This question already has an answer here:

In general my question is related to an approach to make an interpolation of a data which is not on a regular mesh. Another aspect of this question has already been discussed after my question "Numerical integration of a numeric data available as a nested list".

I have data in the form of a list of triads {x, y, z} that came from an external simulation. It may be and may be not ordered. Further, x and y lie within a rectangle R, and are not regularly spaced. I would like to coarse-grain and simultaneously sort this data. The coarse-graining consists in partitioning the rectangle R into multiple small rectangles and constructing a list {xij, yij, zij} where xij and yij are the coordinates of the centers of the rectangles, while zij is the average of all z values for which x and y belong to the rectangle specified by i and j. To be specific, let us build a list mimicking the real one, but small enough:

a = RandomReal[{0, 10}, {10000, 2}];
b = RandomReal[{-1, 1}, {10000}];
lst = 
  MapThread[Insert[#1, #2, 3] &, {a, b}] /. 
   {x_, y_, z_} -> {x, y, 10 Exp[-((x - 5)^2 + (y - 5)^2)/4] + z};

The list lst has 10000 elements with x and y located between 0 and 10. The coarse-grained list may be obtained by the averaging over squares with the sizes 0.5*0.5 like this:

lstCoarseGrained = Flatten[Table[
    (s = Select[
       lst, ((i <= #[[1]] <= i + 0.5) && (j <= #[[2]] <= j + 0.5) &)];
      {{i + 0.25, j + 0.25}, Mean[Transpose[s][[3]]]}),
    {i, 0, 9.75, 0.5}, {j, 0, 9.75, 0.5}], 1];

It has



elements and can be straightforwardly interpolated:

f = Interpolation[lstCoarseGrained, InterpolationOrder -> 3, Method -> "Spline"]


  ListPlot3D[lst, PlotRange -> All, ImageSize -> 250], 
  Plot3D[f[x, y], {x, 0, 10}, {y, 0, 10}, PlotRange -> All, ImageSize -> 250]

enter image description here

and then further post-processed. On the figure above the left image shows the data before, and the right - after the coarse-graining and interpolation.

However, the coarse-graining by this approach requires

        s = Select[lst, ((i <= #[[1]] <= i + 0.5) && (j <= #[[2]] <= j + 0.5) &)];
        {{i + 0.25, j + 0.25}, Mean[Transpose[s][[3]]]}
      {i, 0, 9.75, 0.5}, {j, 0, 9.75, 0.5}],

{13.468750, Null}

13 seconds. My realistic lists have about 10^6 triads and will require about 21 min.

My question is, if you can see a faster way to do this?

In principle, if I could, say, use something like Partition instead of a Table, it would go faster. However, to do this, the list should be first sorted in 2D, and I do not see, how. So the second question is if you see the way to sort the list with elements {x,y,z} with arbitrary order in the (x,y) plane such that it is organized as like, say, the list given by the table:

Table[{i, j}, {i, 1, 5}, {j, 1, 5}]

marked as duplicate by Mr.Wizard Jun 20 '13 at 16:50

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ In your definition of lst, you should use :> (RuleDelayed) rather than -> (Rule). A (faster) alternative here is MapThread[Append[#1,10 Exp[-((#1[[1]]-5)^2+(#[[2]]-5)^2)/4]+#2]&,{a,b}]. Just to show you that we have Append here rather than Insert[_,_,3], I realize is not the step you want to optimize. $\endgroup$ – Jacob Akkerboom Jun 20 '13 at 11:28
  • $\begingroup$ @Jacob Akkerboom My question is not about speeding up making the table returning the list lst. This is a phony table I only generate for the sake of example. The question is how to speed up making the list lstCoarseGrained. $\endgroup$ – Alexei Boulbitch Jun 20 '13 at 11:38
  • 1
    $\begingroup$ Alexei, if you disagree that this question is a duplicate of the one linked by Leonid, please state why. $\endgroup$ – Simon Woods Jun 20 '13 at 13:17
  • $\begingroup$ just realised there is this discussion and the linked both works and is much faster. I'd agree this is a duplicate. $\endgroup$ – gpap Jun 20 '13 at 15:02
  • $\begingroup$ As recommended by Leonid and supported by others I have closed this as a duplicate, and as Simon requested if you disagree please state why. $\endgroup$ – Mr.Wizard Jun 20 '13 at 16:52

Like in the example you link, I'd use Nearest.

Your $ 1000 \times 1000 $ data points:

a = RandomReal[{0, 10}, {1000000, 2}];
b = RandomReal[{-1, 1}, {1000000}];
lst = MapThread[
     10 Exp[-((#1[[1]] - 5)^2 + (#[[2]] - 5)^2)/4] + #2] &, {a, b}];

and a $ 100 \times 100 $ grid

grid = Module[{xres, yres},
   xres = 100;
   yres = 100;
   Flatten[Table[N@{i, j}, {i, 0, 10, 10/xres}, {j, 0, 10, 10/yres}], 

now write a nearest function:

Timing[nf = Nearest[Thread[lst[[All, 1 ;; 2]] -> lst[[All, 3]]]];]
(*OUT=*) {1.88471, Null}

which you can run so that it averages over the $ n $ (here 1000) nearest points in the $ x-y $ plane

Timing[interpolatable = ({Sequence @@ #, Mean@nf[#, 1000]} & /@ 
(*OUT=*) {6.93931, Null}

Much like a moving average you may lose some information if you oversmooth and also the metric in the nearest function isn't binning in rectangles. In fact if you change it, it will become much slower.

The new set, however, is both smooth and regularly spaced and my guess is, if you are averaging over lots of points you won't see much difference:

ListPointPlot3D[interpolatable, PlotRange -> All]

smooth set


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