[ACCEPTED]-Recursive List Flattening-recursion

Accepted answer
Score: 49

Here's an extension that might help. It 11 will traverse all nodes in your hierarchy 10 of objects and pick out the ones that match 9 a criteria. It assumes that each object 8 in your hierarchy has a collection property that holds its child 7 objects.

Here's the extension:

/// Traverses an object hierarchy and return a flattened list of elements
/// based on a predicate.
/// 
/// TSource: The type of object in your collection.</typeparam>
/// source: The collection of your topmost TSource objects.</param>
/// selectorFunction: A predicate for choosing the objects you want.
/// getChildrenFunction: A function that fetches the child collection from an object.
/// returns: A flattened list of objects which meet the criteria in selectorFunction.
public static IEnumerable<TSource> Map<TSource>(
  this IEnumerable<TSource> source,
  Func<TSource, bool> selectorFunction,
  Func<TSource, IEnumerable<TSource>> getChildrenFunction)
{
  // Add what we have to the stack
  var flattenedList = source.Where(selectorFunction);

  // Go through the input enumerable looking for children,
  // and add those if we have them
  foreach (TSource element in source)
  {
    flattenedList = flattenedList.Concat(
      getChildrenFunction(element).Map(selectorFunction,
                                       getChildrenFunction)
    );
  }
  return flattenedList;
}

Examples (Unit Tests):

First we need an object and a nested 6 object hierarchy.

A simple node class

class Node
{
  public int NodeId { get; set; }
  public int LevelId { get; set; }
  public IEnumerable<Node> Children { get; set; }

  public override string ToString()
  {
    return String.Format("Node {0}, Level {1}", this.NodeId, this.LevelId);
  }
}

And 5 a method to get a 3-level deep hierarchy 4 of nodes

private IEnumerable<Node> GetNodes()
{
  // Create a 3-level deep hierarchy of nodes
  Node[] nodes = new Node[]
    {
      new Node 
      { 
        NodeId = 1, 
        LevelId = 1, 
        Children = new Node[]
        {
          new Node { NodeId = 2, LevelId = 2, Children = new Node[] {} },
          new Node
          {
            NodeId = 3,
            LevelId = 2,
            Children = new Node[]
            {
              new Node { NodeId = 4, LevelId = 3, Children = new Node[] {} },
              new Node { NodeId = 5, LevelId = 3, Children = new Node[] {} }
            }
          }
        }
      },
      new Node { NodeId = 6, LevelId = 1, Children = new Node[] {} }
    };
  return nodes;
}

First Test: flatten the hierarchy, no 3 filtering

[Test]
public void Flatten_Nested_Heirachy()
{
  IEnumerable<Node> nodes = GetNodes();
  var flattenedNodes = nodes.Map(
    p => true, 
    (Node n) => { return n.Children; }
  );
  foreach (Node flatNode in flattenedNodes)
  {
    Console.WriteLine(flatNode.ToString());
  }

  // Make sure we only end up with 6 nodes
  Assert.AreEqual(6, flattenedNodes.Count());
}

This will show:

Node 1, Level 1
Node 6, Level 1
Node 2, Level 2
Node 3, Level 2
Node 4, Level 3
Node 5, Level 3

Second Test: Get 2 a list of nodes that have an even-numbered 1 NodeId

[Test]
public void Only_Return_Nodes_With_Even_Numbered_Node_IDs()
{
  IEnumerable<Node> nodes = GetNodes();
  var flattenedNodes = nodes.Map(
    p => (p.NodeId % 2) == 0, 
    (Node n) => { return n.Children; }
  );
  foreach (Node flatNode in flattenedNodes)
  {
    Console.WriteLine(flatNode.ToString());
  }
  // Make sure we only end up with 3 nodes
  Assert.AreEqual(3, flattenedNodes.Count());
}

This will show:

Node 6, Level 1
Node 2, Level 2
Node 4, Level 3
Score: 20

Hmm... I'm not sure exactly what you want here, but 13 here's a "one level" option:

public static IEnumerable<TElement> Flatten<TElement,TSequence> (this IEnumerable<TSequence> sequences)
    where TSequence : IEnumerable<TElement> 
{
    foreach (TSequence sequence in sequences)
    {
        foreach(TElement element in sequence)
        {
            yield return element;
        }
    }
}

If that's not 12 what you want, could you provide the signature 11 of what you do want? If you don't need a 10 generic form, and you just want to do the 9 kind of thing that LINQ to XML constructors 8 do, that's reasonably simple - although 7 the recursive use of iterator blocks is 6 relatively inefficient. Something like:

static IEnumerable Flatten(params object[] objects)
{
    // Can't easily get varargs behaviour with IEnumerable
    return Flatten((IEnumerable) objects);
}

static IEnumerable Flatten(IEnumerable enumerable)
{
    foreach (object element in enumerable)
    {
        IEnumerable candidate = element as IEnumerable;
        if (candidate != null)
        {
            foreach (object nested in candidate)
            {
                yield return nested;
            }
        }
        else
        {
            yield return element;
        }
    }
}

Note 5 that that will treat a string as a sequence 4 of chars, however - you may want to special-case 3 strings to be individual elements instead 2 of flattening them, depending on your use 1 case.

Does that help?

Score: 14

I thought I'd share a complete example with 4 error handling and a single-logic apporoach.

Recursive 3 flattening is as simple as:

LINQ version

public static class IEnumerableExtensions
{
    public static IEnumerable<T> SelectManyRecursive<T>(this IEnumerable<T> source, Func<T, IEnumerable<T>> selector)
    {
        if (source == null) throw new ArgumentNullException("source");
        if (selector == null) throw new ArgumentNullException("selector");

        return !source.Any() ? source :
            source.Concat(
                source
                .SelectMany(i => selector(i).EmptyIfNull())
                .SelectManyRecursive(selector)
            );
    }

    public static IEnumerable<T> EmptyIfNull<T>(this IEnumerable<T> source)
    {
        return source ?? Enumerable.Empty<T>();
    }
}

Non-LINQ version

public static class IEnumerableExtensions
{
    public static IEnumerable<T> SelectManyRecursive<T>(this IEnumerable<T> source, Func<T, IEnumerable<T>> selector)
    {
        if (source == null) throw new ArgumentNullException("source");
        if (selector == null) throw new ArgumentNullException("selector");

        foreach (T item in source)
        {
            yield return item;

            var children = selector(item);
            if (children == null)
                continue;

            foreach (T descendant in children.SelectManyRecursive(selector))
            {
                yield return descendant;
            }
        }
    }
}

Design decisions

I decided to:

  • disallow flattening of a null IEnumerable, this can be changed by removing exception throwing and:
    • adding source = source.EmptyIfNull(); before return in the 1st version
    • adding if (source != null) before foreach in the 2nd version
  • allow returning of a null collection by the selector - this way I'm removing responsibility from the caller to assure the children list isn't empty, this can be changed by:
    • removing .EmptyIfNull() in the first version - note that SelectMany will fail if null is returned by selector
    • removing if (children == null) continue; in the second version - note that foreach will fail on a null IEnumerable parameter
  • allow filtering children with .Where clause on the caller side or within the children selector rather than passing a children filter selector parameter:
    • it won't impact the efficiency because in both versions it is a deferred call
    • it would be mixing another logic with the method and I prefer to keep the logic separated

Sample use

I'm 2 using this extension method in LightSwitch 1 to obtain all controls on the screen:

public static class ScreenObjectExtensions
{
    public static IEnumerable<IContentItemProxy> FindControls(this IScreenObject screen)
    {
        var model = screen.Details.GetModel();

        return model.GetChildItems()
            .SelectManyRecursive(c => c.GetChildItems())
            .OfType<IContentItemDefinition>()
            .Select(c => screen.FindControl(c.Name));
    }
}
Score: 7

Here is a modified Jon Skeet's answer to allow more than "one 2 level":

static IEnumerable Flatten(IEnumerable enumerable)
{
    foreach (object element in enumerable)
    {
        IEnumerable candidate = element as IEnumerable;
        if (candidate != null)
        {
            foreach (object nested in Flatten(candidate))
            {
                yield return nested;
            }
        }
        else
        {
            yield return element;
        }
    }
}

disclaimer: I don't know C#.

The 1 same in Python:

#!/usr/bin/env python

def flatten(iterable):
    for item in iterable:
        if hasattr(item, '__iter__'):
            for nested in flatten(item):
                yield nested
        else:
            yield item

if __name__ == '__main__':
    for item in flatten([1,[2, 3, [[4], 5]], 6, [[[7]]], [8]]):
        print(item, end=" ")

It prints:

1 2 3 4 5 6 7 8 
Score: 6

Isn't that what [SelectMany][1] is for?

enum1.SelectMany(
    a => a.SelectMany(
        b => b.SelectMany(
            c => c.Select(
                d => d.Name
            )
        )
    )
);

0

Score: 4

Function:

public static class MyExtentions
{
    public static IEnumerable<T> RecursiveSelector<T>(this IEnumerable<T> nodes, Func<T, IEnumerable<T>> selector)
    {
        if(nodes.Any() == false)
        {
            return nodes; 
        }

        var descendants = nodes
                            .SelectMany(selector)
                            .RecursiveSelector(selector);

        return nodes.Concat(descendants);
    } 
}

Usage:

var ar = new[]
{
    new Node
    {
        Name = "1",
        Chilren = new[]
        {
            new Node
            {
                Name = "11",
                Children = new[]
                {
                    new Node
                    {
                        Name = "111",
                        
                    }
                }
            }
        }
    }
};

var flattened = ar.RecursiveSelector(x => x.Children).ToList();

0

Score: 2

Okay here's another version which is combined 4 from about 3 answers above.

Recursive. Uses 3 yield. Generic. Optional filter predicate. Optional 2 selection function. About as concise as 1 I could make it.

    public static IEnumerable<TNode> Flatten<TNode>(
        this IEnumerable<TNode> nodes, 
        Func<TNode, bool> filterBy = null,
        Func<TNode, IEnumerable<TNode>> selectChildren = null
        )
    {
        if (nodes == null) yield break;
        if (filterBy != null) nodes = nodes.Where(filterBy);

        foreach (var node in nodes)
        {
            yield return node;

            var children = (selectChildren == null)
                ? node as IEnumerable<TNode>
                : selectChildren(node);

            if (children == null) continue;

            foreach (var child in children.Flatten(filterBy, selectChildren))
            {
                yield return child;
            }
        }
    }

Usage:

// With filter predicate, with selection function
var flatList = nodes.Flatten(n => n.IsDeleted == false, n => n.Children);
Score: 1

The SelectMany extension method does this already.

Projects 3 each element of a sequence to an IEnumerable<(Of 2 <(T>)>) and flattens the resulting 1 sequences into one sequence.

Score: 1

Since yield is not available in VB and LINQ 2 provides both deferred execution and a concise 1 syntax, you can also use.

<Extension()>
Public Function Flatten(Of T)(ByVal objects As Generic.IEnumerable(Of T), ByVal selector As Func(Of T, Generic.IEnumerable(Of T))) As Generic.IEnumerable(Of T)
   If(objects.Any()) Then
      Return objects.Union(objects.Select(selector).Where(e => e != null).SelectMany(e => e)).Flatten(selector))
   Else
      Return objects 
   End If
End Function
public static class Extensions{
  public static IEnumerable<T> Flatten<T>(this IEnumerable<T> objects, Func<T, IEnumerable<T>> selector) where T:Component{
    if(objects.Any()){
        return objects.Union(objects.Select(selector).Where(e => e != null).SelectMany(e => e).Flatten(selector));
    }
    return objects;
  }
}

edited to include:

Score: 1

I had to implement mine from scratch because 7 all of the provided solutions would break 6 in case there is a loop i.e. a child that 5 points to its ancestor. If you have the 4 same requirements as mine please take a 3 look at this (also let me know if my solution 2 would break in any special circumstances):

How 1 to use:

var flattenlist = rootItem.Flatten(obj => obj.ChildItems, obj => obj.Id)

Code:

public static class Extensions
    {
        /// <summary>
        /// This would flatten out a recursive data structure ignoring the loops. The end result would be an enumerable which enumerates all the
        /// items in the data structure regardless of the level of nesting.
        /// </summary>
        /// <typeparam name="T">Type of the recursive data structure</typeparam>
        /// <param name="source">Source element</param>
        /// <param name="childrenSelector">a function that returns the children of a given data element of type T</param>
        /// <param name="keySelector">a function that returns a key value for each element</param>
        /// <returns>a faltten list of all the items within recursive data structure of T</returns>
        public static IEnumerable<T> Flatten<T>(this IEnumerable<T> source,
            Func<T, IEnumerable<T>> childrenSelector,
            Func<T, object> keySelector) where T : class
        {
            if (source == null)
                throw new ArgumentNullException("source");
            if (childrenSelector == null)
                throw new ArgumentNullException("childrenSelector");
            if (keySelector == null)
                throw new ArgumentNullException("keySelector");
            var stack = new Stack<T>( source);
            var dictionary = new Dictionary<object, T>();
            while (stack.Any())
            {
                var currentItem = stack.Pop();
                var currentkey = keySelector(currentItem);
                if (dictionary.ContainsKey(currentkey) == false)
                {
                    dictionary.Add(currentkey, currentItem);
                    var children = childrenSelector(currentItem);
                    if (children != null)
                    {
                        foreach (var child in children)
                        {
                            stack.Push(child);
                        }
                    }
                }
                yield return currentItem;
            }
        }

        /// <summary>
        /// This would flatten out a recursive data structure ignoring the loops. The     end result would be an enumerable which enumerates all the
        /// items in the data structure regardless of the level of nesting.
        /// </summary>
        /// <typeparam name="T">Type of the recursive data structure</typeparam>
        /// <param name="source">Source element</param>
        /// <param name="childrenSelector">a function that returns the children of a     given data element of type T</param>
        /// <param name="keySelector">a function that returns a key value for each   element</param>
        /// <returns>a faltten list of all the items within recursive data structure of T</returns>
        public static IEnumerable<T> Flatten<T>(this T source, 
            Func<T, IEnumerable<T>> childrenSelector,
            Func<T, object> keySelector) where T: class
        {
            return Flatten(new [] {source}, childrenSelector, keySelector);
        }
    }
Score: 0
static class EnumerableExtensions
{
    public static IEnumerable<T> Flatten<T>(this IEnumerable<IEnumerable<T>> sequence)
    {
        foreach(var child in sequence)
            foreach(var item in child)
                yield return item;
    }
}

Maybe like this? Or do you mean that it 1 could potentially be infintly deep?

Score: 0
class PageViewModel { 
    public IEnumerable<PageViewModel> ChildrenPages { get; set; } 
}

Func<IEnumerable<PageViewModel>, IEnumerable<PageViewModel>> concatAll = null;
concatAll = list => list.SelectMany(l => l.ChildrenPages.Any() ? 
    concatAll(l.ChildrenPages).Union(new[] { l }) : new[] { l });

var allPages = concatAll(source).ToArray();

0

More Related questions