similar to: Subclassing lm

I'm trying to create an extension to data.frame with more complex row and column names, and have run into some problems: > setClass("new-data.frame", representation("data.frame")) [1] "new-data.frame" Warning message: old-style ('S3') class "data.frame" supplied as a superclass of "new-data.frame", but no automatic conversion will

Hi Martin, Thankyou for your response. I suspect that we're not going to agree on the main point. Making it trivially simple (as say Java) to set slots to NULL. So, I'll move on to the other points here. ***Note that cited text uses excerpts only.*** > setClassUnion("character_OR_NULL", c("character", "NULL")) > A = setClass("A", slots =

Hello, I have a simple question for you: making: mylm<-lm(y~x) summary(mylm) I get the following results: ****************************************************** Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 16.54087 0.19952 82.91 <2e-16 *** x[1:19] -2.32337 0.04251 -54.66 <2e-16 *** ******************************************************

Hi folks, I am running a very simple regression using mylm <- lm(mass ~ tarsus, na.action=na.exclude) I would like the use the residuals from this analysis for more regression but I'm running into a snag when I try cbind(mylm$residuals, mydata) # where my data is the original data set The error tells me that it cannot use cbind because the length of mylm$residuals is

I'm trying to manipulate/change a formula prior to passing it to another function. A simplified example: User passes formula to my function: y~x My function does: lm(transform(y)~x) Here, transform() is added to the model's response. What is the best way to accomplish this?

Hello list, I have a linear regression: mylm = lm(y~x-1) I've been reading old mail postings as well as the plotmath demo and I came up with a way to print an equation resulting from a linear regression: model = substitute(list("y"==slope%*%"x", R^2==rsq), list(slope=round(mylm$coefficients[[1]],2),rsq=round(summary(mylm)$adj.r.squared, 2))) I have four models and I

Dear all - We perform some measurements with a machine that needs to be recalibrated. The best calibration we get with polynomial regression. The data might look like follows: > true_y <- c(1:50)*.8 > # the real values > m_y <- c((1:21)*1.1, 21.1, 22.2, 23.3 ,c(25:50)*.9)/0.3-5.2 > # the measured data > x <- c(1:50) > # and the x-axes > > # Now I do the following:

Hello, I am randomly generating values and then using an ANOVA table to find the mean square value. I would like to form a loop that extracts the mean square value from ANOVA in each iteration. Below is an example of what I am doing. a<-rnorm(10) b<-factor(c(1,1,2,2,3,3,4,4,5,5)) c<-factor(c(1,2,1,2,1,2,1,2,1,2)) mylm<-lm(a~b+c) anova(mylm) Since I would like to use a loop to

Hi All, I had posted a question on a similar topic, but I think it was not focused. I am posting a modification that I think better accomplishes this. I hope this is ok, and I apologize if it is not. :) I am looping through variables and running several regressions. I have reason to believe that the data is being duplicated because I have been monitoring the memory use on unix. How can I avoid

Hi, I wonder if someone has already figured out a way of making summary(mylm) # where mylm is an object of the class lm() to print the "(Intercept)" at the last line, rather than the first line of the output. I don't know about, say, biostatistics, but in economics the intercept is usually the least interesting of the parameters of a regression model. That's why, say, Stata

DeaR list I would like to predict the values of each column of a matrix A by regressing it on all other columns of the same matrix A. I do this with a for loop: A <- B <- matrix(round(runif(10*3,1,10),0),10) A for (i in 1:length(A[1,])) B[,i] <- as.matrix(predict(lm( A[,i] ~ A[,-i] ))) B It works fine, but I need it to be faster. I've looked at *apply but just can't

Dear R-help, I have a four-part question about regression, matrices, and sandwich package. 1) In the sandwich package, I would like to better understand the meat() function. >From the bread() documentation, for a simple OLS regression, bread() returns (1/n * X'X)^(-1) That is, for a simple regression (per the documentation on bread()): MyLM <- lm(y ~ x) bread(MyLM)

Answering to convey the 'rules' as I know them, rather than to address the underlying issues that I guess you are really after... The S4 practice is to use setOldClass() to explicitly treat an S3 character() vector of classes as an assertion of linear inheritance > x <- structure (sqrt (37), class = c ("sqrt.prime", "numeric") ) > is(x,

Hi, But "any" rises some other problems well known from S3. One has "any" for "free" in S3. You don't need S4. . But I know how "if" polluted functions look like in S3. They are hard to understand and to maintain. Hence I am quite happy to use S4. Type-checking is usefull if you program with data. Also if you are on the "C side" of the

While installing my small package, I met a tricky problem. For clarity, let me explain it with the following simplified example. In ~/pkg/R/aclass.R, setClass("aclass", contains="bclass", representation(i="numeric")) In ~/pkg/R/bclass.R, setClass("bclass", representation(j="numeric")) After building a "pkg" package, the file

I'm interfacing to C code that uses 1-based indexing on arrays -- it ignores the zeroth element. Thus, input vectors from R must be moved up one, and output arrays must be moved down one. What is the best way to deal with this using R internal code? My current approach is: For an input R vector of length n, allocate a new vector(v) of length n+1 and copy input into v[1] to v[1+n]. Call

Dear R-guRus: I am stuck with the following problem in R: Let's say i have a data.frame with columns X,Y & Z. X is a column of numbers, Y is a column of N factors "A", "B", "C", "D", etc and Z is a column of n factors "a", "b", "c", etc if i do MyLm<-lm(X~Y*Z - 1) i get all N coefficients for Y-factors, but only

Is there an easy way to do a linear model with an a priori known slope? In essence, I want to minimize the residuals around a line of known slope. -R -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the

Hi, I'm a biologist, so please forgive me if my question sounds absurd! I have 3 parameters x1, x2, x3 and a response variable y.The sample size is 75. I tried to do the following: mylm<-lm(y~ x1 + x2 + x3, data="mydata") but i can only get stats from anova for the first 2 variables. The third comes up as NA. The degrees of freedom for the third variable are 0. Is there

Hello list, I am used to give a lot of attention to the standardized regression coefficients, which in SPSS are listed automatically. Is there alternative to running the last two lines in the following example to get all the information? ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) summary( lm(ctl ~ trt) )