Skip to main content

Object detection in a low contrast image


I have a small dark object in the upper left corner of an image.


How can I separate it from the noisy rest and determine its IntensityCentroid?


enter image description here



The problem is that the objects intensity is not significantly lower than the intensity in the lower part of the image:


ListPlot3D[ImageData[image] // N]


enter image description here



UPDATE:


Following the question Finding objects in images with inhomogeneous background I tried:


imageN = ColorNegate[image];
background = ImageConvolve[imageN, GaussianMatrix[7]];

subImage = ImageSubtract[imageN, background];
t = FindThreshold[subImage, Method -> "Entropy"];
binImg = DeleteSmallComponents[Binarize[subImage, t], 5];
pts = ComponentMeasurements[ImageMultiply[image, binImg], 
  "IntensityCentroid"]
Show[image, Graphics[{Red, Point[pts[[All, 2]]]}]]

which gives:



{1 -> {77.2279, 397.997}}


enter image description here



Unfortunately the other solutions for the mentioned question cannot detect the object here.


As Rahul mentioned also background = MedianFilter[imagen, 10] can be used to detect the object. But this is much slower:


background = ImageConvolve[imageN, GaussianMatrix[7]]; // RepeatedTiming


{0.005, Null}


background = MedianFilter[imageN, 20]; // RepeatedTiming


{22.8, Null}


Answer



There is a way to simplify all these process:


img=--your image here--;

ComponentMeasurements[

 DeleteSmallComponents@
  ColorNegate@
   LocalAdaptiveBinarize[img, 50, {1, -2, -.01}], {"Centroid", 
  "EquivalentDiskRadius"}]


{1 -> {{77.0441, 397.853}, 4.65243}}



The first is the Centroid of your region, and the second is equivalent disk radius.


The basic idea is to use LocalAdaptiveBinarize. Firstly, I checked the image's size and have a approximate point size, about 10 pixels, so we set the local adaptive range to 50. Then, we want to eliminate the average, so the first argument in {1,-2,-.01} should be 1, representing the elimination of 1*average. then we want to neglect those noises, and from your ListPlot3D, we can easily see that there's almost no noise over 2*sigma, so a proper value for the second part is 2, then the third represents a slight shift in mean, so -0.01 should be proper.



Check the result, it's beautiful!


Then use ComponentMeasurements and everything will be fine.




Note that you need to know the "IntensityCentroid", I made some slight modification to let it show you exactly the "IntensityCentroid" as in previous answer, Binarize will mop out any intensity information.


TakeLargestBy[
 ComponentMeasurements[
  ImageMultiply[
   Dilation[
    DeleteSmallComponents@

     ColorNegate@LocalAdaptiveBinarize[img, 50, {1, -2, -.01}], 
    DiskMatrix[10]], 
   ImageAdjust@
    ImageSubtract[Blur[img, 50], img]], {"IntensityCentroid", 
   "EquivalentDiskRadius"}], #[[2, 2]] &, 1]

The basic idea is to use Blur and ImageSubtract to get out the background information then subtract it. Then we can use the method in the previous part to narrow down the selection. Finally determin the IntensityCentroid in this way.



{1 -> {{76.4188, 397.894}, 10.66}}




A sight difference, but this time the first part: {76.4188, 397.894} tells you the intensity centriod instead of morphological centroid. also, the radius is more accurate~~~ :)


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