I'm trying to blindly detect signals in a spectra.
one way that came to my mind is to detect rectangles in the waterfall (a 2D matrix that can be interpret as an image) .
Is there any fast way (in the order of 0.1 second) to find center and width of all of the horizontal rectangles in an image? (heights of rectangles are not considered for me).
an example image will be uploaded (Note I know that all rectangles are horizontal.
I would appreciate it if you give me any other suggestion for this purpose.
e.g. I want the algorithm to give me 9 center and 9 coordinates for the above image.
Since the rectangle are aligned, you can do that quite easily and efficiently (this is not the case with unaligned rectangles since they are not clearly separated). The idea is first to compute the average color of each line and for each column. You should get something like that:
Then, you can subtract the background color (blue), compute the luminance and then compute a threshold. You can remove some artefact using a median/blur before.
Then, you can just scan the resulting 1D array filled with binary values so to locate where each rectangle start/stop. The center of each rectangle is ((x_start+x_end)/2, (y_start+y_end)/2).
I have images such as below (I am pasting only a sketch here) where I want to calculate the center of symmetry and the displacement between the 2 marked zones in the image(Marked in Red and blue). Could anyone suggest a simple algorithm for this? (Please note the signal is symmetric to 180-degree rotation).
The idea is to calculate the center of symmetry between the red and blue zones
Here is my algorithm approach:
Distinguish the red and blue pixels in the image by thresholding: Lets say set the blue pixels to wihte(255) and set the red pixels(0) and set the rest of image to the (125) in a gray channel image.
Check all of the pixels horizontally in the middle row of the image.
While you are checking, if you hit the red pixel, then try to search blue one. However, if you hit red pixel again, then start searching blue again.
During the search after red pixel if you hit a blue pixel then you can easiy calculate the middle of it. Then re-start the process.
Here is a demonstration of search line:
Note: Since they are dashed lines, luckily you may pass them with gaps. To fix, this you can try couple of random different rows.
I'm trying to show students how the RGB color model works to create a particular color (or moreover to convince them that it really does). So I want to take a picture and convert each pixel to an RGB representation so that when you zoom in, instead of a single colored pixel, you see the RGB colors.
I've done this but for some very obvious reasons the converted picture is either washed out or darker than the original (which is a minor inconvenience but I think it would be more powerful if I could get it to be more like the original).
Here are two pictures "zoomed out":
Here is a "medium zoom", starting to show the RGB artifacts in the converted picture:
And here is a picture zoomed in to the point that you can clearly see individual pixels and the RGB squares:
You'll notice the constant color surrounding the pixels; that is the average RGB of the picture. I put that there so that you could see individual pixels (otherwise you just see rows/columns of shades of red/green/blue). If I take that space out completely, the image is even darker and if I replace it with white, then the image looks faded (when zoomed out).
I know why displaying this way causes it to be darker: a "pure red" will come with a completely black blue and green. In a sense if I were to take a completely red picture, it would essentially be 1/3 the brightness of the original.
So my question is:
1: Are there any tools available that already do this (or something similar)?
2: Any ideas on how to get the converted image closer to the original?
For the 2nd question, I could of course just increase the brightness for each "RGB pixel" (the three horizontal stripes in each square), but by how much? I certainly can't just multiply the RGB ints by 3 (in apparent compensation for what I said above). I wonder if there is some way to adjust my background color to compensate for me? Or would it just have to be something that needs to be fiddled with for each picture?
You were correct to assume you could retain the brightness by multiplying everything by 3. There's just one small problem: the RGB values in an image use gamma correction, so the intensity is not linear. You need to de-gamma the values, multiply, then gamma correct them again.
You also need to lose the borders around each pixel. Those borders take up 7/16 of the final image which is just too much to compensate for. I tried rotating every other pixel by 90 degrees, and while it gives the result a definite zig-zag pattern it does make clear where the pixel boundaries are.
When you zoom out in an image viewer you might see the gamma problem too. Many viewers don't bother to do gamma correction when they resize. For an in-depth explanation see Gamma error in picture scaling, and use the test image supplied at the end. It might be better to forgo scaling altogether and simply step back from the monitor.
Here's some Python code and a crop from the resulting image.
from PIL import Image
im = Image.open(filename)
im2 = Image.new('RGB', (im.size[0]*3, im.size[1]*3))
ld1 = im.load()
ld2 = im2.load()
for y in range(im.size[1]):
for x in range(im.size[0]):
rgb = ld1[x,y]
rgb = [(c/255)**2.2 for c in rgb]
rgb = [min(1.0,c*3) for c in rgb]
rgb = tuple(int(255*(c**(1/2.2))) for c in rgb)
x2 = x*3
y2 = y*3
if (x+y) & 1:
for x3 in range(x2, x2+3):
ld2[x3,y2] = (rgb[0],0,0)
ld2[x3,y2+1] = (0,rgb[1],0)
ld2[x3,y2+2] = (0,0,rgb[2])
else:
for y3 in range(y2, y2+3):
ld2[x2,y3] = (rgb[0],0,0)
ld2[x2+1,y3] = (0,rgb[1],0)
ld2[x2+2,y3] = (0,0,rgb[2])
Don't waste so much time on this. You cannot make two images look the same if you have less information in one of them. You still have your computer that will subsample your image in weird ways while zooming out.
Just pass a magnifying glass through the class so they can see for themselves on their phones or other screens or show pictures of a screen in different magnification levels.
If you want to stick to software, triple the resolution of your image, don't use empty rows and columns or at least make them black to increase contrast and scale the RGB components to full range.
Why don't you keep the magnified image for the background ? This will let the two images look identical when zoomed out, while the RGB strips will remain clearly visible in the zoom-in.
If not, use the average color over the whole image to keep a similar intensity, but the washing effect will remain.
An intermediate option is to apply a strong lowpass filter on the image to smoothen all details and use that as the background, but I don't see a real advantage over the first approach.
I'm trying to align two rectangles using the perspectiveTransform. In the image below there are the two orange rectangles (I know their dimensions and locations) and I want to warp the perspective so that they are of approximately the same size and in line (the green ones in the image). A perspectiveTransform that e.g. makes the small one equal in size with the big one doesn't really do the trick, as the size of the big one changes too. Any help much appreciated!
I'm making a simple camera app for iOS and MAC. After the user snaps a picture it generates a UIimage on iOS (NSImage on MAC). I want to be able to highlight the areas in the image that is over exposed. Basically the overexposed areas would blink when that image is displayed.
Anybody knows the algorithm on how to tell where on the image is overexposed. Do I just add up the R,G,B values at each pixel. And if the total at each pixel is greater than a certain amount, then start blinking that pixel, and do that for all pixels?
Or do I have to do some complicated math from outer space to figure it out?
Thanks
rough
you will have to traverse the image, depending on your desired accuracy and precision, you can combine skipping and averaging pixels to come up with a smooth region...
it will depend on the details of you color space, but imagine YUV space (because you only need to look at one value, the Y or luminance):
if 240/255 is considered white, then a greater value of say 250/255 would be over exposed and you could mark it, then display in an overlay.