blocks|key|74682|text|如果您不反对Math.Net以外的库，可以尝试AlgLib和它们的spline1ddiff函数|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|74683|构建样条线很容易，Akima样条线通过点看起来很好很平滑。如果您想要一个接受一组数据并返回导数的方法，下面是一个使用AlgLib数学库的示例：|74684|public+static+void+CalculateDerivatives(this+Dictionary<double,+double>+inputPoints,+out+Dictionary<double,+double>+firstDerivatives,+out+Dictionary<double,+double>+secondDerivatives)
{
++++++++var+inputPointsXArray+=+inputPoints.Keys.ToArray();
++++++++var+inputPointsYArray+=+inputPoints.Values.ToArray();

++++++++spline1dinterpolant+akimaSplineToDifferentiate;
++++++++alglib.spline1dbuildakima(inputPointsXArray,+inputPointsYArray,+out+akimaSplineToDifferentiate);

++++++++firstDerivatives+=+new+Dictionary<double,+double>();
++++++++secondDerivatives+=+new+Dictionary<double,+double>();
++++++++foreach+(var+pair+in+inputPoints)
++++++++{
++++++++++++var+xPoint+=+pair.Key;
++++++++++++double+functionVal,+firstDeriv,+secondDeriv;
++++++++++++alglib.spline1ddiff(akimaSplineToDifferentiate,+xPoint,+out+functionVal,+out+firstDeriv,+out+secondDeriv);

++++++++++++firstDerivatives.Add(point,+firstDeriv);
++++++++++++secondDerivatives.Add(point,+secondDeriv);
++++++++}
}|code-block|syntax|javascript|74685|警告:+Akima+Spline在您的数据集范围之外有不可预测的行为。|74686|entityMap|0|LINK|mutability|MUTABLE|url|http://www.alglib.net/interpolation/spline3.php#header5|1|http://www.alglib.net/translator/man/manual.csharp.html#sub_spline1ddiff^0|N|6|0|X|C|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@$A|X|B|Y|1|Z]|$A|10|B|11|1|12]]|C|$]]|$1|D|3|E|5|6|7|13|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|14|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|15|8|@]|9|@]|C|$]]|$1|M|3|-4|5|6|7|16|8|@]|9|@]|C|$]]]|N|$O|$5|P|Q|R|C|$S|T]]|U|$5|P|Q|R|C|$S|V]]]]

If you aren't opposed to libraries other than Math.Net, you could try <a href="http://www.alglib.net/interpolation/spline3.php#header5" rel="nofollow noreferrer">AlgLib</a> and their <a href="http://www.alglib.net/translator/man/manual.csharp.html#sub_spline1ddiff" rel="nofollow noreferrer">spline1ddiff</a> function

Building the spline is easy, and the Akima spline looks to be nice and smooth through the points. If you wanted a method that takes in a set of data and returns the derivatives, here's an example using the AlgLib math library:

<pre><code>public static void CalculateDerivatives(this Dictionary&lt;double, double&gt; inputPoints, out Dictionary&lt;double, double&gt; firstDerivatives, out Dictionary&lt;double, double&gt; secondDerivatives)
{
 var inputPointsXArray = inputPoints.Keys.ToArray();
 var inputPointsYArray = inputPoints.Values.ToArray();

 spline1dinterpolant akimaSplineToDifferentiate;
 alglib.spline1dbuildakima(inputPointsXArray, inputPointsYArray, out akimaSplineToDifferentiate);

 firstDerivatives = new Dictionary&lt;double, double&gt;();
 secondDerivatives = new Dictionary&lt;double, double&gt;();
 foreach (var pair in inputPoints)
 {
 var xPoint = pair.Key;
 double functionVal, firstDeriv, secondDeriv;
 alglib.spline1ddiff(akimaSplineToDifferentiate, xPoint, out functionVal, out firstDeriv, out secondDeriv);

 firstDerivatives.Add(point, firstDeriv);
 secondDerivatives.Add(point, secondDeriv);
 }
}
</code></pre>

Be Warned: Akima Spline has unpredictable behavior outside of the range of your dataset.

blocks|key|73695|text|获取数据点并创建一个Math.NET+Numerics+Cubic+Spline对象。然后使用.Differentiate()方法获取每个所需点处的坡度。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|73696|示例|73697|尝试下面的代码：|73698|static+class+Program
{
++++const+int+column_width+=+12;
++++///+<summary>
++++///+The+main+entry+point+for+the+application.
++++///+</summary>
++++[STAThread]
++++static+void+Main(string[]+args)
++++{
++++++++var+xvec+=+new+DenseVector(new+double[]+{+0.0,+1.0,+2.0,+3.0,+4.0+});
++++++++var+yvec+=+new+DenseVector(new+double[]+{+3.0,+2.7,+2.3,+1.6,+0.2+});
++++++++Debug.WriteLine("Input+Data+Table");
++++++++Debug.WriteLine($"{"x",column_width}+{"y",column_width}");
++++++++for(int+i+=+0;+i+<+xvec.Count;+i%2B%2B)
++++++++{
++++++++++++Debug.WriteLine($"{xvec[i],column_width:G5}+{yvec[i],column_width:G5}");
++++++++}
++++++++Debug.WriteLine("+");
++++++++var+cs+=+CubicSpline.InterpolateNatural(xvec,+yvec);

++++++++var+x+=+new+DenseVector(15);
++++++++var+y+=+new+DenseVector(x.Count);
++++++++var+dydx+=+new+DenseVector(x.Count);
++++++++Debug.WriteLine("Interpoaltion+Results+Table");
++++++++Debug.WriteLine($"{"x",column_width}+{"y",column_width}+{"dy/dx",column_width}");
++++++++for(int+i+=+0;+i+<+x.Count;+i%2B%2B)
++++++++{
++++++++++++x[i]+=+(4.0*i)/(x.Count-1);
++++++++++++y[i]+=+cs.Interpolate(x[i]);
++++++++++++dydx[i]+=+cs.Differentiate(x[i]);
++++++++++++Debug.WriteLine($"{x[i],column_width:G5}+{y[i],column_width:G5}+{dydx[i],column_width:G5}");
++++++++}


++++}
}|code-block|syntax|javascript|73699|并查看调试输出：|73700|Input+Data+Table
+++++++++++x++++++++++++y
+++++++++++0++++++++++++3
+++++++++++1++++++++++2.7
+++++++++++2++++++++++2.3
+++++++++++3++++++++++1.6
+++++++++++4++++++++++0.2

Interpoaltion+Results+Table
+++++++++++x++++++++++++y++++++++dy/dx
+++++++++++0++++++++++++3+++++-0.28214
+++++0.28571++++++++2.919+++++-0.28652
+++++0.57143+++++++2.8354+++++-0.29964
+++++0.85714+++++++2.7469++++++-0.3215
++++++1.1429+++++++2.6509+++++-0.35168
++++++1.4286+++++++2.5454+++++-0.38754
++++++1.7143++++++++2.429+++++-0.42864
+++++++++++2++++++++++2.3+++++++-0.475
++++++2.2857++++++++2.154+++++-0.55809
++++++2.5714+++++++1.9746++++++-0.7094
++++++2.8571+++++++1.7422+++++-0.92894
++++++3.1429+++++++1.4382++++++-1.1979
++++++3.4286+++++++1.0646++++++-1.4034
++++++3.7143++++++0.64404++++++-1.5267
+++++++++++4++++++++++0.2++++++-1.5679|73701|​|73702|📷|atomic|73703|73704|entityMap|0|LINK|mutability|MUTABLE|url|http://numerics.mathdotnet.com/api/MathNet.Numerics.Interpolation/CubicSpline.htm|1|IMAGE|IMMUTABLE|imageUrl|https://ask.qcloudimg.com/http-save/yehe-900000/bcd657885351159fdf116bcc1ae3bb11.png|imageAlt^0|1B|G|S|C|0|0|0|0|0|0|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@$9|1F|A|1G|1|1H]]|E|$]]|$1|F|3|G|5|6|7|1I|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|1J|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1K|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1L|8|@]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1M|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1N|8|@]|D|@]|E|$]]|$1|U|3|V|5|W|7|1O|8|@]|D|@$9|1P|A|1Q|1|1R]]|E|$]]|$1|X|3|T|5|6|7|1S|8|@]|D|@]|E|$]]|$1|Y|3|-4|5|6|7|1T|8|@]|D|@]|E|$]]]|Z|$10|$5|11|12|13|E|$14|15]]|16|$5|17|12|18|E|$19|1A|1B|-4]]]]

Take your data points and create a Math.NET Numerics <a href="http://numerics.mathdotnet.com/api/MathNet.Numerics.Interpolation/CubicSpline.htm" rel="noreferrer">Cubic Spline</a> object. Then use the <code>.Differentiate()</code> method to get the slope at each point you want.

<h2>Example</h2>

Try the code below:

<pre><code>static class Program
{
 const int column_width = 12;
 /// &lt;summary&gt;
 /// The main entry point for the application.
 /// &lt;/summary&gt;
 [STAThread]
 static void Main(string[] args)
 {
 var xvec = new DenseVector(new double[] { 0.0, 1.0, 2.0, 3.0, 4.0 });
 var yvec = new DenseVector(new double[] { 3.0, 2.7, 2.3, 1.6, 0.2 });
 Debug.WriteLine("Input Data Table");
 Debug.WriteLine($"{"x",column_width} {"y",column_width}");
 for(int i = 0; i &lt; xvec.Count; i++)
 {
 Debug.WriteLine($"{xvec[i],column_width:G5} {yvec[i],column_width:G5}");
 }
 Debug.WriteLine(" ");
 var cs = CubicSpline.InterpolateNatural(xvec, yvec);

 var x = new DenseVector(15);
 var y = new DenseVector(x.Count);
 var dydx = new DenseVector(x.Count);
 Debug.WriteLine("Interpoaltion Results Table");
 Debug.WriteLine($"{"x",column_width} {"y",column_width} {"dy/dx",column_width}");
 for(int i = 0; i &lt; x.Count; i++)
 {
 x[i] = (4.0*i)/(x.Count-1);
 y[i] = cs.Interpolate(x[i]);
 dydx[i] = cs.Differentiate(x[i]);
 Debug.WriteLine($"{x[i],column_width:G5} {y[i],column_width:G5} {dydx[i],column_width:G5}");
 }


 }
}
</code></pre>

And look at the debug output:

<pre><code>Input Data Table
 x y
 0 3
 1 2.7
 2 2.3
 3 1.6
 4 0.2

Interpoaltion Results Table
 x y dy/dx
 0 3 -0.28214
 0.28571 2.919 -0.28652
 0.57143 2.8354 -0.29964
 0.85714 2.7469 -0.3215
 1.1429 2.6509 -0.35168
 1.4286 2.5454 -0.38754
 1.7143 2.429 -0.42864
 2 2.3 -0.475
 2.2857 2.154 -0.55809
 2.5714 1.9746 -0.7094
 2.8571 1.7422 -0.92894
 3.1429 1.4382 -1.1979
 3.4286 1.0646 -1.4034
 3.7143 0.64404 -1.5267
 4 0.2 -1.5679
</code></pre>

<a href="https://i.stack.imgur.com/nRdAw.png" rel="noreferrer"><img src="https://i.stack.imgur.com/nRdAw.png" alt="plot1"></a>

blocks|key|2729230|text|感谢您的回复。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2729231|我相信我需要遍历数组，从Math.NET调用Differentiate函数，或者简单地编写我自己的(rise+over+run)计算。|2729232|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Thanks for the responses.

I believe I'll need to iterate through the array, calling the Differentiate function from Math.NET, or perhaps simply write my own (rise over run) calculation.

blocks|key|74763|text|这里有函数的派生扩展|type|unstyled|depth|inlineStyleRanges|entityRanges|data|74764|public+static+Func<double[],+double>+Derivative(this+Func<double[],+double>+func,+int+derivativeIndex)
{
++++double+step+=+0.001;
++++return+income+=>+
++++{
++++++++double[]+increasedIncome+=+(double[])income.Clone();
++++++++increasedIncome[derivativeIndex]+%2B=+step;

++++++++double[]+decreasedIncome+=+(double[])income.Clone();
++++++++decreasedIncome[derivativeIndex]+-=+step;

++++++++return+(func(increasedIncome)+-+func(decreasedIncome))+/+(2+*+step);
++++};
}|code-block|syntax|javascript|74765|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Here you have derivative extension for function
<pre><code>public static Func&lt;double[], double&gt; Derivative(this Func&lt;double[], double&gt; func, int derivativeIndex)
{
 double step = 0.001;
 return income =&gt; 
 {
 double[] increasedIncome = (double[])income.Clone();
 increasedIncome[derivativeIndex] += step;

 double[] decreasedIncome = (double[])income.Clone();
 decreasedIncome[derivativeIndex] -= step;

 return (func(increasedIncome) - func(decreasedIncome)) / (2 * step);
 };
}
</code></pre>

I'm looking for a simple function that will accept an array of double values and return the mathematical derivative.

Math.NET appears to have such a function, but it's asking for the following syntax:

<pre><code>double FirstDerivative(Func&lt;double, double&gt; f, double x)
</code></pre>

I'm not sure why I would need to specify a function. I just want a preexisting function that I can pass data to.

Calculating a derivative with Math.NET C#

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在寻找一个简单的函数，它将接受一个双精度值数组并返回数学导数。Math.NET似乎有这样一个函数，但它要求使用以下语法：double FirstDerivative(Func<double, double> f, double x)我不确定为什么我需要指定一个函数。我只想要一个可以传递数据的预先存在的函数。

问使用Math.NET C#计算导数
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