我正在寻找一种更好的方法来对可能具有循环依赖性的项进行递归。目前,我传递已处理项目的列表,以便不再处理它们,但这可能不是最好的方法。
以下是我目前正在做的事情:
/// <summary>
/// caching dependencies in order to increase performance
/// </summary>
private static readonly IDictionary<string, IEnumerable<OwnedItem>> dependencies
= new Dictionary<string, IEnumerable<OwnedItem>>();
/// <summary>
/// recursively find OwnedItem this oi depends upon
/// in order to correctly handle cyclic dependencies, already considered
/// dependencies need to be supplied as well (can be null or empty)
/// </summary>
/// <param name="oi"></param>
/// <param name="parentDeps"></param>
/// <returns></returns>
private static IEnumerable<OwnedItem> GetDependencies(
OwnedItem oi,
IEnumerable<OwnedItem> parentDeps)
{
if (null == oi)
{
return Enumerable.Empty<OwnedItem>();
}
if (dependencies.ContainsKey(oi.UniqueId))
{
return dependencies[oi.UniqueId];
}
var comparer = new TCObjectComparer<OwnedItem>();
var result = new HashSet<OwnedItem>(comparer);
result.Add(oi);
result.UnionWith(parentDeps ?? Enumerable.Empty<OwnedItem>());
foreach (var oi2 in oi.AllUsedOwnedItemsToBeIncluded.Except(
result, comparer))
{
result.UnionWith(GetDependencies(oi2, result));
}
dependencies[oi.UniqueId] = result;
return result;
}
这些项目属于'OwnedItem',并在属性IEnumerable<OwnedItem>
中保留其直接依赖项的列表(AllUsedOwnedItemsToBeIncluded
),但基本上这应该适用于'items'保留'items'列表的地方可能发生循环依赖。
使用字典只是避免多次执行相同的计算;这不是必要的。此外,只需要TCObjectComparer
的单个实例,但这也不是必需的。
有什么建议?我认为必须有一些经典算法来处理这个问题,但我找不到它。
答案 0 :(得分:1)
您要做的是基本上浏览连接图的所有节点。 AllUsedOwnedItemsToBeIncluded属性是连接到当前节点的节点列表。
您可以在此处查找可能有用的graph traversal algorithms。
您的算法是进行图遍历的一种方法。您必须浏览每个节点并保留一个访问过的节点列表,不要再访问他两次。
另一种减少遍历次数的算法可以是:
list nodesToExplore;
list exploredNodes;
nodesToExplore.add(startNode);
for all node in nodesToExplore
{
nodesToExplore.remove(node);
exploredNodes.add(node);
for all child in node.childs
{
if(child not in exploredNodes)
nodesToExplore.add(child);
}
}
当它结束时,探索的节点将包含您需要的内容。使用hashset / dictionnary而不是list会改善性能
答案 1 :(得分:1)
该算法可以提取到一个类,保持你的代码更整洁,摆脱那个臭静态字段。
TV
通用算法的实现如下:
private static IEnumerable<T> GetDependencies(T oi)
{
return new FlattenedCircularTree<OwnedItem>(oi, o => o.AllUsedOwnedItemsToBeIncluded)
.AllNodes();
}
答案 2 :(得分:1)
您可以实现以下内容:
public static partial class LinqGraph {
public static IEnumerable<T> SelectBreadthFirst<T>(this IEnumerable<T> source,
Func<T, IEnumerable<T>> children) {
if (Object.ReferenceEquals(null, source))
throw new ArgumentNullException(nameof(source));
else if (Object.ReferenceEquals(null, children))
throw new ArgumentNullException(nameof(children));
HashSet<T> proceeded = new HashSet<T>();
Queue<IEnumerable<T>> queue = new Queue<IEnumerable<T>>();
queue.Enqueue(source);
while (queue.Count > 0) {
IEnumerable<T> src = queue.Dequeue();
if (Object.ReferenceEquals(null, src))
continue;
foreach (var item in src)
if (proceeded.Add(item)) {
yield return item;
queue.Enqueue(children(item));
}
}
}
}
然后使用它
var items = new OwnedItem[] {startItem} // root nodes
//TODO: provide a rule of returning children on given parent
.SelectBreadthFirst(parent => parent.AllUsedOwnedItemsToBeIncluded);
答案 3 :(得分:0)
这是我对Vincent算法的实现:
private static readonly TCObjectComparer<OwnedItem> comparer
= new TCObjectComparer<OwnedItem>();
/// <summary>
/// caching dependencies in order to increase performance
/// </summary>
private static readonly IDictionary<string, IEnumerable<OwnedItem>> dependencies
= new Dictionary<string, IEnumerable<OwnedItem>>();
/// <summary>
/// recursively find OwnedItems this oi depends upon
/// see http://stackoverflow.com/questions/37614469/how-to-recurse-over-items-having-cyclic-dependencies
/// </summary>
/// <param name="oi"></param>
/// <returns></returns>
private static IEnumerable<OwnedItem> GetDependencies(OwnedItem oi)
{
if (null == oi)
{
return Enumerable.Empty<OwnedItem>();
}
if (dependencies.ContainsKey(oi.UniqueId))
{
return dependencies[oi.UniqueId];
}
var resultExploredNodes = new HashSet<OwnedItem>(comparer);
var nodesToExplore = new Queue<OwnedItem>();
nodesToExplore.Enqueue(oi);
while (nodesToExplore.Count > 0)
{
var node = nodesToExplore.Dequeue();
resultExploredNodes.Add(node);
// add nodes not already visited to nodesToExplore
node.AllUsedOwnedItemsToBeIncluded
.Except(resultExploredNodes, comparer)
.ForEach(n => nodesToExplore.Enqueue(n));
}
dependencies[oi.UniqueId] = resultExploredNodes;
return resultExploredNodes;
}
同样,缓存只是为了提高性能,它对算法来说并不重要。