Say you want to put pixel with a color (R0 G200 B255) in a BMP picture and you have transp option in percents.
How do I determine the new pixel color, considering the transp and the background color?
I actually could figure out a formula that looks promising:
newpixel = newpixel + (bgpixel * transp) / %(transp of 255)
I created it by analyzing the pixel color change in GIMP. Not sure if that is the correct formula. I think it is also rounded up.
The standard formula is pixel = new_pixel * alpha + pixel * (1 - alpha), where alpha is a number between 0 and 1 that describes the opacity of the new (foreground) pixel.
You'll note that if the new pixel is fully transparent (alpha = 0) the pixel is unchanged and that if the new pixel is fully opaque (alpha = 1) the new pixel replaces the old one.
This formula must be applied separately for each pixel components (red, green and blue).
the code needs to change to 24 bit pixels.
where the 4th byte is the transparency factor.
The actual color values do not change
Related
I have an image that I import into Octave 5.2, and I would like to create an outline all the way around the image array using RGB values.
I'm having trouble inserting the RGB values back into the array correctly, inserting / generating the two rows at top, two columns on the left, two columns on the right, and two rows on the bottom all the way around the image / converted double array.
Example of original image:
Example of the image I'm trying to get when done:
My logic was:
To convert the image to a double array so I can do math / insert the RGB values where I wanted them.
Insert the RGB values into the left, right, top, bottom of the array.
Convert the double array back to uint8 to export / view it as image.
My code so far:
pkg load image
img_fn=('https://i.imgur.com/KKxJaOy.png'); %original image
f=imread(img_fn);
[im_r im_c]=size(f);
size_min=min(im_r,im_c); %get minum size from row and col
f_double=double(f); %need to convert to double to do math functions on it
outline_left_of_box=repmat(255,[rows(f_double),2]); %Create left line array of outline red box
f_double_tmp_red(:,:,1)=[outline_left_of_box,f_double];%Create left line of outline red box
red_outline_right_of_box=repmat(255,[rows(f_double),2]); %Create right line array of outline red box
red_outline_top_of_box=repmat(255,[2,columns(f_double)]); %Create top line array of outline red box
red_outline_bottom_of_box=repmat(255,[2,columns(f_double)]); %Create bottom line array of outline red box
%convert back to image
red_outline_img=uint8(f_double);
imshow(red_outline_img); %used to show image in octave plot window
Please note: I'm converting the image array into a double because calculations will be done on the array to get the desired color box around the image, but I'm just trying to get the inserting RGB values into the array issue fixed first.
Maybe it's easier to simply paste the inner part of the input image onto some "background" image with the desired border color, like so:
pkg load image
% Read image, get dimensions, convert to double
f = imread('https://i.imgur.com/KKxJaOy.png');
[im_ro, im_co, im_ch] = size(f);
f_double = double(f);
% Set up width and color of border
bw = 2;
color = ones(1, 1, im_ch);
color(1, 1, :) = [255, 0, 0];
% Create image of same size as input with solid color, and paste inner part of input
red_outline_img = ones(im_ro, im_co, im_ch) .* color;
red_outline_img(bw+1:end-bw, bw+1:end-bw, :) = f_double(bw+1:end-bw, bw+1:end-bw, :);
red_outline_img = uint8(red_outline_img);
imshow(red_outline_img);
That'd be the output:
Another thing you could try is plot the lines as you suggest, which can be done very efficiently with some clever use of xlim/ylim, and then print the whole thing as an -RGBImage to get it back in image form.
The only caveat here though is that you will need to play with the formatting options to get what you're really after (e.g. in case you want higher or lower resolution), since you really are printing what's on your screen at this point.
E.g.
L = imread('leaf.png');
imshow(L)
hold on
plot( [xlim, fliplr(xlim);xlim, xlim], [ylim, ylim;fliplr(ylim), ylim], 'r', 'linewidth', 2 )
hold off
set( gca, 'position', [0, 0, 1, 1] );
set( gcf, 'paperposition', [0, 0, 1, 1] );
R = print( '-RGBImage' );
close
imshow(R); set( gcf, 'color', 'k'); axis off
I am using ggtern to graph some RGB data, where each point refers to the colour % from red, green or blue.
I was wondering if it is possible to change the colour of specific plot points to match the actual colour which they are representing?
I have tried using some ggplot2 commands but it doesn't work.
This is what I am using now:
library(ggplot2)
library(ggtern)
library(readxl)
Image9 <- read_excel("~/PhD/Image9.xlsx")
View(Image9)
ggtern(data=Image9, aes(x = Blue, y = Red, z= Green )) + geom_point( size=2,
shape=16, color="black") + theme_rgbw() + theme_hidegrid_major() +
theme_hidetitles()+ theme_rotate(degrees = 120) +
geom_text(aes(label=Colour), size=3, color="black", hjust=0, vjust=0)
Using this command I can plot the data as a colour intensity percentage in Red, Green and Blue and label the points so that I can map them back to the original image
1; representative image of plot.
Ideally I would like to plot each point as the same shade of color which they represent within the color space, I don't know if this is possible with this package.
Any suggestions would be greatly appreciated.
So I'm no expert in coding: I just have a vague understanding of a few bits and bobs.
I understand that images of pixel dimensions X*Y are stored in array of size 3*X*Y where each vector pulled from a given (x,y) value has 3 elements for the 3 RGB values.
I also understand that one can also store an image in a 4*X*Y array, where each pulled vector now has 4 values, RGBA with A being Alpha, used to represent the opacity of a particular pixel.
Now I'm into animation, and have pencil drawings of white clouds on a black background that I want to import into Flash. I would like the blacker parts of the drawing to be more transparent and the lighter parts to be more opaque. I have the scans of the drawings saved in .png format.
If I had any idea how to manipulate an image at the 'array level', I could have a stab at this myself but I'm at a loss.
I need a program that, given a .png image and a reference RGB value {a b c}, obtains the RGB array of the image and converts it into an RBGA array such that:
a pixel of RGB value {p q r}
...... Becomes ......
a pixel of RGBA value {p q r 1-M[(|p-a|^2 + |q-b|^2 + |r-c|^2)^1/2]}.
Where M is a normalisation factor which makes the largest alpha value = 1.
i.e. M = 1/[(255^2 + 255^2 + 255^2)^1/2]) = 0.0026411...
i.e. the alpha value of the replacement pixel is the 'distance' between the colour of the pixel and some reference colour which can be input.
This then needs to export the new RGBA Array as a png image.
Any ideas or any fellow animators know if this can be done directly with actionscript?
Example: Reference = {250 251 245}
RGB array =
|{250 251 245} {250 250 250}|
|{30 255 22} {234 250 0 }|
...... Becomes ......
RGBA array =
|{250 251 245 1} {250 251 245 0.987}|
|{30 255 22 0.173} {234 250 0 0.352}|
You can do this quite simply, just at the command-line, with ImageMagick which is installed on most Linux distros and is available for free on OSX and Windows.
The "secret sauce" is the -fx operator - described here.
So, let's generate a 300x200 black image and then use -fx to calculate the red channel so that the red varies across the image according to what fraction of the width (w) we are from the left side (i):
convert -size 300x200 xc:black -channel R -fx 'i/w' result.png
Note that I am generating an image "on-the-fly" with -size 300x200 xc:black, whereas if you have a PNG file with your animation frame in it, you can put that in, in its place.
Now let's say we want to vary the opacity/alpha too - according to the distance down the image from the top:
convert -size 300x200 xc:black -alpha on \
-channel R -fx 'i/w' \
-channel A -fx 'j/h' result.png
Ok, we are getting there... your function is a bit more complicated, so, rather than typing it on the command-line every time, we can put it in a script file called RGB2Opacity.fx like this:
convert -size 300x200 xc:black -alpha on -channel A -fx #RGB2Opacity.fx result.png
where RGB2Opacity.fx is simple and looks like this for the moment:
0.2
Now we need to put your "reference" pixel on the command line with your animation frame so that ImageMagick can work out the difference. That means your actual command-line will look more or less exactly like the following:
convert -size 300x200 xc:"rgb(250,251,245)" YourImage.png -alpha on -channel A -fx #RGB2Opacity.fx result.png
And then we need to implement your formula in the -fx script file. Your variable names must be at least 2 letters long with no digits in them, and you should return a single value for the opacity. Variables are all scaled between [0,1.0] so your 255 scaling is a little different. I have no sample image and I am not sure how the answer is supposed to look, but it will be pretty close to this:
MM=1/pow(3,0.5);
pmasq=pow((u.r-v.r),2.0);
qmbsq=pow((u.g-v.g),2.0);
rmcsq=pow((u.b-v.b),2.0);
1-MM*pow((pmasq+qmbsq+rmcsq),0.5)
I don't know if/how to put comments in -fx scripts, so I will explain the variable names below:
pmasq is the square of p minus a.
qmbsq is the square of q minus b.
rmcsq is the square of r minus c.
u.r refers to the red channel of the first image in ImageMagick's list, i.e. the red channel of your reference pixel.
v.g refers to the green channel of the second image in ImageMagick's list, i.e. the green channel of your animation frame.
Let's create your animation frame now:
convert xc:"rgb(250,251,245)" xc:"rgb(250,250,250)" xc:"rgb(30,255,22)" xc:"rgb(234,250,0)" +append frame.png
And check it looks correct:
convert frame.png txt:
Output
# ImageMagick pixel enumeration: 4,1,65535,srgb
0,0: (64250,64507,62965) #FAFBF5 srgb(250,251,245)
1,0: (64250,64250,64250) #FAFAFA grey98
2,0: (7710,65535,5654) #1EFF16 srgb(30,255,22)
3,0: (60138,64250,0) #EAFA00 srgb(234,250,0)
If we apply that to your image and check the results, you can see I have got it slightly wrong somewhere, butI'll leave you (or some other bright spark) to work that out...
convert -size 4x1 xc:"rgb(250,251,245)" frame.png -alpha on -channel A -fx #RGB2Opacity.fx result.png
convert result.png txt:
# ImageMagick pixel enumeration: 4,1,65535,srgba
0,0: (64250,64507,62965,65535) #FAFBF5FF srgba(250,251,245,1)
1,0: (64250,64507,62965,64764) #FAFBF5FC srgba(250,251,245,0.988235)
2,0: (64250,64507,62965,19018) #FAFBF54A srgba(250,251,245,0.290196)
3,0: (64250,64507,62965,29041) #FAFBF571 srgba(250,251,245,0.443137)
I am creating an application in win32 api, which will use a progress bar. This progress bar, should change its color. From red (left end) to green(right), and in the middle some yellow.
I searched a little, and found out, that I should use HSV to achieve this. I just don't know how? I found in this link, two functions, to convert the color, from RGB to HSV and back.
But what should I do if the color has been converted to HSV?
Like RGB coordinates, HSV coordinates define a point in a three dimensional space.
You may find a trajectory form one point (x0, one color) to the second (x1) with a formula like:
x = x0 + alpha * (x1-x0)
with alpha varying form 0.0 to 1.0
You can do this for all three components simultaneaously.
With a trajectory from green to red in HSV space you will mainly modify the H (Hue) value. If you want to see some yellow in the middle of your path (and not violett) you need to define a second or even third color and walk
green -> yellow -> red
Edit: Example
int hue0 = 0; // red
int hue2 = 120; // green
// find 100 colors between red and green
for(double alpha = 0; alpha <= 1.0; alpha += 0.01)
{
hueX = hue0 + alpha * (hue1 - hue0);
// same for value, saturation:
// valX = val0 + alpha * (val1 - val0)
// ...
// plot this color
}
I have a one ambient light with intesity ( 10000,10000, 5000 ). I am trying to color the primitive.
As you know, color values for R,G, and B are between 0 and 255. How can I find color of the pixel according to light intesity ?
platform : linux and programming language c
EDIT :
In ray tracer, we are calculating
for each ambient light in the environment
color . R += Intensity of the light * ambient coefficient for color R
color . G += Intensity of the light * ambient coefficient for color G
color . B += Intensity of the light * ambient coefficient for color B
However, whenever I have tried to emit this pixel color value on the screen with openGL.
set pixel color ( color )
I have taken wrong color because of intensity is high and maximum color value is low.
The question is unclear, as written, so here's some general advice.
Colours in renderers are typically held as values with a nominal range of [0..1] for each of the RGB components.
When those colours are rendered to pixels, they're usually just multiplied by 255 to give a 24-bit colour value (8 bits per component).
If the original values are outside of the [0..1] range they must be "clamped" so that the resulting pixel values fall in the [0..255] range.
That clamping can either be done "per component", which in your case would result in (255, 255, 255), or each component could be divided by the maximum component, giving (255, 255, 127) - i.e. preserving their relative intensities in pseudo-code:
float scale = max(r, g, b);
if (scale < 1) {
scale = 1; // don't normalise colours that are "in range"
}
byte R = 255 * (r < 0 ? 0 : r / scale);
byte G = 255 * (g < 0 ? 0 : g / scale);
byte B = 255 * (b < 0 ? 0 : b / scale);
It's usual for all intermediate calculations to preserve the full dynamic range of intensities. For example, it wouldn't make sense for the Sun to have a brightness of "1", since every other object in the scene would by comparison have an almost infinitesimally small value.
The net result of the clamping will therefore be that light sources which contribute too much light to the scene will produce "saturation" of the image - i.e. the same effect that you get if you leave the shutter open too long on a picture of a bright scene.
It's this one you want:
brightness = sqrt( .241*R*R + .691*G*G + .068*B*B )
Find more here: http://www.nbdtech.com/Blog/archive/2008/04/27/Calculating-the-Perceived-Brightness-of-a-Color.aspx
or here: http://en.wikipedia.org/wiki/Luminance_(relative)