search for: coefficient

...; 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 when we don't add new ones if avoidable) playing a bit more with it, it would make sense to me that the following have the same output: coefficient <- 3 data.frame(value1 = 5) |> transform(coefficient, value2 = coefficient * value1) #> value1 X3 value2 #> 1 5 3 15 data.frame(value1 = 5, coefficient) |> transform(value2 = coefficient * value1) #> value1 coefficient value2 #> 1 5 3 15...

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 s...

...we > don't add new ones if avoidable) > I find silence to be much more unpleasant in practice when debugging, myself, but that may be a personal preference. > > > playing a bit more with it, it would make sense to me that the following > have the same output: > > > coefficient <- 3 > > > data.frame(value1 = 5) |> transform(coefficient, value2 = coefficient * > value1) > > #> value1 X3 value2 > > #> 1 5 3 15 > > > data.frame(value1 = 5, coefficient) |> transform(value2 = coefficient * > value1) > > #>...

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") +...

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. Regards, Peng > x=1:10 > y=1:10 > r=lm(x~y) > class(r) [1] &...

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=...

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 erro...

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 diorit...

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 find how these coefficients are stored, if at all. When...

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 operat...

...psm function in the design package, i.e., all patients who's visits come to an end due to any event other than the event of interest are considered to be censored. Being new to the psm and survreg packages (and to parametric survival modeling) I am not entirely sure how to interpret the coefficient values that psm returns. I have included the following code to illustrate code similar to what I am using on my data. I suppose that the coefficients are somehow rescaled , but I am not sure how to return them to the original scale and make sense out of the coefficients, e.g., estimate the the effe...

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 calculate the % =0 step 4 multiple the % by 2 However, I am new to the package boot. I wrote a co...

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 = rep(0,B) stor.Ind5 = rep(0,B) stor.LNPRI25 = rep(0,B) stor.NPRI10 = rep(0,B) stor.Mine = rep(0,B) for (i in 1:...

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 mnsmkpreve 3.580065e...

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 coefficients table than summary( Dat.nlrq)$coefficients. Below is a series of syntax (mostly from the nlrq sample script) reproducing the error....

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: > fitted(f) 1 2 3 1 4 9 This is very confusing. How should one understand the result of coefficients()? Thanks for any tips, Firas. -- Firas Swidan,...

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 for all three cases. Hoever, using my data...

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( stamp(...

...of (d1(x1,y1), ..., dn(xn,yn), x_class != y_class) rows bound together as a data frame (actually I construct it by columns), and then the obvious thing to try was glm(different.class ~ ., family = binomial(), data = distance.frame) The thing is that this gives me both positve and negative coefficients, whereas the linear combination is only guaranteed to be a metric if the coefficients are all non-negative. There are four fairly obvious ways to deal with that: (1) just force the negative coefficients to 0 and hope. This turns out to work rather well, but still... (2) keep all the coefficie...

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 <- data.frame(y=c(0,25,32,15)...