Some statistical result / background (Link in the picture: Function to calculate R2 (R-squared) in R) Computational details Computations involved here is basically the computation of the variance-covariance matrix. Once we have it, results for all pairwise regression is just element-wise matrix arithmetic. The variance-covariance matrix can be obtained by R function cov, but functions … Read more
A summary.lm object stores these values in a matrix called ‘coefficients’. So the value you are after can be accessed with: a2Pval <- summary(mg)$coefficients[2, 4] Or, more generally/readably, coef(summary(mg))[“a2″,”Pr(>|t|)”]. See here for why this method is preferred.
You’re almost there. You just have to paste the entire formula together, something like this: paste(“roll_pct ~ “,b,sep = “”) coerce it to an actual formula using as.formula and then pass that to lm. Technically, I think lm may coerce a character string itself, but coercing it yourself is generally safer. (Some functions that expect … Read more
When specifying interval and level argument, predict.lm can return confidence interval (CI) or prediction interval (PI). This answer shows how to obtain CI and PI without setting these arguments. There are two ways: use middle-stage result from predict.lm; do everything from scratch. Knowing how to work with both ways give you a thorough understand of … Read more
This issue is raised over and over again, but unfortunately no satisfying answer has been made which can be an appropriate duplicate target. Looks like I need to write one. Most people know this is related to “contrasts”, but not everyone knows why it is needed, and how to understand its result. We have to … Read more
This is a problem of using different names between your data and your newdata and not a problem between using vectors or dataframes. When you fit a model with the lm function and then use predict to make predictions, predict tries to find the same names on your newdata. In your first case name x … Read more
If I don’t get you wrong, you are working with a dataset like this: set.seed(0) dat <- data.frame(y1 = rnorm(30), y2 = rnorm(30), y3 = rnorm(30), x1 = rnorm(30), x2 = rnorm(30), x3 = rnorm(30)) x1, x2 and x3 are covariates, and y1, y2, y3 are three independent response. You are trying to fit three … Read more
There is a special identifier that one can use in a formula to mean all the variables, it is the . identifier. y <- c(1,4,6) d <- data.frame(y = y, x1 = c(4,-1,3), x2 = c(3,9,8), x3 = c(4,-4,-2)) mod <- lm(y ~ ., data = d) You can also do things like this, to … Read more
First, you want to use model <- lm(Total ~ Coupon, data=df) not model <-lm(df$Total ~ df$Coupon, data=df). Second, by saying lm(Total ~ Coupon), you are fitting a model that uses Total as the response variable, with Coupon as the predictor. That is, your model is of the form Total = a + b*Coupon, with a … Read more
Since 2009, dplyr has been released which actually provides a very nice way to do this kind of grouping, closely resembling what SAS does. library(dplyr) d <- data.frame(state=rep(c(‘NY’, ‘CA’), c(10, 10)), year=rep(1:10, 2), response=c(rnorm(10), rnorm(10))) fitted_models = d %>% group_by(state) %>% do(model = lm(response ~ year, data = .)) # Source: local data frame [2 … Read more