search for: coefficiency

Thanks and good point about unspecified behavior. The way it behaves now (when it doesn't ignore) is more consistent with data.frame() though so I prefer that to a "warn and ignore" behaviour: data.frame(a = 1, b = 2, 3) #> a b X3 #> 1 1 2 3 data.frame(a = 1, 2, 3) #> a X2 X3 #> 1 1 2 3 (and in general warnings make for unpleasant debugging so I prefer

Is there a way to get the following code to include NA values where the coefficients are ?NA?? ((summary(reg))$coefficients) explanation: Using a loop, I am running regressions on several ?subsets? of ?data1?. ?reg <- ( lm(lm(data1[,1] ~., data1[,2:l])) )? My regression has 10 independent variables, and I therefore expect 11 coefficients. After each regression, I wish to save the

On Thu, Mar 2, 2023 at 2:02?PM Antoine Fabri <antoine.fabri at gmail.com> wrote: > Thanks and good point about unspecified behavior. The way it behaves now > (when it doesn't ignore) is more consistent with data.frame() though so I > prefer that to a "warn and ignore" behaviour: > > data.frame(a = 1, b = 2, 3) > > #> a b X3 > > #> 1 1 2 3

Anyone know how to get p-values for the t-values from the coefficients produced in vglm? Attached is the code and output ? see comment added to output to show where I need p-values + print(paste("********** Using VGAM function gamma2 **********")) + modl2<- vglm(MidPoint~Count,gamma2,data=modl.subset,trace=TRUE,crit="c") + print(coef(modl2,matrix=TRUE))

Hi, I want to know what is there returned values of 'lm'. 'class' and 'lm' does not show that the returned value has the variable coefficients, etc. I am wondering what is the command to show the detailed information. If possible, I aslo want the lower level information. For example, I want to show that 'coefficients' is a named list and it has 2 elements.

Hi, I thought that 'coefficients' is a named list, but I can not refer to its element by something like r$coefficients$y. I used str() to check r. It says the following. Can somebody let me know what it means? ..- attr(*, "names")= chr [1:2] "(Intercept)" "y" $ Rscript lm.R > x=1:10 > y=1:10 > r=lm(x~y) > class(r) [1] "lm" >

Dear R users, I have problem with extracting coefficients from a object. Here, X (predictor)and Y (response) are two matrix , I am regressing X ( dimensions 10 x 20) on each of columns of Y[,1] (10 x 1) and want to store the coefficient values. I have performed a Elastic Net regression and I want to store the coeffcients in each iteration. I got an error message . I do not

Hello fellow R users, I am trying to extract the coefficient values during a bootstrap operation. Here is the list of my variables that I would like to extract the coefficient values from: (Intercept) LogRds_25k GeoRockbimodal volcanic rocks GeoRockgranodiorite, quartz diorite GeoRockintermediate volcanic

I am writing a resampling program for multiple regression using lm(). I resample the data 10,000 times, each time extracting the regression coefficients. At present I extract the individual regression coefficients using brg = lm(Newdv~Teach + Exam + Knowledge + Grade + Enroll) bcoef[i,] = brg$coef This works fine. But now I want to extract the t tests on these coefficients. I cannot

Dear all, After running a glm, I use the summary ( ) function to extract its coefficients and related statistics for further use. Unfortunately, the screen only displays a small (last) part of the results. I tried to overcome the problem by creating/saving an object "coef" for coefficients of the model and export/save it e.g. as a cvs document. While I succed with this operatiion, I do

I am using psm to model some parametric survival data, the data is for length of stay in an emergency department. There are several ways a patient's stay in the emergency department can end (discharge, admit, etc..) so I am looking at modeling the effects of several covariates on the various outcomes. Initially I am trying to fit a survival model for each type of outcome using the psm

Hello, I want to know if two quadratic regressions are significantly different. I was advised to make the test using step 1 bootstrapping both quadratic regressions and get their slope coefficients. (Let's call the slope coefficient *â*^1 and *â*^2) step 2 use the slope difference *â*^1-*â*^2 and bootstrap the slope coefficent step 3 find out the sampling distribution above and

Hello, I am doing a simple linear regression analysis that includes few variables. I am using a bootstrap analysis to obtain the variation of my variables to replacement. I am trying to obtain the coefficients 95% confidence interval from the bootstrap procedure. Here is my script for the bootstrap: N = length (data_Pb[,1]) B = 10000 stor.r2 = rep(0,B) stor.r2 = rep(0,B) stor.inter =

summary(x), where x is the output of lm, produces the expectedd display, including standard errors of the coefficients. summary(x)$coefficients produces a vector (x is r$individual[[2]]): > r$individual[[2]]$coefficients tX(Intercept) tXcigspmkr tXpeld tXsmkpreve mn -2.449188e+04 -4.143249e+00 4.707007e+04 -3.112334e+01 1.671106e-01 mncigspmkr mnpeld

I have been quantreg library for a number of projects but have just hit a snag. I am using nlrq to examine an asymptotic relationship between 2 variables at the 99th percentile. It performs as expected, however when I try to extract the coefficients along with se and significance I am running into problems. The problem is that for the nlrq regression Dat.nlrq, summary(Dat.nlrq) reports a different

Hi, I can not comprehend the linear fitting results of polynoms. For example, given the following data (representing y = x^2): > x <- 1:3 > y <- c(1, 4, 9) performing a linear fit > f <- lm(y ~ poly(x, 2)) gives weird coefficients: > coefficients(f) (Intercept) poly(x, 2)1 poly(x, 2)2 4.6666667 5.6568542 0.8164966 However the fitted() result makes sense: >

Dear all, I did a non-linear least square model fit y ~ a * x^b (a) > nls(y ~ a * x^b, start=list(a=1,b=1)) to obtain the coefficients a & b. I did the same with the linearized formula, including a linear model log(y) ~ log(a) + b * log(x) (b) > nls(log10(y) ~ log10(a) + b*log10(x), start=list(a=1,b=1)) (c) > lm(log10(y) ~ log10(x)) I expected coefficient b to be identical

Hi I have the following code which I would like to simplify. Id does linear regressions and returns the r-squares, and the coefficients. It runs slow, as it is doing the regressions for each - is it possible to get the values in a dataframe which looks as follow: expert | xx | seeds | r.squared | slope | intercept Thanks in advance, Rainer library(reshape) rsqs <- as.data.frame(

I am trying to automatically construct a distance function from a training set in order to use it to cluster another data set. The variables are nominal. One variable is a "class" variable having two values; it is kept separate from the others. I have a method which constructs a distance matrix for the levels of a nominal variable in the context of the other variables. I want to

Dear all, I would like to use predict.lm() with an existing lm object but with new arbitrary coefficients. I modify 'fit$coef' (see example below) "by hand" but the actual model in 'fit' used for prediction does not seem to be altered (although fit$coef is!). Can anyone please help me do this properly? Thanks in advance, J?r?mie > dat <-