我自己可以写这个，但是我想要完成它的具体方法就是把我扔掉。我正在尝试编写一个类似于.NET 3.5中引入的其他通用扩展方法，它将采用IEnumerables的嵌套IEnumerable（依此类推）并将其展平为一个IEnumerable。有人有什么想法吗？
具体来说，我遇到了扩展方法本身的语法问题，因此我可以使用展平算法。
答案 0 :(得分：42)
这是一个可能有帮助的扩展。它将遍历对象层次结构中的所有节点，并选择符合条件的节点。它假定层次结构中的每个对象都有一个包含其子对象的集合属性。
/// 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;
}
首先，我们需要一个对象和一个嵌套的对象层次结构。
一个简单的节点类
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);
}
}
一种获得3级深度节点层次结构的方法
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;
}
第一次测试：展平层次结构，不过滤
[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());
}
这将显示：
Node 1, Level 1
Node 6, Level 1
Node 2, Level 2
Node 3, Level 2
Node 4, Level 3
Node 5, Level 3
第二次测试：获取具有偶数NodeId的节点列表
[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());
}
这将显示：
Node 6, Level 1
Node 2, Level 2
Node 4, Level 3
答案 1 :(得分：19)
嗯......我不确定完全你想要什么，但这里是“一级”选项：
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;
}
}
}
如果那不是你想要的，你能提供你想要的签名吗？如果您不需要通用表单，并且您只想做LINQ to XML构造函数所做的事情，那就相当简单 - 尽管迭代器块的递归使用效率相对较低。类似的东西：
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;
}
}
}
请注意，这会将字符串视为字符序列，但是 - 您可能希望特殊字符串是单个元素而不是展平它们，具体取决于您的用例。
这有帮助吗？
答案 2 :(得分：7)
我认为我会分享一个完整的错误处理示例和单逻辑应用程序。
递归展平非常简单：
LINQ版
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>();
}
}
非LINQ版本
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;
}
}
}
}
设计决策
我决定：
IEnumerable，可以通过删除异常抛出来改变它：
source = source.EmptyIfNull();之前添加return if (source != null)之前添加foreach .EmptyIfNull() - 请注意，如果选择器返回null，则SelectMany将失败if (children == null) continue; - 请注意，foreach参数<{1}} IEnumerable将失败
.Where子句的子项，而不是传递子过滤器选择器参数：
使用示例
我在LightSwitch中使用此扩展方法来获取屏幕上的所有控件：
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));
}
}
答案 3 :(得分：6)
这不是[SelectMany] [1]的用途吗？
enum1.SelectMany(
a => a.SelectMany(
b => b.SelectMany(
c => c.Select(
d => d.Name
)
)
)
);
答案 4 :(得分：5)
这是一个修改后的Jon Skeet's answer，允许超过“一个级别”：
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;
}
}
}
Python中的相同内容：
#!/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=" ")
打印：
1 2 3 4 5 6 7 8
答案 5 :(得分：2)
功能：
public static class MyExtentions
{
public static IEnumerable<T> RecursiveSelector<T>(this IEnumerable<T> nodes, Func<T, IEnumerable<T>> selector)
{
if(nodes.Any())
return nodes.Concat(nodes.SelectMany(selector).RecursiveSelector(selector));
return nodes;
}
}
用法：
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();
答案 6 :(得分：1)
SelectMany扩展方法已经这样做了。
将序列的每个元素投影到 IEnumerable＆lt;（＆lt;（T＆gt;）＆gt;）和 将得到的序列展平成 一个序列。
答案 7 :(得分：1)
由于在VB中没有yield，LINQ提供了延迟执行和简洁的语法，你也可以使用。
<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)
Return objects.Union(objects.SelectMany(selector).Flatten(selector))
End Function
答案 8 :(得分：1)
我必须从头开始实施我的，因为所有提供的解决方案都会中断，以防有一个循环，即指向其祖先的孩子。如果您有与我相同的要求，请看一下（如果我的解决方案在任何特殊情况下都会中断，也请告诉我）：
使用方法：
var flattenlist = rootItem.Flatten(obj => obj.ChildItems, obj => obj.Id)
代码：
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);
}
}
答案 9 :(得分：1)
好的，这是另一个版本，由上面的3个答案组合而成。
递归。使用产量。通用。可选的过滤谓词。可选的选择功能。关于我能做到的简洁。
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;
}
}
}
用法：
// With filter predicate, with selection function
var flatList = nodes.Flatten(n => n.IsDeleted == false, n => n.Children);
答案 10 :(得分：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;
}
}
也许是这样的？或者你的意思是它可能无限深？
答案 11 :(得分：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();
答案 12 :(得分：-1)
基本上，你需要有一个超出递归函数的主IENumerable，然后是你的递归函数（Psuedo-code）
private void flattenList(IEnumerable<T> list)
{
foreach (T item in list)
{
masterList.Add(item);
if (item.Count > 0)
{
this.flattenList(item);
}
}
}
虽然我真的不确定IEnumerable嵌套在IEnumerable中你的意思...那是什么意思？嵌套多少级？什么是最终类型？显然我的代码不正确，但我希望它让你思考。