similar to: 4D Plot ??

Hi, I would like to add contour lines to a (trellis/lattice-) levelplot. Sure, there is the "contour=TRUE" argument, but this uses "cuts=..." (which is usually chosen very high for my plots. I guess cuts=99 is the best you can do (?)) for plotting the contour lines. Furthermore, I do not like the numbering of the contour lines this way. Therefore, I tried to add a

Hello there, I am creating a series of images using levelplot but I also want to overlay a contour for a particular value as reference. Here is the levelplot command for the image: print(levelplot(d~x+y,data=t,cuts=20,scales=list(draw=F),xlab=NULL,ylab= NULL,col.regions=heat.colors(100)[100:1]),split=c(1,1,1,1),more=T) and then to add the contour plot (I only want a contour at level 5):

I have a complex function that I want to maximize (I have multiplied this function by -1 so that it becomes a minimization problem in the code below). This function has two equality constraints. I get the programs to run but the answer isn't correct because, when it does converge, at least one of the constraints is violated. Any suggestions? Code below Violated constraint (an easy check):

Hi everybody, I want to add a single contourline to a levelplot. While everything works fine if the value at which the line should be drawn is positive, there is an error if the value is negative: library(lattice) my.panel <- function(..., at, contour=FALSE, labels=NULL) { panel.levelplot(..., at=at, contour=contour, labels=labels) panel.contourplot(..., contour=TRUE,

I'm solving 4 complex equations simultaneously. Code is below. The code returns only zero's for the solution though there should also be a non-zero result. I'm pretty confident that the equations are correct because they are straight from a published paper and I checked them pretty thoroughly. The parameter values I used are from the published paper as well. Any suggestions for how

Estimados Usuarios-R: Estoy convirtiendo un programa en Matlab a R. El original lo saqu de: Lai, M., & Poltera, Y. (2009). Lecture with computer exercises: Modelling and simulating social systems with matlab. Tech. rep., Swiss Federal Institute of Technology (December 2009). 27. Ahora estoy convirtiendo la siguiente funcin: function sizes = gridsizes(N,varargin) % gridsizes(N) calculates

Hello, I want to add a single contour line to a levelplot but can't figure out how to do it 'on-the-fly'. When I include the last line in the code below, I get the following error: Error in NextMethod("[") : Argument "subscripts" is missing, with no default Any tips on how to fix this are greatly appreciated! Ian Jonsen

Hi, I'm doing a function that describe two populations in competition. that's the function that i wrote: exclusao<-function(n10, n20, k1, k2, alfa, beta, t){ n1<-k1-(alfa*n20) n2<-k2-(beta*n10) if(t==0){plot(t, n10, type='b', xlim=range(c(1:t),c (1:t)), ylim=range(n10, n20), xlab='tempo', ylab='tamanho populacional') points(t, n20, type='b',

Hi, Here are some new library functions for klibc. Some of them are required for udev, which currently has a klibc_fixups.c file that implements these functions. mh -- Martin Hicks Wild Open Source Inc. mort@wildopensource.com 613-266-2296 # This is a BitKeeper generated patch for the following project: # Project Name: The kernel C library # This patch format is intended

Hello...i want to find the empirical rate for type 1 error using fixed trimmed mean. To make it easy, i'm referring to journal given by this website http://www.academicjournals.org/ajmcsr/PDF/pdf2011/Yusof%20et%20al.pdf. I already run the programme and there is no error in it but i got zero for the empirical rate of type 1 error. The empirical rate for the type 1 error given in the journal

By accident, I left out the lines defining n1 and n2. Here it is as a function. Peter B. hotelling <- function(d1,d2){ k <- ncol(d1) n1 <- nrow(d1) n2 <- nrow(d2) xbar1 <- apply(d1,2,mean) xbar2 <- apply(d2,2,mean) dbar <- xbar2-xbar1 v <- ((n1-1)*var(d1)+(n2-1)*var(d2))/(n1+n2-2) t2 <- n1*n2*dbar%*%solve(v)%*%dbar/(n1+n2) f <-

Hello, I haven't found errors in your code. I implemented the test in the paper (the first, fixed symetric mean) and it also gives me zero Type I errors, when alpha = 0.05. Try to see the value of min(pv) or to plot the histogram of 'pv', hist(pv) and you'll see that there are no significant p-values, at that level. Anyway I'll continue to look at it, but my first

Hi to all the people (again), I'm doing some simulations with the memisc package of an own function, but I've the problem that I didn't know how to read the result of such simulations. My function is: > Torre<-function(a1,N1,a2,N2) + {Etorre<-(a1*N1)/(1+a1*N1) + Efuera<-(a2*N2)/(1+a2*N2) + if(Etorre>Efuera)Subir=TRUE + if (Etorre<Efuera)Subir=FALSE +

Thank you David. Using xtabs operation simplifies the code very much, many thanks ;) On Tue, Jun 6, 2017 at 7:44 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > > > Hi Bogdan, > > Kinda messy, but: > > > > N <-

Hi R-Developers, After my message about a month ago concerning attributes being lost during simple operations, Martin Maechler invited me to fix the code if I was in a hurry (especially for attributes in general). We've just made the following changes here, which appear to implement the rules in the Blue Book (pg. 257) Enclosed is the output from "diff -c" from /R-0.61.1/src

Hi, Erin: A cleaner way is to pass "n2" to "outer" as a "..." argument, as in the following modification of your code: boot1 <- function(y,method="f",p=1) { n1 <- length(y) n2 <- n1*p n3 <- n2 - 1 a <- 0.5*(outer(1:n3,1:n3,function(x,y, n2.){n2. - pmax(x,y)}, n2.=n2)) return(a) } y1 <- c( 9, 8, 7, 3, 6) boot1(y=y1,p=4)

Simple matrix indexing suffices without any fancier functionality. ## First convert M and N to character vectors -- which they should have been in the first place! M <- sort(as.character(M[,1])) N <- sort(as.character(N[,1])) ## This could be a one-liner, but I'll split it up for clarity. res <-matrix(NA, length(M),length(N),dimnames = list(M,N)) res[as.matrix(C[,2:1])] <-

Hi I have a couple of functions that work on the object created by another R command and then print out or summarise the results of this work. The main function is defined as: hotelling.t <- function(obj) { #internal commands } I then have print.hotelling.t() that takes the list returned by hotelling.t and prints it with some extra significance calculations, formatting, etc. I want to

Full_Name: Martha Nason Version: 2.0.1 OS: Windows XP Submission from: (NULL) (137.187.154.154) I am running simulations using fisher's test on 2 x c tables and a very small p.value from fisher's test (<2.2e-16) is returned as a negative number. Code follows. > set.seed(0) > nreps.outer <-7 > pvalue.fisher <- rep(NA,nreps.outer) > > population1 <- c(

Here's another approach: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400)) # Rebuild the factors using M and N C$m <-