Matlab:如何弯曲图像中的线条

问题描述

我手头有一张图片(png 格式).椭圆的边界线(代表核)过直，这是不切实际的.我怎样才能从图像中提取线条并使它们弯曲，前提是它们仍然包围着细胞核.

I have a image(png format) in hand. The lines that bound the ellipses (represent the nucleus) are over straight which are impractical. How could i extract the lines from the image and make them bent, and with the precondition that they still enclose the nucleus.

图片如下:

弯曲后

编辑:如何将 answer2 中的 Dilation And Filter 部分翻译成 Matlab 语言?我想不通.

EDIT: How can i translate the Dilation And Filter part in answer2 into Matlab language? I can't figure it out.

推荐答案

好的，这里有一个方法，需要几个随机化步骤来获得自然"的非对称外观.

Ok, here is a way involving several randomization steps needed to get a "natural" non symmetrical appearance.

我在 Mathematica 中发布了实际代码，以防万一有人愿意将其翻译成 Matlab.

I am posting the actual code in Mathematica, just in case someone cares translating it to Matlab.

(* A preparatory step: get your image and clean it*)
i = Import@"http://i.stack.imgur.com/YENhB.png";
i1 = Image@Replace[ImageData[i], {0., 0., 0.} -> {1, 1, 1}, {2}];
i2 = ImageSubtract[i1, i];
i3 = Inpaint[i, i2]

(*Now reduce to a skeleton to get a somewhat random starting point.
The actual algorithm for this dilation does not matter, as far as we
get a random area slightly larger than the original elipses *)
id = Dilation[SkeletonTransform[
             Dilation[SkeletonTransform@ColorNegate@Binarize@i3, 3]], 1]

(*Now the real random dilation loop*)
(*Init vars*)
p = Array[1 &, 70]; j = 1;
(*Store in w an image with a different color for each cluster, so we
can find edges between them*)
w = (w1 =
      WatershedComponents[
       GradientFilter[Binarize[id, .1], 1]]) /. {4 -> 0} // Colorize;
(*and loop ...*)
For[i = 1, i < 70, i++,
 (*Select edges in w and dilate them with a random 3x3 kernel*)
 ed = Dilation[EdgeDetect[w, 1], RandomInteger[{0, 1}, {3, 3}]];
 (*The following is the core*)
 p[[j++]] = w =
   ImageFilter[  (* We apply a filter to the edges*)
    (Switch[
          Length[#1],  (*Count the colors in a 3x3 neighborhood of each pixel*)
          0, {{{0, 0, 0}, 0}},          (*If no colors, return bkg*)
          1, #1,                        (*If one color, return it*)
          _, {{{0, 0, 0}, 0}}])[[1, 1]] (*If more than one color, return bkg*)&@
      Cases[Tally[Flatten[#1, 1]],
       Except[{{0.`, 0.`, 0.`}, _}]] & (*But Don't count bkg pixels*),
    w, 1,
    Masking -> ed,       (*apply only to edges*)
    Interleaving -> True (*apply to all color chanels at once*)]
 ]

结果是:

编辑

对于面向 Mathematica 的读者来说，最后一个循环的功能代码可能更容易(也更短):

For the Mathematica oriented reader, a functional code for the last loop could be easier (and shorter):

NestList[
 ImageFilter[
   If[Length[#1] ==  1, #1[[1, 1]], {0, 0, 0}] &@
     Cases[Tally[Flatten[#1, 1]], Except[{0.` {1, 1, 1}, _}]] & , #, 1,
   Masking      -> Dilation[EdgeDetect[#, 1], RandomInteger[{0, 1}, {3, 3}]],
   Interleaving -> True ] &,
 WatershedComponents@GradientFilter[Binarize[id,.1],1]/.{4-> 0}//Colorize,
5]

这篇关于Matlab:如何弯曲图像中的线条的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！