[英]SelectMany to flatten a nested structure
I am parsing an XML structure and my classes look like the following: 我正在解析XML结构,我的类看起来像如下:
class MyXml
{
//...
List<Node> Content { get; set; }
//...
}
class Node
{
// ...
public List<Node> Nodes { get; set; }
public string Type { get; set; }
//...
}
MyXml represents the XML file I am parsing, whose elements are all called <node>
. MyXml表示我正在解析的XML文件,其元素都称为
<node>
。 Each node has a type attribute, which can have different values. 每个节点都有一个type属性,可以有不同的值。
The type of the node is not connected to its depth. 节点的类型未与其深度相关联。 I can have any node type at any depth level.
我可以在任何深度级别拥有任何节点类型。
I can parse the structure correctly, so I get a MyXml object whose content is a list of Nodes, where ever node in the List can have subnodes and so on (I used recursion for that). 我可以正确地解析结构,所以我得到一个MyXml对象,其内容是节点列表,其中List中的节点可以有子节点等等(我使用了递归)。
What I need to do is flatten this whole structure and extract only the nodes of a certain type. 我需要做的是展平整个结构并仅提取某种类型的节点。
I tried with: 我尝试过:
var query = MyXml.Content.SelectMany(n => n.Nodes);
but it's taking only the nodes with a structure depth of 1. I would like to grab every node, regardless of depth, in the same collection and then filter what I need. 但它只采用结构深度为1的节点。我想在同一个集合中抓取每个节点,无论深度如何,然后过滤我需要的东西。
This is a naturally recursive problem. 这是一个自然递归的问题。 Using a recursive lambda, try something like:
使用递归lambda,尝试类似于:
Func<Node, IEnumerable<Node>> flattener = null;
flattener = n => new[] { n }
.Concat(n.Nodes == null
? Enumerable.Empty<Node>()
: n.Nodes.SelectMany(flattener));
Note that when you make a recursive Func
like this, you must declare the Func
separately first, and set it to null. 请注意,当您像这样生成递归
Func
,必须首先单独声明Func
,并将其设置为null。
You could also flatten the list using an iterator-block method: 您还可以使用迭代器块方法展平列表:
public static IEnumerable<Node> Flatten(Node node)
{
yield return node;
if (node.Nodes != null)
{
foreach(var child in node.Nodes)
foreach(var descendant in Flatten(child))
yield return descendant;
}
}
Either way, once the tree is flattened you can do simple Linq queries over the flattened list to find nodes: 无论哪种方式,一旦树被展平,您可以在展平列表上执行简单的Linq查询以查找节点:
flattener(node).Where(n => n.Type == myType);
Response adapted from: https://stackoverflow.com/a/17086572/1480391 响应改编自: https : //stackoverflow.com/a/17086572/1480391
You should implement a method Node.GetFlattened
, which returns the node itself and then calls itself on all subnodes: 您应该实现
Node.GetFlattened
方法,该方法返回节点本身,然后在所有子节点上调用自身:
public IEnumerable<Node> GetFlattened()
{
yield return this;
foreach (var node in this.Nodes.SelectMany(n => n.GetFlattened()))
yield return node;
}
You would then be able to call this method and it recursively returns all nodes regardless of their depth. 然后,您就可以调用此方法,并以递归方式返回所有节点,而不管其深度如何。 This is a depth-first search, if you want a breadth-first search, you will have to try another approach.
这是深度优先搜索,如果您想要广度优先搜索,则必须尝试其他方法。
class MyXml
{
public List<Node> AllNodes()
{
List<Node> allNodes = new List<Node>();
foreach (var node in Content)
AddNode(node, nodes);
}
public void AddNode(Node node, List<Node> nodes)
{
nodes.Add(node);
foreach (var childNode in node.Nodes)
AddNode(childNode, nodes);
}
public List<Node> AllNodesOfType(NodeType nodeType)
{
return AllNodes().Where(n => n.NodeType == nodeType);
}
}
First flatten the list with a function and query on that. 首先使用函数和查询对列表进行展平。
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.