How to speedup calling the interpolation function?

Question

Consider the following data:

dat={{0.00001, 7000.}, {0.0001, 7000.}, {0.001, 7000.}, {0.002, 
  6999.999999999987}, {0.0045000000000000005, 
  6999.999999999987}, {0.008, 6999.999999999987}, {0.02, 
  6999.999999999987}, {0.045, 6999.999999999987}, 
   {0.08, 4313.535758081632}, {0.15000000000000002, 
  3853.703772545988}, {0.30000000000000004, 3128.349789547097}, {0.5, 
  2800.9698445932686}, {0.7, 2437.7100097749117}, {0.9, 
  2363.8265106983717}, 
   {1.1, 2393.576853673258}, {1.2999999999999998, 
  2383.170945978633}, {1.4849999999999999, 2302.8502915536897}};

I would like to obtain the interpolation int[x] from this data and generate the table {x, f[x]} for x belonging to some dataset xrand. This is my code:

xrand=RandomReal[MinMax[dat[[All,1]]],10^6];
int[x_]=Interpolation[dat,InterpolationOrder->1][x];
finaldata=int[xrand];//AbsoluteTiming

However, it is slow:

{1.04,Null}

The main problem is in the slow calling of int. In principle, instead of generating the interpolation, I may find a fit to dat, using e.g. NonlinearModelFit. However, if dealing with an arbitrary dataset, it would not be possible to make the procedure automatic since then one would need to change the fit function for NonlinearModelFit. Therefore, I am interested in a speedup of calling the interpolated function.

Is there any way to improve the performance?

10^-6 s per value seems pretty good for software evaluation (select interpolant poly and then evaluate it). But I’m not sure what the real lower limit is. — Michael E2, Mar 10 '23 at 18:37

Ulrich Neumann · Answer 1 · 2023-03-10T10:43:07.627

1

Try purefunctionas Interpolation object

xrand = RandomReal[MinMax[dat[[All, 1]]], 10^6];
ip = Interpolation[dat, InterpolationOrder -> 1] ;
finaldata = ip[xrand]; // AbsoluteTiming
({1.19475, Null})

which is much faster than your attempt

int[x_] = Interpolation[dat, InterpolationOrder -> 1][x];  
finaldata = int[xrand]; // AbsoluteTiming
(*{1.49481, Null}*)

edited Mar 10 '23 at 10:43

answered Mar 10 '23 at 10:31

Ulrich Neumann

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On my machine (Mathematica 12.1), both ip[xrand] and int[xrand] require 1 second. – John Taylor Mar 10 '23 at 10:34
What is your Mathematica version? Mine: v12.2 Windows11 – Ulrich Neumann Mar 10 '23 at 10:35
Sorry, I have just updated my comment, it is 12.1@Windows 11. – John Taylor Mar 10 '23 at 10:38
Only if I call int and ip multiple times (more than 1) I get the same execution times too! – Ulrich Neumann Mar 10 '23 at 10:46
I got ~1.03 s for int and 1.15 s for ip if launching them separately on a clean kernel. – John Taylor Mar 10 '23 at 10:50

How to speedup calling the interpolation function?

1 Answers1