Hi Max,

In the end, I want to convert the line to a vector object, ...

I am not sure about your intention, but maybe this helps:

I start with an object like this (coming from the new and useful function Canvas):

Then I basically use ComponentMeasurements to calculate the data needed for a vector representation:

img = Binarize[Image @@ obj];
cm = ComponentMeasurements[ColorNegate[img], {"Centroid", "Orientation"}];
arrows = With[{v = {Cos[#2], Sin[#2]}, l = 25}, {#1 + l v, #1 - l v}] & @@@ cm[[All, 2]];
Show[img, Graphics[{Red, Arrow /@ arrows}]]

Liebe Gruesse ins schoene Freiburg! -- Henrik

Hello Max,

ahh, I see what you meant! Now - just as a footnote - here is another attempt (using your example array arr from above):

img = Image[arr, ImageSize -> 400];
pts = ImageValuePositions[img, 1];
endPts = ImageCorners[img, MaxFeatures -> 2];
{start, end} = Flatten[Position[pts, #] & /@ endPts];
order = Last@FindShortestTour[pts, start, end];
GraphicsColumn[{Show[img, Graphics[{Red, Thick, Line[pts[[order]]]}]],
   Show[img, Graphics[{Red, Thick, BSplineCurve[pts[[order]]]}]]}]

Maybe this is in a way helpful, regards -- Henrik

Hello Henrik,

the ImageGraphics function is generating the outline, but I want to trace the line itself from one end to the other. And although FindShortestTour is elegant, it is quite slow. I have tens of images with around 1000-2000 lines each. So ideally it should vectorize one line within tens of milliseconds. Less than a second is still acceptable.

In the end, the goal is to have a distribution of angles present in all the lines and see if there is a preference for a certain angle; so it is essential to smoothen the lines a bit to not only get 0°; 45° and 90°.

Max

Good evening Henrik, your code looks quite good and is very fast. I think the two prominent directions are artefacts though. If I rotate the image by a random value, the peaks remain where they are. Probably the gradient function treats certain angles with a preference. This is also true for my algorithm but not to that extent. If I play with the cutoff for the refinement of the subdivision (the value 0.5 in the function "splt"), the algorithm also allows bigger deviations from the exact line, therefore also more variations in the angle of the line. I assume it is indeed a difficult issue to get a true representation of the real angular distribution, i.e. the ground truth of the underlying data, which is what we try to obtain.

We use a SEM to look at structures on a cell surface and want to test the hypothesis that they are aligned with the orientation of the tissue. First, I obtained the outline of the structures with U-Net, then I determined their center line with SkeletonTransform and Thinning. Actually I did not really phrase it correctly previously; it is not so much about finding a preferential direction but more about determining if there is an alignment with the orientation of the tissue. For this image, the anterior/posterior axis of the sample is at 18 deg, counted from the horizontal direction counterclockwise. We would assume a preference of the structures to align either parallel or perpendicular to the axis.

Good night,

Max