search for: vaiable

Hi, if I know the colnames x and y in the following example, I can easily to do lm. tmp <- data.frame(x=c(1,1.2),y=c(1,2)) lm(y ~ x, data=tmp) when the colnames are variable, what should I do? for example colnames(tmp)[1] <- paste("aa",1,sep="_") lm(y ~ paste("aa",1,sep="_"), data = tmp) it gives me error. [[alternative HTML version

I have a vaiable named NAM having value : 1,2,3,4,5,6,7,8,9. I want to make an indicator variable that will take value 1 if NAM=7 or NAM=8 or NAM=9. How can I do that? I usually do: Var001<- ifelse(NAM==7,1,0) for the simplest case. [[alternative HTML version deleted]]

Hi folks, I have a vaiable Y and an associated factor Z at several (13) levels. boxplot(Y~Z) produces a nice array of boxplots, one for each level of Z, and each duly labaelled with its level of Z. I would like to superpose on each boxplot the actual data points which it represents, i.e. do something conceptually (thoug...

hello everybody, here I have a question regarding printing of the constants( like 2 .63 etc.,) present in the instruction while iterating over the instructions within a basic block . I am able to print the vaiables but not the constants. Can you please tell me how to get the constants printed out while iterating over the instructions because the constants do not have names as the variables do( like temp12,temp131 etc.,). thanking you, yours sincerely anubham suresh TU-Darmstadt Germany Anubham Suresh M...

...MASS library to do this, but examining the output I notice that the higher-order orthogonal polynomials are getting larger coefficients than the more important lower-order ones. This is clearly because some scaling of the variables is being done by lda(), and because the higher-order polynomial vaiable values are smaller, they are scaled up. I would like to turn off this scaling as it is not what is needed in this problem and will cause the tail to "wag the dog". There is no obvious parameter to do this in lda(x, grouping, prior = proportions, tol = 1.0e-4, su...

...lo del modelo no sería necesario procesar por cada ejecución de variable (A y B). Expresado de otra forma, (1) es el procesamiento del modelo, (2.1) y (2.2) son procesos sobre el modelo, por ejemplo ranef para solo A y ranef para solo B. El resultado de ranef para A, es decir filtrando solo la vaiable A se escribe en el archivo A. Por lo que tendría dos archivos, uno A y otro B, donde ambos tienen una parte en común (1) y la correspondiente (2.1) para A y (2.2) para B. Mi deseo es no tener que procesar muchas veces la parte (1). ¿Alguna sugerencia? Javier Marcuzzi [[alternative HTML ver...

...flowed > > On Thu, 2 Mar 2006, anubham suresh wrote: > > here I have a question regarding printing of the > > constants( like 2 .63 etc.,) present in the > > instruction while iterating over the instructions > > within a basic block . > > I am able to print the vaiables but not the > > constants. > > Can you please tell me how to get the constants > > printed out while iterating over the instructions > > because the constants do not have names as the > > variables do( like temp12,temp131 etc.,). > > I'm not really sure wha...

...por cada ejecución de variable (A y B). > > Expresado de otra forma, (1) es el procesamiento del modelo, (2.1) y > (2.2) son procesos sobre el modelo, por ejemplo ranef para solo A y > ranef para solo B. > > El resultado de ranef para A, es decir filtrando solo la vaiable A se > escribe en el archivo A. > > Por lo que tendría dos archivos, uno A y otro B, donde ambos > tienen una > parte en común (1) y la correspondiente (2.1) para A y (2.2) para > B. Mi > deseo es no tener que procesar muchas veces la parte (1). > &gt...

...s follows: response ~ (intercept) + value + group OR: glm( response ~ (intercept) + value + group , family=binomial(link='logit')) lrm( response ~ (intercept) + value + group ) ROC( from = response ~ (intercept) + value + group , plot='ROC') the response is a binary vaiable, the independent predictor 'value' is a continuous variable, and the grouping factor is a ordered factor (with 5 levels (25,50,100,200,400)) When I compare the GLM model with the ROC model and the LRM model setting 'group' as factor variable, the resulting coefficients are s...

Dear R developers and users: I have the following data, x is the response vaiable, nsample(individual) nested within trt, and subsample nested within nsample, I want to fit trt as fixed effect, and block, nsample(trt) as random effects using lme, is the following coding correct? dat$vgrp <- getGroups(dat, form = ~ 1|trt/nsample, level = 2) ge.lme1 <- lme(fixed=x~trt, dat...

...*** This all seems to be as it should. I then decided to try and confim this result by running a glm without any of the 5 potential cubic terms ( note - TRI:I(TRI^2) was the only one that made it into the final model but there were 5 potential). After entering the 73 potential terms (12 primary vaiables and now 66 minus 5 interactions = 73 total), the glm and stepAIC produces a completely different final model. It has 8 variables that were not in the model that was chosen with scope statement and manually removing TRI:TRI^2, and it is missing 7 variables that were in the model chosen with the sc...

Hi all, I''m sure this is really simple, but I can''t seem to figure it out. I want to let users preview their messages before submitting. Something in the form of: def new @post = Post.new end def preview #preview message text here #@post = Post.new(params[:post]) <--? end def create #save message on user''s confirmation ... @post.save end Have any

dear all, I am a student in cs college. I would like to know how to plot infinte number of genes after using the svm. the data set i have consists of x which is a matrix of 39 cancer patients [rows] and 2000 gene names [colmns]. each cell is the value of the gene for a particular patient. there are two types of cancer people representedas factor y. here is the code: library(e1071) #load