【问题梗概】
求一个函数的一阶导数。
【代码方案】
- using System;
- using System.Linq.Expressions;
- namespace Derivative
- {
- class Program
- {
- // 求一个节点表达的算式的导函数
- static Expression GetDerivative(Expression node)
- {
- if (node.NodeType == ExpressionType.Add
- || node.NodeType == ExpressionType.Subtract)
- { // 该节点在做加减法,套用加减法导数公式
- BinaryExpression binexp = (BinaryExpression)node;
- Expression dleft = GetDerivative(binexp.Left);
- Expression dright = GetDerivative(binexp.Right);
- BinaryExpression resbinexp;
- if (node.NodeType == ExpressionType.Add)
- resbinexp = Expression.Add(dleft, dright);
- else
- resbinexp = Expression.Subtract(dleft, dright);
- return resbinexp;
- }
- else if (node.NodeType == ExpressionType.Multiply)
- { // 该节点在做乘法,套用乘法导数公式
- BinaryExpression binexp = (BinaryExpression)node;
- Expression left = binexp.Left;
- Expression right = binexp.Right;
- Expression dleft = GetDerivative(left);
- Expression dright = GetDerivative(right);
- return Expression.Add(Expression.Multiply(dleft, right),
- Expression.Multiply(left, dright));
- }
- else if (node.NodeType == ExpressionType.Parameter)
- { // 该节点是x本身(叶子节点),故而其导数即常数1
- return Expression.Constant(1.0);
- }
- else if (node.NodeType == ExpressionType.Constant)
- { // 该节点是一个常数(叶子节点),故其导数为零
- return Expression.Constant(0.0);
- }
- throw new NotImplementedException(); // 其余的尚未实现
- }
- static Func<double, double> GetDerivative(Expression<Func<double, double>> func)
- {
- // 从Lambda表达式中获得函数体
- Expression resBody = GetDerivative(func.Body);
- // 需要续用Lambda表达式的自变量
- ParameterExpression parX = func.Parameters[0];
- Expression<Func<double, double>> resFunc
- = (Expression<Func<double, double>>)Expression.Lambda(resBody, parX);
- // 打印看一下导数解析式,在这个实现中结果是1+1*x+x*1(因为缺乏优化)
- Console.WriteLine("diff function = {0}", resFunc);
- // 编译成CLR的IL表达的函数
- return resFunc.Compile();
- }
- static double GetDerivative(Expression<Func<double, double>> func, double x)
- {
- Func<double, double> diff = GetDerivative(func);
- return diff(x);
- }
- static void Main(string[] args)
- {
- // 举例:求出函数f(x) = x*x+x 在x=32处的导数
- double y = GetDerivative(x => x * x + x, 32);
- Console.WriteLine("f'(x) = {0}", y);
- }
- }
- }