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I'm aware of the solution to a similar problem by Hayashi Yoshiaki. Is there a way of extending the solution to break out the chromosomes from this example I have a mass of data in this format that needs to be converted to a training set. Note the there are holes in the chromosome that the human ignores when segmenting. Doing this manually for hundreds of images is a challenge. Thanks in advance

Example subset of images of chromosome

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  • $\begingroup$ Please give a link to Yoshiaki's solution. $\endgroup$
    – C. E.
    Jul 31, 2017 at 17:30
  • $\begingroup$ mathematica.stackexchange.com/questions/109154/… $\endgroup$
    – Boris
    Jul 31, 2017 at 17:47
  • $\begingroup$ Are all your images aligned and roughly equally spaced like this one? $\endgroup$
    – yohbs
    Jul 31, 2017 at 18:55
  • $\begingroup$ Yes, thats right. They are autogenerated. some are long and bent but all are essentially equidistant. $\endgroup$
    – Boris
    Jul 31, 2017 at 19:00
  • $\begingroup$ This image looks pretty easy to divide, to the extent that a very straightforward algorithm can be written (basically - detecting columns that are almost totally white). It would be helpful if you could provide a few more samples so that we can check the robustness. $\endgroup$
    – yohbs
    Jul 31, 2017 at 19:14

1 Answer 1

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Here's the very straightforward method I suggested in my comment:

tot = Total@ImageData[Binarize[im, 0.9]];
columnsSequence = Flatten@Position[tot, x_ /; x >= 103];
columns = Round@(Mean /@ Split[columnsSequence, #2 - #1 == 1 &]);
imd = ImageData[im];
imd[[All, columns, All]] = 0;
Image[imd]

This is just drawing the lines that separate the chromosomes. From here you can split the image or do any other further manipulation. For calculating the best separating "white column" I used this trick.

enter image description here

Here's how to divide the image to subimages:

imd=ImageData[ims]
images = Table[
  imd[[All, columns[[i]] ;; columns[[i + 1]], All]], {i, 
   Length[columns] - 1}]
Image /@ images

enter image description here

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  • $\begingroup$ OK, its a good start - how do I generate a list of images from this? I'll post a page of data from one example in the morning. $\endgroup$
    – Boris
    Jul 31, 2017 at 22:31
  • $\begingroup$ Hi yohbs, Ive tested your routine on other sets and find that I need to adjust the position parameter but thats OK I can automate that. Im having trouble breaking out the individual components due to my limited familiarity with array handling. I shall persevere. $\endgroup$
    – Boris
    Aug 1, 2017 at 10:28
  • $\begingroup$ this seems to work- not elegant Im afraid: $\endgroup$
    – Boris
    Aug 1, 2017 at 13:59
  • $\begingroup$ pt1 = Drop[Partition[Riffle[columns, 0], 2], -1]; pt2 = Drop[Partition[Riffle[columns, 116], 2], 1]; components = Partition[Partition[Flatten[Transpose[{pt1, pt2}]], 2], 2]; images = ImageTrim[im, #] & /@ components $\endgroup$
    – Boris
    Aug 1, 2017 at 14:00
  • $\begingroup$ @Boris You can do it more cleanly. I'll edit my answer. $\endgroup$
    – yohbs
    Aug 1, 2017 at 14:04

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