Skip to main content

correlation - How to align images?


I have two grayscale images (img1, img2) which contain objects seen with two cameras. Some objects are the same (not same shape and intensity) and are seen in both images. Some objects are only seen in img1 or img2.


I would like to align the images in such a way that the objects seen in both images are overlapping.


How can I determine the vertical and horizontal shift between the two images?


Here are the images:



img1:


enter image description here


img2:


enter image description here


What I tried:


pts = ImageCorrespondingPoints[img1, img2, KeypointStrength -> 0.0002]

{{{34.6035, 72.9785}}, {{48.1733, 82.9132}}}

xshift = Mean[pts[[All, All, 1]][[2]] - pts[[All, All, 1]][[1]]]


13.5698

yshift = Mean[pts[[All, All, 2]][[2]] - pts[[All, All, 2]][[1]]]

9.93468

This seems to be correct. When I look only at the vertically elongated object in the center of img2 then I find manually roughly: xshift=10, yshift=13.


What confuses me:


The found points pts do not correspond to img1 or img2:



HighlightImage[img1, pts]

enter image description here


HighlightImage[img2, pts]

enter image description here


Where is the error in HighlightImage?


Can ImageCorrelate or ImageAlign be used to find the shift or do you have another solution?



Answer



My solution is the following:



Determine the shift between both images:


{merit, trans} = 
FindGeometricTransform[img2, img1,
TransformationClass -> "Translation"];


enter image description here



Applying the shifts


imgt = ImageTransformation[img2, trans, DataRange -> Full]


Combine images


Blend[{ColorNegate[img1], imgt}, {0.8, 0.2}]


enter image description here



Comments

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}]

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.