Skip to main content

Detecting and correcting gradational brightness variations in a grayscale image


I am trying to quantify the proportions of various components in a greyscale image (backscattered electron image of a polished rock sample). Here is the original image:


bse = Import@"http://i.stack.imgur.com/eQece.png";


enter image description here


The image is comprised of components (minerals) that should have the same intensity response. For example, the abundant 'mid-grey' component (plagioclase) should have a fixed grey-level across the entire image. It doesn't.


By masking the other components of the image and adjusting the contrast, the gradations in brightness become more obvious:


bsek = ImageSubtract[bse, ColorNegate@Binarize[bse, {0.16, 0.24}]];
ImageAdjust[bsek, {3, 1.1}]

enter image description here


There seems to be two effects: a first order decrease in brightness towards the bottom corners and a second-order periodic vertical striping (comprised of short wavelength gradients).


Why bother? - Correcting for variations in brightness is an important step prior to segmentation analysis of the image



By approximating the variation in grey-values for the plagioclase, the difference in birghtness between the top and bottom of the image can be modelled (albeit crudely).


ksim[im_] := Module[{kdf, kdl},
  kdf = SmoothKernelDistribution[First@im];
  kdl = SmoothKernelDistribution[Last@im];
  Plot[{PDF[kdf, x], PDF[kdl, x]}, {x, 0, 0.3}, Filling -> Axis, 
   Exclusions -> None]]
ksim[ImageData[bsek]]

enter image description here


The following creates a gradational filter that causes the grey-level peaks to overlap (by adjusting the value of shift).



bsedim = ImageDimensions[bse]
shift = 0.2;
grad = 1 + 
   Transpose@
    Array[Table[x, {x, -shift, 0., shift/bsedim[[1]]}] &, bsedim[[1]]];
gradi = Image[grad];

bsekm = ImageMultiply[bsek, gradi];
ksim[ImageData[bsekm]]


enter image description here


This filter can then be applied to the original image:


bsem = ImageMultiply[bse, gradi]

enter image description here


This clumsy method does an OK job of levelling the first-order brightness defect. However, this effect looks to have a spherical gradient (dimmer in the bottom corners than the middle) which is not accounted for.


As yet, I have not found a way to correct the second order periodic effect.


Can anyone suggest a more straightforward way to detect and correct these kinds of brightness variation defects? Extra bonus gratitude to anyone who can help to correct the second-order periodic aberration.



Update



Here is a contrast stretched image showing the vertical aberrations:


 ImageAdjust[bse, {4.5, 3.85}]

enter image description here



Answer



Does this help? I've treated the error as additive, though you could easily change it to a multiplicative correction.


Starting with your bsek image, find the mean offset for rows and for columns:


id = ImageData[bsek];
horiz = (# - Mean[#]) &[Total[id]/Total[Unitize[id]]];
vert = (# - Mean[#]) &[Total[id, {2}]/Total[Unitize[id], {2}]];


ListLinePlot[horiz]
ListLinePlot[vert]

enter image description here enter image description here


The horizontal plot shows an overall darkening at the left and right sides of the image, plus the periodic component. The vertical plot shows an overall darkening towards the bottom of the image, with a periodic component also.


Combining the horizontal and vertical offsets into an overall offset:


a = Outer[Plus, vert, horiz];
Image[a] // ImageAdjust


enter image description here


Now subtract the offset from the original data:


bse2 = Image[ImageData[bse] - a]

enter image description here


Using s.s.o.'s ClusteringComponents visualisation appears to show an improvement:


ClusteringComponents[bse, 10] // Colorize
ClusteringComponents[bse2, 10] // Colorize

enter image description here enter image description here



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