regression

I am fitting data with weights using scipy.odr but I don't know how to obtain a measure of goodness-of-fit or an R squared. Does anyone have suggestions for how to obtain this measure using the output stored by the function? The res_var attribute of the Output is the so-called reduced Chi-square value for the fit, a popular choice of goodness-of-fit statistic. It is somewhat problematic for non-linear fitting, though. You can look at the residuals directly ( out.delta for the X residuals and out.eps for the Y residuals). Implementing a cross-validation or bootstrap method for determining

This Q & A arises from How to make group_by and lm fast? where OP was trying to do a simple linear regression per group for a large data frame. In theory, a series of group-by regression y ~ x | g is equivalent to a single pooled regression y ~ x * g . The latter is very appealing because statistical test between different groups is straightforward. But in practice doing this larger regression is not computationally easy. My answer on the linked Q & A reviews packages speedlm and glm4 , but pointed out that they can't well address this problem. Large regression problem is difficult,

I want to do a piecewise linear regression with one break point, where the 2nd half of the regression line has slope = 0 . There are examples of how to do a piecewise linear regression, such as here . The problem I'm having is I'm not clear how to fix the slope of half of the model to be 0. I tried lhs <- function(x) ifelse(x < k, k-x, 0) rhs <- function(x) ifelse(x < k, 0, x-k) fit <- lm(y ~ lhs(x) + rhs(x)) where k is the break point, but the segment on the right is not a flat / horizontal one. I want to constrain the slope of the second segment at 0. I tried: fit <- lm(y ~ x * (x < k) + x *

I have the following code for minimizing the sum of deviation using optim() to find beta0 and beta1 but I am receiving the following errors I am not sure what I am doing wrong: sum.abs.dev<-function(beta=c(beta0,beta1),a,b) { total<-0 n<-length(b) for (i in 1:n) { total <- total + (b[i]-beta[1]-beta[2]*a[i]) } return(total) } tlad <- function(y = "farm", x = "land", data="FarmLandArea.csv") { dat <- read.csv(data) #fit<-lm(dat$farm~dat$land) fit<-lm(y~x,data=dat) beta.out=optim(fit$coefficients,sum.abs.dev) return(beta.out) } Here's the error and warnings are receive: Error in `contrasts<-`(`

I have data on revenue of a company from sales of various products (csv files), one of which looks like the following: > abc Order.Week..BV. Product.Number Quantity Net.ASP Net.Price 1 2013-W44 ABCDEF 92 823.66 749 2 2013-W44 ABCDEF 24 898.89 749 3 2013-W44 ABCDEF 243 892.00 749 4 2013-W45 ABCDEF 88 796.84 699 5 2013-W45 ABCDEF 18 744.80 699 Now, I'm fitting a multiple regression model with Net.Price as Y and Quantity, Net.ASP as x1 and x2. There are more than 100 such files and I'm trying to do it using the following code: fileNames <- Sys.glob("*.csv") for (fileName in fileNames) { abc <-

I would like to run repeated measure anova in R using regression models instead an 'Analysis of Variance' ( AOV ) function. Here is an example of my AOV code for 3 within-subject factors: m.aov<-aov(measure~(task*region*actiontype) + Error(subject/(task*region*actiontype)),data) Can someone give me the exact syntax to run the same analysis using regression models? I want to make sure to respect the independence of residuals, i.e. use specific error terms as with AOV. In a previous post I read an answer of the type: lmer(DV ~ 1 + IV1*IV2*IV3 + (IV1*IV2*IV3|Subject), dataset)) I am really not