search for: hair

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 f...

...ike 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 it. It seems to only work with independence models: > data(HairEyeColor) > mosaicplot(~ Hair + Eye + Sex, data=HairEyeColor) > mosaicplot(~ Hair * Eye + Sex, data=HairEyeColor) Error in aperm(a, perm, resize) : `perm' is of wrong length here's a traceback: > debug(mosaicplot) > mosaicplot(~Hair + Eye + Sex + Hair:Eye, data=HairEyeColor) de...

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 all factors [hair][eye][age][sex]? So does > mosaicplot(frequency ~ hair + eye) mean that the model is of conditional independence [hairAgeSex...

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...ailed. Here's a reproducer using only gnm: > dat <- structure(c(326, 688, 343, 98, 38, 116, 84, 48, 241, 584, 909, + 403, 110, 188, 412, 681, 3, 4, 26, 85), .Dim = 4:5, .Dimnames = structure(list( + Eye = c("Blue", "Light", "Medium", "Dark"), Hair = c("Fair", + "Red", "Medium", "Dark", "Black")), .Names = c("Eye", "Hair" + )), class="table") > > f <- Freq ~ Eye + Hair + Mult(Eye, Hair) > gnm::gnm(f, family=poisson, data=dat) Error in which(sappl...

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 => _X.,1,NoOp(my...

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, relapse Dark,male, relapse Fair, female, relapse Dark, female, relapse   Fair, male, TRM Dark,male, TRM Fair, female, TRM Dark, female, TRM   Fair, male, censored Dark,male, censored...

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, relapse Dark,male, relapse Fair, female, relapse Dark, female, relapse Fair, male, TRM Dark,male, TRM Fair, female, TRM Dark, female, TRM Fair, male, censored Dark,male, censored...

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...

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 nationality. By the way, which is the simplest command to use to study the independence in a matrix like the one that I wrote? Using > chisq.test(a)...

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 fax: 217-244-6678 Champaign, IL 61820

> 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 experience with the Panasonics, but not the EuroISD...

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-/JYPxA39Uh5TLH3...

...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 before. In the sequence below, drop1 didn't produce what I expected it to produce. data(HairEyeColor); lab<-dimnames(HairEyeColor) HairEye<-cbind(expand.grid(Hair=lab$Hair,Eye=lab$Eye, Sex=lab$Sex),Fr=as.vector(HairEyeColor)) HairEye.fit<-glm(Fr~.^2,poisson,HairEye) drop1(HairEye.fit) The following sequence however works. HairEye.fit<-glm(Fr~(Hair+Eye+...

...else { require(plyr) result <- aaply(f, stratum, fun2way) ## order of dimensions screwed up! } result } For example, by hand (or with a loop) I can calculate the pieces and combine them as I want using abind(): > # apply separately to strata > t1<-fun2way(HairEyeColor[,,1]) > t2<-fun2way(HairEyeColor[,,2]) > > library(abind) > abind(t1, t2, along=3) , , 1 Brown Blue Hazel Black 32 11 10 Brown 53 50 25 Red 10 10 7 , , 2 Brown Blue Hazel Black 36 9 5 Brown 66 34 29 Red 16...

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 would like to make the labels on the x-axis diagonal, so I tried the following: barplot_reference<-barplot(a, beside = T...

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 it's likely not to be brown. L...

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, at 05:57 PM, Milan Bouchet-Valat wrote: > Hi! > > Some time ago, I replaced calls to library() with calls to > requireNamespace() in my package logmult, in order to follow the new...

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 -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Street http://www....