Does np.interp Implemented? #132
Comments
|
It is not implemented yet and if you look closely you will notice that the code in the test case (which is auto-generated by the way) everything is commented out using I am looking into it if this can be implemented easily. |
|
Got it,I've use DeepSeek to trying implement this,hope it works as expected. Temp Close. |
|
Can you share your solution please? |
using System;
using System.Collections.Generic;
using System.Text;
namespace ShtolaUtils
{
/// <summary>
/// 用于模拟Numpy.interp的1D线性插值类
/// </summary>
public class Interpolator
{
/// <summary>
/// 用于模拟Numpy.interp的1D线性插值
/// </summary>
/// <param name="x">用于计算插值值的x坐标。</param>
/// <param name="xp">X轴上的数据点。<br></br>
/// 当周期为true时则必定要求为单调递增，反之XP将在标准化后进行内部排序（用不上period）</param>
/// <param name="fp">Y轴上的数据点，和XP同长</param>
/// <param name="left">FP的左极值点</param>
/// <param name="right">FP的右极值点</param>
/// <returns>插值后的1D Array</returns>
/// <exception cref="ArgumentException"></exception>
public static double[] Interp(double[] x, double[] xp, double[] fp, double left = double.NaN, double right = double.NaN)
{
// 参数校验
if (xp.Length != fp.Length)
throw new ArgumentException("xp和fp长度必须一致");
if (x.Length == 0)
return new double[0];
// 处理边界默认值
if (double.IsNaN(left)) left = fp[0];
if (double.IsNaN(right)) right = fp[^1]; // 使用 ^1 表示最后一个元素
// 缓存边界值
double minXP = xp[0];
double maxXP = xp[^1];
double[] result = new double[x.Length];
int xpIndex = 0;
for (int i = 0; i < x.Length; i++)
{
double currentX = x[i];
// 处理左边界
if (currentX <= minXP)
{
result[i] = left;
continue;
}
// 处理右边界
if (currentX >= maxXP)
{
result[i] = right;
continue;
}
// 找到当前 x 在 xp 中的位置
while (xpIndex < xp.Length - 1 && xp[xpIndex + 1] < currentX)
{
xpIndex++;
}
// 计算线性插值
double x0 = xp[xpIndex];
double x1 = xp[xpIndex + 1];
double y0 = fp[xpIndex];
double y1 = fp[xpIndex + 1];
// 提前计算斜率
double slope = (y1 - y0) / (x1 - x0);
result[i] = y0 + slope * (currentX - x0);
}
return result;
}
}
}Not testing so much with performance,but i got same result as original numpy implementation |
|
Ok, I see. Of course numpy will be faster if the arrays are really big. If they are small your c# code may be faster than Numpy because there is an overhead of calling into python. Anyway, I'll put this on my todo list and let you know when it's done. |
Thanks a lot buddy,hope everything in progress and peachy |
Hi bud,is that everything going as progressed or not?Is that feature published? |
|
I got it working. The following testcase shows the usage: [TestMethod]
public void interpTest()
{
// >>> xp = [1, 2, 3]
// >>> fp = [3, 2, 0]
// >>> np.interp(2.5, xp, fp)
// 1.0
// >>> np.interp([0, 1, 1.5, 2.72, 3.14], xp, fp)
// array([ 3. , 3. , 2.5 , 0.56, 0. ])
// >>> UNDEF = -99.0
// >>> np.interp(3.14, xp, fp, right=UNDEF)
// -99.0
//
var xp = new float[] { 1, 2, 3 };
var fp = new float[] { 3, 2, 0 };
Assert.AreEqual(1.0f, np.interp(2.5f, xp, fp));
var given = np.interp(new float[] { 0, 1, 1.5f, 2.72f, 3.14f }, xp, fp);
var expected =
"array([3. , 3. , 2.5 , 0.55999994, 0. ])";
Assert.AreEqual(expected, given.repr);
var UNDEF = -99.0f;
Assert.AreEqual(-99.0f, np.interp(3.14f, xp, fp, right: UNDEF));
// Interpolation with periodic x-coordinates:
// >>> x = [-180, -170, -185, 185, -10, -5, 0, 365]
// >>> xp = [190, -190, 350, -350]
// >>> fp = [5, 10, 3, 4]
// >>> np.interp(x, xp, fp, period=360)
// array([7.5, 5., 8.75, 6.25, 3., 3.25, 3.5, 3.75])
//
//var x = [-180, -170, -185, 185, -10, -5, 0, 365];
var x = new float[] { -180, -170, -185, 185, -10, -5, 0, 365 };
xp = new float[] { 190, -190, 350, -350 };
fp = new float[] { 5, 10, 3, 4 };
given = np.interp(x, xp, fp, period: 360);
expected =
"array([7.5 , 5. , 8.75, 6.25, 3. , 3.25, 3.5 , 3.75])";
Assert.AreEqual(expected, given.repr);
// Complex interpolation:
// >>> x = [1.5, 4.0]
// >>> xp = [2,3,5]
// >>> fp = [1.0j, 0, 2+3j]
// >>> np.interp(x, xp, fp)
// array([ 0.+1.j , 1.+1.5j])
x = new float[] { 1.5f, 4.0f };
xp = new float[] { 2, 3, 5 };
var fp2 = new [] { new Complex(0, 1), new Complex(0, 0), new Complex(2, 3) };
given = np.interp(x, xp, fp2);
expected =
"array([0.+1.j , 1.+1.5j])";
Assert.AreEqual(expected, given.repr);
} |
|
Released to nuget as as Numpy v3.11.1.35 (1f35173) |
Nice work,Since it's 1:00 AM in China right now,I will bring your reply when I wake up and fire up my workstation,good night: ) |
|
Hi.I'm back here. I've met a new problem with new version of Numpy.NET is that:when i running up the program,it will throw an warning message: Could not find platform dependent libraries <exec_prefix> And then,when i call a numpy array in program,it may cause: PythonEngine is not initialized |
|
It won't happend on 3.10.1.30 release |
|
I am pretty sure that you are doing something wrong. I updated the MatmulExample in the Numpy.NET source code to use the latest numpy library and ran it and it works just fine: Output: Sure it also prints the I don't have time to debug your app so here is my advice. Use the MatmulExample as a base and add your code. Maybe that'll work for you. |
Oh i see,My Anaconda base is Py3.10. Is that a related reason? |
Yes, that would explain it. I can backport the change onto a 3.10 version of numpy. |
Hi:
i've there's a testing code at:
Numpy.NET/test/Numpy.UnitTest/NumPy_math.tests.cs
Line 3388 in 5955d0c
Yet when i wanna using this function in project,it raised:
It raise it's not defined.
The text was updated successfully, but these errors were encountered: