similar to: What does model.matrix() return?

Hi, I run the following commands. 'A' has 3 levels and 'B' has 4 levels. Should there be totally 3+4 = 7 coefficients (A1, A2, A3, B1, B2, B3, B4)? > a=3 > b=4 > n=1000 > A = rep(sapply(1:a,function(x){rep(x,n)}),b) > B = as.vector(sapply(sapply(1:b, function(x){rep(x,n)}), function(x){rep(x,a)})) > Y = A + B + rnorm(a*b*n) > > fr =

The model matrix for the code at the end the email is shown below. Since the model matrix doesn't have -1, I think that it is made of dummy regressors rather than deviation regressors. I'm wondering how to make a model matrix using deviation regressors. Could somebody let me know? > model.matrix(aaov) (Intercept) A2 B2 B3 A2:B2 A2:B3 1 1 0 0 0 0 0 2

Hi all, I believe this is a bug in the model.matrix function. I'd like a second opinion before filing a bug report. If I have a nested covariate B with multiple values for just one level of A, I can not get a non-singular design matrix out of model.matrix > df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x",

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Dear all, I have problems to code contrasts for performing an asymmetrical anova with aov(). I am using aov() because I want to get the Mean Squares for further analyses. I didn't find any solution to my problem in the help files of functions aov(), contrasts(), C(), etc. Let's say I have three locations, one with treatment P, and two with treatment C: >

Hello, I've fitted a parametric survival model by > survreg(Surv(Week, Cens) ~ C(Treatment, srmod.contr), > data = poll.surv.wo3) where srmod.contr is the following matrix of contrasts: prep auto poll self home [1,] 1 1 1.0000000 0.0 0 [2,] -1 0 0.0000000 0.0 0 [3,] 0 -1 0.0000000 0.0 0 [4,] 0 0 -0.3333333 1.0 0 [5,] 0 0

This works: > model.matrix(~I(pos>3),data=data.frame(pos=c(1:5))) (Intercept) I(pos > 3)TRUE 1 1 0 2 1 0 3 1 0 4 1 1 5 1 1 attr(,"assign") [1] 0 1 attr(,"contrasts") attr(,"contrasts")$"I(pos > 3)" [1] "contr.treatment"

Dear R-Community! The example "oats" in MASS (2nd edition, 10.3, p.309) is calculated for aov and lme without interaction term and the results are the same. But I have problems to reproduce the example aov with interaction in MASS (10.2, p.301) with lme. Here the script: library(MASS) library(nlme) options(contrasts = c("contr.treatment", "contr.poly")) # aov: Y ~

Dear R-devel list members, I'd like to suggest a more flexible procedure for generating contrast names. I apologise for a relatively long message -- I want my proposal to be clear. I've never liked the current approach. For example, the names generated by contr.treatment paste factor to level names with no separation between the two; contr.sum simply numbers contrasts (I recall an

Hi all, I have been trying to calculate Type III SS in R for an unbalanced two-way anova. However, the Type III SS are lower for the first factor compared to type I but higher for the second factor (see below). I have the impression that Type III are always lower than Type I - is that right? And a clarification about how to fit Type III SS. Fitting model<-aov(y~a*b) in the base package and

I was surprised by this (in R 2.0.1): > a <- ordered(-1:1) > a [1] -1 0 1 Levels: -1 < 0 < 1 > model.matrix(~ a) (Intercept) a.L a.Q 1 1 -7.071068e-01 0.4082483 2 1 -9.073800e-17 -0.8164966 3 1 7.071068e-01 0.4082483 attr(,"assign") [1] 0 1 1 attr(,"contrasts") attr(,"contrasts")$a [1]

On 6/23/05, RenE J.V. Bertin <rjvbertin at gmail.com> wrote: > Hello, > > I was just having a look at the aov function source code, and see that when the model used does not have an Error term, Helmert contrasts are imposed: > > if (is.null(indError)) { > ... > } > else { > opcons <- options("contrasts") >

Hi, I recently ran into an inconsistency in the way model.matrix.default handles factor encoding for higher level interactions with categorical variables when the full hierarchy of effects is not present. Depending on which lower level interactions are specified, the factor encoding changes for a higher level interaction. Consider the following minimal reproducible example: -------------- >

I read somewhere that t.test is equivalent to a hypothesis testing for one way ANOVA. But I'm wondering how they are equivalent. In the following code, the p-value by t.test() is not the same from the value in the last command. Could somebody let me know where I am wrong? > set.seed(0) > N1=10 > N2=10 > x=rnorm(N1) > y=rnorm(N2) > t.test(x,y) Welch Two Sample t-test data:

useRs, A no doubt simple question, but I am baffled. Indeed, I think I once knew the answer, but can't recover it. The default contrasts for aov (and lm, and...) are contr.treatment and contr.poly for unordered and ordered factors, respectively. But, how does one invoke the latter? That is, in a data.frame, how does one indicate that a factor is an *ordered* factor such that

Hi, I need to build a function to generate one column for each level of a factor in the model matrix created on an arbitrary formula (instead of using the available contrasts options such as contr.treatment, contr.SAS, etc). My approach to this was first to use the built-in function for contr.treatment but changing the default value of the contrasts argument to FALSE (I named this function

Hi, I found that using different contrasts (e.g. contr.helmert vs. contr.treatment) will generate different fitted probabilities from multinomial logistic regression using multinom(); while the fitted probabilities from binary logistic regression seem to be the same. Why is that? and for multinomial logisitc regression, what contrast should be used? I guess it's helmert? here is an example

Hi! I would like to suggest to make it possible, in one way or another, to get meaningful contrast names when using contr.sum(). Currently, when using contr.treatment(), one gets factor levels as contrast names; but when using contr.sum(), contrasts are merely numbered, which is not practical and can lead to mistakes (see code at the end of this message). This issue was discussed quickly in 2005

Hi useRs, Perhaps I am having a senior moment? I have a nested variable situation to model, toy example: > df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x", "a")), + B = factor(c("b", "x", "x", "x"), levels = c("x", "b"))) > >

Hi, I am trying to find out a collinearity in explanatory variables with alias(). I creat a dataframe: dat <- ds[,sapply(ds,nlevels)>=2] dat$Y <- Response Explanatory variables are factor and response is continuous random variable. When I run a regression, I have the following error: fit <- aov( Y ~ . , data = dat) Error in "contrasts<-"(`*tmp*`, value =