ToArray()));
Console.WriteLine(string.Join(", ", Seq.SkipLastN(3).Select(x => x.ToString()).ToArray()));
}