我可能可以自己写这个代码,但是我尝试编写一个通用的扩展方法,类似于.NET 3.5中引入的其他扩展方法,它将获取一个嵌套的IEnumerable集合(等等)并将其展开为一个IEnumerable。有人有什么想法吗?
具体来说,我在扩展方法的语法上遇到了问题,以便我可以编写展平算法。
我可能可以自己写这个代码,但是我尝试编写一个通用的扩展方法,类似于.NET 3.5中引入的其他扩展方法,它将获取一个嵌套的IEnumerable集合(等等)并将其展开为一个IEnumerable。有人有什么想法吗?
具体来说,我在扩展方法的语法上遇到了问题,以便我可以编写展平算法。
这里有一个可能会有所帮助的扩展。它将遍历对象层次结构中的所有节点,并挑选出符合条件的节点。它假设您的层次结构中的每个对象都有一个集合属性,用于保存其子对象。
/// 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);
}
}
并且提供一种获取三级节点深度层次结构的方法
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,不进行过滤
[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
第二个测试: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
nodes.Map(i => true, n => n.Children)
。我还建议不要将这个扩展方法命名为“Map”,“Flatten”更清晰地描述了它的功能。 - Amicable嗯...我不确定您想要什么内容,但这是一个“单层次”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;
}
}
}
如果这不是你想要的,请提供你所需要的签名。如果你不需要一个通用的表单,而只是想做LINQ到XML构造函数所做的事情,那么这是相当简单的 - 尽管迭代器块的递归使用相对较低效。可以尝试以下代码:
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;
}
}
}
我想分享一个完整的例子,包括错误处理和单一逻辑方法。
递归展开就像这样简单:
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");
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
进行展开,可以通过删除异常抛出并执行以下操作来更改:
return
之前添加source = source.EmptyIfNull();
foreach
之前添加if (source != null)
.EmptyIfNull()
- 请注意,如果选择器返回null,则SelectMany
将失败if (children == null) continue;
- 请注意,如果IEnumerable
参数为null,则foreach
将失败.Where
子句过滤子项,而不是传递一个子项过滤选择器参数:
示例用法
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));
}
}
!source.Any() ? source
部分。 - FindOutIslamNow这是一个修改过的Jon Skeet的答案,允许多级:
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;
}
}
}
声明:我不懂C#。
同样的Python代码:
#!/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=" ")
1 2 3 4 5 6 7 8
enum1.SelectMany(
a => a.SelectMany(
b => b.SelectMany(
c => c.Select(
d => d.Name
)
)
)
);
功能:
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);
}
}
使用方法:
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();
好的,这里是另一个版本,结合了上面大约3个答案。
递归。使用yield。通用的。可选的过滤器谓词。可选的选择函数。尽可能简洁。
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;
}
}
}
使用方法:
// With filter predicate, with selection function
var flatList = nodes.Flatten(n => n.IsDeleted == false, n => n.Children);
<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;
}
}
编辑 包括:
SelectMany
扩展方法已经实现了这个功能。
将序列的每个元素投影到一个 IEnumerable<(Of <(T>)>) 中,并将生成的序列平坦化为一个序列。
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);
}
}