regression

I am trying to build a rolling regression function based on the example here , but in addition to returning the predicted values, I would like to return the some rolling model diagnostics (i.e. coefficients, t-values, and mabye R^2). I would like the results to be returned in discrete objects based on the type of results. The example provided in the link above sucessfully creates thr rolling predictions, but I need some assistance packaging and writing out the rolling model diagnostics: In the end, I would like the function to return three (3) objects: Predictions Coefficients T values R^2

I have the following data.frame d from an experiment: - Variable y (response, continuous) - Factor f (500 levels) - Time t (posixct) In the last 8 years, y was measured roughly once a month (exact date in t) for each level of f. Sometimes there are 2 measures per month, sometimes a couple of month passed without any measures. Sorry for not providing example data, but making up unregular time series goes beyond my R knowledge. ;) I'd like to do the following with this data: make a regression using the loess() function (y ~ t) , for each level of f make a prediction of y for the first day of

I have an example dataset below. train<-data.frame(x1 = c(4,5,6,4,3,5), x2 = c(4,2,4,0,5,4), x3 = c(1,1,1,0,0,1), x4 = c(1,0,1,1,0,0), x5 = c(0,0,0,1,1,1)) Suppose I want to create separate models for column x3 , x4 , x5 based on column x1 and x2 . For example lm1 <- lm(x3 ~ x1 + x2) lm2 <- lm(x4 ~ x1 + x2) lm3 <- lm(x5 ~ x1 + x2) I want to then take these models and apply them to a testing set using predict, and then create a matrix that has each model outcome as a column. test <- data.frame(x1 = c(4,3,2,1,5,6), x2 = c(4,2,1,6,8,5)) p1 <- predict(lm1, newdata = test) p2 <- predict(lm2,

I used regsubsets to search for models. Is it possible to automatically create all lm from the list of parameter selections? library(leaps) leaps<-regsubsets(y ~ x1 + x2 + x3, data, nbest=1, method="exhaustive") summary(leaps)$which (Intercept) x1 x2 x3 1 TRUE FALSE FALSE TRUE 2 TRUE FALSE TRUE TRUE 3 TRUE TRUE TRUE TRUE Now i would manually do model_1 <- lm(y ~ x3) and so on. How can this be automated to have them in a list? I don't know why you want a list of all models. summary and coef methods should serve you well. But I will first answer your question from a pure programming aspect, then

How can I calculate and plot a confidence interval for my regression in r? So far I have two numerical vectors of equal length (x,y) and a regression object(lm.out). I have made a scatterplot of y given x and added the regression line to this plot. I am looking for a way to add a 95% prediction confidence band for lm.out to the plot. I've tried using the predict function, but I don't even know where to start with that :/. Here is my code at the moment: x=c(1,2,3,4,5,6,7,8,9,0) y=c(13,28,43,35,96,84,101,110,108,13) lm.out <- lm(y ~ x) plot(x,y) regression.data = summary(lm.out) #save regression

Closed. This question is off-topic . It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 3 years ago . I am working on a desktop application which uses Infragistic grids. We need to automate the regression tests for same. QTP alone does not support this, we need to buy new plug in for same which my company is not very much interested in. Do we have any open source tool for automating regression testing of desktop application? Application is in Dot net but i do not think it makes much of a difference. Please

In the code for the main OLS class in Python Pandas , I am looking for help to clarify what conventions are used for the standard error and t-stats reported when weighted OLS is performed. Here's my example data set, with some imports to use Pandas and to use scikits.statsmodels WLS directly: import pandas import numpy as np from statsmodels.regression.linear_model import WLS # Make some random data. np.random.seed(42) df = pd.DataFrame(np.random.randn(10, 3), columns=['a', 'b', 'weights']) # Add an intercept term for direct use in WLS df['intercept'] = 1 # Add a number (I picked 10) to