similar to: Format of numbers in plotmath expressions.

Hi, i'm trying to put a legend on some figures and they're coming out a bit wonky. here's an example: a <- c(1:10) par(mfrow=c(2,1)) plot(a,type="s",lwd=3) leg <- c(expression(paste("data1 (",rho,"=1)")), expression(paste("data2 (",rho,"=0.0)"))) legend("bottomright",legend=leg,col=c(1,2),lwd=3)

Dear all, I have a vector of expressions and would like to "paste" some string to it before using it in a plot: vars <- vector("expression", 2) vars[1] <- expression(alpha) vars[2] <- expression(beta) plot(0, 0, main=substitute(bold("Foo" ~~ VAR), list(VAR=vars[2]) )) Although I tried hard, I just can't figure out how to solve this. The title should be

A colleague asked me if there is a way to specify with a ***variable*** (say ``cflag'') whether there is an intercept in a linear model. She had in mind something like lm(y ~ x - cflag) where cflag could be 0 or 1; if it's 0 an intercept should be fitted, if it's 1 then no intercept. This doesn't work ``of course''. The cflag just gets treated as another predictor

I have factors with levels ``Unit", "Achieved", and "Scholarship"; I wish to replace these with "U", "A", and "S". So I do fff <- factor(fff,labels=c("U","A","S")) This works as long as all of the levels are actually present in the factor. But if ``Scholarship'' is absent (as if often is) then

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 had such good luck with my previous question to r-help, (a few minutes ago) that I thought I would try again with the following query: > Suppose I have > > xNm <- "gamma" > > I would like to be able to do > > plot(1:10,xlab = <something involving xNm">) > > and get the x axis label to be the Greek letter gamma > (rather than the

Another question on functions - I have something that looks like plotter<-function(i){ temp.i<-rwb[rwb$vector1 <=(i*.10),] with(temp.i, plot(vector2, vector3, main=(i*.10),)) mod<-lm(vector3~vector3-1,data=temp.i) r2<-summary(mod)$adj.r.squared rsqrd[i]<-r2 legend("bottomright", legend=signif(r2), col="black") abline(mod) rsqrd<<-rsqrd } I'd

Hi: I'm trying to plot dates on the x-axis of a code, but the legend is not being displayed. I receive the following error: Error in match.arg(x, c("bottomright", "bottom", "bottomleft", "left", : 'arg' should be one of bottomright, bottom, bottomleft, left, topleft, top, topright, right, center In addition: Warning message: longer

Hi Rolf, this topic is probably already saturated, but here is a tidyverse solution: ``` library(purrr) x <- list( ? `1` = c(7, 13, 1, 4, 10), ? `2` = c(2, 5,? 14, 8, 11), ? `3` = c(6, 9, 15, 12, 3) ) x |> ? pmap(~ c(..1, ..2, ..3)) |> ? reduce(c) #> [1]? 7? 2? 6 13? 5? 9? 1 14 15? 4? 8 12 10 11? 3 ``` Here, we map over the elements of the list in parallel (hence pmap),

Hello, I've written a small function that's supposed to save me some time, and it's ending up killing it- the intention is to iteratively subset a dataset fram on framevec, fit a model (either lm or nls depending on type) and return the r2 or AIC from the model, respectively. Although as far as I can tell in my code the plots are dependent on the fit of the model to the data and the

I am trying to create a plot that has multiple plot characters for each point (e.g. a point within a triangle, a triangle within a square, etc). The workaround I have found to do this is by plotting twice, as in this example: x <- c(1.1, 2.3, 4.6) y <- c(2.0, 1.6, 3.2) plot(x, y) points(x,y, pch=20, col="red", cex=0.5) This works, but perhaps there is a better way to do it in one

I want to write some text in a corner of my plot. Is it possible to get the xlim and ylim of an open window? Or is there anything similar like legend(x="bottomright", inset=0.01,legend=...) for text(x=1,y=2, "test") Thomas

Hello, I seem unable to construct a legend which contains a substitution as well as math symbols. I'm trying to do the following: strain2 <- "YJG48" legend.txt <- c( substitute( strain * %==% * "YJG45, rpn10" * %Delta%, list(strain=strain2) ), "Verhulst/Logistic", "Malthus" ) legend( 100,2.5, legend.txt, cex=0.75,

Sorry to append, but I just realised that of course ``` x |> ? pmap(c) |> ? reduce(c) |> ? unname() ``` also works and is a general solution in case your list has more than three elements. Here, we map in parallel over all elements of the list, always combining the current set of elements into a vector, and then reduce the resulting list into a vector by combining the elements

For reasons best known only to myself ( :-) ) I wish to create a data frame with 0 rows and 9 columns. The best I've been able to come up with is: junk <- as.data.frame(matrix(0,nrow=0,ncol=9)) Is there a sexier way? cheers, Rolf ###################################################################### Attention:\ This e-mail message is privileged and confid...{{dropped:9}}

Hello, I need to change the font for one of the items (C. elegans) in my legend to italic. Can someone suggest how to accomplish this? legend('bottomright', bty='n', c('C. elegans range', 'Study area'), cex=0.8, fill=c('light gray', 'white'), border=c('black','black')) I tried using lab.font=c(1,3) but R ignored and did not write

Hola a todos, Os envío una consulta que considero sencilla pero me está resultando imposible de resolver. Si ejecutáis el siguiente código, obtendréis la gráfica que os adjunto: library(ltm) modelo <- rasch(LSAT) plot(modelo, main="Curva probabilidad pregunta 1",legend = TRUE, cx = "bottomright", items=1,xlab="Conocimiento",ylab="Probabilidad") Resulta

Hi all, Does anyone have any tips for extracting chunks of data from a distance matrix? For instance, if one was interested in only a subset of distance comparisons (i.e., that of rows 4 thru 6, and no others), is there a simple way to pull that data out? >From some playing around with an example (below), I've been able to figure out that a distance matrix in R is stored as a single

I see a book coming: "666 ways to do the same thing in R ranked by sexiness." Kidding aside, if you look under the covers of some of the functions we are using, we may find we are taking steps back as some of them use others and perhaps more functionality than we need. But for a new reader , looking at many approaches may open up other ways and ideas and see the problem space as quite

Hi, I'm running this command: # upsrw -s "battery.date=2013Jun7" -u ups-mon-IT "IT-APC-BOTTOMRIGHT at 127.0.0.1" Password: OK but I'm getting this listing: # upsc IT-APC-BOTTOMRIGHT at 127.0.0.1 battery.alarm.threshold: 0 battery.charge: 100.0 battery.charge.restart: 15 battery.date: 01/10/06 battery.packs: 000 battery.runtime: 1440 battery.runtime.low: 120