similar to: Formula with no intercept

Esteemed R-forum subscribers, I'm having a tough time configuring contrasts for my 3-way ANOVA. In short: I don't know how to configure (all) my contrasts correctly in order to specify (all) my comparisons of interest. I succeeded getting my contrasts of interest set up for a simpler 2-way ANOVA based on the fairly intuitive logic of the design col.names. But i'm not able to

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Trying to work out a model formula that will do what I want ... suppose I want to model y = b_i x + epsilon (i.e. a linear model with zero intercept and with slopes differing by groups), and I want to parameterize the slopes in the "usual" way of having a baseline slope value for the first level of the factor b and (n-1) values

Full_Name: Adriano Azevedo Filho Version: 1.9.1 OS: Windows, Linux Submission from: (NULL) (200.171.246.212) R-squared and Adjusted R-squared appear to be wrong when the formula in lm() is specified without intercept. Problem present in both Windows and Linux 1.9.1 version. Also in the 1.8.1 version for Windows (other versions not checked). Possible example which reproduces the problem:

I have some data with two categorises plus/minus (p53) and a particular time (Time) and the outcome is a continuous vairable (Result). I set up a maximum model. ancova <- lm(Result~Time*p53) > summary(ancova) .. Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.05919 0.55646 0.106 0.916 Time -0.02134 0.01785 -1.195 0.241 p53plus

For a pedagogical purpose, I was trying to show how the formula for a simple regression line (~1+x) could be crossed with a factor (~1:group + x:group) to fit separate regressions by group. For example: set.seed(201108) dat <- data.frame(x=1:15, y=1:15+rnorm(15), group = sample(c('A','B'), size=15, replace=TRUE)) m1 <- lm(y~ 1 + x, data=dat)

I'm trying to write a formula method for canonical correlation analysis, that could be called similarly to lm() for a multivariate response: cancor(cbind(y1,y2,y3) ~ x1+x2+x3+x4, data=, ...) or perhaps more naturally, cancor(cbind(y1,y2,y3) ~ cbind(x1,x2,x3,x4), data=, ...) I've adapted the code from lm() to my case, but in this situation, it doesn't make sense to include an

Hi, I'm a new user of R and I'm trying to make a linear model from this kind of dataset x [1] 16.87 19.93 25.85 20.94 17.06 19.49 19.93 25.45 27.74 20.15 25.81 21.06 17.17 20.03 25.50 27.79 20.44 16.88 19.93 25.79 z<-x-10 y [1] 0.80 1.27 2.22 1.32 0.90 1.18 1.84 2.41 2.97 1.25 2.07 1.41 1.14 1.66 2.59 3.51 1.53 0.81 1.26 2.30 plot(x,y) I want to be able to force the line of

Hi, I'm trying to create a linear model for a dataset that has a breakpoint e.g. # dummy dataset x <- 1:20 y <- c(1:10,seq(10.5,15,0.5)) plot(x,y) I've modelled this using the following formula: temp <- lm(y ~ x*(x<=10)+x*(x>10)) I want to be able to omit the intercept (i.e. force the line through zero) from the first of these segments (x<=10) so that I'm only

I need to use Alt-F2 shortcut in Far Manager running with wineconsole. But Alt-F2 invokes standard Linux "run command" dialog. How can I intercept the Alt-F2 so that it is processed by Wine application?

I realize that the following has been talked about on this list many times before in some related way but I am going to ask for help anyway because I still don't know what to do. Suppose I have no intercept models such as the following : Y = B*X_1 + error Y = B*X_2 + error Y = B*X_3 + error Y = B*X_4 + error and I run regressions on each ( over the same sample of Y ) and now I want to

Hi all, Dear Dr. Wheeler, I am trying to use the lmPerm package to perform multiple regression on microarray data with certain empirical variables associated with treatments of the experiment. In order the circumvent the very conservative multiple test corrections such as Bonferroni and BH, I try to use permutated probabilities to assess associations. In addition to mu previous posting I

Hello, I'm implementing a function using non uniform B-Splines. Looking at the code of the bs() function, I realized that if the intercept was set to FALSE, the behavior of the function was the following (df is the number of degrees of freedom that I believe can be interpreted as the number of control points): - Compute df- ord + 1 _internal_ knots (ord is the order of the spline) - Add ord

Dear list, I'm trying to fit a glm model using family=poisson(link = "identity"). The problem is that the glm function fits a model with a negative intercept, which sounds like a nonsense to me, being the response a Poisson variable. >From a previous discussion on this list I've understood that the glm function uses IRLS for the fitting without any constraint so it is

It's easy to run a linear regression on a simple model without an intercept just by doing this: lm(y ~ x1 + x2 -1) Is there a similar trick to suppress the intercept when your model is in a large dataframe and you don't want to write out the names of individual columns? -- View this message in context:

I would appreciate if someone could tell me how to fit a non-intercept model using lrm (and not glm). The -1 in the formula of the glm does not work with lrm. Thanks, Clarissa [[alternative HTML version deleted]]

Dear R users, glm.fit() gave me the same AIC's regardless of TRUE or FALSE intercept option. > myX <- as.matrix(1:10) > myY <- 3+5*myX > foo <- glm.fit(x=myX, y=myY, family = gaussian(link = "identity"), intercept=TRUE) > foo$aic [1] 38.94657 > foo <- glm.fit(x=myX, y=myY, family = gaussian(link = "identity"), intercept=FALSE) > foo$aic [1]

Hi, dear R-help, I want to fit a Buckley-James censored regression without intercept. The function bj() inside the Design library have to have an intercept, I try Surv(y,d)~ 0 + x , or -1+x, or x-1, none works. Any suggestions? Is there another BJ() code out there that do not have to have an intercept? Or may be it is easier to modify the bj()? I need the estimator only, no need of var estimate

SVM patch to add add init intercept handler. Applies cleanly to 9622. Please apply. Signed-off-by: Tom Woller <thomas.woller@amd.com> _______________________________________________ Xen-devel mailing list Xen-devel@lists.xensource.com http://lists.xensource.com/xen-devel

Hi, This might be a silly question, but how do I create a pooled model (binomial family) as a baseline for a likelihood-ratio test for a random-intercept lmer model? R won't compare a glm model with an lmer model (or glmmPQL ). Thanks in advance. -Bobby

This very likely falls in the category of an unexpected result due to user ignorance. I generated the following data: time <- 0:10 set.seed(4302007) y <- 268 + -9*time + .4*(time^2) + rnorm(11, 0, .1) I then fit models using both orthogonal and raw polynomials: fit1 <- lm(y ~ poly(time, 2)) fit2 <- lm(y ~ poly(time, degree=2, raw=TRUE)) > predict(fit1, data.frame(time =