similar to: Problem with mathematical expression and loop

Recent questions about using plotmath have renewed my interest in this question I want to have expressions take values of variables from the environment. I am able to use expressions, and I am able to use paste to put text and values of variables into plots. But the two things just won't work together. Here is some example code that shows what I mean. plot(NA,xlim=c(0,100),ylim=c(0,100))

I have a problem, that the plot of mathematical formula sometimes crashes R on Linux (SuSE 6.2 and 7.0 on a Dual PIII/500 machine with 1GB RAM). The problem occurs for example in demo("graphics"), i.e. in the last part were the mathematical symbols are plotted. There are several circumstances to reproduce this segmentation fault, but the following is one of the most simple examples. The

I want to write something like (LaTeX style) b_{norm}=\frac{F\prime(0)}{R\prime(0)} how do I add the "prime" (first derivative) to a R-plot? The help of "plotmath" just talks about "partialdiff". Can you complete this command? text( 30,0.05,labels=expression(b[plain("norm")]==frac(F(0),R(0))) ) Thanks, Thomas

Hi, I try to make a histogram from a variable that contains the number of shoots from about 1000 individuals from a specific plant species (the range is 1-110). Those numbers are highly skewed to the right. My question is: how can I make my own classes with the lattice "histogram"? I tried it with "breaks=c(0,5,10,15,20,25,110)" but my "25-110"-class is presented

Is it possible to use "paste" to write out an expression and evaluate it? Suppose I want to add two vectors X1 and X2, defined as follows: X1 <- 1:6 X2 <- 6:1 If I write the following it looks like what I want but is a character: noquote(paste(paste("X", 1, sep = ""), paste("X", 2, sep = ""), sep = "+")) Is there a way to tell R

Hi everyone I have the following problem: My dataframe has 3 variables: ID, Year and and an outcome variable. The dataframe contains repeated measurements because the subjects filled out a questionnaire every year. The time span covers 2 years. Now I want to check if there is a significant change in the outcome over the 2 years with a paired wilcox.test. The problem: Not every subject

The source to the noquote() function looks like this: noquote <- function(obj, right = FALSE) { ## constructor for a useful "minor" class if(!inherits(obj,"noquote")) class(obj) <- c(attr(obj, "class"), if(right) c(right = "noquote") else "noquote") obj } Notice what happens with right =

It is very wierd... Can some of you confirm the following behavior ? It is a new bug (feature ?) which was not yet in 0.49 ... noquote <- function(obj) { ## constructor for a useful "minor" class if(!inherits(obj,"noquote")) class(obj) <- c(class(obj),"noquote") obj } "[.noquote" <- function (x, subs) structure(unclass(x)[subs], class =

Hello, I just wanted to save a graphic in the pdf-format. But id failed: Fehler: Invalid font type Zus?tzlich: Warning messages: 1: font family not found in PostScript font database 2: font family not found in PostScript font database I use R 2.6.0 with all packages updated recently. Any idea? Thank you very much! Best, Denis __ Denis Aydin mailto:Denis.Aydin at stud.unibas.ch Montag, 22.

hello, I have the following problem setting parameter 'frac' in the sigest function of the kernlab package. ## executing the ?sigest example: library(kernlab) data(spam) srange <- sigest(type~.,data = spam) ## works fine... ## setting 'frac' explicitly ## (in this case even to the default of .25) options(error=recover) srange <- sigest(type~.,data = spam, frac = .25) ##

I currently have a program that automates 2-way ANOVA on a series of endpoints, but before the ANOVA is carried out I want the code to test the assumptions of normality and equal variance and report along with each anova result in the output file.  How can I do this? I have pasted below the code that I currently use.   library(car) numFiles = x #

Hello, I'm having difficulty passing an object name to a lapply function. Can somebody tell me the trick to make this work? #Works T13702 <- TRACKDATA[["13702.xls"]][["data"]] min(unlist(lapply(list(T13702), function(x) mdy.date(x[1, 2], x[1, 1], x[1, 3])))) 16553 #Works d<-2 assign(paste("T",substr(names(TRACKDATA)[d],1,(nchar(names(TRACKDATA)[d]

... or if I understand correctly, simply expression(frac(additive ~ HCO[3]^"-", true ~ HCO[3]^"-" ))) Cheers, Bert On Fri, Feb 2, 2024 at 3:06?AM Rui Barradas <ruipbarradas at sapo.pt> wrote: > ?s 10:01 de 02/02/2024, Troels Ring escreveu: > > Hi friends - I'm plotting a ratio of bicarbonates i ggplot2 and > > > >

The following line of code seems fairly straight forward, yet it is kicking back an error message: for (i in 1:length(mean.natveg.frac)){month.observed[i]=as.numeric(names(mean.natveg.frac[i]))%%12} Error message: Error in month.observed[i] = as numeric(names(mean.natveg.frac[i]))%%12 : object 'month.observed' not found Seems like its not found because I'm trying to define it. Also

?s 10:01 de 02/02/2024, Troels Ring escreveu: > Hi friends - I'm plotting a ratio of bicarbonates i ggplot2 and > > ylab(expression(paste(frac("additive BIC","true BIC")))) worked OK - but > now I have been asked to put the chemistry instead - so I wrote > >

Hi, my proposal is based on a mathematical proof (see mail before) and yours is correct too. They cannot but match. Let's inspect your examples: Ady wrote: > ISO size: 1'085'736'960 bytes ( > 1GiB) This are exactly 132 cylinders of 255x63. 132 is divisible by 4. So there is no padding needed. > Cylinders (1st attempt)=132 > [...] > 2120580 / 4 = 530145, so

I haven't been able to get anywhere tracking this down. It seems that c.noquote() does something strange with its third (and subsequent) parameters: R-2.7.0 under NetBSD, R-2.6.0 under Solaris, and R-2.8.0 (unstable) (2008-06-10 r45893) under WinXP, I get: > c(noquote('z'), 'y', 'x', '*') [1] z y x * x * > or: > c(noquote('z'),

Hi friends - I'm plotting a ratio of bicarbonates i ggplot2 and ylab(expression(paste(frac("additive BIC","true BIC")))) worked OK - but now I have been asked to put the chemistry instead - so I wrote ?ylab(expression(paste(frac("additive",HCO[3]^"-","true",HCO[3]^"-")))) - and frac saw that as additive = numerator and HCO3- =

Hi all, I've been trying (unsuccessfully) to modify an xyplot I created using the lattice package. I would like to change default strip colors and locations. I started with numeric data in 4 columns, which look like this: 0.25 2 1 32 0.25 2 2 30 0.25 2 3 27 0.25 2 4 23 0.25 2 5 17 0.25 3 1 30 0.25

Hi R, When I use the below to write the text file try=data.frame(rep("a",5), rep("b",5)) write.table(try,"z:\\try.txt",row.names=F,col.names=F,sep="\t") the output contains two columns with quotes! Is there a way to write without quotes? I tried try[,1]=noquote(try[,1]) try[,2]=noquote(try[,2]) Thank you, Regards, Ravi Shankar