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I am doing 3D interpolation and I would like to get the function definition as output so that I can use it in gradient descent method optimization. My data set is of the form

x      y     z     loss
val1  val2  val3   val4

I would like to get a function definition from my interpolation of the form f(x,y,z)= loss. I need it in an explicit form like this so that I can do optimization using the function. Any assistance is very greatly appreciated.

So far my code is as follows:

from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
import pandas as pd
from scipy.interpolate import griddata as gd

x = electrode2["Base Gap"].values
y = electrode2["Top Gap"].values
z = electrode2["SiO2 Thickness"].values
out = electrode2["Loss"].values

X, Y, Z=np.mgrid[0:8:20j, 0:7:20j, 0:2:20j]
V = gd((x,y,z), out, (X.flatten(),Y.flatten(),Z.flatten()), method='linear')
#Plotting original data
fig1 = plt.figure()
ax1=fig1.gca(projection='3d')
sc1=ax1.scatter(x, y, z, c=out, cmap=plt.hot())
plt.colorbar(sc1)
ax1.set_xlabel('X')
ax1.set_ylabel('Y')
ax1.set_zlabel('Z')

#Plotting interpolated data 
fig2 = plt.figure()
ax2=fig2.gca(projection='3d')
sc2=ax2.scatter(X, Y, Z, c=V, cmap=plt.hot())
plt.colorbar(sc2)
ax2.set_xlabel('X')
ax2.set_ylabel('Y')
ax2.set_zlabel('Z')

How can I get a function definition f(x,y,z) = loss from this? I found some suggestions here Get the formula of a interpolation function created by scipy but they are all for simple 1D interpolation and they don't seem to suggest that I can find a simple formula definition. I'm also open to other suggestions on how I can perform optimization without an explicit function definition. Any thoughts are very appreciated.

Also, how can I check the validity of my interpolation? My original data was nearly grid structured but not entirely so I'm hoping this method still works. I'm having some trouble visualizing it because it is a 4D result. My scatter plots gave the following which I think looks good but I'm not entirely sure: Scatter Plot of Original Data Scatter Plot of Interpolated Data

Also, here is some of my data (not all) so that you can get a better feel for the type of values I'm using:

|x     |         y  |      z |         val|
|------|------------|---------|------------|
| 4.0 |     2.5 |     0.7 |     0.7503418140000001 |
| 4.0 |     2.75 |     0.7 |     0.64151014 |
| 4.0 |     3.0 |     0.7 |     0.6229984510000001 |
| 4.5 |     2.3 |     0.7 |     0.730188891 |
| 4.5 |     2.6 |     0.7 |     0.236956251 |
| 4.5 |     3.2 |     0.7 |     0.092038571 |
| 4.5 |     3.5 |     0.7 |     0.08848320300000001 |
| 5.0 |     2.25 |     0.7 |     0.9488792770000001 |
| 5.0 |     2.6 |     0.7 |     0.190624075 |
| 5.0 |     2.95 |     0.7 |     0.135275036 |
| 5.0 |     3.3 |     0.7 |     0.032763743 |
| 5.0 |     3.65 |     0.7 |     0.029430211 |
| 5.0 |     4.0 |     0.7 |     0.027537075 |
| 5.5 |     2.5 |     0.7 |     0.281232137 |
| 5.5 |     2.9 |     0.7 |     0.054980707000000004 |
| 5.5 |     3.3 |     0.7 |     0.023085487999999998 |
| 5.5 |     3.7 |     0.7 |     0.031317836 |
| 5.5 |     4.1 |     0.7 |     0.010820878 |
| 5.5 |     4.5 |     0.7 |     0.01016186 |
| 6.0 |     2.3 |     0.7 |     0.397089577 |
| 6.0 |     2.75 |     0.7 |     0.074292346 |
| 6.0 |     3.2 |     0.7 |     0.015918433 |
| 6.0 |     3.65 |     0.7 |     0.004494633 |
| 6.0 |     4.1 |     0.7 |     0.002195262 |
| 6.0 |     4.55 |     0.7 |     0.00175018 |
| 6.0 |     5.0 |     0.7 |     0.0017010179999999999 |
| 6.5 |     2.5 |     0.7 |     0.32217322 |
| 6.5 |     3.0 |     0.7 |     0.037894300000000006 |
| 6.5 |     3.5 |     0.7 |     0.005178418 |
| 6.5 |     4.0 |     0.7 |     0.001170227 |
| 6.5 |     4.5 |     0.7 |     0.00104462 |
| 6.5 |     5.0 |     0.7 |     0.00022831099999999998 |
| 6.5 |     5.5 |     0.7 |     0.000204873 |
| 7.0 |     2.15 |     0.7 |     0.799562745 |
| 7.0 |     2.7 |     0.7 |     0.169379167 |
| 7.0 |     3.25 |     0.7 |     0.019800399 |
| 7.0 |     3.8 |     0.7 |     0.001902681 |
| 7.0 |     4.35 |     0.7 |     0.000501842 |
| 7.0 |     4.9 |     0.7 |     0.000126206 |
| 7.0 |     5.45 |     0.7 |     2.6700000000000002e-05 |
| 7.0 |     6.0 |     0.7 |     2.0600000000000003e-05 |
| 7.5 |     2.3 |     0.7 |     1.21559407 |
| 7.5 |     2.9 |     0.7 |     0.050740772 |
| 7.5 |     3.5 |     0.7 |     0.009225684 |
| 7.5 |     4.1 |     0.7 |     0.000752695 |
| 7.5 |     4.7 |     0.7 |     0.000144944 |
| 7.5 |     5.3 |     0.7 |     1.62e-05 |
| 7.5 |     5.9 |     0.7 |     3.72e-06 |
| 7.5 |     6.5 |     0.7 |     2.67e-06 |
| 8.0 |     2.45 |     0.7 |     0.31635749999999996 |
| 8.0 |     3.1 |     0.7 |     0.049589355 |
| 8.0 |     3.75 |     0.7 |     0.0016244529999999999 |
| 8.0 |     4.4 |     0.7 |     0.000321852 |
| 8.0 |     5.05 |     0.7 |     4.3799999999999994e-05 |
| 8.0 |     5.7 |     0.7 |     3.32e-06 |
| 8.0 |     6.35 |     0.7 |     1.04e-06 |
| 8.0 |     7.0 |     0.7 |     4.25e-07 |
| 4.0 |     2.0 |     0.83 |     1.123030977 |
| 4.0 |     2.25 |     0.83 |     0.643862594 |
| 4.0 |     2.5 |     0.83 |     0.494527641 |
| 4.0 |     2.75 |     0.83 |     0.460772203 |
| 4.0 |     3.0 |     0.83 |     0.449952666 |
| 4.5 |     2.0 |     0.83 |     1.177597462 |
| 4.5 |     2.3 |     0.83 |     0.24628630399999998 |
| 4.5 |     2.6 |     0.83 |     0.126303219 |
| 4.5 |     2.9 |     0.83 |     0.8058431640000001 |
| 4.5 |     3.2 |     0.83 |     0.064124542 |

Also, I should mention that I have tried using scipy's Rbf for interpolation because I read it would be useful when the data is not in grid format. I don't think I trust the results though because the resulting graph looked like this Rbf Interpolation Results

mathemagician617
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  • Well, there doesn't exist a simple function definition. With the default parameter, `griddata` subdivides the 3D space into tetrahedra. Each vertex is one of the points xyz, with an associated `out` value. To find the value for an arbitrary point, first find the tetrahedron which contains the point, and then interpolate the `out` values depending on the distances of the 4 vertices. (If the point is outside all tetrahedra, the function is ill-defined). You will never get an interpolated value that is lower than the minimum `out` value. – JohanC Nov 27 '21 at 16:24
  • @JohanC Thank you for your response! Should I be looking into curve fitting then? My original hesitation with using this was that I wanted the function to be forced to go through my original data points(hence the use of interpolation). Also, I'm not sure how to use curvefit from scipy optimize because I don't have an original function that produced the data, just discrete data points that I collected. Do you have any tips that can help? – mathemagician617 Nov 27 '21 at 16:26
  • To find the general minimum, you need the xyz corresponding to that minimum `out` value. `electrode2[electrode2['loss'] ==electrode2['loss'].min()]` selects the row(s) corresponding to the minimum. Curve fitting is only useful if you are certain that the real function corresponds to a certain type of curve. Otherwise, you'd just be optimizing a dummy function, not related to your real function. To get closer to the "real minumum", you'd need to do more measurements, probably in the environment of the current minima. – JohanC Nov 27 '21 at 16:28
  • @JohanC Thank you for your suggestion. My goal is to find the minimum with respect to a continuous function that describes the data. This is why I want to use the gradient descent method on a continuous function definition describing the data. I've already found the minimum value from my discrete data. I have about 1331 data points within the region of the minima. Would this not be enough to create a good functional approiximation? – mathemagician617 Nov 27 '21 at 16:31
  • `griddata` or similar functions won't help for gradient descent. You need to implement one of the existing gradient descent approaches to select new candidates. Depending on the real function, and how smooth it is near the minima, you need strategies to avoid local minima. – JohanC Nov 27 '21 at 16:36
  • @JohanC Thank you! So, to clarify, do you believe that I can get the real function that you are referring to from some sort of curve fitting? I don't have an original function to work from since the data points were gathered from a simulation. – mathemagician617 Nov 27 '21 at 16:39

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