详尽搜索/生成表达式树的每个组合

Question

我正在玩一个基本的表达式树优化器来构建查询计划。在解析树时，我可以根据我可以分配给每个操作的权重来决定如何构建它。

如果我有一个简单的树，有2个关于如何执行动作的选择，我希望能够生成树的两个变体，然后可以比较每个的权重以查看哪些是最高效。

例如，下面的代码会允许我来构建表达式树的两个变化加入操作：一个带有MergeJoinExpression，一个具有NestedLoopJoinExpression

class Customer 
{ 
     public int Id { get; set; } 
} 
class Orders 
{ 
     public int Id { get; set; } 
     public int CustomerId { get; set; } 
} 

class MergeJoinExpresion : JoinExpression 
{ 
} 

class NestLoopJoinExpresion : JoinExpression 
{ 
} 

class Visitor : ExpressionVisitor 
{ 
    public List<Expression> GetPlans(Expression expr) 
    { 
     // ??? 
    } 

    override VisitJoin(JoinExpression join) 
    { 
     // For this join, I can return the following (trite example) 
     // return MergeJoinExpresion 
     // return NestLoopJoinExpresion 

     return base.VisitJoin(join); 
    } 
} 

我如何构建，将产生每个方法树的变种并将它们还给我？

class Program 
{ 
     static void Main(string[] args) 
     { 
      var query = from c in customers 
         join o in orders on c.Id equals o.CustomerId 
         select new 
         { 
          CustomerId = c.Id, 
          OrderId = o.Id 
         }; 


      var plans = new Visitor().GetPlans(query); 
     } 
} 

谁能告诉我怎样才能修改VisitorGetPlans类的方法来产生这些变化？

编辑 - 是这样的：

class Visitor : ExpressionVisitor 
{ 
    private List<Expression> exprs = new List<Expression>(); 

    public List<Expression> GetPlans(Expression expr) 
    { 
     Visit(expr);  
     return exprs; 
    } 

    override VisitJoin(JoinExpression join) 
    { 
     // For this join, I can return the following (trite example) 
     // return MergeJoinExpresion 
     // return NestLoopJoinExpresion  
     var choices = new Expression[] { MergeJoinExpresion.Create(join), NestLoopJoinExpresion.Create(join) }; 

     foreach(var choice in choices) 
     { 
      var cloned = Cloner.Clone(choice); 
      var newTree = base.VisitJoin(cloned); 
      exprs.Add(newTree); 
     } 

     return base.VisitJoin(join); 
    } 
} 

Answer 1

所以下手，我们将创建一个访问者，这将只是帮助我们从一个Expression提取JoinExpression对象的列表：

internal class FindJoinsVisitor : ExpressionVisitor 
{ 
    private List<JoinExpression> expressions = new List<JoinExpression>(); 
    protected override Expression VisitJoin(JoinExpression join) 
    { 
     expressions.Add(join); 
     return base.VisitJoin(join); 
    } 
    public IEnumerable<JoinExpression> JoinExpressions 
    { 
     get 
     { 
      return expressions; 
     } 
    } 
} 
public static IEnumerable<JoinExpression> FindJoins(
    this Expression expression) 
{ 
    var visitor = new FindJoinsVisitor(); 
    visitor.Visit(expression); 
    return visitor.JoinExpressions; 
} 

下一步，我们将使用下面的方法，从this blog post拍摄，以获得序列的序列的笛卡尔乘积：

static IEnumerable<IEnumerable<T>> CartesianProduct<T>(
    this IEnumerable<IEnumerable<T>> sequences) 
{ 
    IEnumerable<IEnumerable<T>> emptyProduct = new[] { Enumerable.Empty<T>() }; 
    return sequences.Aggregate( 
     emptyProduct, 
     (accumulator, sequence) => 
      from accseq in accumulator 
      from item in sequence 
      select accseq.Concat(new[] {item})); 
} 

接下来，我们将创建一个访问者是需要表达对的序列，并替换第一个表达式的所有实例对与第二：

internal class ReplaceVisitor : ExpressionVisitor 
{ 
    private readonly Dictionary<Expression, Expression> lookup; 
    public ReplaceVisitor(Dictionary<Expression, Expression> pairsToReplace) 
    { 
     lookup = pairsToReplace; 
    } 
    public override Expression Visit(Expression node) 
    { 
     if(lookup.ContainsKey(node)) 
      return base.Visit(lookup[node]); 
     else 
      return base.Visit(node); 
    } 
} 

public static Expression ReplaceAll(this Expression expression, 
    Dictionary<Expression, Expression> pairsToReplace) 
{ 
    return new ReplaceVisitor(pairsToReplace).Visit(expression); 
} 

public static Expression ReplaceAll(this Expression expression, 
    IEnumerable<Tuple<Expression, Expression>> pairsToReplace) 
{ 
    var lookup = pairsToReplace.ToDictionary(pair => pair.Item1, pair => pair.Item2); 
    return new ReplaceVisitor(lookup).Visit(expression); 
} 

最后，我们通过寻找所有在我们表达了加入表情把一切融合在一起，项目的出来对的序列，其中的JoinExpression是在对中的第一项，第二是每一个可能的替代值。从那里我们可以得到它的笛卡尔乘积以获得表达式替换对的所有组合。最后，我们可以将每个组合的替换投影到实际替换原始表达式中所有这些对的表达式中：

public static IEnumerable<Expression> AllJoinCombinations(Expression expression) 
{ 
    var combinations = expression.FindJoins() 
     .Select(join => new Tuple<Expression, Expression>[] 
     { 
      Tuple.Create<Expression, Expression>(join, new NestLoopJoinExpresion(join)), 
      Tuple.Create<Expression, Expression>(join, new MergeJoinExpresion(join)), 
     }) 
     .CartesianProduct(); 

    return combinations.Select(combination => expression.ReplaceAll(combination)); 
} 

Answer 2

你需要一成不变的树木肯定。

创建一个类：

class JoinOptionsExpression: JoinExpression { 
    public IEnumerable<JoinExpression> Options {get; private set;} 
    private JoinOptionsExpression(){} 
    public static JoinOptionsExpression Create(IEnumerable<JoinExpression> options){ 
     return new JoinOptionsExpression{Options = options.ToList().AsReadOnly()}; // you can improve this probably 
    } 
} 

然后在您的VisitJoin方法返回的选项，并返回所有的选择：

private List<Dictionary<JoinOptionsExpression,int>> selections = new List<Dictionary<JoinOptionsExpression,int>>{new Dictionary<JoinOptionsExpression,int>()}; 
override VisitJoin(JoinExpression join) 
{ 
    var choices = new Expression[] { MergeJoinExpresion.Create(join), NestLoopJoinExpresion.Create(join) }; 
    List<Expression> exprs = new List<Expression>(); 
    foreach(var choice in choices) 
    { 
     var cloned = Cloner.Clone(choice); 
     var newTree = base.VisitJoin(cloned); 
     exprs.Add(newTree); 
    } 
    var result = JoinOptionsExpression.Create(exprs); 
    // now add all choices 
    if (exprs.Count > 0) 
     foreach (selection in selections.ToList()) // to make sure your don't modify during enumeration, you can improve this too 
     { 
      selection.Add(result, 0); 
      for (i=1; i<exprs.Count; i++) 
      { 
       var copy= new Dictionary<JoinOptionsExpression, int>(selection); 
       copy[result] = i; 
       selections.Add(copy); 
      } 
     } 
    return result; 
} 

然后你需要一个第二访问者，从框架游客派生，并且没有其他原因，只是提取您的选项：

class OptionsExtractor:ExpressionVisitor 
{ 
    public IEnumerable<Expression> Extract(Expression expression, List<Dictionary<JoinOptionsExpression,int>> selections) 
    { 
     foreach(var selection in selections) 
     { 
      currentSelections = selection; 
      yield return Visit(expression); 
     } 
    } 
    private Dictionary<JoinOptionsExpression,int> currentSelections; 
    override Expression Visit(Expression node) 
    { 
     var opts = node as JoinOptionsExpression; 
     if (opts != null) 
      return base.Visit(opts.Options.ElementAt(currentSelections[opts]); 
     else 
      return base.Visit(node); 
    } 
} 

反正一个详尽的搜索可以迅速爆炸在你的脸上，我想你知道这一点。 声明：我只是在这个编辑器中输入了它，它甚至不会编译，但你应该能够明白。