I'm trying to detect the monospaced-grid in a possibly slightly deformed image of a receipt. Example input (full size here)

enter image description here

My idea is to obtain the centroids or bounding boxes of all components and do a kind of Hough transform on these points, but I wouldn't know how to do this in Mathematica.

The output could have different formats, but should look something like this.

In this particular example, equi-spaced lines horizontally and vertically are enough, but It would be a pro if the algorithm can also handle a little perspective distortion.

  • $\begingroup$ You can do a Hough transform using the command Radon reference.wolfram.com/language/ref/Radon.html Look under the Details section and choose the Hough option. $\endgroup$ – bill s Nov 29 '14 at 16:22
  • $\begingroup$ @bill, thank you! I found Radon, but the input for Radon is an image. It would be preferable if I can just input a list of 2D-points. $\endgroup$ – Thijs Nov 29 '14 at 16:32
  • $\begingroup$ If you have a 2D list, then you can turn it into an image using Image[list], then apply Radon. $\endgroup$ – bill s Nov 29 '14 at 17:14
  • 3
    $\begingroup$ The input you mention in your question is an image. In the comment above you state the input is a list. IMHO you have to be a bit clearer about your intentions. $\endgroup$ – Sjoerd C. de Vries Nov 29 '14 at 17:31
  • $\begingroup$ @sjoerd, yeah you're right. Sorry I was confused. In my own approach, I use the centroids of the components as points. belisarius' answer is exactly what i'm looking for though. $\endgroup$ – Thijs Nov 29 '14 at 19:46

Simple enough, and fails only on one scarcely populated column. Easily fixed if you enforce the "monospaced" property

i = Binarize@Import@"http://i.stack.imgur.com/XWhHv.png";
(* delete noisy border*)
i1 = DeleteSmallComponents[ImageTake[i, {1, 1650}, {30, -1}], 10];

id = ImageDimensions@i1;
cm = ComponentMeasurements[i1, "Centroid"][[All, 2]];
bb = Abs@Mean[ Subtract @@@ ComponentMeasurements[i1, "BoundingBox"][[All, 2]]];

(* The following is basically a MeanShiftFilter[], but I couldn't make
   MeanShiftFilter[] to behave properly after a cursory try. Probably you should try it *)

yc = Mean /@ Split[Sort@cm[[All, 2]], Abs[#1 - #2] <= bb[[1]]/4 &];
xc = Mean /@ Split[Sort@cm[[All, 1]], Abs[#1 - #2] <= bb[[2]]/4 &];
a = 1.3;
Show[i1, Graphics[{{Red, {Line[{{1, # - a bb[[1]]}, {id[[1]], # - a bb[[1]]}}]} & /@ yc}, 
                  {Yellow, {Line[{{# - a bb[[2]], 1}, {# - a bb[[2]], id[[2]]}}]} & /@ xc}}]]

Mathematica graphics

| improve this answer | |
  • $\begingroup$ This is great, thanks! $\endgroup$ – Thijs Nov 29 '14 at 19:49
  • $\begingroup$ Belisarius, do you by any chance also have ideas on how to adapt this method to support slight perspective distortion? $\endgroup$ – Thijs Nov 29 '14 at 20:53
  • $\begingroup$ @Thijs Usually done fitting a FindGeometricTransform[]. Butyou need to provide a few sample images to be sure $\endgroup$ – Dr. belisarius Nov 29 '14 at 21:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.