similar to: rgb and col2rgb color conversion/modification/shading

Hello I have a palette vector of colour blind colours (in hexadecimal) which I?m using for plots, but they are not see-through, and as I wanted to overlay some histograms I wanted to convert these colours to rgb, when you can set the opacity. I have found the function col2rgb(), which works in the sense that it gives a vector of numbers but these don?t work directly in rgb because they are too

Does adjustcolor() help? cb8<- c("#000000", "#E69F00", "#56B4E9", "#009E73","#F0E442", "#0072B2", "#D55E00", "#CC79A7") plot(0,0,xlim=c(1,8),ylim=c(0,1)) points(1:8,rep(0.5,8),col=cb8,pch=19,cex=2) points(1:8,rep(0.75,8),col=adjustcolor(cb8, alpha.f = 0.3), pch=19,cex=2) On 2023-07-23 2:15 p.m., Nick Wray

Hello Core Folk. Quite by accident I discovered today that col2rgb, when fed factors, acts on the integer representation of those factors: td <- as.factor(rainbow(5)) # not how I discovered the issue! td res <- col2rgb(td) res # but col2rgb("#FF0000FF") # this is td[1] but does not produce the same answer res[,1] # moreover as.integer(td) col2rgb(5) # the integer value of the

Thanks That works nicely Nick On Sun, 23 Jul 2023 at 19:26, Ben Bolker <bbolker at gmail.com> wrote: > Does adjustcolor() help? > > cb8<- c("#000000", "#E69F00", "#56B4E9", "#009E73","#F0E442", "#0072B2", > "#D55E00", "#CC79A7") > plot(0,0,xlim=c(1,8),ylim=c(0,1)) >

Dear r-helpers, I wonder whether, given the "#rrggbb" representation of a colour, there is a simple way to select the complementary colour, also expressed as a "#rrggbb" string. Any suggestions would be appreciated. John -------------------------------- John Fox, Professor Department of Sociology McMaster University Hamilton, Ontario Canada L8S 4M4 905-525-9140x23604

Just one addition which may or may not be useful: The color palette you use is also known as "Okabe-Ito" and it is the default set of colors in the palette.colors() function. This function also has an optional alpha argument. So if you want to generate these colors with an alpha of 0.3 you can also do: palette.colors(8, alpha = 0.3) or more explicitly palette.colors(8, palette =

Nick, I've also made colors transparent by pasting the hex equivalent of, say, 0.3*256 = 76.9 to the hex color code. e.q. for black it might be "#0000004d" and the 4d is 77 in hex. That way you don't need to convert back and forth so much. If col is "#000000" the transparent version is tcol <- paste0(col,"4d") This would work in one step on a whole

The release notes for R 2.0.0 states: It is now possible to specify colours with a full alpha transparency channel via the new 'alpha' argument to the rgb() and hsv() functions, or as a string of the form "#RRGGBBAA". NOTE: most devices draw nothing if a colour is not opaque, but PDF and Quartz devices will render semitransparent colours. A new argument

-----Original Message----- From: Duncan Murdoch [mailto:murdoch at stats.uwo.ca] Sent: ned 2004-10-17 15:34 To: Gorjanc Gregor Cc: r-help at stat.math.ethz.ch Subject: Re: [R] Plotcorr: colour the ellipses to emphasize the differences On Sun, 17 Oct 2004 02:51:58 +0200, "Gorjanc Gregor" <Gregor.Gorjanc at bfro.uni-lj.si> wrote: [removed old stuff] >>I would also like to

Hello List Dwellers: I?ve looked around quite a bit, but don?t quite see an answer that I understand. I?m looking for a way to take any kind of color specification (rgb, hsv, hcl, hex) and match it to the n-nearest R official color names. Clearly it is easy to interconvert different specification schemes and color spaces, but matching to the name seems a bit trickier. Seems like if one has a

Hi Group, I have a matrix, and I would like to replace numbers less than 0 by the smallest minimum number. Below is an small matrix, and the loop I used. I would like to get suggestions on the "R way" to do this. Thanks, Juliet # example data set mymat <- structure(c(-0.503183609420937, 0.179063475173256, 0.130473004669938, -1.80825226960127, -0.794910626384209, 1.03857280868547,

Based on reading 'rgb' documentation, I would have thought the following would have produced identical results. Can someone explain how to make this happen? I need to be able to specify an array of rgb values for the 'col' parameter. colnames.col <- c("black", "red", "blue", "green") colnames.rgb <- apply(as.matrix(colnames.col), 1,

Hola, Por si lo quieres con colores rellenando cada punto: #---------------- library(data.table) library(lattice) dat <- read.table("pba.csv", header=TRUE, dec=",", as.is=TRUE) row.names(dat) <- NULL dat <- as.data.table(dat) dat$mycol <- ifelse(dat$sol =="ControlAE", "red", dat$sol) dat$mycol <- ifelse(dat$mycol

I would really like some help with understanding the panel function, in zoo. Thank you. R 15.1 and zoo 1.7-7. library(zoo) x = seq(0,3*pi,length.out=100) y = sin(x) zobj = zoo(y, x) ########################################################### ## EXAMPLE 1 - GLOBAL ARGUMENT ## This panel function works ## But, it relies on mycol, which is a global variable

This is probably easy for experienced users but I could not find a solution. I have several R scripts that process several columns of a dataframe (several dataframes and columns actually, but simplified for my question). References such as: myDF$myCol are all over. I like to automate this for other dataframes and columns by defining a reference only once in the beginning of the script. One

Hi, I'm trying to draw a 2D plot using multiple tints of red. The (simplified) setup is the following: || year | x | y || My idea is that each year is plotted with a different tint of red. Older year (lightest) -> Later year (darkest). I've managed to plot this with different scales of grays simply by doing: palette(gray(length(years):0/length(years))) before the plot and for each

Hi all, I used image.plot() to create a heat map of a matrix: as.matrix(read.table("Matrix.txt", sep="\t"))->x HeatBrk<-seq(5,25,2.5) MyCol= gray((7:0)/7) library(fields) image.plot(x, col=MyCol, breaks=HeatBrk, legend.shrink=0.3) dev.copy(device=pdf, file="HEAT4!.pdf", height=8, width=8) dev.off() There are a few things that I would like to do that I

Hi All, I'm having a problem with barplot: mydata [1,] 2 108 0 0 0 1 3 0 0 0 0 0 7 18 3 4 8 20 26 20 19 7 1 1 mycol = c(rep('yellow', 2), rep('white', 3), rep('orange',2), rep('white', 5), rep('orange',3), rep('red',9)) barplot(mydata, col = mycol) gives me an uniformly yellow barplot. How do I solve this? bw Federico

Dear R users, I am currently trying to create my first personnal function and use it with the apply function. The purpose of this function is to create a vector summarizing the number of levels in a given selection of data.frame columns. I tried to transpose the indexation method used by the nlevels function but it doesn't seem to work. I did not find anything uesful in the archives so

Dear users, I have a double for loop that does exactly what I want, but is quite slow. It is not so much with this simplified example, but IRL it is slow. Can anyone help me improve it? The data and code for foo_reg() are available at the end of the email; I preferred going directly into the problematic part. Here is the code (I tried to simplify it but I cannot do it too much or else it