similar to: model.matrix bug? Nested factor yields singular design matrix.

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 all, What are current methods people use in R to identify mis-spelled column names when selecting columns from a data frame? Alice Johnson recently tackled this issue (see [BioC] posting below). Due to a mis-spelled column name ("FileName" instead of "Filename") which produced no warning, Alice spent a fair amount of time tracking down this bug. With my fumbling fingers

Hi, When I use Anova(car) to produce type III SS, 'Sum Sq' is reported in integers: > Anova(bot.lm3, type ="III") Anova Table (Type III tests) Response: bottemp Sum Sq Df F value Pr(>F) (Intercept) 45295 1 29436.4440 < 2e-16 fungroup 3 2 0.8259 0.44006 numsp.fun 11 2

Full_Name: Steven McKinney Version: 2.9.0 OS: Mac OS X 10.5.6 Submission from: (NULL) (142.103.207.10) A corrupt data frame can be constructed as follows: foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4]

Hi useRs, A recent coding infelicity along these lines yielded a corrupt data frame. foo <- matrix(1:12, nrow = 3) bar <- data.frame(foo) bar$NewCol <- foo[foo[, 1] == 4, 4] bar lapply(bar, length) > foo <- matrix(1:12, nrow = 3) > bar <- data.frame(foo) > bar$NewCol <- foo[foo[, 1] == 4, 4] > bar X1 X2 X3 X4 NewCol 1 1 4 7 10 <NA> 2 2 5 8 11

Hi all, I have been having problems generating png files on a Mac running OS X 10.6.8. Here's a simple example > png("foo.png", type = "cairo"); plot(1:10); dev.off(); null device 1 libpng warning: Application built with libpng-1.2.26 but running with 1.5.2 The resultant file is of size 0 Kb. Is this the proper place to report this issue? Or should

R version 2.10.1 (2009-12-14) Copyright (C) 2009 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with

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, 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

Full_Name: Lixin Han Version: 2.4.0 OS: Windows 2000 Submission from: (NULL) (155.94.110.222) A character vector c('a','b') is supplied to rm(). As a result, 'c' is deleted unintentionally. > a <- 1:5 > b <- 'abc' > c <- letters > ls() [1] "a" "b" "c" > rm(c('a','b')) > ls() character(0)

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

Dears Sirs During my computational work I encountered unexpected behavior when calling "ar" function, namely # time series x<-ts(c(-0.2052083,-0.3764986,-0.3762448,0.3740089,0.2737568,2.8235722,- 1.7783313,0.2728676,-0.3273164),start=c(1978,3),frequency=4,end=c(1980,3)) # ar function res.ar<-ar(x,aic=TRUE,demean=F) # call "ar" again and ............

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

Dear experts: Is it possible to create a new function based on stats:::model.matrix.default so that an alternative factor coding is used when the function is called instead of the default factor coding? Basically, I'd like to reproduce the results in 'mat' below, without having to explicitly specify my desired factor coding (identity matrices) in the 'contrasts.arg'. dd

For ordered factor the natural contrast coding would be to parametrize by the succsessive differences between levels, which does not assume equal spacing of factor levels as does the polynomial contrasts (implicitly at least). This requires the contr.cum, which could be: contr.cum <- function (n, contrasts = TRUE) { if (is.numeric(n) && length(n) == 1) levs <- 1:n

Dear R-users, I want to fit a linear model with Y as response variable and X a categorical variable (with 4 categories), with the aim of comparing the basal category of X (category=1) with category 4. Unfortunately, there is another categorical variable with 2 categories which interact with x and I have to include it, so my model is s "reg3: Y=x*x3". Using fit.contrast to make the

Hi, I don't understand what the meaning of the following lines returned by model.matrix(). Can somebody help me understand it? What can they be used for? attr(,"assign") [1] 0 1 2 2 attr(,"contrasts") attr(,"contrasts")$A [1] "contr.treatment" attr(,"contrasts")$B [1] "contr.treatment" Regards, Peng > a=2 > b=3 > n=4

Hi Folks, I'm a bit puzzled by the following (example): N<-factor(sample(c(1,2,3),1000,replace=TRUE)) unique(N) # [1] 3 2 1 # Levels: 1 2 3 So far so good. Now: contrasts(N)<-contr.treatment(3, base=1, contrasts=FALSE) contrasts(N) # 1 2 # 1 1 0 # 2 0 1 # 3 0 0 whereas: contr.treatment(3, base=1, contrasts=FALSE) # 1 2 3 # 1 1 0 0 # 2 0 1 0 # 3 0 0 1 contr.treatment(3, base=1,

Hi all, I am trying to do a ordered probit regression using polr(), replicating a result from SAS. >polr(y ~ x, dat, method='probit') suppose the model is y ~ x, where y is a factor with 3 levels and x is a factor with 5 levels, To get coefficients, SAS by default use the last level as reference, R by default use the first level (correct me if I was wrong), The result I got is a

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