Skip to main content

Image correlation


I have two images from motor scan, Image1 is shifted by MxN pixels from Image2. How can I find the shift between Image1 and Image2 ?


Image -1 1 Image -2 2


If we compare the shift between the images in terms of {X,Y} I will be able to tell how many steps of the motor scan corresponds to the pixel shift on the image array sensor.




Answer



There are many different image registration algorithms, all with different advantages and disadvantages. But I usually try the simplest choice first: I simply pass the two images to ImageCorrelate.


img1 = Import["http://i.stack.imgur.com/laZeT.png"];
img2 = Import["http://i.stack.imgur.com/LAHyF.png"];
corr = ImageAdjust[ImageCorrelate[img1, img2, EuclideanDistance]]

enter image description here


Intuitively, ImageCorrelate shifts img2 by every possible offset, calculates the difference with img1 with that offset, and stores the difference in the output image. So the pixel location with the lowest value in this image corresponds to the offset that minimizes the difference:


minOffset = PixelValuePositions[corr, Min[ImageData[corr]]][[1]] - 
ImageDimensions[img1]/2;

FlipView[{ImageTransformation[img1, TranslationTransform[minOffset ],
PlotRange -> Full], img2}]

enter image description here


If you look at the result of ImageCorrelate, you can see that there's one global minimum. That's the offset where the "U" shapes in the two images are aligned. There's a second local minimum left of it; That's probably where the left "leg" of one U is aligned with the right leg of the other. The shape of the correlation function around the global minimum and the relative correlation values at the local minima give a qualitative impression of the accuracy and reliability of the alignment.


EuclideanDistance compares brightness values. So the different overall brightness in the two images can "move" the minimum. If you don't want that, you can apply a filter to the images that basically removes the low-frequency background lighting changes:


filter = LaplacianGaussianFilter[#, 25] &;
corr = ImageAdjust[
ImageCorrelate[filter[img1], filter[img2], CosineDistance]]


enter image description here


(where 25 is roughly the width of the U-shape. Details smaller than that and low-frequency changes larger than that are removed by the LoG filter.)


Comments

Popular posts from this blog

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...

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}]