Does anyone know of a package in R that allows me to give several photographs the same average luminance?
The photographs I have are generally low definition. I have seen the package 'magick' but there is no mention of luminance as something that can be altered.
Although I am pretty new to the physics of light as I understand it luminance and brightness are different so altering the brightness is not quite enough.
Using imager, you can change the color space of an image (from RGB to HSL/HSV/HSI/YUV/YCbCR). In one color space, you can manipulate each channel as you like.
Here, I convert im from RGB to YUV (corresponding to one luma (Y) and two chrominance (UV) channels). Then, as an example, I incrementally increase the luma:
library(imager)
im <- load.image(system.file('extdata/parrots.png',package='imager'))
im.yuv <- RGBtoYUV(im) # convert from RGB to YUV
par(mfrow=c(1,3))
plot(im)
for (i in rep(0.2, 2)){
im.yuv[,,1,1] <- im.yuv[,,1,1] + i # the last dimension gives the channel; 1:Y 2:U 3:V
im.cor <- YUVtoRGB(im.yuv) # convert back from YUV to RGB
plot(im.cor)
}
Related
I really really need some advice. I have a Raster with many pixels. Each pixel has one value. Now I want to do a spatial analysis of these pixels. I want to see in which region have the most pixels and were not. Sounds simple, but it's not.
I had an idea to do this with the kernal density but it does not work with rasterlayer. It doesn't work either with ppp, because you can't transform a raster into this data type. I'm really lost. I don't know what could work. So I would be very grateful if I could get some help.
My Pixels looks like this:
There must be a way to show the regions with the most pixels and so on. But I don't know how I can do that.
Short answer: convert your raster object to a pixel image of class im in the spatstat package. Then use Smooth.im. Example:
library(spatstat)
Z <- as.im(my_raster_data)
S <- Smooth(Z)
plot(S)
Long answer: you're using the term "pixel" in a nonstandard sense. The pixels are the small squares which make up the image. Your illustration shows a pixel image in which the majority of the pixels have the value 0 (represented by white colour), but a substantial number of individual pixels have values greater than 0 (ranging from 0 to 0.3).
If I understand correctly, you would like to generate a colour image or heat map which has a brighter/warmer colour in those places where more of the pixels have positive values.
The simplest way is to use Gaussian smoothing of the pixel values in the image. This will calculate a spatially-varying average of the values of the nearby pixels, including the zero pixels. To do this, convert the raster to a pixel image of class im in the spatstat package
Z <- as.im(my_raster_object)
then apply Smooth.im
S <- Smooth(Z)
plot(S)
Look at the help for Smooth.im for options to control the degree of smoothing.
If you wanted to ignore the actual colours (pixel values) in the input data, you could just transform them to binary values before smoothing:
B <- (Z > 0)
SB <- Smooth(B)
plot(SB)
I understand that domain or color wheel plotting is typical for complex functions.
Incredibly, I can't find a million + returns on a web search to easily allow me to reproduce some piece of art as this one in Wikipedia:
There is this online resource that reproduces plots with zeros in black - not bad at all... However, I'd like to ask for some simple annotated code in Octave to produce color plots of functions of complex numbers.
Here is an example:
I see here code to plot a complex function. However, it uses a different technique with the height representing the Re part of the image of the function, and the color representing the imaginary part:
Peter Kovesi has some fantastic color maps. He provides a MATLAB function, called colorcet, that we can use here to get the cyclic color map we need to represent the phase. Download this function before running the code below.
Let's start with creating a complex-valued test function f, where the magnitude increases from the center, and the phase is equal to the angle around the center. Much like the example you show:
% A test function
[xx,yy] = meshgrid(-128:128,-128:128);
z = xx + yy*1i;
f = z;
Next, we'll get its phase, convert it into an index into the colorcet C2 color map (which is cyclic), and finally reshape that back into the original function's shape. out here has 3 dimensions, the first two are the original dimensions, and the last one is RGB. imshow shows such a 3D matrix as a color image.
% Create a color image according to phase
cm = colorcet('C2');
phase = floor((angle(f) + pi) * ((size(cm,1)-1e-6) / (2*pi))) + 1;
out = cm(phase,:);
out = reshape(out,[size(f),3]);
The last part is to modulate the intensity of these colors using the magnitude of f. To make the discontinuities at powers of two, we take the base 2 logarithm, apply the modulo operation, and compute the power of two again. A simple multiplication with out decreases the intensity of the color where necessary:
% Compute the intensity, with discontinuities for |f|=2^n
magnitude = 0.5 * 2.^mod(log2(abs(f)),1);
out = out .* magnitude;
That last multiplication works in Octave and in the later versions of MATLAB. For older versions of MATLAB you need to use bsxfun instead:
out = bsxfun(#times,out,magnitude);
Finally, display using imshow:
% Display
imshow(out)
Note that the colors here are more muted than in your example. The colorcet color maps are perceptually uniform. That means that the same change in angle leads to the same perceptual change in color. In the example you posted, for example yellow is a very narrow, bright band. Such a band leads to false highlighting of certain features in the function, which might not be relevant at all. Perceptually uniform color maps are very important for proper interpretation of the data. Note also that this particular color map has easily-named colors (purple, blue, green, yellow) in the four cardinal directions. A purely real value is green (positive) or purple (negative), and a purely imaginary value is blue (positive) or yellow (negative).
There is also a great online tool made by Juan Carlos Ponce Campuzano for color wheel plotting.
In my experience it is much easier to use than the Octave solution. The downside is that you cannot use perceptually uniform coloring.
While I use R quite a bit, just started an image analysis project and I am using the EBImage package. I need to collect a lot of data from circular/elliptical images. The built-in function computeFeatures gives the maximum and minimum radius. But I need all of the radii it computes.
Here is the code. I have read the image, thresholded and filled.
actual.image = readImage("xxxx")
image = actual.image[,2070:4000]
image1 = thresh(image)
image1 = fillHull(image1)
As there are several objects in the image, I used the following to label
image1 = bwlabel(image1)
I generated features using the built in function
features = data.frame(computeFeatures(image1,image))
Now, computeFeatures gives max radius and min radius. I need all the radii of all the objects it has computed for my analysis. At least if I get the coordinates of boundaries of all objects, I can compute the radii through some other code.
I know images are stored as matrices and can come up with a convoluted way to find the boundaries and then compute radii. But, was wondering if there a more elegant method?
You could try extracting each object + some padding, and plotting the x and y axis intensity profiles for each object. The intensity profiles is simply the sum of rows / columns which can be computed using rowSums and colSums in R
Then you could find where it dropps by splitting each intensity profiles in half and computing the nearest minimum value.
Maybe an example would help clear things up:
Hopefully this makes sense
At the moment I'm doing a colourizing effect using additive blending in HSV space. Have a diff value in HSV space which is added to an image texture's individual pixels to get the desired color effect. But this is turning out to be expensive as the fragment shader has to do two costly conversions to do the addition
RGB -> HSV
HSV addition
HSV -> RGB
Is there a better way to do this? The diff value will be provided in HSV only. And the final color representation is in RGB to draw.
Many Thanks,
Sak
You can get a similar effect to HSV manipulations by using a color matrix in RGB. For example, a rotation around the r=g=b axis is similar to a hue addition. (Adding x degrees in the hue channel is similar to a rotation of x degrees around r=g=b in RGB.) A translation along the r=g=b axis is similar to a value addition. (I believe that adding x to the value channel should be similar to adding x to all of r, g, and b.) And a uniform scale perpendicular to the r=g=b axis is similar to a saturation addition. I don't know off the top of my head the exact translation between adding x to saturation and scaling in RGB, but it shouldn't be too hard to work out. You should be able to compose these matrixes into a single matrix, and implement it as a single matrix multiply by the RGB value.
I use the R colorspace package to convert a three-dimensional point into a LAB color. The LAB color is defined with three coordinates, the first one ranges from 0 to 100 and the two other ones range from -100 to 100.
But searching with Google I do not find a cuboidal representation of the LAB color space. Why ?
Short answer
The LAB color space, a.k.a. gamut, contain colors that are impossible to reproduce in nature or on a screen (according to this page).
Elaboration on converting RGB to LAB
I guess the reason you ask is that you want to make some kind of printed material and want to be sure the colors turn out right. I am merely an enthusiastic amateur in this field, but think this paragraph from the wikipedia article on lab color space explains some of the complications.
There are no simple formulas for conversion between RGB or CMYK values
and L*a*b*, because the RGB and CMYK color models are device
dependent. The RGB or CMYK values first need to be transformed to a
specific absolute color space, such as sRGB or Adobe RGB. This
adjustment will be device dependent, but the resulting data from the
transform will be device independent, allowing data to be transformed
to the CIE 1931 color space and then transformed into L*a*b*.
That is, in order to create a lab color cube, you must first find the transformation from your monitor specific color space into absolute color space. This is surprisingly difficult since the mapping is not linear or on any other simple form. The transformation is not likely to be perfect either since the RGB and LAB spaces do not span the same subspace (speculating here). I once talked to a printmaker about this and he said altough the human eye only has 4 types of color receptors (RGB + light intensity) you need about 17 color components on generate the full spectrum of visible colors on paper. Both RGB and LAB compromises on that, optimized for different purposes.
Bottom line
You can calibrate your screen to set up the transformation needed to convert the RGB of the screen to the LAB colors of human eyes, and then go on to make a color cube. However, it will only apply to your very monitor and not be perfect. You are best off test printing different color profiles and choose the one you like best.
Because there is no such thing. The CIELAB colour space has a Cartesian representation (of infinite size), but the (finite) gamut that we can perceive is not cubic, it has a complicated shape. Varying the two coordinates a* and b* independently in a pre-defined range may seem convenient, but this is fundamentally not the way human perception works.