Skip to main content

image processing - How to create a 1D vector by selecting a straight line in a given angle of 2D array


Suppose I have an image created by the code provided below. I want to select a line (1D vector) (y−y0)=n(x−x0) from the output image (2D matrix) by specifying n and (x0,y0). Can anyone explain how to do this?


Code:


w1 = 600;
w2 = 600;
mask = Sum[
  RotateRight[
    Exp[I k] DiskMatrix[20, {w1, w2}], {k 100, k 200 }], {k, 2}] + 
    Exp[I 5] DiskMatrix[20, {w1, w2}] + 

    Sum[RotateLeft[Exp[I k] DiskMatrix[20, {w1, w2}], {k 200, k  }], {k, 1}]
<< Developer`
Image[RotateRight[
Re[Fourier[mask]], {w1/2, w2/2}]\[TensorProduct]ToPackedArray[{1.0, 0.3, 0.1}], Magnification -> 0.4]

Mathematica graphics



Answer



I'm sure you know that a line will almost never run through exact pixel positions. Therefore, you have two choices. First, you interpolate your image matrix and then you can sample as many points along the line as you like. In this case, I probably wouldn't recommend it because the values depend on the interpolation itself. Another, very easy way is to use an algorithm which is used draw an approximate line in a pixel grid.


Lucky for you we already had a post about his. Therefore, you can read in this answer how it is done and with the provided function, you will get the pixel coordinates along a line between points p0 and p1.


bresenham



bresenham[p0_, p1_] := 
 Module[{dx, dy, sx, sy, err, newp}, {dx, dy} = Abs[p1 - p0];
  {sx, sy} = Sign[p1 - p0];
  err = dx - dy;
  newp[{x_, y_}] := 
   With[{e2 = 2 err}, {If[e2 > -dy, err -= dy; x + sx, x], 
     If[e2 < dx, err += dx; y + sy, y]}];
  NestWhileList[newp, p0, # =!= p1 &, 1]]

and then you take your image and use ImageData to get the pixel matrix. The extraction of the diagonal image is as simple as



img = Import["http://i.stack.imgur.com/FQthc.png"];
Extract[ImageData[img], bresenham[{1, 1}, ImageDimensions[img]]]

Final note: pay attention that the image matrix is reversed to what you see in the image. I guess you can do the transformation of a line in two-point form and yours by yourself.


Comments

Post a Comment

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...
Post a Comment
Read more

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...
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