How to take all but the last element in a sequence using LINQ?

Question

Let\'s say I have a sequence.

IEnumerable sequence = GetSequenceFromExpensiveSource();
// sequence now contains: 0,1,2,3,...,999999,1000000
         


        

Answer 1

I don't know a Linq solution - But you can easily code the algorithm by yourself using generators (yield return).

public static IEnumerable TakeAllButLast(this IEnumerable 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 SkipLastN(this IEnumerable source, int n) {
    var  it = source.GetEnumerator();
    bool hasRemainingItems = false;
    var  cache = new Queue(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()));
}