linear-regression

I am trying to implement batch gradient descent on a data set with a single feature and multiple training examples ( m ). When I try using the normal equation, I get the right answer but the wrong one with this code below which performs batch gradient descent in MATLAB. function [theta] = gradientDescent(X, y, theta, alpha, iterations) m = length(y); delta=zeros(2,1); for iter =1:1:iterations for i=1:1:m delta(1,1)= delta(1,1)+( X(i,:)*theta - y(i,1)) ; delta(2,1)=delta(2,1)+ (( X(i,:)*theta - y(i,1))*X(i,2)) ; end theta= theta-( delta*(alpha/m) ); computeCost(X,y,theta) end end y is the

I want to calculate a linear regression using the lm() function in R. Additionally I want to get the slope of a regression, where I explicitly give the intercept to lm() . I found an example on the internet and I tried to read the R-help "?lm" (unfortunately I'm not able to understand it), but I did not succeed. Can anyone tell me where my mistake is? lin <- data.frame(x = c(0:6), y = c(0.3, 0.1, 0.9, 3.1, 5, 4.9, 6.2)) plot (lin$x, lin$y) regImp = lm(formula = lin$x ~ lin$y) abline(regImp, col="blue") # Does not work: # Use 1 as intercept explicitIntercept = rep(1, length(lin$x)) regExp = lm

I am building a quadratic model with lm in R: y <- data[[1]] x <- data[[2]] x2 <- x^2 quadratic.model = lm(y ~ x + x2) Now I want to display both the predicted values and the actual values on a plot. I tried this: par(las=1,bty="l") plot(y~x) P <- predict(quadratic.model) lines(x, P) but the line comes up all squiggely. Maybe it has to do with the fact that it's quadratic? Thanks for any help. 李哲源 You need order() : P <- predict(quadratic.model) plot(y~x) reorder <- order(x) lines(x[reorder], P[reorder]) My answer here is related: Problems displaying LOESS regression line and confidence

I've trained a Linear Regression model with R caret. I'm now trying to generate a confusion matrix and keep getting the following error: Error in confusionMatrix.default(pred, testing$Final) : the data and reference factors must have the same number of levels EnglishMarks <- read.csv("E:/Subject Wise Data/EnglishMarks.csv", header=TRUE) inTrain<-createDataPartition(y=EnglishMarks$Final,p=0.7,list=FALSE) training<-EnglishMarks[inTrain,] testing<-EnglishMarks[-inTrain,] predictionsTree <- predict(treeFit, testdata) confusionMatrix(predictionsTree, testdata$catgeory) modFit<-train(Final~UT1+UT2

My understanding of orthogonal polynomials is that they take the form y(x) = a1 + a2(x - c1) + a3(x - c2)(x - c3) + a4(x - c4)(x - c5)(x - c6)... up to the number of terms desired where a1 , a2 etc are coefficients to each orthogonal term (vary between fits), and c1 , c2 etc are coefficients within the orthogonal terms, determined such that the terms maintain orthogonality (consistent between fits using the same x values) I understand poly() is used to fit orthogonal polynomials. An example x = c(1.160, 1.143, 1.126, 1.109, 1.079, 1.053, 1.040, 1.027, 1.015, 1.004, 0.994, 0.985, 0.977) #

I'm trying hard to add a regression line on a ggplot. I first tried with abline but I didn't manage to make it work. Then I tried this... data = data.frame(x.plot=rep(seq(1,5),10),y.plot=rnorm(50)) ggplot(data,aes(x.plot,y.plot))+stat_summary(fun.data=mean_cl_normal) + geom_smooth(method='lm',formula=data$y.plot~data$x.plot) But it is not working either. In general, to provide your own formula you should use arguments x and y that will correspond to values you provided in ggplot() - in this case x will be interpreted as x.plot and y as y.plot . More information about smoothing methods and