lm

问题 coefplot from library(coefplot) has a variable decreasing which when set to to TRUE the coefficients should be plotted in descending order But when I run a toy example: data(tips, package = "reshape2") mod1 <- lm(tip ~ day + sex + smoker, data = tips) coefplot.glm(mod2, decreasing = TRUE) the coefficients aren't in descending order. What am I missing? EDIT I was missing sort = "magnitude" . However, this doesn't work with multiplot : data(tips, package = "reshape2") mod1 <- lm(tip ~ day + sex

问题 This question already has answers here : How to use formula in R to exclude main effect but retain interaction (3 answers) Why do I get NA coefficients and how does `lm` drop reference level for interaction (1 answer) Closed 2 years ago . Setting aside the debate about Type III ANOVA and the Principle of Marginality and all that... I've set up two models whose sum of squares should be different (and Type III ANOVA would test that difference). Here's the code: library(car) library(openintro)

问题 I have a simple data set for which I applied a simple linear regression model. Now I would like to use fixed effects to make a better prediction on the model. I know that I could also consider making dummy variables, but my real dataset consist of more years and has more variables so I would like to avoid making dummies. My data and code is similar to this: data <- read.table(header = TRUE, stringsAsFactors = FALSE, text="CompanyNumber ResponseVariable Year ExplanatoryVariable1

问题 When building the linear regression model using lm , the data set has about 20 independent variables. Do I need to explicitly clarify them as factor ? If I have to, how can I do that? It can be very tedious to declare one by one. 回答1: First, check which variables R has automatically converted into factors with the commande str(mydata) Then if you want to convert several variable into factors easily, you can do something like this: create a "mycol" variable with the No of columns you want to

问题 I have soil moisture data with x-, y- and z-coordinates like this: gue <- structure(list(x = c(311939.1507, 311935.4607, 311924.7316, 311959.553, 311973.5368, 311953.3743, 311957.9409, 311948.3151, 311946.7169, 311997.0803, 312017.5236, 312006.0245, 312001.5179, 311992.7044, 311977.3076, 311960.4159, 311970.6047, 311957.2564, 311866.4246, 311870.8714, 311861.4461, 311928.7096, 311929.6291, 311929.4233, 311891.2915, 311890.3429, 311900.8905, 311864.4995, 311870.8143, 311866.9257, 312002.571,

问题 I am trying to run many regression models with all possible combinations of a set of independent variables. In this example, I am interested in the coefficients of cyl with all possible combinations of other variables listed in xlist . df <- mtcars md <- "mpg ~ cyl" xlist <- c("disp", "hp", "am") n <- length(xlist) # get a list of all possible combinations of xlist comb_lst <- unlist(lapply(1:n, function(x) combn(xlist, x, simplify=F)), recursive = F) # get a list of all models md_lst <-

问题 I know that the support for linear models with multiple LHS is limited. But when it is possible to run a function on an "mlm" object, I would expect the results to be trusty. When using rstudent , strange results are produced. Is this a bug or is there some other explanation? In the example below fittedA and fittedB are identical, but in the case of rstudent the 2nd column differs. y <- matrix(rnorm(20), 10, 2) x <- 1:10 fittedA <- fitted(lm(y ~ x)) fittedB <- cbind(fitted(lm(y[, 1] ~ x)),

问题 This question was asked in stackoverflow.com/q/38378118 but there was no satisfactory answer. LASSO with λ = 0 is equivalent to ordinary least squares, but this does not seem to be the case for glmnet() and lm() in R. Why? library(glmnet) options(scipen = 999) X = model.matrix(mpg ~ 0 + ., data = mtcars) y = as.matrix(mtcars["mpg"]) coef(glmnet(X, y, lambda = 0)) lm(y ~ X) Their regression coefficients agree by at most 2 significant figures, perhaps due to slightly different termination

问题 I have this code: library(biglm) library(ff) myData <- read.csv.ffdf(file = "myFile.csv") testData <- read.csv(file = "test.csv") form <- dependent ~ . model <- biglm(form, data=myData) predictedData <- predict(model, newdata=testData) the model is created without problems, but when I make the prediction... it runs out of memory: unable to allocate a vector of size xx.x MB some hints? or how to use ff to reserve memory for predictedData variable? 回答1: I have not used biglm package before.

问题 I have data and I expect several linear correlations of the form y_i = a_i + b_i * t_i, i = 1 .. N where N is a priori unknown. The short version of the question is: Given a fit how can I extract N ? how can I extract the equations? In the reproducible example below, I have data (t,y) with corresponding parameters p1 (levels p1_1 , p1_2 ) and p2 (levels p2_1 , p2_2 , p2_3 ). Thus the data looks like (t, y, p1, p2) which has at most 2*3 different best-fit lines and the linear fit from has then