How to access sequential elements in an IQueryable <T> object?

Question

I need to access the current and previous item in an IQueryable object. If I had an int array, I would do the following:

var array = new int[]{0,1,2,3,4};
for(var i = 1; i<array.Length ; i++)
{
    method1(array[i-1], array[i]);
}

I do not know what to do with IQueryable, since it does not implement IList.

Answer 1

Using extension methods makes this pretty easy.

public static class IEnumerableExtensions
{
  public static IEnumerable<ValueWithPrevious<T>> WithPrevious<T>(this IEnumerable<T> @this)
  {
    using (var e = @this.GetEnumerator())
    {
      if (!e.MoveNext())
        yield break;

      var previous = e.Current;

      while (e.MoveNext())
      {
        yield return new ValueWithPrevious<T>(e.Current, previous);
        previous = e.Current;
      }
    }
  }
}

public struct ValueWithPrevious<T>
{
  public readonly T Value, Previous;

  public ValueWithPrevious(T value, T previous)
  {
    Value = value;
    Previous = previous;
  }
}

Using:

var array = new int[] { 1, 2, 3, 4, 5 };
foreach (var value in array.WithPrevious())
{
  Console.WriteLine("{0}, {1}", value.Previous, value.Value);
  // Results: 1, 2
  //          2, 3
  //          3, 4
  //          4, 5
}

Answer 2

EDIT

A bit of a concern. This code will give you sequential elements

public static IEnumerable<Pair<T,T>> GroupIntoConsequetive(this IEnumerable<T> enumerable) {
  using ( var e = enumerable.GetEnumerator() ) {
    if ( !e.MoveNext() ) { 
      yield break;
    }
    var last = e.Current;
    while ( e.MoveNext() ) {
      yield return new Pair<T,T>(last, e.Current);
      last = e.Current;
    }
  }
}

I'm not sure there is a default way, but writing an extension method for this should not be difficult. I assume there is a simple Pair implementation

public static IEnumerable<Pair<T,T>> Window(this IEnumerable<T> enumerable) {
  using ( var e = enumerable.GetEnumerator() ) {
    while ( e.MoveNext() ) { 
      var first = e.Current;
      if ( !e.MoveNext() ) {
        throw new InvalidOperationException("Need even number");
      }
      var second = e.Current;
      yield return new Pair<T,T>(first,second);
   }
  }
}

var col = GetQueryableItem();
col.Window().Select(pair => method1(pair.First, pair.Second));

public struct Pair<T1,T2> {
  public readonly T1 First;
  public readonly T2 Second;
  public Pair(T1 first, T2 second) {
    First = first;
    Second = second;
  }
}

Answer 3

IQueryable < > List < > ToList < > ().

Answer 4

IQueryable<T>, . ToArray() ToList(). , .

Answer 5

IQueryable IEnumerable. - :

var a = new [] {1, 2, 3, 4}.AsQueryable();

if (a.Count() < 2) {
    return;
}

var prev = a.First();
var isFirst = true;

foreach (var i in a) {
    if (isFirst) {
        isFirst = false;
        continue;
    }

    method1(prev, i);
    prev = i;
}

IQueryable IList:

var list = a.ToList();