穷举搜索/生成表达式树的每个组合

Question

我正在使用基本的表达式树优化器来构建查询计划。在解析树时，我可以决定如何“最好”地构造它，这取决于我可以分配给每个操作的权重。

如果我有一个简单的树，有两个关于如何执行操作的选择，我希望能够生成树的两种变体，然后可以比较每个变体的权重，看看什么是最有效的.

例如，下面的代码将允许我构造表达式树连接操作的两种变体：一种带有MergeJoinExpression，另一种带有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);
    }
}

我如何构造一个方法来生成树的每个变体并将它们返回给我？

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);
        }
}

谁能告诉我如何修改Visitor 类GetPlans 方法来生成这些变体？

编辑 - 类似：

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

首先我们将创建一个访问者，它可以帮助我们从Expression 中提取JoinExpression 对象列表：

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;
}

接下来我们将使用以下方法（取自this blog post）来获取序列序列的笛卡尔积：

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，它接受一系列表达式对，并将该对中第一个表达式的所有实例替换为第二个：

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，我们通过查找表达式中的所有连接表达式将所有内容放在一起，将它们投影到一系列对中，其中JoinExpression 是对中的第一项，第二项是每个可能的替换值。从那里我们可以使用它的笛卡尔积来获得表达式替换对的所有组合。最后，我们可以将每个替换组合投影到表达式中，该表达式实际上是替换原始表达式中的所有这些对：

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

你肯定需要不可变的树。

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 方法中返回选项，并返回所有选项：

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;
}

然后您将需要第二个访问者，它源自框架访问者，并且没有其他原因，只需提取您的选项：

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);
    }
}

无论如何，详尽的搜索很快就会在您的脸上爆炸，我想您知道这一点。 免责声明：我只是在这个编辑器中输入了这个，它甚至可能无法编译，但你应该能够明白。