search for: mylm

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...

I'd trying to subclass the "lm" class to produce a "mylm" class whose instances behave like lm objects (are accepted by methods like summary.lm) but have additional data or slots of my own design. For starters: setClass("mylm", "lm") produces the somewhat cryptic: Warning message: Old-style (``S3'') class "mylm&q...

...<-IDeffect[as.character(longdat$ID)] longdat$y<- (longdat$x + longdat$x^3 + (longdat$x-1)^4 / 5 + 1/(abs(longdat$x/50) + 0.02) + longdat$IDeffect + rnorm(1:nrow(longdat)) * 2 ) longdat<-longdat[order(longdat$x),] library(splines) # Calling ns within lm works fine: mylm<- lm( y ~ ns(x,df=splineDF), data=longdat) longdat$lmfit<-predict(mylm) library(ggplot2) print( ggplot(longdat, aes(x, y)) + geom_point(shape=1) + geom_line(aes(x=x, y=lmfit), color="red") ) cat("Enter to attempt lme.") readline() library(nlme) if(WhichApproach==1...

...Deffect[as.character(longdat$ID)] longdat$y<- (longdat$x + longdat$x^3 + (longdat$x-1)^4 / 5 + 1/(abs(longdat$x/50) + 0.02) + longdat$IDeffect + rnorm(1:nrow(longdat)) * 2 ) longdat<-longdat[order(longdat$x),] library(splines) # Calling ns within lm works fine: mylm<- lm( y ~ ns(x,df=splineDF), data=longdat) longdat$lmfit<-predict(mylm) library(ggplot2) print( ggplot(longdat, aes(x, y)) + geom_point(shape=1) + geom_line(aes(x=x, y=lmfit), color="red") ) cat("Enter to attempt lme.") readline() library(nlme) if(WhichApproach...

...ating 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 generate this several times it would be helpful to know how to extract the mean square value from ANOVA. Thanks [[alternative HTML version deleted]]

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(...

...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: > > m_y_2 <- m_y^2 > m_y_3 <- m_y^3 > mylm <- lm(true_y ~ m_y + m_y_2 + m_y_3) ; mylm Call: lm(formula = true_y ~ m_y + m_y_2 + m_y_3) Coefficients: (Intercept) m_y m_y_2 m_y_3 1.646e+00 1.252e-01 2.340e-03 -9.638e-06 Now I can get the real result with > calibration <- 1.646e+00 + 1.252e-01*m_y +...

...3 <- rnorm(50) myData <- data.frame(response,var1,var2,var3) var.names <- names(myData)[2:4] numVars <- length(var.names) betas <- rep(-1,numVars) names(betas) <- var.names #run regression on var1 through var3. for (Var_num in 1:numVars) { col.name <- var.names[Var_num] mylm <- lm(response ~ get(col.name),data=myData,model=FALSE) betas[Var_num] <- coef(mylm)[2] }

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...

...and $deviance elements as in glm objects... I tried to find out an answer on R-help & Pineihro & Bates (2000). Partial success only: - null deviance: Response: possibly yes: see http://tolstoy.newcastle.edu.au/R/help/05/12/17796.html (Spencer Graves). The (null?) deviance is -2*logLik(mylme), but a personnal trial with some glm objects did not led to the same numbers that the one given by the print.glm method... - the deviance due to the the random effect(s). I was supposing that the coefficients given by ranef(mylme) may be an entry... but beyond this, I guess those coefficient...

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) )

...<- 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 seem to figure it out. Maybe the solution could look somewhat like this: mylm <- function(y,ci) { x <- A[,-ci] b <- lm(y~x) } B <- apply(A,2,mylm,ci=current_column_index(A)) Is there a way to pass the index of the current column in apply to my function? Am I on the right path at all? Thanks for your help. Regards, Stefan

...gression, 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) solve(crossprod(cbind(1, x))) * length(y) (The last two terms above produce the same output, the matrix described above.) In terms of the basic data matrix and coefficients, what does meat() return for a simple OLS regression? (I don't know the term "empirical...

Hello, I've had no luck finding an R package that has the ability to estimate a Tobit model allowing for heteroskedasticity (multiplicative, for example). Am I missing something in survReg? Is there another package that I'm unaware of? Is there an add-on package that will test for heteroskedasticity? Thanks for your help. Cheers, Alan Spearot -- Alan Spearot Department of Economics

...0L)) ## this is what I really want to be able to do, but without contrast ## complaining about the imbalance in the number of rows cat("### With scanner\n") fixedFormula=as.formula("fmri ~ group * task + scanner") randomFormula = as.formula("random = ~ 1 | subject") mylme = lme(fixed=fixedFormula, random=randomFormula, data=my.model) ## now look at the risky choices (40outcome and 80outcome) versus the ## safe choices (20outcome) con=contrast( mylme, a=list(group="Group1", task=c("40outcome", "80outcome"), scanner=levels(my.model$...

Hello, I am trying to generate a confidence interval (90 or 95%) of a regression line. This is primarily just for illustration on a scatter plot (i.e. I am trying to make this http://www.ast.cam.ac.uk/~rgm/scratch/statsbook/graphics/anima4.gif). I have been trying to use the predict.lm function, with interval set as "confidence", but this still seems to be giving me a prediction

...20, rpois(n, lam= 9))), > f3 = factor(pmin(32, rpois(n, lam= 12)))) > with(ldat, > ldat$y <<- 10 + 4*x1 + 2*x2 + rnorm(n) + > ## no rounding here: > + 10 * rnorm(nlevels(f1))[f1] + > + 100* rnorm(nlevels(f2))[f2]) > str(ldat) > > mylm <- lm(y ~ .^2, data = ldat) > proc.time() ## (~= 100 sec on P4 1.6 GHz "lynne") > str(mm <- model.matrix(mylm)) > smlm <- summary(mylm) > > p1 <- predict(mylm) > p2 <- predict(mylm, type = "terms") > > str(myim <- influence.measures(m...

...will be " , inNumberOfOutputBriks, " stats briks\n") outStats <- vector(mode="numeric", length=inNumberOfOutputBriks) ##cat (inData, "\n") if ( ! all(inData == 0 ) ) { ##try( ##if( inherits( print(inModelFormula) print(inRandomFormula) mylme <- lme(fixed=inModelFormula, random=inRandomFormula, data = inModel) print(mylme) ##, ## silent=FALSE), ##"try-error") ) { ##temp <- 0 ## cat (paste("Error on slice", inZ, "\n")) ##} else { temp <- as.vector(unlis...

Hi, I want to plot the residuals of a least-squares regression. plot(lm(y~x), which=1) does this, but it plots the y-axis of my data on the x-axis of the residuals plot. That is, it plots the residual for each y-value in the data. Can I instead use the x-axis of my data as the x-axis of the residuals plot, showing the residual for a given x? Thanks! Jason Priem University of North