similar to: Hopefully an easy error bar question

All, I've been using "sub" (subtitle) instead of "main" such that captions are below figures produced by xyplot. This works fine and captions are on a single line. However, when I try this for bar plots with error bars (altering the error.bars function form Crawley's The R Book, see below), the captions are split on more than 1 line. Is there a way to get the

I like to use complex numbers for 2-dimensional geometry. E.g., > polyAngles2 function (xV, yV) { stopifnot((length(xV) == length(yV)) && (length(xV) >= 3)) z <- complex(re = xV, im = yV) c(NA, diff(Arg(diff(z))), NA) # radians, positive is counter-clockwise } > x <- c(0:3) > y <- c(0,0,1,1) > polyAngles2(x,y) / pi * 180 [1] NA 45 -45 NA Bill

Assuming your polyline is defined by two vectors, one for the x coordinates, one for the y coordinates, you can try the following library(NISTunits) polyangles <- function(xV,yV) { stopifnot( (length(xV)==length(yV)) && (length(xV) >= 3)) v <- function(i) { c( xV[i]-xV[i-1], yV[i]-yV[i-1])} vlen <- function(v) { sqrt(sum(v*v)) } lV <- rep(NA_real_,length(xV)) for

Dear R-help Can anyone tell me why my curve appears the wrong way round on a rug plot? I am using the same code as on pg 596 of the Crawley R-book. mod<-glm(mort~logBd,binomial) par(mfrow=c(2,2)) xv<-seq(0,8,0.01) yv<-predict(mod,list(logBd=xv),type="response") plot(logBd,mort) lines(xv,yv) I've tried swapping xv and yv around but no luck. Thanks, Pete

Hi All, I'm having some trouble assigning arguments inside a function to produce a plot from a model Can anyone help me? Below I've outlined the situation and examples of failing and working code. Regards Mike ## data ## decay.data <- ... behaviors lift reach.uu estimated.conversions.uu total.reach 1 1 432.0083 770 770 0.00 2

Hello, I am trying to fit a model to some "death over time" data that does not fit the criteria for the usual LD50 type models (the counts are too large). I am using a simple linear model in an attempt to plot a nice line on a scatter plot and calculate some LD values to use in designing an experiment. Here is the basic idea of what I'm doing: head(mort) Time Density 0

Dear all, I am quite new with R and I have a problem with plotting a binomial regression line in a plot. This is what I type in: > model<-glm(Para~Size,binomial) > par(mfrow=c(1,1)) > xv<-seq(3.2,4.5,0.01) > yv<-predict(model,list(area=xv),type="response") > plot(Size,Para) > lines(xv,yv) The error message that I get is: > Error in xy.coords(x, y) :

Hello there I have a list, Y, and each component of that list is a real-valued function (that is, Y[[i]](u) returns a number). I was wishing to build the mean function and the first thing I thought of was Ybar<-function(u){ mean(Y[[1:n]](u)) } but obviously this doesn't work, since Y[[1:n]] is not allowed at all. Any elegant suggestions? I really would like to avoid something like

On 15/06/17 05:29, David Winsemius wrote: > >> On Jun 14, 2017, at 10:18 AM, David Winsemius <dwinsemius at comcast.net> wrote: >> >> >>> On Jun 14, 2017, at 9:46 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: >>> >>> I don't see a question. If your question is whether R supports pattern fills, AFAIK it does not. If that is

Hello all, I'm a beginner in R working on a script that will produce a set of models (linear, polynomial and logistic) for each location in a dataset. However, the self-starting logistic model often fails - if this happens I would like to just skip to the next iteration of the loop using tryCatch. I've looked at a few examples and read the help file, but didn't understand tryCatch

> On Jun 14, 2017, at 10:18 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > >> On Jun 14, 2017, at 9:46 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: >> >> I don't see a question. If your question is whether R supports pattern fills, AFAIK it does not. If that is not your question, ask one. >> -- >> Sent from my

Please excuse me for asking such basic questions: Here is my code > y=c(1.21,0.51,0.14,1.62,-0.8,0.72,-1.71,0.84,0.02,-0.12) > ybar=mean(y) > ll=length(y); > ybarv=rep(ybar,ll) > x=1:ll > plot(x,ybarv,pch=1) > segments(x[1],ybar,x[ll],ybar) What I get is a collection of small circles, with a segment "on top" of the circles, which is almost what I want. But I

Dear forum, Does anyone know any functions for plotting vertical error bars, if the x-axis variable is a date? I can produce a pretty plot with dates on the x-axis, if I don't want to include error bars. I can also produce a nice plot with error bars, if I just use Julian days for the x-axis, but I'd much rather have the actual dates. I have tried writing my own function, plotCI

#See David Williams' book "Weighing the odds", p286 y <- c(1.21, 0.51, 0.14, 1.62, -0.8, 0.72, -1.71, 0.84, 0.02, -0.12) ybar <- mean(y) ylength <- length(y) ybarv <- rep(ybar, ylength) x <- 1:ylength plot(x,y,asp=1,xlab="position",ylab="ybar",type="n",ylim=c(-1,1)) segments(x[1], ybar, x[ylength], ybar) segments(x,ybarv,x,y)

Hi I had to use a glm instead of my basic lm on some data due to unconstant variance. now, when I plot the model over the data, how can I easily get the 95% confidence interval that sormally coming from: > yv <- predict(modelVar,list(aveLength=xv),int="c") > matlines(xv,yv,lty=c(1,2,2)) There is no "interval" argument to pass to the

> On Jun 14, 2017, at 1:53 PM, Rolf Turner <r.turner at auckland.ac.nz> wrote: > > On 15/06/17 05:29, David Winsemius wrote: >>> On Jun 14, 2017, at 10:18 AM, David Winsemius <dwinsemius at comcast.net> wrote: >>> >>> >>>> On Jun 14, 2017, at 9:46 AM, Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: >>>>

Sigh. I never load packages in .Rprofile to avoid the irreproducibility trap. Might seem drastic to some, but I don't feel much pain because I almost always edit my code in a file rather than on the fly at the console, and re-run it frequently from a fresh R process to check my progress. -- Sent from my phone. Please excuse my brevity. On June 14, 2017 3:27:15 PM PDT, David Winsemius

Envoy? depuis mon appareil Samsung -------- Message d'origine -------- De : Rolf Turner <r.turner at auckland.ac.nz> Date : 14/06/2017 22:53 (GMT+01:00) ? : David Winsemius <dwinsemius at comcast.net> Cc : r-help at r-project.org Objet : Re: [R] [FORGED] Re:&nbsp; draw stripes in a circle in R On 15/06/17 05:29, David Winsemius wrote: > >> On Jun 14, 2017, at

dear community, I am a beginer in R , and can't predict with logistic model in package logistf, could anyone help me ? thanks ! the following is my command and result : >library(logistf) >data(sex2) >fit<-logistf(case ~ age+oc+vic+vicl+vis+dia, data=sex2) >predict(fit,newdata=sex2) Error in predict(fit, newdata = sex2) : no applicable method for "predict"

Hello: If anyone could guide me with this I would greatly appreciate it. Thanking you in advance for your assistance. Using a 3-level input factor alternative so that a function(below) can compute both a two-sided and one-sided p-values. Making the two-sided test the default. And produce output information about which alternative was tested. Where would I place the ifelse statement?