I am creating an animated gif with Water and Land. The Land part has shores, so the water will in some cases be overlaying those shores. However, I wish the shores to still be slightly visible underneath the water. Basically, imagine 1 transparent png which represents water, over another that has sand. The 2 images combined can be viewed as a third image.
My question is - how would I calculate the RGB value of the new pixels, based on the top ones RGBA (with transparency) and the bottom ones RGB values, to mimic a natural look.
I will change the transparency level myself to see what suits me best, but I'm missing a formula.
EDIT:
atm I do have 1 idea - take the percentage of the transparency level of the water, and based on that percentage calculate a new RGB. Do the same for the sand pixels, but with the "remaining" percentage (100-transparency%)
ie (A part is in %, for simplicity)
(R1,G1,B1, 40) (R2,G2,B2) = 40% of first group + 60% of second
Alpha blending is either dead simple (alpha is on 0-255, so the colour is destination * (255 - a) + overlay * a, all over 255) or quite subtle, when you allow the destination to also have alpha. Also it is necessary to implement efficiently for many applications, which is done by pre-multiplying, then hardcoding the division by 255 using
((x+1) * 257) >> 16;
Malcolm McLean gives the correct answer for the case where destination (aka background) is completely opaque, which happens to be the case we are discussing.
Just for completeness, I'd like you give the formula for the general case and the intuition behind it.
First of all, assume our colors are 4-component floating-point vectors, with each channel having values between 0 and 1. Also assume our colors are alpha-premultiplied, that is the r, g and b channels are pre-multiplied by the alpha channel. Then the alpha-blending formula is:
blended = front + back * (1.0 - front.a)
The intuition is: we are taking the front color and adding to it a bit of the back color, namely as much as we can see through the front color.
Note that the blended color is also alpha-premultiplied.
Related
I apologize in advance if a question like this was already answered. All of my searches for adding filters resulted in how to add dog faces. I wasn't sure what the proper terminology is.
What techniques do phone apps (such as Snapchat's text overlay or a QR code program for android) use to "darken" a section of the image? I am looking to replace this functionality in OpenCV. Is it possible to do this with other colors? (Such as adding a blue accent)
Example: Snapchat text overlay
https://i.imgur.com/9NHfiBY.jpg
Another Example: Google Allo QR code search
https://i.imgur.com/JnMzvWT.jpg
Any questions or comments would be appreciated
In General:
Change of brightness can be achieved via Addition/Subtraction.
If you want to brighten your Image, you can add a specific amount (e.g. 20) to each channel of the image. The other way around for darkening (Subtraction).
If you would subtract 50 from each channel of the image, you would get:
To darken pixel dependent you could also use Division. This is how a division with 1.5 would change the image:
Another way would be to use the Exponential Operator. This operator takes the value of each channel and will then calculate pixel^value. The resulting value will be then scaled back to the 0-255 range (for 8 bit RGB) via looking the highest value and then calculating the scaling factor via 255/resulting value.
If use it with values > one, it will darker the image. This is because
Here a chart how the exponential operator will change the value of each pixel. As you can see, values for the operator above 1 will darken the image (meaning the channels will be shifted towards lower values), whilst values below 0 will shift all pixels towards white and thus increase brightness.
Here is an example image for application of the operator using the value 0.5, meaning you take each pixel^0.5 and scale it back to the range of 0-255:
For a value of 2 you get:
Sadly i can not help you further, because i am not familiar with OpenCV, but it should be easy enough to implement yourself.
For your question about tinting: Yes, that is also possible. Instead of shifting towards white, you would have to shift the values of each pixel towards the respective color. I recommend to inform you about blending.
Original image taken from here
Update: I was able to darken an image by blending an image matrix with a black matrix. After that, it was just a matter of darkening certain parts of the image to replicate an overlay.
The lower the alpha value is, the darker the image.
Result
void ApplyFilter(cv::Mat &inFrame, cv::Mat &outFrame, double alpha)
{
cv::Mat black = cv::Mat(inFrame.rows, inFrame.cols, inFrame.type(), 0.0);
double beta = (1.0 - alpha);
cv::addWeighted(inFrame, alpha, black, beta, 0.0, outFrame);
}
https://docs.opencv.org/2.4/doc/tutorials/core/adding_images/adding_images.html
Thank you for the help everyone!
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 am extracting primitives from pixel-based line diagrams and wish select by colour. Thus in the following
I wish to extract the "blue", the "green" and the "black" primitives. (I am prepared to try to reconstruct primitives which have been split by primitives of another colour).
However the "blues" have a varying amount of white added (similar to a gray scale for black). Thus the commonest colours (rounded to 12-bit for simplicity) with their counts might be
000 881 // black
88f 1089 // white-blue
fff 70475 // white
but there are other degrees of whiteness at lower frequency
// other white-blue
99f 207
// other grey
ddd 196
I believe that the authors will have used only a very limited number of pure colours (e.g. 3-6) in many diagrams and that various rendering tools will have added the white. IOW the colours can be expressed by (0 =< x =< 1)
000 + x(FFF)
00F + x(FF0) // blue
0F0 + x(F0F) // green
However there is no requirement to use primary colours and the set could be any colour with arbitrary amounts of white.
How can I reconstruct the (small) set of different colours? If this is possible I can then select those regions, transform to grey, and binarize in the normal way.
I'd prefer source in Java but I suspect that any code will be adequate;
I have read two useful SO questions
"Rounding" colour values to the nearest of a small set of colours
HCL color to RGB and backward
which use H-C-L and might be a way forward although they don't directly answer my requirements.
You could try using region growing. I think it should fit your needs well. Just change the threshold for when it's the same color. I think it should work well here since there seems to be a big difference between any two colors that are connected as objects.
If your intuition is correct (all pixels being a linear mixture of some color and pure white), in the RGB cube all colors will be aligned on line segments originating from the white corner.
If you pick one representative pixel per different color (as far as possible from white, for better accuracy), you can identify the color of any other pixel by finding the best alignment formed by this pixel, by white and by the representative pixels.
Alignment is tested by computing the cosine of the angle formed (use 3D vectors, the cosine is the dot product over the product of the norms; drop the sign). In theory the cosine should be exactly 1, but due to numerical errors it can be smaller, so just consider the representative color that maximizes the cosine.
Take special care of the white pixels (short distance to the white corner), otherwise they will be randomly assigned to some representative color.
Depending on the number of colors involved and their similarity, a simple threshold of the R, G, and B values would quickly reduce everything to one of 8 colors (black, red, green, blue, cyan, magenta, yellow, or white).
I would like to subtract color from another. For example, I have two image 100X100 pixel, one with color R:236 G:226 B:43, and another R:63 G:85 B:235. I would like to cut color R:236 G:226 B:43 from R:63 G:85 B:235. But I know it can't subtract like the mathematically method, by layer R:236-63, G:226-85, B:43-235 because i found that the color that less than 0 and more than 255 can't define.
I found another color space in RYB color space.but i don't know how it really work.
Thank you for your help.
You cannot actually subtract colors. But you surely can detect their difference. I suppose this is what you need, anyway.
Here are some thoughts and remarks:
Convert your images to HSV colorspace which transforms RGB values to
Hue, Saturation and Brightness (Value).
All your images should be around a yellowish color (near 60 deg. on
the Hue circle) so they should all have about the same Hue with
minor differences.
Typically if all images are taken at constant lighting conditions
they should have the same Value (brightness).
Saturation, which corresponds to the mixture of white in a color,
typically represents how intense you perceive a color to be. This
would typically be of about the same value for all your images in
constant lighting conditions.
According to your first description, the main difference should be detected in the Hue channel.
A good thing about HSV is that H (hue) is represented by a counterclockwise circle and colors are just positions on this circle, so positive and negative values all make sense (search google for a description of HSV colorspace to get a view of how it looks and works).
You may either detect differences by a subtraction that will lead you to a value either positive either negative, or by taking the absolute value of the subtraction, which will just give a measure of the difference of the two values of Hue (but without any information on the direction of the difference). If you need the direction of the difference you should just stick to a plain subtraction.
For example:
Hue_1 - Hue_2 = Hue_3 (typically a small value for your problem)
if Hue_3 > 0 this means that Hue_1 is a bit towards Green if
Hue_3 < 0 this means that Hue_1 is a bit towards Red
Of course you may also need to take a look at the differences in the other channels, S and V to see if colors are more saturated or more bright, but I cannot be sure you need to do this since we haven't seen any images here.
Of course you can do a lot more sophisticated things...Like apply clustering or classification techniques on the detected hues and classify them to classes according to your problem needs...
I am trying to apply histogram normalization to create a dense color histogram.
Split the channels into R, G, B
Normalize the individual histogram
Merge
I think that this is the normal step perhaps if I am wrong please let me know. Now,
for a rainbow image as shown below I get I get max of 255 for all 3 channel and 0 as min. Using the formula for
Pixel - Min / (Max - min) * 255
I will get the same image as the original one back. What is the critical step that I am missing. Please advise me.Thank you!
REf: http://www.roborealm.com/help/Normalize.php.. I used this reference
White = (255,255,255). Black = (0,0,0). So your program finds the white background, and the black line in the bottom right.
Remove the white and change it to black. Then make your program ignore black.
images having white and black pixels cannot be normalized as such. your formula is giving you the same value. try ignoring all white and black pixels and normalize the pixels one by one.
as i see here you have a well distributed image for all channels already, so normalizing this one may not work well anyways..