similar to: Relating Tables

Hi, I would like to select the most frequent value level in a set of three variables. Three different observators have judged hair color in study subjects. Mostly they judge the same color, sometimes there is a slight difference. I want to know what most of the observators have chosen (so at least 2) from the 3 observations. E.g. If two out of three observators decide the hair is black, then

Hello there! I'd just like to say in advance, "Thank you," for any help and/or advice. My problem is as follows: I have a dataset that is made up of percentages. I've assigned my "error" percentages a value of '-100', my "non-existent" percentages a value of '0', and all my other percentages are normal values that range from the high

Hi everyone I am using barchart to make my graphs. Here is my code. barchart(percent_below ~ factor(Year)| factor(Season, levels=unique(Season)), data= .season_occurrence, origin = 0, layout = c(4, 1), scales=list(tick.number=ticknum,labels=NULL), ylim=c(0, ymax), group = factor(Year), xlab= "Year", auto.key = list(points = FALSE, rectangles =

Hi , I am new to R. I am trying to run a simple R script as shown below: aov.R ------ data1<-c(49,47,46,47,48,47,41,46,43,47,46,45,48,46,47,45,49,44,44,45,42,45,45,40 ,49,46,47,45,49,45,41,43,44,46,45,40,45,43,44,45,48,46,40,45,40,45,47,40) matrix(data1, ncol= 4, dimnames = list(paste("subj", 1:12), c("Shape1.Color1", "Shape2.Color1", "Shape1.Color2",

Dear All, Consider the following trivial code snippet rm(list=ls()) name_vec <- c("color1", "color2") pdf("test_color.pdf") plot(seq(5), seq(5), main=paste(name_vec[1]," and ",name_vec[2], sep="")) dev.off() What I would like to achieve is rather simple to explain, but it is giving me a headache: how can I have two colors in main? Let us

If you are not using an anonymous function and say you had written the function out The below gives me the error > 'f(colordata2$color1)' is not a function, character or symbol' But then how is the anonymous function working? f <- function(col){ ifelse(col == 'blue', 1, 0) } responses <- lapply(colordata2[ -1 ], f(colordata2$color1) )

Lapply is not a vectorized function. It is compact to read, but it would not be worth using for this calculation. However, if your data frame had multiple color columns in your data frame that you wanted to make responses for then you might want to use lapply as a more compact version of a for loop to repeat this operation. colordata2 <- data.frame(id = c(1,2,3,4,5), color1 =

lapply(colordata2[ -1 ], f ) When you put the parentheses on, you are calling the function yourself before lapply gets a chance. The error pops up because you are giving a vector of numbers (the answer f gave you) to the second argument of lapply instead of a function. -- Sent from my phone. Please excuse my brevity. On April 7, 2016 7:31:18 AM PDT, Michael Artz <michaeleartz at

Thaks so much! And how would you incorporate lapply() here? On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbarron at gmail.com> wrote: > ifelse is vectorised, so just use that without the loop. > > colordata$response <- ifelse(colordata$color == 'blue', 1, 0) > > David > > On 7 April 2016 at 12:41, Michael Artz <michaeleartz at gmail.com> wrote: >

hi all, i have a simple scatter plot, and i'd like to make it so the scatter plot colors are slightly transparent. i see in a previous post that someone mentioned the "alpha" parameter, but i am not sure how it can be used with the 'plot' function [*]. for example, suppose i have: plot(mydata$column1, mydata$column2, col="red", cex=1) i now want to make it so

Hello R usuer, The code given below superimposes a pie diagram on another plot containing some points. However, I would like to center the pie diagram on the xy location on the plot, but not on the center. is there any way to re-center pic diagram. Any suggestion or better alternative are highly appreciated. Thank you in advance for your help. Regards, Bibke library(visualFields) library(car)

There are a lot of ways to specify colours in R plot functions (number from the palette, by name, etc.). I'd like to pass a colour from an R function to an external function, and I'd like it to have the same flexibility. To avoid having to interpret all possible colour specifications myself, I need a function to convert a general colour specification into a standard form (e.g. r,g,b

I couldn't find this in online help or the archives: Is there any R function or object giving the mapping of the colornames as given by colors() into the hsv() model? Regards -- Dr. Jens Oehlschl?gel-Akiyoshi MD FACTORY GmbH Bayerstrasse 21 80335 M?nchen Tel.: 089 545 28-27 Fax.: 089 545 28-10 http://www.mdfactory.de

I couldn't find this in online help or the archives: Is there any R function or object giving the mapping of the colornames as given by colors() into the hsv() model? Regards -- Dr. Jens Oehlschl?gel-Akiyoshi MD FACTORY GmbH Bayerstrasse 21 80335 M?nchen Tel.: 089 545 28-27 Fax.: 089 545 28-10 http://www.mdfactory.de

Good afternoon, This is my first post, so please be gentle ;) Anyway, here's my question....I've recently put together a Linux Server for our company, replacing an old Novel server. I have the box set up with Samba, as a PDC on our network. So far, just about everything is working correctly. One issue I'm coming up against, though, is printing from DOS. I have Samba

Dear All: I tried to use heatmap.2 to generate hierarchical clustering using the following command: heatmap.2(datamatrix, scale="row", trace="none", col=greenred(256), labRow=genelist[,1], margins=c(10,10), Rowv=TRUE, Colv=TRUE) datamatrix is subset of a RMA normalized data subset by a genelist. The problem is a lot of times, the z-score in key are from, like -5 to 15 or

Hi, I have a question about exporting interactive 3D plots. I use the following code to plot a contour of a trivariate normal distribution: library(mvtnorm) library(rgl) library(misc3d) n=25 x=seq(-3,3,length=n) X=cbind(rep(x,each=n**2),rep(rep(x,each=n),n),rep(x,n**2)) p=array(dmvnorm(X,sigma=diag(3)*0.5+0.5),c(n,n,n)) contour3d(p,x,x,x,level=mean(p)) lim=c(-3,3)

Hi, I''m using drbfire to send a "DRbUndumped" ActiveRecord object from the server to the client, through an ADSL line of 4Mbits/s download, and 600Kbits/s upload. I was afraid there would be too many round-trips between the client and the server, and unfortunately this is true: Server code: ------------ #!/usr/local/bin/ruby require ''rubygems'' require

Hi , I would like to use ggplot2 to plot a matrix as an image. You can copy paste the following Data<-matrix(data=rnorm(900,80,20),nrow=30,ncol=30) lengthOut<-5 Lengths<- 15 library(reshape2) library(ggplot2) tdm <- melt(Data) ggplot(tdm, aes(x = Var2, y = Var1, fill = factor(value)),levels=seq(0,1,by=0.1)) +                   labs(x = "MHz", y =

Hi Martin, Great that you follow this. My original intention was to be able to translate colornames to hsv because this would allow using colornames to cut out a certain part of the colorwheel for colorcoding (HSV component H, see my code below) I think internally we might have colors represented as Colornames, ColorIntegers, ColorHexcodes, ColorRGBs, ColorHSVs (ColorCMYs?) however the R-user