In Processing, how can I save part of the window as an image? - image

Question

I am using Processing under Fedora 20, and I want to display an image of the extending tracks of objects moving across part of the screen, with each object displayed at its current position at the end of the track. To avoid having to record all the co-ordinates of the tracks, I usesave("image.png"); to save the tracks so far, then draw the objects. In the next frame I use img = loadImage("image.png"); to restore the tracks made so far, without the objects, which would still be in their previous positions.. I extend the tracks to their new positions, then usesave("image.png"); to save the extended tracks, still without the objects, ready for the next loop round. Then I draw the objects in their new positions at the end of their extended tracks. In this way successive loops show the objects advancing, with their previous positions as tracks behind them.
This has worked well in tests where the image is the whole frame, but now I need to put that display in a corner of the whole frame, and leave the rest unchanged. I expect that createImage(...) will be the answer, but I cannot find any details of how to to so.
A similar question asked here has this recommendation: "The PImage class contains a save() function that exports to file. The API should be your first stop for questions like this." Of course I've looked at that API, but I don't think it helps here, unless I have to create the image to save pixel by pixel, in which case I would expect it to slow things down a lot.
So my question is: in Processing can I save and restore just part of the frame as an image, without affecting the rest of the frame?

Answer 1

I have continued to research this. It seems strange to me that I can find oodles of sketch references, tutorials, and examples, that save and load the entire frame, but no easy way of saving and restoring just part of the frame as an image. I could probably do it using Pimage but that appears to require an awful lot of image. in front of everything to be drawn there.
I have got round it with a kludge: I created a mask image (see this Processing reference) the size of the whole frame. The mask is defined as grey areas representing opacity, so that white, zero opacity (0), is transparent and black, fully opaque (255) completely conceals the background image, thus:
{ size (1280,800);
background(0); // whole frame is transparent..
fill(255); // ..and..
rect(680,0,600,600); // ..smaller image area is now opaque
save("[path to sketch]/mask01.jpg");
}
void draw(){}
Then in my main code I use:
PImage img, mimg;
img = loadImage("image4.png"); // The image I want to see ..
// .. including the rest of the frame which would obscure previous work
mimg = loadImage("mask01.jpg"); // create the mask
//apply the mask, allowing previous work to show though
img.mask(mimg);
// display the masked image
image(img, 0, 0);
I will accept this as an answer if no better suggestion is made.

Answer 2

void setup(){
size(640, 480);
background(0);
noStroke();
fill(255);
rect(40, 150, 200, 100);
}
void draw(){
}
void mousePressed(){
PImage img =get(40, 150, 200, 100);
img.save("test.jpg");
}

Answer 3

Old news, but here's an answer: you can use the pixel array and math.
Let's say that this is your viewport:
You can use loadPixels(); to fill the pixels[] array with the current content of the viewport, then fish the pixels you want from this array.
In the given example, here's a way to filter the unwanted pixels:
void exportImage() {
// creating the image to the "desired size"
PImage img = createImage(600, 900, RGB);
loadPixels();
int index = 0;
for(int i=0; i<pixels.length; i++) {
// filtering the unwanted first 200 pixels on every row
// remember that the pixels[] array is 1 dimensional, so some math are unavoidable. For this simple example I use the modulo operator.
if (i % width >= 200) { // "magic numbers" are bad, remember. This is only a simplification.
img.pixels[index] = pixels[i];
index++;
}
}
img.updatePixels();
img.save("test.png");
}
It may be too late to help you, but maybe someone else will need this. Either way, have fun!