使用表达式计算欧几里得距离的通用方法

Question

我想要一种可以使用表达式计算欧几里得距离并订购 IQueryable 的方法：

sqrt[(q1 - p1)^2 + (q2 - p2)^2 + ... + (qn - pn)^2]

这是我想出的方法签名：

public static IOrderedQueryable<T> EuclideanDistanceOrder<T>(
    this IQueryable<T> query, IEnumerable<Expression<Func<T, double>>> expressions)
    {
        var orderedQuery = query.OrderBy(i => Math.Sqrt(expressions.Aggregate((total, item) => total + Math.Pow(item, 2))));
        return orderedQuery;
    }

我不确定如何处理 item 和 total（因为它们是 Expression<Func<T, double>>）。 我已经尝试了几种不同的方法，包括使用Expression.Power 和Expression.Add。我尝试定义要单独组合的表达式：

Expression<Func<double, double>> power = i => Math.Pow(i, 2);
Expression<Func<List<Expression<Func<T, double>>>, double>> dist = (items) => Math.Sqrt(items.Sum(power));

但我仍然不知道如何处理power。

有没有更好的方法来解决这个问题？

Answer 1

要使其与 EF 或 LinqToSQL 一起使用，您必须将所有信息作为表达式传递，甚至是 P 和 Q 的属性访问器。这就是我修改您的方法声明的原因：

public static class Extension
{
    public static IOrderedQueryable<T> EuclideanDistanceOrder<T>(
        this IQueryable<T> query, 
        IEnumerable<Expression<Func<T, double>>> pExpressions, 
        IEnumerable<Expression<Func<T, double>>> qExpressions)
    {
        var parameter = Expression.Parameter(typeof(T));
        var pBodies = pExpressions
            .Select(x => ReplaceParameter(x.Body, parameter))
            .ToArray();

        var qBodies = qExpressions
            .Select(x => ReplaceParameter(x.Body, parameter))
            .ToArray();

        var distances = pBodies
            .Select((x, i) => CreateDistance(x, qBodies[i]))
            .ToArray();

        var squers = distances
            .Select(x => CreateSquerExpression(x))
            .ToArray();

        var sum = squers.First();
        for (int i = 1; i < squers.Count(); i++)
        {
            sum = Expression.Add(sum, squers[i]);
        }
        var funcExpression = Expression.Lambda<Func<T, double>>(sum, parameter);
        //the sqrt is irrelevant to order of this sequence
        return query.OrderBy(funcExpression);
    }

    private static Expression CreateDistance(Expression p, Expression q)
    {
        return Expression.Subtract(q, p);
    }

    private static Expression CreateSquerExpression(Expression x)
    {
        var method = typeof(Math).GetMethod("Pow", BindingFlags.Static | BindingFlags.Public);
        return Expression.Call(method, x, Expression.Constant(2.0));
    }

    private static Expression ReplaceParameter(Expression expression, ParameterExpression parameter)
    {
        var unaryExpression = expression as UnaryExpression;
        MemberExpression memberExpression;
        if (unaryExpression != null)
        {
            memberExpression = unaryExpression.Operand as MemberExpression;
        }
        else
        {
            memberExpression = expression as MemberExpression;
        }

        if (memberExpression == null)
            throw new NotImplementedException();

        if (!(memberExpression.Expression is ParameterExpression) || !(memberExpression.Member is PropertyInfo))
            throw new NotImplementedException();

        return Expression.Property(parameter, (PropertyInfo)memberExpression.Member);
    }
}

这样调用时：

    var list = new[]{ new Item
    {
        P1 = 0,
        Q1 = 0,
        P2 = 3,
        Q2 = 1,
    },
    new Item
    {
        P1 = 0,
        Q1 = 0,
        P2 = 2,
        Q2 = 1,
    }
};

var query = list.AsQueryable();

var result = query.EuclideanDistanceOrder(new Expression<Func<Item, double>>[]{
    x => x.P1,
    x => x.P2
},
new Expression<Func<Item, double>>[]{
    x => x.Q1,
    x => x.Q2
}).ToArray();

internal class Item
{
    public double P1 { get; set; }
    public double Q1 { get; set; }
    public double P2 { get; set; }
    public double Q2 { get; set; }
}

它适用于 liq to objects。我只是不确定EF或linqtoSql是否会将Math.Power方法映射到sql。如果不是，则很容易更改为乘法。

Answer 2

我无法对此进行测试，但它似乎应该可以工作。最后没有平方根，但顺序应该是一样的。

public static IOrderedQueryable<T> EuclideanDistanceOrder<T>(this IQueryable<T> query, IEnumerable<Expression<Func<T, double>>> expressions)
{
    var parameter = Expression.Parameter(typeof(T), "item");
    var seed = Expression.Lambda<Func<T, double>>(Expression.Constant((double)0), parameter);
    return query.OrderBy(expressions.Aggregate(seed, GetAggregateExpression));
}

private static Expression<Func<T, double>> GetAggregateExpression<T>(Expression<Func<T, double>> sum, Expression<Func<T, double>> item)
{
    var parameter = Expression.Parameter(typeof(T), "item");
    return Expression.Lambda<Func<T, double>>(Expression.Add(Expression.Invoke(sum, parameter), Expression.Power(Expression.Invoke(item, parameter), Expression.Constant((double)2))), parameter);
}

编辑：

由于您不能使用Expression.Invoke()，您需要内联传递给EuclideanDistanceOrder 的表达式的主体。似乎没有任何“好”的方法可以做到这一点，所以我写了一个Replace 方法来做到这一点。我只为一些更常见的Expression 类型实现了Replace，希望这足以涵盖您的使用，但您可能需要为其他Expression 类型实现它。

public static IOrderedQueryable<T> EuclideanDistanceOrder<T>(this IQueryable<T> query, IEnumerable<Expression<Func<T, double>>> expressions)
{
    var parameter = Expression.Parameter(typeof(T), "item");
    var seed = Expression.Constant((double)0);
    var agg = expressions.Aggregate((Expression)seed, (s, item) => Expression.Add(s, Expression.Power(Replace(item.Body, item.Parameters[0], parameter), Expression.Constant((double)2))));
    return query.OrderBy(Expression.Lambda<Func<T, double>>(agg, parameter));
}

private static Expression Replace(Expression expression, ParameterExpression original, ParameterExpression replacement)
{
    if (expression is BinaryExpression)
    {
        var binaryExpression = (BinaryExpression)expression;
        return Expression.MakeBinary(expression.NodeType, Replace(binaryExpression.Left, original, replacement), Replace(binaryExpression.Right, original, replacement), binaryExpression.IsLiftedToNull, binaryExpression.Method, binaryExpression.Conversion);
    }
    if (expression is ConditionalExpression)
    {
        var conditionalExpression = (ConditionalExpression)expression;
        return Expression.Condition(Replace(conditionalExpression.Test, original, replacement), Replace(conditionalExpression.IfTrue, original, replacement), Replace(conditionalExpression.IfFalse, original, replacement), conditionalExpression.Type);
    }
    if (expression is ConstantExpression)
    {
        return expression;
    }
    if (expression is MemberExpression)
    {
        var memberExpression = (MemberExpression)expression;
        return Expression.MakeMemberAccess(Replace(memberExpression.Expression, original, replacement), memberExpression.Member);
    }
    if (expression is ParameterExpression)
    {
        var parameterExpression = (ParameterExpression)expression;
        return parameterExpression == original ? replacement : parameterExpression;
    }
    if (expression is UnaryExpression)
    {
        var unaryExpression = (UnaryExpression)expression;
        return Expression.MakeUnary(unaryExpression.NodeType, Replace(unaryExpression.Operand, original, replacement), unaryExpression.Type, unaryExpression.Method);
    }
    throw new Exception(string.Format("Unsupported expression type: {0}", expression.NodeType));
}

例如，如果我们的输入表达式是：

p => p.X1 - p.X2
p => p.Y1 - p.Y2

最初的实现应该是这样构建的：

i => 0 + expressions[0](i) ^ 2 + expressions[1](i) ^ 2

新实现采用原始表达式，并将输入参数（上面的p）替换为将传递给最终lambda的参数（i），并直接使用表达式的主体输出：

i => 0 + (i.X1 - i.X2) ^ 2 + (i.Y1 - i.Y2) ^ 2