interpolation - Interpolating 2D data with missing values

I have a list (21 x 21) containing values. I want to eliminate slots containing zeros by interpolating the nearest values and overwriting the zeros. How do I use the Mathematica's ListInterpolation function to do that?

Answer

You have to use Interpolation because there it is possible to give values which do not lie on a structured grid. Let's create some test data which has zeroes in it

data = Table[ Sin[x] + Cos[y], {x, 0, Pi/2, Pi/2/29.}, {y, 0, Pi/2, Pi/2/29.}]*
   RandomInteger[{0, 1}, {30, 30}];
ListPlot3D[data]

Mathematica graphics

Now you attach the position of every value to it so you get tuples of the form {{x,y},value} and Select only those, where value is not zero. This can then be interpolated

data2 = Select[Flatten[MapIndexed[{#2, #1} &, data, {2}], 1], 
   Last[#] != 0 &];
ip = Interpolation[data2];
Plot3D[ip[y, x], {x, 1, 30}, {y, 1, 30}]

Mathematica graphics

Note, that with such data you always have an InterpolationOrder of 1.

Update regarding your comment

Yes, this is indeed a problem and you have to decide what to do. First you have to understand the issue. Let's assume the following grid with values, where all white cells are zero and have to be interpolated

Mathematica graphics

When you look at cell {3,1}, you see that such a situation is not a problem, because the value can be interpolated since it is place in between the cells {1,2} and {1,4}.

On the other hand, the cell {5,5} is a problem, because no surrounding cells with values are left which can be used for interpolation. Therefore, what you have to ensure is, that you have values on all 4 corners of your matrix. When they got selected out because they were zero, then you have to fill them with values. When they are indeed zero, and got kicked out accidentally, then you have to fill them back in.

You can ignore the corners when you Select non-zero values by changing the condition in the end

{ny, nx} = Dimensions[data];
data2 = Select[Flatten[MapIndexed[{N@#2, #1} &, data, {2}], 1], 
   Last[#] != 0 && 
   (First[#] =!= {1, 1} || First[#] =!= {1, nx} || 
    First[#] =!= {ny, 1} || First[#] =!= {ny, nx}) &];

Comments

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}]

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}]

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.

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