英文:
How to generate a 3D surface function to fit given 3D points and interpolate 3rd coordinate if I have other 3 coordinates
问题
我是Python的初学者。
我有一些3D点数据，如下所示，
x = [600, 450, 302, 151, 600, 450, 302, 151, 599, 450, 302, 150, 599, 449, 301, 150]
y = [0.4, 0.2, 0.1, 0.0, 2.1, 2, 2, 2, 4, 4.1, 4, 4.1, 6.1, 6.1, 6, 6]
z = [23.77, 17.74, 11.93, 5.97, 23.60, 17.60, 11.78, 5.82, 23.09, 17.24, 11.39, 5.39, 21.89, 16.18, 10.36, 4.26]
我想要拟合这些点的曲面，然后创建一个函数来表示这个曲面，这样，如果我想在这个曲面上绘制一个新点，并且我有它的两个坐标（x，y），我应该能够找到它的z坐标。
请问有人能帮助我，我该如何解决这个问题？
我尝试使用scipy.linalg.lstsq函数来拟合二次曲线，但我没有得到任何输出。
我只是期望得到一个图表和一个能够给我提供Z值的曲面方程，如果我有X和Y的话。
英文:
I am a beginer in Python.
I have some 3D point data as below,
x= [600, 450, 302, 151, 600, 450, 302, 151, 599, 450, 302, 150, 599, 449, 301, 150]
y= [0.4, 0.2, 0.1, 0.0, 2.1, 2, 2, 2, 4, 4.1, 4, 4.1, 6.1, 6.1, 6, 6]
z =[23.77, 17.74, 11.93,5.97, 23.60, 17.60, 11.78, 5.82, 23.09, 17.24, 11.39, 5.39, 21.89, 16.18, 10.36, 4.26]
I want to fit a surface to these points and then create a function of this surface such that if I want to plot a new point on this surface and I have two coordinates (x, y) of it then I should be able to find the z coordinate of the it.
Could someone please assist me, how can I proceed with this problem?
I tried using scipy.linalg.lstsq function to fit quadratic curve, but I am not getting any output.
I am simply expecting a plot and a surface equation which can give me the value of Z if i have X, and Y.
答案1
得分: 0
如果您想定义一个最佳拟合平面，而不必从零开始实现数学计算，您可以使用scikit-spatial库，如下所示（如示例中所解释的）。
安装该库（例如使用pip，但您也可以使用conda）：
pip install scikit-spatial
导入库：
from skspatial.objects import Plane
定义您的点：
points = [[0, 0, 0], [1, 0, 0], [0, 1, 0], [0, 0, 1]]
计算最佳拟合平面：
plane = Plane.best_fit(points)
您还可以使用plot_3d方法绘制平面。
此外，您可以使用cartesian方法获取平面的系数：
coefficients = plane.cartesian()
系数包括a、b、c和d，平面方程为ax + by + cz + d = 0。因此，给定x和y坐标，您可以如下计算z坐标：
x = 3
y = 4
z = -(a*x + b*y + d) / c
print(z)
最后，如果您需要拟合二次曲面，您可以参考我的这个答案（不幸的是，它既未被接受也未被投票支持，因此我无法建议它作为重复问题）。
英文:
If you want to define a best-fit plane without implementing the math from zero, you can use the scikit-spatial library in the following way (as explained in the example).
Install the library (for instance using pip, but you can also use conda)
pip install scikit-spatial
Import the library:
from skspatial.objects import Plane
Define your points:
points = [[0, 0, 0], [1, 0, 0], [0, 1, 0], [0, 0, 1]]
Compute the best-fit plane:
plane = Plane.best_fit(points)
You can also plot the plane using the plot_3d method.
Moreover, you can get the plane's coefficients using the cartesian method:
coefficients = plane.cartesian()
The coefficients are a, b, c and d and the plane equation is ax + by + cz + d = 0. Thus, given the x and y coordinates, then you can compute the z coordinate as follows:
x = 3
y = 4
z = -(a*x + b*y + d) / c
print(z)
Finally, if you need to fit a quadratic surface, you can take a look at this answer of mine (unfortunately, it's neither accepted nor upvoted, so I cannot suggest it as a duplicate).
答案2
得分: -1
线性回归与平面方程：
线性回归与二次曲面方程：
RLMSE：根最小均方误差。
注意：根据上述结果，因为A和C都很小，可以预期仅使用四个参数（而不是六个）的二次曲面方程 z = B y^2 + D x + E y + F 也能获得几乎同样好的结果。
计算细节：
英文:
If you have no clue for an equation model try several equations until you obtain an acceptable result.
Linear regression with the equation of plane :
Linear regression with equation of quadratic surface :
RLMSE : Root Least Mean Square Error.
Note : From the above result and because A and C are very small one can expect an almost as good result with the only four parameters (instead of six) equation of quadratique surface : z = B y^2 + D x + E y + F .
Details of the calculus :
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