search for: hairs

Dear R-helpers, Suppose I have dataset like this below: data(HairEyeColor) dfHEC <- as.data.frame(as.table(HairEyeColor)) my.dfHEC <- data.frame(Hair=rep(dfHEC$Hair,dfHEC$Freq), Eye=rep(dfHEC$Eye,dfHEC$Freq), Sex=rep(dfHEC$Sex,dfHEC$Freq)) my.dfHEC my.dfHEC$Hair my.dfHEC$Eye my.dfHEC$Sex and I know all levels for Hair, Eye and Sex. In my case, I

I have been tinkering with mosaicplot() and friends as a way of learning R. As part of this, I've written a pair.table() method for mosaic matrices, and would like to extend mosaicplot to work with loglin and logln (MASS) objects. I'm using R 1.4.0 on Win 98. I've been trying to figure out the formula interface, and think there's a bug, but not sure how to find it, yet alone fix

Hi, Can someone set me straight as to how to write formulas in R to indicate: complete independence [A][B][C] joint independence [AB][C] conditional independence [AC][BC] nway interaction [AB][AC][BC] ? For example, if I have 4 factors: hair colour, eye colour, age, sex does > mosaicplot( frequency ~ hair + eye + age + sex) mean that the model fitted is of complete independence of

*Hair Treatment & Transplantation Tips* [image: Hair Treatment and Transplantation]<http://info-center.50webs.com/hair-transplant/> *Prevent and Reversing Your Hair Loss* *Make your Hair Silky and Healthy* *Hair Treatment* *and* *Transplantation Safe Sites*. *Tips for your Hairs. <http://info-center.50webs.com/hair-transplant/>* --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To post to this group, send email to rubyonrails-talk-/JYPxA39Uh5TLH3MbocF...

Hi! Some time ago, I replaced calls to library() with calls to requireNamespace() in my package logmult, in order to follow the new CRAN policies. But I just noticed it broke jackknife/bootstrap using several workers via package parallel. The reason is that I'm running model replicates on the workers, and the formula includes non-standard terms like Mult() which are provided by gnm. If gnm

I'm looking for a way to transmit a user specific variable to my dialplan If we use the example of the hair color, I was thinking of having something like: [bob] context=users host=dynamic secret=password type=friend username=bob hair=brown [lary] context=users host=dynamic secret=password type=friend username=lary hair=black And be able to access from the dialplan: [users] Exten =>

Hello everyone,   I am a very new user of R and I have a query about the cuminc function in the package cmprsk. In particular I would like to verify that I am interpreting the output correctly when we have a stratification variable.   Hypothetical example:   group : fair hair, dark hair fstatus: 1=Relapse, 2=TRM, 0=censored strata: sex (M or F)   Our data would be split into:   Fair, male,

Hello everyone, I am a very new user of R and I have a query about the cuminc function in the package cmprsk. In particular I would like to verify that I am interpreting the output correctly when we have a stratification variable. Hypothetical example: group : fair hair, dark hair fstatus: 1=Relapse, 2=TRM, 0=censored strata: sex (M or F) Our data would be split into: Fair, male,

The data is in the attachment. What I wanna get is: , , Sex = Male Eye Hair Brown Blue Hazel Green Black 32 11 10 3 Brown 53 50 25 15 Red 10 10 7 7 Blond 3 30 5 8 , , Sex = Female Eye Hair Brown Blue Hazel Green Black 36 9 5 2 Brown 66 34 29 14 Red 16 7 7 7 Blond 4

Hello I have the matrix a<-matrix(c(2,1,0,1,2,2,1,5,7,2,5,12),nrow=6) a [,1] [,2] [1,] 2 1 [2,] 1 5 [3,] 0 7 [4,] 1 2 [5,] 2 5 [6,] 2 12 Suppose that in the first row we have 3 men of England, 2 with hair, and 1 no In the second we have 6 italian men, 1 with hair and 5 no ... I want to find if there is a dependence between men withouth hair and

It being friday, I would like to call your attention to an innovative data analysis by Leslie Lamport available from: http://research.microsoft.com/users/lamport/pubs/hair.pdf I particularly liked the graphics. url: www.econ.uiuc.edu/~roger Roger Koenker email rkoenker at uiuc.edu Department of Economics vox: 217-333-4558 University of Illinois

> Date: Thu, 23 Jun 2005 08:50:50 +0200 > From: "Robert Rozman" <rozman@fri.uni-lj.si> > Subject: [Asterisk-Users] Loosing hair on connecting Panasonic PBX- * > - Euroisdn Italy > > I'm pulling my hair down and getting bold :-) ..... I have Asterisk between > Panasonic KXTD816 and Euroisdn in Italy (beronet octobri and bristuff > Asterisk).... Plenty of

if you have hair loss problem this will help you http://www.yenisaclarim.com --~--~---------~--~----~------------~-------~--~----~ You received this message because you are subscribed to the Google Groups "Ruby on Rails: Talk" group. To post to this group, send email to rubyonrails-talk-/JYPxA39Uh5TLH3MbocFFw@public.gmane.org To unsubscribe from this group, send email to

I am using R to teach, and here are a couple of things that I thought would work didn't work. 1. I noticed the utility data(***,package=***) recently and like it very much, but unless I type in the whole word "package" I'll get an error in 1.4.0. For example, data(cats,package=MASS) works fine but data(cats,pac=MASS) doesn't. 2. drop1 doesn't seem to be as smart as

I have an application where I have a function to calculate results for a 2-way table or matrix, which returns a matrix with one less row and column. To keep this short, the function below captures the structure: fun2way <- function(f){ if (!length(dim(f)) ==2) stop("only for 2-way arrays") R <- dim(f)[1] C <- dim(f)[2] f[1:(R-1), 1:(C-1)] } Now, I want to

My question is based on an example provided in the following: Referencing: Statistics with R Vincent Zoonekynd <zoonek at math.jussieu.fr> 6th January 2007 URL: http://zoonek2.free.fr/UNIX/48_R/all.html data(HairEyeColor) a <- as.table( apply(HairEyeColor, c(1,2), sum) ) # Provided Example barplot(a, beside = TRUE, legend.text = attr(a, "dimnames")$Hair) # I

Hi, I would like to select the most frequent value level in a set of three variables. Three different observators have judged hair color in study subjects. Mostly they judge the same color, sometimes there is a slight difference. I want to know what most of the observators have chosen (so at least 2) from the 3 observations. E.g. If two out of three observators decide the hair is black, then

Hi Milan, I expect I may be able to do something about the way the terms are evaluated, to ensure the evaluation is done in the gnm namespace (while still ensuring the variables can be found!). In the meantime, I think the following will work: Mult <- gnm::Mult f <- Freq ~ Eye + Hair + Mult(Eye, Hair) gnm::gnm(f, family=poisson, data=dat) Hope that helps, Heather On Wed, Apr 29, 2015,

I have been pulling my hairs out about this for a while now. Running the latest version of SuSE with all patches applied, I have my machine joined to the domain no problem. But when I go to use mount -t smbfs I get this error. Mounting share failed, smbmnt must be installed suid root for direct user mounts (1000,1000) smbmnt...

In ?title I see the plot(cars, main = "") title(main = list("Stopping Distance versus Speed", cex=1.5, col="red", font=3)) I can't seem to generalize this to use several colors in a single title. What I'd like is in latex-ish \red{Hair color} \black{ and } \blue{Eye color} to serve also as an implicit legend for points that are plotted. -Michael --