similar to: barplot as histogram

Dear R-users Idea: Plot a dnorm line using specific mean/sd to complete a histogram (skewed). xs:range of y-values, ys: dnorm function Problem: I expected to multiply the ys function with the sample size (n=250-300). I was wondering about a factor between 12'000 and 30'000 to match the size of the dnorm line with the specific histogram. Thanks Sibylle hist(Biotree[Ld,]$Height2008,

Here's some code to make lissajous dance. I've attached a small sample GIF. Cheers, Rob Steele robsteele at yahoo dot com plot.lissajous = function(omega.x, omega.y, delta = 0, num.thetas = 200) { thetas = seq(0, 2 * pi, length = num.thetas) xs = sin(omega.x * thetas + delta) ys = cos(omega.y * thetas) plot(xs, ys, type = 'l', lwd = 3, ann = FALSE, axes = FALSE) }

Dear R-List users, Can anyone explain exactly the difference between Weights options in lm glm and gls? I try the following codes, but the results are different. > lm1 Call: lm(formula = y ~ x) Coefficients: (Intercept) x 0.1183 7.3075 > lm2 Call: lm(formula = y ~ x, weights = W) Coefficients: (Intercept) x 0.04193 7.30660 > lm3 Call:

I've written a simple function to draw a regression line in a plot and annotate the line showing the slope with a label. It works, as I'm using it, when the horizontal variable is 'x', but gives incorrect results otherwise. What's wrong? # simple function to show the slope of a line show.beta <- function(model, x="x", x1, x2, label, col="black", ...)

Pervasives is deprecated since Ocaml 4.08 and has been removed in OCaml 5.0. https://github.com/ocaml/ocaml/pull/1605 --- mlstdutils/std_utils.ml | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/mlstdutils/std_utils.ml b/mlstdutils/std_utils.ml index 0d2fa22..86b21a7 100644 --- a/mlstdutils/std_utils.ml +++ b/mlstdutils/std_utils.ml @@ -341,12 +341,12 @@ module List =

Full_Name: Stephen Eglen Version: 1.4 OS: Redhat Linux 7.1 Submission from: (NULL) (128.252.204.36) abline() produces a spurious line in addition to the correct line, at least with the X11 driver. The postscript file generated also has nan values in it, which causes an error under ghostscript. xs <- c(0, 150, 300) ys <- c(75, 40, 23) par(mfrow=c(1,2)) plot(xs,ys, log="")

On Thursday 07 July 2016 17:30:03 Richard W.M. Jones wrote: > This adds imperative list manipulation functions inspired by Perl. > The functions are passed list refs which get updated in place. > > This allows us to replace some awkward pure functional code like: > > let xs = ys in > let xs = if foo then xs @ zs else xs in > > with: > > let xs = ref ys in

Hi everyone, i've been trying to make a special plot with ggplot2, but I can't get it to fill the polygon I'd like to see filled so very very much. I want to display the difference or change in the distribution of the modified Rankin Scale between two groups. mRS is a scale for disability or daily activities competence. It looks like this.

On Thu, Jul 07, 2016 at 06:08:43PM +0100, Richard W.M. Jones wrote: > On Thu, Jul 07, 2016 at 07:00:46PM +0200, Pino Toscano wrote: > > On Thursday 07 July 2016 17:30:03 Richard W.M. Jones wrote: > > > This adds imperative list manipulation functions inspired by Perl. > > > The functions are passed list refs which get updated in place. > > > > > > This

Hi, It is most probably just my R-ignorance, but I have following problem with using predict(). I train the model using 164 cases and then I try to use it on the data set with 35 cases, but I am getting 164 predictions ? R-code below illustrates in more detail what I am doing. Truly yours, R train = read.csv("train.csv", header = TRUE, row.names = "mol",

Dear all, I considered an ordinary ridge regression problem. I followed three different ways: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-t(as.matrix(cbind(2,3,4,5))) n<-nrow(X) p<-ncol(X) #Without

Hello all I'm trying to do a simulation that involves identifying the minimum point between two peaks of a (usually) bimodal distribution. I can do this easily if there are only two peaks: CnBdens<-density(Ys/Xs) #probability density function for ratio of Ys to Xs for(p in 1:512) ifelse(CnBdens$y[p]>CnBdens$y[p-1],peak1<-p,break) #identifies first peak in probability

Dear all, For an ordinary ridge regression problem, I followed three different approaches: 1. estimate beta without any standardization 2. estimate standardized beta (standardizing X and y) and then again convert back 3. estimate beta using lm.ridge() function X<-matrix(c(1,2,9,3,2,4,7,2,3,5,9,1),4,3) y<-as.matrix(c(2,3,4,5)) n<-nrow(X) p<-ncol(X) #Without standardization

Nicolas Capens wrote: > Here's essentially what I try to generate: > > void add(float z[4], float x[4], float y[4]) > { > z[0] = x[0] + y[0]; > z[1] = x[1] + y[1]; > z[2] = x[2] + y[2]; > z[3] = x[3] + y[3]; > } This is the vectorized llvm-assembly equivalent: ----- define void @add(<4 x float>* %z, <4 x float>* %x, <4 x float>* %y) {

I am using following codes to plot map library(sp) library(rgdal) library(maps) library(gplots) library(clim.pact) library(fields) source("/R/PlotGridded2DMap.R") source("/R/image.plot.fix.R") source("/R/image.plot.plt.fix.r") seasonal_plot<-function(input,lonll=-180,latll=-90,lonres=5.,latres=3.75,write_file=TRUE,The_title=NULL){ if(is.null(The_title)){

Dear R users/developers I requested help to solve the problem of formulating Multivariate Sample selection model by using Full Information Maximum Likelihood (FIML)estimation method. I could not get any response. I formulated the following code of FIML to analyse univariate sample selection problem. Would you please advise me where is my problem library (sem) library(nrmlepln) Selection

Dear all, is there any function to calculate the center of a polygon mass in R? Actually I need to find the best location within polygons to place labels. Thanks for any hint Jens Oehlschl?gel -- COMPUTERBILD 15/03: Premium-e-mail-Dienste im Test\ --------...{{dropped}}

Hello All, I wanted to perform a fourier transform on high frequency financial data. I have searched and have not found much on this topic for R. I was wondering if anyone has used any libraries for it or have come across any papers I may read. Many Thanks, Neil Gupta [[alternative HTML version deleted]]

Dear R helpers, I would like to have abline, for a lm model for example, lying within data range. Do you know how to get it? Thank in advance Nguyen D Nguyen #CODE x<- rnorm(200, 35,5) y<- rnorm(200, 0.87,0.12) plot(y~x, xlim=c(0,50), pch=17, bty="l") abline(lm(y~x)) # I would like abline is between min(x) and max(x) [[alternative HTML version deleted]]

Hello I have a model with several hundred Y variables, and also several 1000 X variables. The model is linear lm(Y ~ X). My questions are: 1.- how to avoid writing all Xs variables? is list() the right function? 2.- about the multiple Ys with dependence among some of them, how to incorporate that information in the linear model? Thank you Rosario