linear-regression

I know there is a small difference between $sigma and the concept of root mean squared error . So, i am wondering what is the easiest way to obtain RMSE out of lm function in R ? res<-lm(randomData$price ~randomData$carat+ randomData$cut+randomData$color+ randomData$clarity+randomData$depth+ randomData$table+randomData$x+ randomData$y+randomData$z) length(coefficients(res)) contains 24 coefficient, and I cannot make my model manually anymore. So, how can I evaluate the RMSE based on coefficients derived from lm ? Residual sum of squares: RSS <- c(crossprod(res$residuals)) Mean squared error:

I am trying to use a linear regression on a group by pandas python dataframe: This is the dataframe df: group date value A 01-02-2016 16 A 01-03-2016 15 A 01-04-2016 14 A 01-05-2016 17 A 01-06-2016 19 A 01-07-2016 20 B 01-02-2016 16 B 01-03-2016 13 B 01-04-2016 13 C 01-02-2016 16 C 01-03-2016 16 #import standard packages import pandas as pd import numpy as np #import ML packages from sklearn.linear_model import LinearRegression #First, let's group the data by group df_group = df.groupby('group') #Then, we need to change the date to integer df['date'] = pd.to_datetime(df['date']) df['date_delta

This R code throws a warning # Fit regression model to each cluster y <- list() length(y) <- k vars <- list() length(vars) <- k f <- list() length(f) <- k for (i in 1:k) { vars[[i]] <- names(corc[[i]][corc[[i]]!= "1"]) f[[i]] <- as.formula(paste("Death ~", paste(vars[[i]], collapse= "+"))) y[[i]] <- lm(f[[i]], data=C1[[i]]) #training set C1[[i]] <- cbind(C1[[i]], fitted(y[[i]])) C2[[i]] <- cbind(C2[[i]], predict(y[[i]], C2[[i]])) #test set } I have a training data set (C1) and a test data set (C2). Each one has 129 variables. I did k means cluster analysis on the C1 and then split my data set

Here is what I am doing: $ python Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54) [GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin >>> import statsmodels.api as sm >>> statsmodels.__version__ '0.5.0' >>> import numpy >>> y = numpy.array([1,2,3,4,5,6,7,8,9]) >>> X = numpy.array([1,1,2,2,3,3,4,4,5]) >>> res_ols = sm.OLS(y, X).fit() >>> res_ols.params array([ 1.82352941]) I had expected an array with two elements?!? The intercept and the slope coefficient? behzad.nouri Try this: X = sm.add_constant(X) sm.OLS(y,X) as in the documentation : An intercept is not included by default and

what is the benefit of using Gradient Descent in the linear regression space? looks like the we can solve the problem (finding theta0-n that minimum the cost func) with analytical method so why we still want to use gradient descent to do the same thing? thanks When you use the normal equations for solving the cost function analytically you have to compute: Where X is your matrix of input observations and y your output vector. The problem with this operation is the time complexity of calculating the inverse of a nxn matrix which is O(n^3) and as n increases it can take a very long time to

Consider the following table : DB <- data.frame( Y =rnorm(6), X1=c(T, T, F, T, F, F), X2=c(T, F, T, F, T, T) ) Y X1 X2 1 1.8376852 TRUE TRUE 2 -2.1173739 TRUE FALSE 3 1.3054450 FALSE TRUE 4 -0.3476706 TRUE FALSE 5 1.3219099 FALSE TRUE 6 0.6781750 FALSE TRUE I'd like to explain my quantitative variable Y by two binary variables (TRUE or FALSE) without intercept. The argument of this choice is that, in my study, we can't observe X1=FALSE and X2=FALSE at the same time, so it doesn't make sense to have a mean, other than 0, for this level. With intercept m1 <- lm(Y~X1+X2, data=DB) summary(m1)

I need to run a multiple regression in R, with the variables X1, X2 and X3, where there is a variable θ = β2 + β3. So instead of β2, for the coefficient of X2 I need to use (θ - β3). How could I do this? Note that Y = b1 * x1 + (t - b3) * x2 + b3 * x3 is equivalent to Y = b1 * x1 + t * x2 - b3 * x2 + b3 * x3 = b1 * x1 + t * x2 + b3 * (x3 - x2) So, you can continue from there easily. 来源： https://stackoverflow.com/questions/53561699/linear-combination-of-regression-coefficients-in-r