similar to: Test for multiple contrasts?

Hi R-people, I am wanting to run Factorial ANOVA followed by Scheffe tests on some spatial subjective data. I'm comparing X-Y independent coordinates against x-y dependent coordinates. There are only four independent spatial coordinates that form a square. I am wondering whether I am doing the right thing, because there doesn't seem to be a simple way of doing this. I have attempted to

Hello everyone I'm trying to use Scheffe's complete cubic model (mixture design). In the bibliographies, they indicate that the term is of the type: A * B * (A-B). But I see that trying to adjust the three cubic terms results in singularities. I know this implies not having the inverse matrix: solve (t (X)% *% X) does not exist. The bibliographies show all three cubic terms. So my

Hello all, to obtain a list of all objects in all search paths, I've found the following to work: > biglist <- sapply(1:length(search()), objects) This more obvious one, however, does not work: > biglist <- sapply(search(), objects) Error in pos.to.env(pos) : invalid argument Still, search() gives [1] ".GlobalEnv" "package:ctest" "Autoloads"

I think this is pretty certainly a bug, so I'm cc'ing this to r-bugs. -p Kaspar Pflugshaupt <pflugshaupt@geobot.umnw.ethz.ch> writes: > Hello, > > first, I'd like to congratulate the core team to the new R version 1.1. I > think it's a great update, with glimpses into an even greater future > (tcltk!). > > While playing around with the new functions

This is not a bug. There is no way one can scale the plot axes equally when no indication has been given of the scaling expected for either! It has always seemd to be that plot(x=1, y=1) shoul dnot work: it has not supplied enough information! On Thu, 8 Nov 2001 pflugshaupt@geobot.umnw.ethz.ch wrote: > I found the following in eqscplot, library MASS: > > > When given just one

Currently (R-1.4.1 as well as R-devel, according to http://stat.ethz.ch/R-alpha/R-devel/library/base/html/par.html), the html version of help(par) shows [...] lab A numerical vector of the form c(x, y, len) which modifies the way that axes are annotated. The values of x and y give the (approximate) number of tickmarks on the x and y axes and len specifies the label size. The default

I found the following in eqscplot, library MASS: When given just one point, it fails: > eqscplot(x=1, y=1) Error in plot.window(xlim, ylim, log, asp, ...) : need finite xlim values After inserting browser() just before the last line in eqscplot() (in which plot() gets called), I found the following: > eqscplot(x=1, y=1) Called from: eqscplot(x = 1, y = 1) Browse[1]> xlim [1] NaN

Hello, I've a question concerning the behaviour of the "scale" function in the base package. I'm using R 1.1.1 on Windows 95. If I take a matrix with NA values, such as > tm <- matrix(c(2,1,0,1,0,NA,NA,NA,0), nrow=3) > tm [,1] [,2] [,3] [1,] 2 1 NA [2,] 1 0 NA [3,] 0 NA 0 and scale it, the columns containing NAs come out all NA: >

Hello, I'd like to propose an extension to the function summary.aov. In Splus (2000, I don't know about other versions), summary.aov allows a parameter ssType to be set to 1 or 3 (defaults to 1) to choose the type of Sums of Squares. I know I can get Type III SS in R with drop1(model), but including the functionality into summary.aov would, in my opinion, - yield a more usable table

I had run into problems when compiling some contributed packages to an installation of R 1.40 by fink on MacOS X. Namely, packages KernSmooth (2.22-7) and cluster (1.4-0) would not find required libraries, though those were present on the system. Jeff Whitaker, the maintainer of the R fink packages, kindly sent me this: > Kaspar: It's not looking in /sw/lib, where the libs are (the

Hello, first, I'd like to congratulate the core team to the new R version 1.1. I think it's a great update, with glimpses into an even greater future (tcltk!). While playing around with the new functions (on Win 95), I found the following: As the documentation states, when I generate a plot and save it with recordPlot, I can regenerate it by printing the variable: > plot(1:10) >

Hello, does anyone know of a quick way to create a data frame "like" another, but with more rows? What I'd like to do is this: if mydata is a data.frame like a b c 1 TRUE yes 2 FALSE no 3 TRUE yes I'd like to get mydata2 with the same column names and column types, but without the values and with more rows. All I could think of was to manually do

Hello, to save searching time in the S-news archive, here is the simple solution to merge data frames vertically (cols must be equivalent, of course). The frames must be components of a list, such as produced by framelist<-split(bigframe,factor). bigframe2<-do.call("rbind",framelist) # posted to S-news by Bill Venables, found in summary message

(message from 22.2.2000 13:04 Uhr): > > > Why don't you inverse the modelling instead: > > t.m.i <- lm((x~y) > Jan, thanks for the tip, but it's not just the same. The coefficients come out differently, since the squared y residuals are minimized. Orthogonal regression would be symmetric, but least squares is not, I'm afraid. And, what's more, I have to

Hello, after some searching in the mailing list archives, R-Help and S-Plus-Help, I dare asking the question here: How can I do inverse prediction from a lm model with R? What I want to achieve is this: predict term values for given new response values. If I've got a model t.m<-lm(y?x) I would type something like inverse.predict(t.m, newdata=data.frame(y=1:10)) which would give me new

Hello, after I've read a few answers from the list, I realised that I stated my problem incorrectly: my x values are not _fixed_, as I wrote (I did not set them). They just have no errors to them, but, apart from that, are random. I think I shall be reading up on calibration next. Many thanks for the references! Kaspar -- Kaspar Pflugshaupt Geobotanisches Institut Zuerichbergstr. 38

Hello all, here's a collection of answers I got on my question concerning partial correlation coefficients: Some people gave a simple formula for the three-variable-case, as did Dave Lucy: pcor <- function(v1, v2, v3) { c12 <- cor(v1, v2) c23 <- cor(v2, v3) c13 <- cor(v1, v3) partial <- (c12-(c13*c23))/(sqrt(1-(c13^2)) * sqrt(1-(c23^2)))

Hello, this is probably more a statistical question than an R-specific problem, but I'll risk it. I've fitted a Cox Proportional hazard model with one factor Treatment (seven levels) as a predictor variable. The general Null hypothesis (all groups show the same survival behaviour) is clearly rejected. Now, is there any (statistically sensible) way of doing pairwise comparisons and/or

Hello, I'm using R 1.40 on MacOS X X.1.2 (installed via the fink package manager). To upgrade my installed packages, I tried to use update.packages() today. All went well for most packages, with the exception of KernSmooth and cluster. In both cases, libraries were not found although I think they are present. Here's what happened: ---------------------------------- >

Hello, is there a quick way of estimating Weibull parameters for some data points that are assumed to be Weibull-distributed? I guess I'm just too lazy to set up a Maximum-Likelihood estimation... ...but maybe there is a simpler way? Thanks for any hint (and yes, I've read help(Weibull) ;) Kaspar Pflugshaupt -- Kaspar Pflugshaupt Geobotanical Institute ETH Zurich, Switzerland