09-26-2024, 03:16 PM
(09-22-2024, 09:32 AM)Ofnuts Wrote:(09-22-2024, 08:51 AM)tomatoSauce23 Wrote:(09-22-2024, 08:13 AM)Ofnuts Wrote: Ok, so the question you are asking is a typical one for computer vision, so maybe you should learn Python (which is very easy if you already have programming experience) and check the openCV library.
With Gimp, I would try this:
Now out to hide the pool in my backyard.
- The color selection can use a RGB color but also tint, hue, saturation, lightness, or any of the RGB channels, all with a threshold.
- From a selection you can obtain a path. This will rarely be exactly 4 points in a square, even if the shape is rectangular
- From the path anchors you can determine corners (extremums on X/Y)
- From these 4 corners you can create a selection a bucket-fill a layer
- From the initial path you can obtain a selection and bucket-fill another layer
- You put the top layer in Difference mode and merge the layer. The resulting layer is a map of the differences
- With the histogram function you can have the pixel count of the difference and see how big that is relative to the square
Thank you. So, I hadn't ever heard of computer vision before, and a quick look suggests to me that the openCV library is a library of programmes written in the common programming languages for various tasks that people have use for. So, am I right to think that you are advising that I go there and see if there is already a programme available to do the task that I'm asking about? and if so, then I guess that all would happen outside of Gimp and inside whichever programming language I choose (Java, for example). So I'd find a programme in the openCV library that would find those squares in my images and crop those images to those squares.
As for those Gimp instructions, I reckon I could follow them manually... and then are you saying that it would then be possible to write a Python script that does all of those steps as a plugin?
Yes to both.
Thanks. As it happens I just wrote a Java programme to do it for me in the end which was relatively straight-forward to do.