similar to: "formula plotting" -> substitute pecularity

Hi: I am looking for some help in making two boxplots next to each other. I have a data like this: N1 T1 N2 T2 N3 T3 N4 T4 ... Nn Tn 7 8.2 4 5 8 10 4 5 ..... 10 11 I want to have box plot for all Normal samples (N1,N2,N3,N4,,,,Nn) and another box plot for all tumors (T1,T2,T3,T4,...Tn). I have data in a numeric class. If data is represented as N1

Hi Folks, I've got a question concerning the calculation of the Schwarz-Criterion (BIC) done by summary.regsubsets() of the leaps-package: Using regsubsets() to perform subset-selection I receive an regsubsets object that can be summarized by summary.regsubsets(). After this operation the resulting summary contains a vector of BIC-values representing models of size i=1,...,K. My problem

This is doing my head in. Staying away from R for too long is bad for the health! I have two vectors of character names where there may be repetition and from which I want to form two factors with the same levels but only if there are more than N instances of each name in each vector. I can get the list of common names quite easily, using:

Hi wizards, I need to know how can I generate this numbers I have n numbers N1,N2 , Nn , but Nn = 1- sum(N1+N2+...+Nn-1) and sum(N1+N2+..Nn)=1 and N1,N2,..Nn with 0<N<1 . Does somebody know how to generate it ? Some distribution or algorithm. Thanks in advance. -- Web Page http://www.geocities.com/ricardo_rios_sv/index.html

I don't have time now to investigate myself, and I'm not feeling like deciding myself if the following is a bug: myplot <- function(dat, cex = 1.2, ...) { if(!is.data.frame(dat <- as.data.frame(dat))) stop("`dat' must be a data.frame") if(any(is.na(match(c("x","y"), names(dat))))) stop("`dat' must have a `x' and a

Hi. I have a machine with a local DHCP server and a couple of virtual networks and I've configured the server for each virtual interface, so that I would be able to install VMs on the corresponding subnets using PXE. The problem is that the two DHCP servers (my local server and dnsmasq) are conflicting with each other causing the boot process to either fails or takes ages untill a VM can

I have the following problem. I have an automatically generated named list with "stringified" names: a <- list("A"=..., "B"=..., "C"=..., ) then I want to refer to the elements of the list, stored as an vector of names: nn <- c("A", "B", "C"), so that I could get list elements like a$nn[1], a$nn[2], etc. Obviously it

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

Hi Martin, When I attended the LinuxWorld Expo in NYC back in January, I chatted with some folks at the AMD booth, as well as guys from Penguin Computing (where we bought our Opteron box). I was told that the Operton has this somewhat strange setup that the memory is controlled by one CPU. The net effect of this being that when both CPUs are running, one might only be running at around 90%

Hi, Here is a listw object z corresponding to the matrix W. I understand n, nn, S0, S1 and S2 in the weights constants summary. Is it simply so that n1 = n-1, n2 = n-2 and n3 = n-3? If this is true where they are needed? Just wondering Osmo > W [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [1,] 0 1 0 1 0 0 0 0 0 [2,] 1 0 1 0 1 0

Kia Ora > vec_size(data.frame(x = 1:10)) > #> [1] 10 > vec_size(array(dim = c(10, 4, 1))) > #> [1] 10 I think that "length" (in the R way) or "size" should incorporate the number of columns (or other dimension information). How about: > vec_npoints (x) > vec_nrecords (x) > vec_nobs (x) Depending on whether you want to promote geometry, databases

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

Hello r helpers! Below is the whole coding for my programme. Before proceed more further, let me explain for you. First of all, I need to compute trimmed mean. Till that step is ok. Then I need to compute ssdw which is sum of square deviation. If I do equal trimming at both tail of distribution that I chose, I will use the first ssd formulae which is "a". But if I am doing unequal

Hi, final<-data.frame() ?? for (m1 in 4:10) { ?????? for (n1 in 4:10){? ?????????? for (x1 in 0: m1) { ????????????? for (y1 in 0: n1) { final<- rbind(final,c(m1,n1,x1,y1)) res}}}} ?final1<-within(final,{flag<-ifelse(x1/m1>y1/n1, 1,0)}) ?head(final1) #? m1 n1 x1 y1 flag #1? 4? 4? 0? 0??? 0 #2? 4? 4? 0? 1??? 0 #3? 4? 4? 0? 2??? 0 #4? 4? 4? 0? 3??? 0 #5? 4? 4? 0? 4??? 0 #6? 4? 4?

I would be more strong on this advice: learn to think in R, rather than thinking in Python, when programming in R. R has atomic vectors... Python does not (until you import a package that implements them). I find that while it is possible to import R thinking into Python, Python programmers seem to object for stylistic reasons even though such thinking speeds up Python also. A key step in that

The one difference I see, is the necessity to pass the dots to the function dotlength : dotlength <- function(...) nargs() myfun <- function(..., someArg = 1){ n1 <- ...length() n2 <- dotlength() n3 <- dotlength(...) return(c(n1, n2, n3)) } myfun(stop("A"), stop("B"), someArg = stop("c")) I don't really see immediately how one can

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',

Dear R-users, I am trying to translate a matlab code for calculating the Local Whittle estimator in time series with long memory originally written by Shimotsu and available free in his webpage ( http://www.econ.queensu.ca/pub/faculty/shimotsu/ ) The Matlab code is ======================================================================================= function[r] = whittle(d,x,m) % WHITTLE.M

Windows 7, R 2.12.1 Colleagues, I am trying to understand the n.for.2means function. The code below is a copy of the function (renamed to n.for.2means.js). I have inserted a single line of code towards the bottom of the function which uses the cat function to print the value of n1. You will note the value (preceded by stars) is printed as 1. The function (1) prints a lot of output without any

Dear all, Here I come with another stupid question. Suppose I want to use nls() to fit a series of data (here modelled by generated points), then plot the points and the fitting curve. I figured out some way of doing it: x <- runif(1:20, 0, 10) y <- 0.1*x^2 - rep(3, length(x)) + rnorm(length(x), sd = 0.5) yfit <- nls(y ~ a*x^2 + b*x + c, start = list(a = 1, b = 1, c = 1),