使用表达式树构建 Func<int, double> 多项式

Question

TL;DR

如何使用系数数组构建表达式并将其转换为 Func<int, double>？有没有比表达式树更好的方法？

我有一个不可变的 Sequence 类型，它是使用 Func<int, double> formula 构造的，用于为序列 A 生成项 An。我开始构建一个助手 class 来构造常用的数学公式一些简单的参数：

public static Sequence CreateLinearSequence (double m, double b)
{ return new Sequence (n => m * n + b); }

我为常数序列、对数和简单多项式（线性、二次、三次和四次）构建了标准方法，但我想使用 params 关键字扩展它以支持任意数量的项。

这是我的方法：

 public static Sequence CreatePolynomialSequence (params double[] coeff)
 {
     Expression<Func<int, double>> e = x => 0;
     double pow = 0;

     for (int i = coeff.Length - 1; i >= 0; i--)
     {
         double c = coeff[i];
         var p = Expression.Parameter (typeof (int), "x");
         e = Expression.Lambda<Func<int, double>> (
             Expression.Add (
                 e, 
                 (Expression<Func<int, double>>)(x => c * Math.Pow (x, pow))
             ), 
             p);
         pow++; 
     }
     return new Sequence (e.Compile ());
 }

你们可能很明显我做错了什么；我搞砸了一点，直到我得到一些我觉得应该工作的东西，但它没有。

目标是让序列像这样对数组工作double[] coeff = {a,b,c,d,e,f,g,h}

x => h + gx + fx^2 + ex^3 + dx^4 + cx^5 + bx^6 + ax^7 使用适当的 Math.Pow(x, exponent) 调用。

运行

var s2 = SequenceHelper.CreatePolynomialSequence (new[] { 1d, 2 });
Console.WriteLine ("s2: " + s2);

结果

Unhandled Exception: System.InvalidOperationException: The binary operator Add is not defined for the types 'System.Func2[System.Int32,System.Double]' and 'System.Func2[System.Int32,System.Double]'. at System.Linq.Expressions.Expression.GetUserDefinedBinaryOperatorOrThrow (ExpressionType binaryType, System.String name, System.Linq.Expressions.Expression left, System.Linq.Expressions.Expression right, Boolean liftToNull) [0x0004a] in /private/tmp/source-mono-mac-4.2.0-branch/bockbuild-mono-4.2.0-branch/profiles/mono-mac-xamarin/build-root/mono-4.2.1/mcs/class/dlr/Runtime/Microsoft.Scripting.Core/Ast/BinaryExpression.cs:658 at System.Linq.Expressions.Expression.Add (System.Linq.Expressions.Expression left, System.Linq.Expressions.Expression right, System.Reflection.MethodInfo method) [0x00057] in /private/tmp/source-mono-mac-4.2.0-branch/bockbuild-mono-4.2.0-branch/profiles/mono-mac-xamarin/build-root/mono-4.2.1/mcs/class/dlr/Runtime/Microsoft.Scripting.Core/Ast/BinaryExpression.cs:1409 at System.Linq.Expressions.Expression.Add (System.Linq.Expressions.Expression left, System.Linq.Expressions.Expression right) [0x00000] in /private/tmp/source-mono-mac-4.2.0-branch/bockbuild-mono-4.2.0-branch/profiles/mono-mac-xamarin/build-root/mono-4.2.1/mcs/class/dlr/Runtime/Microsoft.Scripting.Core/Ast/BinaryExpression.cs:1390 at Sequence.SequenceHelper.CreatePolynomialSequence (System.Double[] coeff) [0x00110] in /Users/Knoble/MonoProjects/Sequences/Sequence/SequenceHelper.cs:88
at Sequence.Test.Main () [0x0001f] in /Users/Knoble/MonoProjects/Sequences/Sequence/Test.cs:53 [ERROR] FATAL UNHANDLED EXCEPTION: System.InvalidOperationException: The binary operator Add is not defined for the types 'System.Func2[System.Int32,System.Double]' and 'System.Func2[System.Int32,System.Double]'. at System.Linq.Expressions.Expression.GetUserDefinedBinaryOperatorOrThrow (ExpressionType binaryType, System.String name, System.Linq.Expressions.Expression left, System.Linq.Expressions.Expression right, Boolean liftToNull) [0x0004a] in /private/tmp/source-mono-mac-4.2.0-branch/bockbuild-mono-4.2.0-branch/profiles/mono-mac-xamarin/build-root/mono-4.2.1/mcs/class/dlr/Runtime/Microsoft.Scripting.Core/Ast/BinaryExpression.cs:658 at System.Linq.Expressions.Expression.Add (System.Linq.Expressions.Expression left, System.Linq.Expressions.Expression right, System.Reflection.MethodInfo method) [0x00057] in /private/tmp/source-mono-mac-4.2.0-branch/bockbuild-mono-4.2.0-branch/profiles/mono-mac-xamarin/build-root/mono-4.2.1/mcs/class/dlr/Runtime/Microsoft.Scripting.Core/Ast/BinaryExpression.cs:1409 at System.Linq.Expressions.Expression.Add (System.Linq.Expressions.Expression left, System.Linq.Expressions.Expression right) [0x00000] in /private/tmp/source-mono-mac-4.2.0-branch/bockbuild-mono-4.2.0-branch/profiles/mono-mac-xamarin/build-root/mono-4.2.1/mcs/class/dlr/Runtime/Microsoft.Scripting.Core/Ast/BinaryExpression.cs:1390 at Sequence.SequenceHelper.CreatePolynomialSequence (System.Double[] coeff) [0x00110] in /Users/Knoble/MonoProjects/Sequences/Sequence/SequenceHelper.cs:88
at Sequence.Test.Main () [0x0001f] in /Users/Knoble/MonoProjects/Sequences/Sequence/Test.cs:53 The application was terminated by a signal: SIGHUP

Answer 1

您需要解决三件事：

在 Add 内使用 e.Body 而不是 e。
对所有内容使用相同的参数对象。这有点棘手：Expression.Parameter(typeof (int), "x");中的x、e = x => 0中的x和x => c * Math.Pow (x, pow)中的x是不同个参数。
在循环内创建 pow 的副本。否则，pow 被捕获并且 pow 的相同（最终）值用于所有系数。

在下面的代码示例中，我通过使用内部表达式的参数调用新表达式来解决第二个问题。另一种选择是手动构建 x => c * Math.Pow(x, pow) 而不是使用 C# lambda 表达式或统一参数，如 this question.

中所述

static void Main(string[] args)
{
    var seq = CreatePolynomialSequence(1, 2, 3);
    Console.WriteLine(seq.Invoke(1)); // yields 6 = 1 + 2 + 3
    Console.WriteLine(seq.Invoke(2)); // yields 11 = 1*4 + 2*2 + 3
}

public static Func<int, double> CreatePolynomialSequence(params double[] coeff)
{
    Expression<Func<int, double>> e = x => 0;
    double pow = 0;

    for (int i = coeff.Length - 1; i >= 0; i--)
    {
        var p = e.Parameters[0];
        double c = coeff[i];
        var _pow = pow; // avoid closing over the outer variable
        var next = (Expression<Func<int, double>>)(x => c * Math.Pow(x, _pow));
        var nextInvoked = Expression.Invoke(next, p);

        e = Expression.Lambda<Func<int, double>>(Expression.Add(e.Body, nextInvoked), p);
        pow++;
    }
    return e.Compile();
}

Answer 2

基于@Heinzi 的回答，这里是您如何使用 CreatePolynomialExpression 手动构建整个表达式树的方法：

public static Expression<Func<int, double>> CreatePolynomialExpression(params double[] coeff)
{
    if (coeff.Length == 0)
        return x => 0;

    double pow = 1;

    var x_param = Expression.Parameter(typeof(int), "x");

    Expression expression = Expression.Constant(coeff[coeff.Length - 1]);

    for (int i = coeff.Length - 2; i >= 0; i--)
    {
        Expression sub_expression =
            Expression.Multiply(
                Expression.Constant(coeff[i]),
                Expression.Power(
                    Expression.Convert(x_param, typeof(double)),
                    Expression.Constant(pow)));

        expression = Expression.Add(expression , sub_expression);

        pow++;
    }

    return Expression.Lambda<Func<int, double>>(expression, x_param);
}

Answer 3

另一个构建器 with fiddle，它不使用捕获的变量和 Math.Pow 因此工作速度更快

 public static Func<int, double> CreatePolynomialSequence (params double[] coeff)
 {              
     // polynom builder
     // double argument
     var y = Expression.Variable(typeof(double), "y");  

     // func result
     var res = Expression.Variable(typeof(double), "res");       

     var expr = Expression.Assign(res, Expression.Constant(coeff[0]));

     // build polynom in format: ((a*x+b)*x+c)*x+d  <=>  a*x^3 + b*x^2 + c*x + d
     for (int i = 1; i < coeff.Length; i++)
     {
         expr = Expression.Add
                (
                    Expression.Multiply(expr, y), 
                    Expression.Constant(coeff[i]) 
                );
     }       

     // function body
     var x = Expression.Variable(typeof(int), "x"); 
     var block = Expression.Block
         (
             new ParameterExpression[]{ y, res },  // local variables
             new Expression[]
             {
                 // cast int argument to double
                 Expression.Assign(y, Expression.Convert(x, typeof(double))),
                 //compute result
                 expr
             }
        );

     return Expression.Lambda<Func<int, double>>(block, x).Compile();
 }

Answer 4

我对这个问题和所有三个答案感到困惑；如果您打算做的只是将它们编译成委托，为什么还要搞乱表达式树？直接 return 代表！

public static Func<double, double> CreatePolynomialFunction (params double[] coeff)
{
    if (coeff == null) throw new ArgumentNullException("coeff");
    return x => 
    {
        double sum = 0.0;
        double xPower = 1;
        for (int power = 0; power < coeff.Length; power += 1)
        {
            sum += xPower * coeff[power];
            xPower *= x;
        }
        return sum;
    };
}

完成。不需要弄乱表达式树。

（我注意到我假设数组中的第 n 项是第 n 个系数；显然你在数组中向后列出了你的系数。这似乎容易出错，但如果那是你想要的那么它并不困难将此答案修改为运行循环 down 从 Length-1 到零。）

使用表达式树构建 Func<int, double> 多项式

Building a Func<int, double> Polynomial using Expression Trees

c#

math

expression-trees

func

polynomials

TL;DR