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) $ \left(y-y_{0}\right)=n\left(x-x_{0}\right) $ from the output image (2D matrix) by specifying $n$ and $\left(x_{0},y_{0}\right)$. 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 $p_0$ and $p_1$.


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

How to thread a list

I have data in format data = {{a1, a2}, {b1, b2}, {c1, c2}, {d1, d2}} Tableform: I want to thread it to : tdata = {{{a1, b1}, {a2, b2}}, {{a1, c1}, {a2, c2}}, {{a1, d1}, {a2, d2}}} Tableform: And I would like to do better then pseudofunction[n_] := Transpose[{data2[[1]], data2[[n]]}]; SetAttributes[pseudofunction, Listable]; Range[2, 4] // pseudofunction Here is my benchmark data, where data3 is normal sample of real data. data3 = Drop[ExcelWorkBook[[Column1 ;; Column4]], None, 1]; data2 = {a #, b #, c #, d #} & /@ Range[1, 10^5]; data = RandomReal[{0, 1}, {10^6, 4}]; Here is my benchmark code kptnw[list_] := Transpose[{Table[First@#, {Length@# - 1}], Rest@#}, {3, 1, 2}] &@list kptnw2[list_] := Transpose[{ConstantArray[First@#, Length@# - 1], Rest@#}, {3, 1, 2}] &@list OleksandrR[list_] := Flatten[Outer[List, List@First[list], Rest[list], 1], {{2}, {1, 4}}] paradox2[list_] := Partition[Riffle[list[[1]], #], 2] & /@ Drop[list, 1] RM[list_] := FoldList[Transpose[{First@li...
Post a Comment
Read more

front end - keyboard shortcut to invoke Insert new matrix

I frequently need to type in some matrices, and the menu command Insert > Table/Matrix > New... allows matrices with lines drawn between columns and rows, which is very helpful. I would like to make a keyboard shortcut for it, but cannot find the relevant frontend token command (4209405) for it. Since the FullForm[] and InputForm[] of matrices with lines drawn between rows and columns is the same as those without lines, it's hard to do this via 3rd party system-wide text expanders (e.g. autohotkey or atext on mac). How does one assign a keyboard shortcut for the menu item Insert > Table/Matrix > New... , preferably using only mathematica? Thanks! Answer In the MenuSetup.tr (for linux located in the $InstallationDirectory/SystemFiles/FrontEnd/TextResources/X/ directory), I changed the line MenuItem["&New...", "CreateGridBoxDialog"] to read MenuItem["&New...", "CreateGridBoxDialog", MenuKey["m", Modifiers-...
Post a Comment
Read more