regression

I didn't find a function to calculate the orthogonal regression (TLS - Total Least Squares). Is there a package with this kind of function? Update: I mean calculate the distance of each point symmetrically and not asymmetrically as lm() does. You might want to consider the Deming() function in package MethComp [ function info ]. The package also contains a detailed derivation of the theory behind Deming regression. The following search of the R Archives also provide plenty of options: Total Least Squares Deming regression Your multiple questions on CrossValidated, here and R-Help imply that

In order to correct heteroskedasticity in error terms, I am running the following weighted least squares regression in R : #Call: #lm(formula = a ~ q + q2 + b + c, data = mydata, weights = weighting) #Weighted Residuals: # Min 1Q Median 3Q Max #-1.83779 -0.33226 0.02011 0.25135 1.48516 #Coefficients: # Estimate Std. Error t value Pr(>|t|) #(Intercept) -3.939440 0.609991 -6.458 1.62e-09 *** #q 0.175019 0.070101 2.497 0.013696 * #q2 0.048790 0.005613 8.693 8.49e-15 *** #b 0.473891 0.134918 3.512 0.000598 *** #c 0.119551 0.125430 0.953 0.342167 #--- #Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0

I am trying to replicate a logit regression from Stata to R. In Stata I use the option "robust" to have the robust standard error (heteroscedasticity-consistent standard error). I am able to replicate the exactly same coefficients from Stata, but I am not able to have the same robust standard error with the package "sandwich". I have tried some OLS linear regression examples; it seems like the sandwich estimators of R and Stata give me the same robust standard error for OLS. Does anybody know how Stata calculate the sandwich estimator for non-linear regression, in my case the logit regression?

To make clear what I'm asking I've created an easy example. Step one is to create some data: gender <- factor(rep(c(1, 2), c(43, 41)), levels = c(1, 2),labels = c("male", "female")) numberofdrugs <- rpois(84, 50) + 1 geneticvalue <- rpois(84,75) death <- rpois(42,50) + 15 y <- data.frame(death, numberofdrugs, geneticvalue, gender) So these are some random dates merged to one data.frame . So from these dates I'd like to plot a cloud where I can differ between the males and females and where I add two simple regressions (one for females and one for males). So I've started, but I couldn't get to

I use a tensorflow to implement a simple multi-layer perceptron for regression. The code is modified from standard mnist classifier, that I only changed the output cost to MSE (use tf.reduce_mean(tf.square(pred-y)) ), and some input, output size settings. However, if I train the network using regression, after several epochs, the output batch are totally the same. for example: target: 48.129, estimated: 42.634 target: 46.590, estimated: 42.634 target: 34.209, estimated: 42.634 target: 69.677, estimated: 42.634 ...... I have tried different batch size, different initialization, input

I have the following information: Height Weight 170 65 167 55 189 85 175 70 166 55 174 55 169 69 170 58 184 84 161 56 170 75 182 68 167 51 187 85 178 62 173 60 172 68 178 55 175 65 176 70 I want to construct quadratic and cubic regression analysis in Excel. I know how to do it by linear regression in Excel, but what about quadratic and cubic? I have searched a lot of resources, but could not find anything helpful. Ian Boyd You need to use an undocumented trick with Excel's LINEST function: =LINEST(known_y's, [known_x's], [const], [stats]) Background A regular linear regression is calculated

This question relates to a machine learning feature selection procedure. I have a large matrix of features - columns are the features of the subjects (rows): set.seed(1) features.mat <- matrix(rnorm(10*100),ncol=100) colnames(features.mat) <- paste("F",1:100,sep="") rownames(features.mat) <- paste("S",1:10,sep="") The response was measured for each subject ( S ) under different conditions ( C ) and therefore looks like this: response.df <- data.frame(S = c(sapply(1:10, function(x) rep(paste("S", x, sep = ""),100))), C = rep(paste("C", 1:100, sep = ""), 10), response = rnorm(1000),