sergiomsilva / alpr-unconstrained

License Plate Detection and Recognition in Unconstrained Scenarios
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Description of data inside "*_lp.txt" #6

Closed amit2319 closed 5 years ago

amit2319 commented 6 years ago

licence-plate-detection.py returns *_lp.png* & _lp.txt**, can you describe data inside *_lp.txt.

sample data: 4,0.864626,0.946707,0.937679,0.855598,0.571571,0.515678,0.573691,0.629584,,

inamesion commented 5 years ago

4 is class 0.864626,0.946707,0.937679,0.855598,0.571571,0.515678,0.573691,0.629584, is the axis of four point of license plate after Normalizition

amit2319 commented 5 years ago

I want to map these 4 points on the input image of licence-plate-detection.py. Any help, I have tried x*width(img) and y*height(img) but points doesn't match the actual license plate.

sergiomsilva commented 5 years ago

@amit2319 these points are related to the cropped car image, not to the original image. You need to multiply by the width and height of the car bounding-box, and sum the top-left offset to retrieve the points in the original image space.

amit2319 commented 5 years ago

@sergiomsilva actually I was trying to retrive license plate from car_bounding_box only, but x.width(img) and y.height(img) where img refers to image of car_bounding-box doesn't reproduce the results.

In following data: 4,0.864626,0.946707,0.937679,0.855598,0.571571,0.515678,0.573691,0.629584,, (x1,y1) = (0.864626,0.946707) (x2,y2) = (0.937679,0.855598) (x3,y3) = (0.571571,0.515678) (x4,y4) = (0.573691,0.629584)

Am I interpreting it right?

sergiomsilva commented 5 years ago

Ah ok... now I understood. Its not the correct intepretation, considering your notation, the correct should be:

(x1,y1) = (0.864626, 0.571571) (x2,y2) = (0.946707, 0.515678) (x3,y3) = (0.937679, 0.573691) (x4,y4) = (0.855598, 0.629584)

In summary, the first 4 values are the x's, and the next 4, the y's.

amit2319 commented 5 years ago

@sergiomsilva Thanks! it worked.