similar to: Change cex.axis in biplot()

Hi, I'm not sure if this can be done but.. I know that with ifelse() I can do something like: ifelse(x <= 3, 1, 2) to go through each element in my vector x, and if x_i <= 3 substitute the number with 1 else with 2. Essentially I'll get a vector with 2 levels. Can I tweak it so I can get 3-levels? For example: if(x <= 3) then 1 elseif(3 < x <= 4) then 2

Hi, I'm doing a presentation on Neural Networks and Tree-Based Models in two weeks, at the moment I'm looking for a data set to use in the presentation. What I would like to use is a good old data, like the Iris data, that is already known by every statisticians. MASS4 uses the cpus data in Chapter 8.10 and the Cushing's syndrome in Chapter 12.4. These two data sets plus the

Hi, If I have: foo <- survfit(y ~ x) where y is a survival object and x is a n-level factor. The documentation says when I plot(foo), the confidence intervals will not be plotted (which I guess is understandable as otherwise the plot will get really messy). I tried to plot with confidence intervals by using: plot(foo, conf.int = TRUE) and indeed the resulting plot is messy. However

Hi, I've been looking through the documentation for sample(), but can only get it to work with vectors. Is it possible to sample from a dataframe? -- Cheers, Kevin ------------------------------------------------------------------------------ /* Time is the greatest teacher, unfortunately it kills its students */ -- Ko-Kang Kevin Wang Master of Science (MSc) Student SLC Tutor and Lab

Hi, I have a column in a dataframe in the form of: > as.vector(SLDATX[1:20]) [1] "1/6/1986" "1/17/1986" "2/2/1986" "2/4/1986" "2/4/1986" [6] "2/21/1986" "3/6/1986" "3/25/1986" "4/6/1986" "4/10/1986" [11] "4/23/1986" "4/30/1986" "5/8/1986"

Hi, Say I have a factor with 20-levels: 1, 2, 3, ..., 20, called foo. If I do plot(foo) it will draw a barplot. But the x-axis is sorted alphanumerically, i.e. 1, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 2, 3, ..., 9, which is not what I want. I'd like to x-axis to be in the order of 1 ~ 20 numerically. So, the question is, how do I change the order on the x-axis in this case? --

Hi, I hope I haven't got this wrong, but I don't think one needs Perl installed if they install the "precompiled" packages for Windows right? I'm just wondering if the first sentence in the "R Installation and Administration" can be re-worded a bit so newbies won't get confused.... ---------- Forwarded message ---------- Date: Mon, 2 Jun 2003 08:56:37 +1200

Hi, Suppose I have: # Fit a base model d1.ph <- coxph(Surv(start, stop, event)~ ejec + diavol + score + smoking + beta + surg.done, data = data.frame(foo)) summary(update(d1.ph, . ~ . + td1)) summary(update(d1.ph, . ~ . + td2)) As I have many columns in my data frame, foo, called td's. e.g. td1, td2, td3, .... And I'd like to

Hi, What does this mean when I have something like: > qda.boot <- boot(train, qda.bootstrap, R = 500) Error in qda.default(structure(data.matrix(x), class = "matrix"), ...) : Rank deficiency in group M with my qda.bootstrap() looks something like: > qda.bootstrap <- function(data, index) { + boot.qda <- qda(x = data[index, 2:9], group = data[index, 1]) + qda.pred

Hi, Suppose I have: > summary(manova(plank.man)) Df Pillai approx F num Df den Df Pr(>F) plankton.new[, 1] 1 0.5267 9.8316 6 53 2.849e-07 *** Residuals 58 --- Signif. codes: 0 `***' 0.001 `**' 0.01 `*' 0.05 `.' 0.1 ` ' 1 My understanding is the MANOVA summary returns a list.

Hi, Does anyone know what the error message mean? > Boot2.Pillai <- function(x, ind) { + x <- as.matrix(x[,2:ncol(x)]) + boot.x <- as.factor(x[ind, 1]) + boot.man <- manova(x ~ boot.x) + summary(manova(boot.man))[[4]][[3]] + } > > man.res <- manova(as.matrix(pl.nosite) ~ + as.factor(plankton.new[,1]))$residuals > boot2.plank <-

Hi, Suppose I use the following codes to read in a data set. ############################################### > rating <- scan("../Data/Rating.csv", + what = list( + usage = "", + mileage = 0, + sex = "", + excess = "", + ncd = "", +

Hi, This may actually be a theoretical question. When I tried to do the following: ########################################################## > colnames(rating.adclms) [1] "usage" "mileage" "sex" "excess" "ncd" [6] "primage" "minage" "drivers" "district" "cargroup" [11]

Hey, R-listers I was recommended to try using smooth.spline function for estimating 2-Dimensinal curve given a data set. So will you please tell me where to get this R function? Or which package provides this function? Thanks for your point. Fred

> biplot standardGeneric for "biplot" defined from package "stats" function (x, ...) standardGeneric("biplot") <environment: 0x0d4444d8> Methods may be defined for arguments: x Use showMethods("biplot") for currently available ones. > > > showMethods("biplot") Function: biplot (package stats) x="ANY"

I'm attempting to modify how biplot draws its red vectors (among other things). This is how I've started: Biplot <- function(xx, comps = c(1, 2), cex = c(.6, .4)) { ## Purpose: Makes a biplot with princomp() object to not show arrows ## ---------------------------------------------------------------------- ## Arguments: xx is an object made using princomp() ##

Hi, I am trying to understand how the biplot.prcomp is constructed so I can manipulate it to emphasise particular observations and reduce the number of variables shown. The prcomp model I have ran has cor=TRUE and scale=TRUE I have worked out from looking at str(prcomp.model) that... prcomp.model$x = the observations ploted in the biplot prcomp.model$rotation = the variables that form the

Hello all, I just started in on R today. I am at the point where I am trying to distinguish different groups of data (hydrochemical data) in biplot. I search and found a post about converting the sample number in biplot to a symbol like x or + "> temp <- matrix(runif(50), nrow=10) > temp.pca <- princomp(temp) > biplot(temp.pca, xlabs=c("A", "A",

Hi everybody. I'm a new R user and i've been searching a tool for construction of regression tree... I found function "tree()" written by a certain Mr. Ripley, and seems to be just what i'm looking for, but when i try to use it in R replies me:"Object not found". So I was wandering if I should include one special library or something like that? Thanks a lot, ana

Hi everybody, Excuse me for this silly question, but after searching the help archives I'm still unable to find my way to read the source code of the "biplot" function in R. I have installed the mvbutils package, and tried: > fixr(biplot) which only gives me: function (x, ...) UseMethod("biplot") with no further details.... I'd like to read the code to see if