image processing - Measure perimeter of black edge in Mathematica

I have a pattern like the one below. How could I measure the total length of the perimeter of the black/white edge? This would give me a measure of spatial frequency. Any tips would be great!

Thanks!

Answer

Edit

You should be aware that there are 2 basic ways of counting perimeter pixels: 4-ways neighbors or 8-ways neighbors. The difference is about whether counting or not the yellow pixels in this figure:

Mathematica graphics

If you want to count them, then almost manually:

i = Import["http://i.stack.imgur.com/7jt0I.gif"]; 
mask = Complement[Partition[#, 3] & /@ ({0, 0, 0, 0, 1, 0, 0, 0, 0} - 
       RotateLeft[{1, 0, 0, 0, 0, 0, 0, 0, 0}, #] & /@ Range[9]), {Array[0 &, {3, 3}]}];
ip = ImagePad[i, 10, RGBColor[0.4980 {1, 1, 1}]];
is = Binarize@

  ImageSubtract[ HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .1], 
   ImageSubtract[HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .1], 
                 HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .8]]]

ImageMeasurements[is, "Total"]

14715

If you don't want to count the diagonal neighbors, then we change the mask to consider only four neighbors per pixel:

i = Import["http://i.stack.imgur.com/7jt0I.gif"];
mask = {{{0, -1, 0}, {0, 1,  0}, {0, 0, 0}}, {{0, 0, 0}, {-1, 1, 0}, {0, 0, 0}},
        {{0,  0, 0}, {0, 1, -1}, {0, 0, 0}}, {{0, 0, 0}, { 0, 1, 0}, {0,-1, 0}}};

ip = ImagePad[i, 10, RGBColor[0.4980 {1, 1, 1}]];
is = Binarize@
  ImageSubtract[ HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .1], 
   ImageSubtract[HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .1], 
                 HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .8]]];


ImageMeasurements[is, "Total"]

10411

Comments

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 - 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 - 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....

Blog

Search This Blog