Skip to main content

performance tuning - Fast Spherical Linear Interpolation of list of quaternions


An accurate way to interpolate between two quaternions is to use Spherical Linear Interpolation (Slerp) because it preserves the unit length, whereas straightforward linear interpolation does not, as shown by this example:


Clear[slerp];
slerp[q1_, q2_, f_] := Module[{omega},

  omega = ArcCos[Dot[Normalize@q1, Normalize@q2]]];
  If[PossibleZeroQ[omega], 
    q1,
    (Sin[(1 - f) omega] q1 + Sin[f omega] q2)/Sin[omega]
  ]
]

q1 = {1,0,0,0};
q2 = {0,1,0,0};
q3 = {0,0,1,0};


(* Linear interpolation using Interpolation *)
qint = Interpolation[{{0,q1},{1,q2}},InterpolationOrder->1];

Plot[{Norm@qint[t], Norm@slerp[{1, 0, 0, 0}, {0, 1, 0, 0}, t]}, {t, 0, 1}, 
  AxesLabel -> {"time", "length"}, PlotRange -> {0.7, 1.01}, 
  PlotStyle -> {Automatic, Dashed}, 
  Epilog -> {Text["Linear Interpolation", {0.5, 0.75}], 
  Text["Spherical Linear Interpolation", {0.5, 0.98}]}
]


Mathematica graphics


What's nice thing about Interpolation, however, is that it is easy to give it a list of vectors to get an interpolating function valid over the specified range, and it works fast:


Interpolation[{{0,q1},{1,q2},{2,q3}},InterpolationOrder->1]


InterpolatingFunction[{{0,2}},<>]



How can I create something like an InterpolatingFunction that preserves the unit-length property?


This is what I came up with, but it seems kludgy and it is very slow (this code does not check input bounds):



Clear[slerpInterpolation]; 
slerpInterpolation[q_List] := Function[{t}, 
  Module[{times, quaternions, pos, u, dt},
    times = q[[All, 1]];
    quaternions = q[[All, 2]];
    pos = Last@Flatten@Position[t - times, x_ /; x >= 0];
    dt = times[[pos + 1]] - times[[pos]];
    u = (t - times[[pos]])/dt;
    slerp[quaternions[[pos]], quaternions[[pos + 1]], u]
  ]

]

It is too slow in practice, giving about 10 evaluations per second on my laptop. Compare that with regular interpolation:


qlist = Transpose[{Range[100000] - 1, Normalize /@ RandomReal[{-1, 1}, {100000, 4}]}];
sint = slerpInterpolation[qlist];
lint = Interpolation[qlist, InterpolationOrder -> 1];

AbsoluteTiming[Do[sint[541.236], {10}]][[1]]/10



0.0967707



AbsoluteTiming[Do[lint[541.236], {10000}]][[1]]/10000


7.2522*10^-6




Answer



Here's a somewhat complete implementation of Shoemake's spherical linear interpolation that functions completely analogously to Interpolation[] and InterpolatingFunction[]. As already noted, much of the slowness is due to your use of a sequential search. In any event, if you prefer, you could use the built-in interpolation as suggested in the other answer, but I think using a bisection routine is a bit more instructive.


Anyway...



SphericalLinearInterpolation::inddp = "The point `1` is duplicated.";

SphericalLinearInterpolation[data_] := 
 Module[{dtr = Transpose[SortBy[data, Composition[N, First]]], diffs, times},
        SphericalInterpolatingFunction[data[[{1, -1}, 1]], dtr] /; 
        If[MemberQ[diffs = Chop[Differences[times = First[dtr]]], 0], 
           Message[SphericalLinearInterpolation::inddp, 
                   First[Extract[times, Position[diffs, 0]]]]; False, True]]

SphericalInterpolatingFunction::dmval = 

  "Input value `1` lies outside the domain of the interpolating function.";

MakeBoxes[SphericalInterpolatingFunction[range_, rest__], _] ^:= 
 InterpretationBox[RowBox[
    {"SphericalInterpolatingFunction", "[", "{", #1, ",", #2, "}", ",", "\"<>\"", "]"}], 
    SphericalInterpolatingFunction[range, rest]] & @@ Map[ToBoxes, range]

SphericalInterpolatingFunction[stuff__][l_List] :=
 SphericalInterpolatingFunction[stuff] /@ l


SphericalInterpolatingFunction[range_, __]["Domain"] := range

SphericalInterpolatingFunction[{r_, s_}, data_][t_?NumericQ] :=
  (Message[SphericalInterpolatingFunction::dmval, t]; $Failed) /; ! (r <= t <= s)

slerp = Compile[{{q1, _Real, 1}, {q2, _Real, 1}, {f, _Real}}, 
   Module[{n1 = Norm[q1], n2 = Norm[q2], omega, so},
    (* vector angle formula by Velvel Kahan *)
    omega = 2 ArcTan[Norm[q1 n2 + n1 q2], Norm[q1 n2 - n1 q2]];
    If[Chop[so = Sin[omega]] == 0, q1, Sin[{1 - f, f} omega].{q1, q2}/so]]];    


SphericalInterpolatingFunction[range_, data_][t_?NumericQ] := Module[{times, quats, k},
        {times, quats} = data;
        k = GeometricFunctions`BinarySearch[times, t];
        slerp[quats[[k]], quats[[k + 1]], Rescale[t, times[[{k, k + 1}]], {0, 1}]]]


See J. Blow's article on why spherical linear interpolation might not always be the best method for interpolating quaternions.


Comments

Post a Comment

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]
Post a Comment
Read more

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]
Post a Comment
Read more

plotting - Mathematica: 3D plot based on combined 2D graphs

I have several sigmoidal fits to 3 different datasets, with mean fit predictions plus the 95% confidence limits (not symmetrical around the mean) and the actual data. I would now like to show these different 2D plots projected in 3D as in but then using proper perspective. In the link here they give some solutions to combine the plots using isometric perspective, but I would like to use proper 3 point perspective. Any thoughts? Also any way to show the mean points per time point for each series plus or minus the standard error on the mean would be cool too, either using points+vertical bars, or using spheres plus tubes. Below are some test data and the fit function I am using. Note that I am working on a logit(proportion) scale and that the final vertical scale is Log10(percentage). (* some test data *) data = Table[Null, {i, 4}]; data[[1]] = {{1, -5.8}, {2, -5.4}, {3, -0.8}, {4, -0.2}, {5, 4.6}, {1, -6.4}, {2, -5.6}, {3, -0.7}, {4, 0.04}, {5, 1.0}, {1, -6.8}, {2, -4.7}, {3, -1.
Post a Comment
Read more