similar to: Colors in interaction plots

Hi, Try either: tolerance <- read.csv("http://www.ats.ucla.edu/stat/r/examples/alda/data/tolerance1.txt") ?aggregate(exposure~male,data=tolerance,mean) ?# male exposure #1??? 0 1.246667 #2??? 1 1.120000 #or ?library(plyr) ?ddply(tolerance,.(male),summarize,exposure=mean(exposure)) #? male exposure #1??? 0 1.246667 #2??? 1 1.120000 #or

Dear Madame, Dear Sir, I am able to obtain the coefficients from a 'summary' of 'lm', but NOT from a 'summary' of 'aov'. The following example shows my steps. ## Initialize rm(list = ls()) # remove (almost) everything in the working environment utils::data(npk, package="MASS") # get data model <- yield ~ block + N*P*K ## Using lm npk.lm <-

Hi all, Perhaps this in an inappropriate post, but I've found a bug in proj I'd like to track down a bit further before making a formal bug report. The example below shows the problem, if you change the rownames proj fails. The problem seems to be that there is a mismatch in the rownames in the qr objects constructed by aov and the rownames that proj is expecting them to have.

Hi Ruser I'm trying to replicate some SAS code. I have to add a spline to my longitudinal spaghetti plot. I have the plot, but I can't add the spline, a overall trend line. In the SAS code they use the command 'I=SM50S' and I would prefer something similar. I?m using R 2.10.1 on windows XP? I have made this working example. tolerance.pp <-

Hi, This might be a minor bug, or it could be that I'm just mis-using the tools (again). If you modify the example for model.tables to introduce an NA, N <- c(0,1,0,1,1,1,0,0,0,1,1,0,1,1,0,0,1,0,1,0,1,1,0,0) P <- c(1,1,0,0,0,1,0,1,1,1,0,0,0,1,0,1,1,0,0,1,0,1,1,0) K <- c(1,0,0,1,0,1,1,0,0,1,0,1,0,1,1,0,0,0,1,1,1,0,1,0) yield <-

Dear listmembers, I would like to define a class with a slot that takes either an object of class aov or NULL. I have been reading "S4 Classes in 15 pages more or less" and "Lecture: S4 classes and methods" #First I tried with list and NULL setClass(listOrNULL") setIs("list", "listOrNULL") setIs("NULL", "listOrNULL") #doesn't

I'm baffled. When I run make check after installing from source, I get a Error 2. From my understanding of how these things work, it would appear to be coming from this (as at the end of base-Ex.Rout.fail: > has.VR <- require(MASS, quietly = TRUE) Attaching package 'MASS': The following object(s) are masked from package:base : confint confint.lm nclass.FD nclass.scott

Dear statistitians / R users, I was told to analyse the effects of the four factors using lm or update for aov. The following is the result from lm. As you can see that in the last few lines enclosed coefficients. 1)Does this mean I can write a formula, like yield =0.26 + 0.03*H + 1.48*T + 0.04*L + 0.004*C 2) in the two levels design, is there any difference between I use (-1, 1) to

The following R commands were written: >help.search("factorial") >data(npk) >npk >coef(npk.aov) In the output of coef command, please explain me the interpretation of coefficients of block1 to block 6 in this 2^3 confounded factorial experiment. Thanks. [[alternative HTML version deleted]]

Dear DevelopeRs, in continuing with my suite of packages on experimental design, I am stuck with an issue that appears to be related to package AlgDesign - I have tried to get it solved by Bob Wheeler, but he seems to be stuck as well. Whenever AlgDesign is loaded, some of my code does not work any more. For example, in a fresh R session: require(DoE.base) fac.design(nlevels=c(2,6,2))

Hello, I noticed that model.tables fails when applied to an aov() fit if called inside a function. The problem seems to occur when as.formula is used inside a function on a string containing "<formula> + Error( x / y )" The reason I tried to use as.formula is to generate dynamic calls to aov(). Here is a minimal example illustrating the problem: ## Example test <-

Hi all, I have been trying to reproduce an analysis from Douglas Montgomery?s book on design and analysis of experiments. Table 6.10 of example 6.2 on page 246, gives a table as follows: > NPK <- expand.grid(A=mp,B=mp,C=mp,D=mp) > Rate <- c(45,71,48,65,68,60,80,65,43,100,45,104,75,86,70,96) > filtration <- cbind(NPK,Rate) > filtration A B C D Rate 1 - - - - 45 2

Hi, I am analyzing a data set with greater than 1000 independent cases (collected in an unrestricted manner), where each case has 3 variables associated with it: one, a factor variable with 0/1 levels (called XX), another factor variable with 8 levels (X) and a third response variable with two levels (Y: 0/1). I am trying to see if X1 has an effect on the relationship between X2 and the

mingw, llvm 2.6 (buid with llvm-gcc) Example source code: http://npcontemplation.blogspot.com/2008/06/secret-of-llvm-c-bindings.html I change LLVMCreateJITCompiler(&engine, provider, &error); to LLVMCreateJITCompiler(&engine, provider, 3, &error); $ llvm-gcc `llvm-config --cflags` -c fac.c $ g++ `llvm-config --libs --cflags --ldflags core analysis executionengine jit

Hi, I need to resample rows in a data frame by subsets L3 <- LETTERS[1:3] d <- data.frame(cbind(x=1, y=1:10), fac=sample(L3, 10, repl=TRUE)) x y fac 1 1 1 A 2 1 2 A 3 1 3 A 4 1 4 A 5 1 5 C 6 1 6 C 7 1 7 B 8 1 8 A 9 1 9 C 10 1 10 A I have seen this used to sample rows with replacement d[sample(nrow(d), replace=T), ] x y fac 7 1 7 B 2

Hi all, I noticed something strange when I ran aov and anova. vtot=c(7.29917, 7.29917, 7.29917) #identical values fac=as.factor(c(1,1,2)) #group 1 has first two elements, group 2 has the 3rd element When I run: > anova(lm(vtot~fac)) Analysis of Variance Table Response: vtot Df Sum Sq Mean Sq F value Pr(>F) fac 1 1.6818e-30 1.6818e-30 0.3333 0.6667 Residuals 1

Dear all, I'm using the mgcv library by Simon Wood to fit gam models with interactions and I have been reading (and running) the "factor 'by' variable example" given on the gam.models help page (see below, output from the two first models b, and b1). The example explains that both b and b1 fits are similar: "note that the preceding fit (here b) is the same as

Suppose I have a dataframe 'd' defined as L3 <- LETTERS[1:3] d0 <- data.frame(cbind(x = 1, y = 1:10), fac = sample(L3, 10, replace = TRUE)) (d <- d0[d0$fac %in% c('A', 'B'),]) x y fac 2 1 2 B 3 1 3 A 4 1 4 A 5 1 5 A 6 1 6 B 8 1 8 A Even though factor 'fac' in 'd' only has 2 levels, but it seems to bear the birthmark

[I originally posted this on the R-help mailing list, and it was suggested that R-devel would be a better place to dicuss it.] Running ?table? on a factor with levels containing non-ASCII characters seems to result in extremely bad performance on Windows. Here?s a simple example with benchmark results (I?ve reduced the number of replications to make the function finish within reasonable time):

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