这个 TreeBuilder 函数可以不引入 Tuple 就写出来吗?
Can this TreeBuilder function be written without introducing a Tuple?
我刚刚将 this answer 中的代码改编为一个通用的树生成器,我的想法是可行的,但是,我最终引入了一个 Tuple 来使其工作。我很确定它可以写得更直接,但我很难看到它..
我引入元组的根本原因是因为在我迭代子节点时,我还没有将这些子节点转换为树节点。这个问题是我用我的方法造成的还是它固有的我正在用代码解决的问题?
下面是树生成器的用法:
// USAGE
var document = DocumentModel.Load(@"Part CareAlerts Page 1 Part 1.docx", LoadOptions.DocxDefault);
var tree = BuildTree<Element, Node>(
document,
elt => elt.GetChildElements(false),
elt => new Node(elt.ElementType.ToString()) {Content = elt.Content.ToString().Trim()},
(parent, child) => parent.Children.Add(child));
// TREE BUILDER FUNCTION
public static TNode BuildTree<TIn, TNode>(TIn root, Func<TIn, IEnumerable<TIn>> childSelector,
Func<TIn, TNode> createNode, Action<TNode, TNode> connectNodes)
where TIn : class
where TNode : class
{
var stack = new Stack<Tuple<TNode, TIn>>(new[] {Tuple.Create(default(TNode), root)});
TNode tree = null;
while (stack.Any())
{
var next = stack.Pop();
var result = createNode(next.Item2);
if (tree == null)
{
tree = result;
}
if (next.Item1 != null)
{
connectNodes(next.Item1, result);
}
foreach (var child in childSelector(next.Item2).Reverse())
{
stack.Push(Tuple.Create(result, child));
}
}
return tree;
}
我认为这样做可以:
public static TNode BuildTree<TIn, TNode>(TIn root, Func<TIn, IEnumerable<TIn>> childSelector,
Func<TIn, TNode> createNode, Action<TNode, TNode> connectNodes)
where TIn : class
where TNode : class
{
var tree = createNode(root);
var children =
childSelector(root)
.Select(c => BuildTree(c, childSelector, createNode, connectNodes))
.ToArray();
children
.ForEach(child => connectNodes(tree, child));
return tree;
}
我通过删除 TIn
和 运行 上的约束来测试这个代码:
var tree = BuildTree<int, Tree<int>>(
1,
elt => Enumerable.Range(2, 2).Select(n => elt * n).Where(x => x < 100),
elt => new Tree<int>() { Value = elt },
(parent, child) => parent.Add(child));
public class Tree<T> : List<Tree<T>>
{
public T Value { get; set; }
}
或者,更好的是,这个:
var tree = BuildTree<int, XElement>(
1,
elt => Enumerable.Range(2, 3).Select(n => elt * n).Where(x => x < 16),
elt => new XElement("Node", new XAttribute("Value", elt)),
(parent, child) => parent.Add(child));
这给出了这个:
<Node Value="1">
<Node Value="2">
<Node Value="4">
<Node Value="8" />
<Node Value="12" />
</Node>
<Node Value="6">
<Node Value="12" />
</Node>
<Node Value="8" />
</Node>
<Node Value="3">
<Node Value="6">
<Node Value="12" />
</Node>
<Node Value="9" />
<Node Value="12" />
</Node>
<Node Value="4">
<Node Value="8" />
<Node Value="12" />
</Node>
</Node>
我刚刚将 this answer 中的代码改编为一个通用的树生成器,我的想法是可行的,但是,我最终引入了一个 Tuple 来使其工作。我很确定它可以写得更直接,但我很难看到它..
我引入元组的根本原因是因为在我迭代子节点时,我还没有将这些子节点转换为树节点。这个问题是我用我的方法造成的还是它固有的我正在用代码解决的问题?
下面是树生成器的用法:
// USAGE
var document = DocumentModel.Load(@"Part CareAlerts Page 1 Part 1.docx", LoadOptions.DocxDefault);
var tree = BuildTree<Element, Node>(
document,
elt => elt.GetChildElements(false),
elt => new Node(elt.ElementType.ToString()) {Content = elt.Content.ToString().Trim()},
(parent, child) => parent.Children.Add(child));
// TREE BUILDER FUNCTION
public static TNode BuildTree<TIn, TNode>(TIn root, Func<TIn, IEnumerable<TIn>> childSelector,
Func<TIn, TNode> createNode, Action<TNode, TNode> connectNodes)
where TIn : class
where TNode : class
{
var stack = new Stack<Tuple<TNode, TIn>>(new[] {Tuple.Create(default(TNode), root)});
TNode tree = null;
while (stack.Any())
{
var next = stack.Pop();
var result = createNode(next.Item2);
if (tree == null)
{
tree = result;
}
if (next.Item1 != null)
{
connectNodes(next.Item1, result);
}
foreach (var child in childSelector(next.Item2).Reverse())
{
stack.Push(Tuple.Create(result, child));
}
}
return tree;
}
我认为这样做可以:
public static TNode BuildTree<TIn, TNode>(TIn root, Func<TIn, IEnumerable<TIn>> childSelector,
Func<TIn, TNode> createNode, Action<TNode, TNode> connectNodes)
where TIn : class
where TNode : class
{
var tree = createNode(root);
var children =
childSelector(root)
.Select(c => BuildTree(c, childSelector, createNode, connectNodes))
.ToArray();
children
.ForEach(child => connectNodes(tree, child));
return tree;
}
我通过删除 TIn
和 运行 上的约束来测试这个代码:
var tree = BuildTree<int, Tree<int>>(
1,
elt => Enumerable.Range(2, 2).Select(n => elt * n).Where(x => x < 100),
elt => new Tree<int>() { Value = elt },
(parent, child) => parent.Add(child));
public class Tree<T> : List<Tree<T>>
{
public T Value { get; set; }
}
或者,更好的是,这个:
var tree = BuildTree<int, XElement>(
1,
elt => Enumerable.Range(2, 3).Select(n => elt * n).Where(x => x < 16),
elt => new XElement("Node", new XAttribute("Value", elt)),
(parent, child) => parent.Add(child));
这给出了这个:
<Node Value="1">
<Node Value="2">
<Node Value="4">
<Node Value="8" />
<Node Value="12" />
</Node>
<Node Value="6">
<Node Value="12" />
</Node>
<Node Value="8" />
</Node>
<Node Value="3">
<Node Value="6">
<Node Value="12" />
</Node>
<Node Value="9" />
<Node Value="12" />
</Node>
<Node Value="4">
<Node Value="8" />
<Node Value="12" />
</Node>
</Node>