Skip to main content

image processing - Measure perimeter of black edge in Mathematica


I have a pattern like the one below. How could I measure the total length of the perimeter of the black/white edge? This would give me a measure of spatial frequency. Any tips would be great!


Thanks!





Answer



Edit


You should be aware that there are 2 basic ways of counting perimeter pixels: 4-ways neighbors or 8-ways neighbors. The difference is about whether counting or not the yellow pixels in this figure:


Mathematica graphics


If you want to count them, then almost manually:


i = Import["http://i.stack.imgur.com/7jt0I.gif"]; 
mask = Complement[Partition[#, 3] & /@ ({0, 0, 0, 0, 1, 0, 0, 0, 0} - 
       RotateLeft[{1, 0, 0, 0, 0, 0, 0, 0, 0}, #] & /@ Range[9]), {Array[0 &, {3, 3}]}];
ip = ImagePad[i, 10, RGBColor[0.4980 {1, 1, 1}]];
is = Binarize@

  ImageSubtract[ HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .1], 
   ImageSubtract[HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .1], 
                 HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .8]]]

ImageMeasurements[is, "Total"]



14715




If you don't want to count the diagonal neighbors, then we change the mask to consider only four neighbors per pixel:


i = Import["http://i.stack.imgur.com/7jt0I.gif"];
mask = {{{0, -1, 0}, {0, 1,  0}, {0, 0, 0}}, {{0, 0, 0}, {-1, 1, 0}, {0, 0, 0}},
        {{0,  0, 0}, {0, 1, -1}, {0, 0, 0}}, {{0, 0, 0}, { 0, 1, 0}, {0,-1, 0}}};

ip = ImagePad[i, 10, RGBColor[0.4980 {1, 1, 1}]];
is = Binarize@
  ImageSubtract[ HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .1], 
   ImageSubtract[HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .1], 
                 HitMissTransform[ColorConvert[ip, "Grayscale"], mask, .8]]];


ImageMeasurements[is, "Total"]


10411



Comments

Post a Comment

Popular posts from this blog

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

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

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more