lm

I'm executing lm() with arguments formula , data , na.action , and weights . My weights are stored in a numeric variable. When I specify formula as a character (i.e. formula = "Response~0+." ), I get an error that weights is not of the proper length (even though it is). When I specify formula without the quotes (i.e. formula = Response~0+. ), the function works fine. I stumbled upon this sentence in the lm() documentation: All of weights, subset and offset are evaluated in the same way as variables in formula, that is first in data and then in the environment of formula. This is difficult for

I'm using a set of points which go from (-5,5) to (0,0) and (5,5) in a "symmetric V-shape". I'm fitting a model with lm() and the bs() function to fit a "V-shape" spline: lm(formula = y ~ bs(x, degree = 1, knots = c(0))) I get the "V-shape" when I predict outcomes by predict() and draw the prediction line. But when I look at the model estimates coef() , I see estimates that I don't expect. Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 4.93821 0.16117 30.639 1.40e-09 *** bs(x, degree = 1, knots = c(0))1 -5.12079 0.24026 -21.313 2.47e-08 *** bs(x, degree = 1, knots = c(0))2 -0

I'm using lm on a time series, which works quite well actually, and it's super super fast. Let's say my model is: > formula <- y ~ x I train this on a training set: > train <- data.frame( x = seq(1,3), y = c(2,1,4) ) > model <- lm( formula, train ) ... and I can make predictions for new data: > test <- data.frame( x = seq(4,6) ) > test$y <- predict( model, newdata = test ) > test x y 1 4 4.333333 2 5 5.333333 3 6 6.333333 This works super nicely, and it's really speedy. I want to add lagged variables to the model. Now, I could do this by augmenting my original training set: > train$y_1 <- c(0

In R, I get the content of a binary file as a string (because of design issues, I can't access the file directly). This file was originally an lm model. How do I convert that string back into the lm model? Thanks I'm assuming you used base::dput() according to the following example (based on this answer ): # Generate some model over some data data <- sample(1:100, 30) df <- data.frame(x = data, y = 2 * data + 20) model <- lm(y ~ x, df) # Assuming this is what you did you have the model structure inside model.R dput(model, control = c("quoteExpressions", "showAttributes"), file = "model.R") # -

This question already has an answer here: How to run linear model in R with certain data range? 1 answer I have a dummy variable black where black==0 is White and black==1 is Black. I am trying to fit a linear model lm for the black==1 category only, however running the code below gives me the incorrect coefficients. Is there a way in R to run a model with the if statement, similar to Stata? library(foreign) df<-read.dta("hw4.dta") attach(df) black[black==0]<-NA model3<-lm(rent~I(income^2)+income+black) If looks like there are a few issues here. First, you've stored all your data in separate

This question already has an answer here: Linear Regression and group by in R 10 answers I'm trying to run anova() in R and running into some difficulty. This is what I've done up to now to help shed some light on my question. Here is the str() of my data to this point. str(mhw) 'data.frame': 500 obs. of 5 variables: $ r : int 1 2 3 4 5 6 7 8 9 10 ... $ c : int 1 1 1 1 1 1 1 1 1 1 ... $ grain: num 3.63 4.07 4.51 3.9 3.63 3.16 3.18 3.42 3.97 3.4 ... $ straw: num 6.37 6.24 7.05 6.91 5.93 5.59 5.32 5.52 6.03 5.66 ... $ Quad : Factor w/ 4 levels "NE","NW","SE",..: 2 2 2 2 2 2 2 2 2 2 ... Column r

I used lmList to fit 480 relationships and I would like the R2 of each of these. Here is an example dataset and model which are pretty close to what it really looks like, except I have 480 eu (experimental units): eu mass day 11 .02 1 11 .03 2 11 .04 3 11 .06 4 12 .01 1 12 .03 2 12 .04 3 12 .05 4 fit<-lmList(mass ~ day | eu, data=df) Printing fit or summary does not give me the information I want. I am ultimately trying to make a new dataframe that will look like: eu intercept slope R2 11 .01 .95 .98 12 .01 .96 .98 I've got the coefficients through coef , now I need the R-squared. Here you go:

I am trying to do fixed effects linear regression with R. My data looks like dte yr id v1 v2 . . . . . . . . . . . . . . . I then decided to simply do this by making yr a factor and use lm : lm(v1 ~ factor(yr) + v2 - 1, data = df) However, this seems to run out of memory. I have 20 levels in my factor and df is 14 million rows which takes about 2GB to store, I am running this on a machine with 22 GB dedicated to this process. I then decided to try things the old fashioned way: create dummy variables for each of my years t1 to t20 by doing: df$t1 <- 1*(df$yr==1) df$t2 <- 1*(df$yr==2) df$t3 <- 1

So I'm trying to compare different linear models in order to determine if one is better than another. However I have several models, so I want to create an list of models and then call on them. Is that possible? Models <- list(lm(y~a),lm(y~b),lm(y~c) Models2 <- list(lm(y~a+b),lm(y~a+c),lm(y~b+c)) anova(Models2[1],Models[1]) Thank you for your help! If you have two lists of models, and you want to compare each pair of models, then you want Map : models1 <- list(lm(y ~ a), lm(y ~ b), lm(y ~ c) models2 <- list(lm(y ~ a + b), lm(y ~ a + c), lm(y ~ b + c)) Map(anova, models1, models2) This is

This question already has an answer here: Linear Regression and group by in R 10 answers I am running a linear regression on some variables in a data frame. I'd like to be able to subset the linear regressions by a categorical variable, run the linear regression for each categorical variable, and then store the t-stats in a data frame. I'd like to do this without a loop if possible. Here's a sample of what I'm trying to do: a<- c("a","a","a","a","a", "b","b","b","b","b", "c","c","c","c","c") b<- c(0.1,0.2,0.3,0.2,0.3, 0.1,0.2,0.3,0.2,0.3, 0.1,0.2,0.3,0.2,0.3) c<- c(0.2,0.1,0.3,0.2,0.4, 0.2,0.5