similar to: predict with model (rms package)

How can I extract the adjusted R^2 value from an ols object (using rms package)? library(rms) x <- rnorm(10) y <- x + rnorm(10) ols1 <- ols(y ~ x) Typing "ols1" displays adjusted R^2 among other things, but how can I assign it to a variable? I tried str(ols1) but couldn't see where to go from there. Thanks, Mark Seeto

Hi all, I have a vector of proportions (post_op_prw) such that >summary(amb$post_op_prw) Min. 1st Qu. Median Mean 3rd Qu. Max. NA's 0.0000 0.0000 0.0000 0.3985 0.9134 0.9962 1.0000 > summary(cut2(amb$post_op_prw,0.0001)) [0.0000,0.0001) [0.0001,0.9962] NA's 1904 1672 1

Hello ALL! I have a problem to plot factor (lets say gender) as a line, or at least both line and point, from ols model: ols1 <- ols(Y ~ gender, data=dat, x=T, y=T) plot(ols1, gender=NA, xlab="gender", ylab="Y", ylim=c(5,30), conf.int=FALSE) If I convert gender into discrete numeric predictor, and use forceLines=TRUE, plot is not nice and true, since it shows values

Dear R help, I'm having some trouble with model formulas for the ols function in the rms package. I want to have two variables represented as restricted cubic splines, and also include an interaction as a product of linear terms, but I get an error message. library(rms) d <- data.frame(x1 = rnorm(50), x2 = rnorm(50), y = rnorm(50)) ols(y ~ rcs(x1,3) + rcs(x2,3) + x1*x2, data=d) Error in

Using hccm() I got a heteroscedasticity correction factor on the diagonal of the return matrix, but I don't know how to incorporate this into my linear model: METHOD 1: > OLS1 <- lm(formula=uer92~uer+low2+mlo+spec+degree+hit) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.0623377 0.0323461 -1.927 0.057217 . uer 0.2274742 0.0758720

Dear R-help, I don't understand why Gls gives me an error when trying to fit a model with AR(2) errors, while gls (from nlme) does not. For example: library(nlme) library(rms) set.seed(1) d <- data.frame(x = rnorm(50), y = rnorm(50)) gls(y ~ x, data=d, correlation = corARMA(p=2)) #This works Gls(y ~ x, data=d, correlation = corARMA(p=2)) # Gives error # Error in

Hi, I have a question about the usage of eval(). Wonder if any experienced user can help me out of it. I use eval() in the following function: semireg.pwl <- function(coef.s=rnorm(1),coef.a=rnorm(1),knots.pos=knots.x,knots.ini.val=knots.val){ knotn <- length(knots.pos) def.par.env <- sys.frame(1) print(def.par.env) print(environment(coef.s)) tg <- eval( (parse(text=

Hello, I have a general question about combining imputations as well as a question specific to the rms and Hmisc packages. The situation is multiple regression on a data set where multiple imputation has been used to give M imputed data sets. I know how to get the combined estimate of the covariance matrix of the estimated coefficients (average the M covariance matrices from the individual

I?ve been using Frank Harrell?s rms package to do bootstrap model validation. Is it the case that the optimum penalization may still give a model which is substantially overfitted? I calculated corrected R^2, optimism in R^2, and corrected slope for various penalties for a simple example: x1 <- rnorm(45) x2 <- rnorm(45) x3 <- rnorm(45) y <- x1 + 2*x2 + rnorm(45,0,3) ols0 <- ols(y

Dear Mac users, Hi, as you might have probably read the thread of "[R] Rsquared in summary(lm)" on May 10, it seems that Mac version of lm() seem to be working incorrectly. I enclose the script to produce the result both for lm() and manual calculation for a simple regression. Could you run the script and report with the version of R, so I don't have to go through every builds

Hi, I have fit a series of ols() models, by group, in this manner: l <- ols(y ~ rcs(x, 4)) ... where the series of 'x' values in each group is the same, however knots are not always identical between groups. The result is a table of 'coefs' derived from the ols objects, by group: group Intercept top top' top'' 1 6.864 0.01 2.241 -2.65

Hello, I have a couple of questions about the ols function in Frank Harrell's rms package. Is there any way to specify variables by their column number in the data frame rather than by the variable name? For example, library(rms) x1 <- rnorm(100, 0, 1) x2 <- rnorm(100, 0, 1) x3 <- rnorm(100, 0, 1) y <- x2 + x3 + rnorm(100, 0, 5) d <- data.frame(x1, x2, x3, y) rm(x1, x2, x3,

Hi r-helpers. Please forgive my ignorance, but I would like to plot a smoothing spline (smooth.spline) from package "stats", and show the knots in the plot, and I can't seem to figure out where smooth.spline has located the knots (when I use nknots). Unfortunately, I don't know a lot about splines, but I know that they provide me an easy way to estimate the location of local

Goodmorning, This is a documentation related question about the B-spline function in R. In the help file it is stated that: "df degrees of freedom; one can specify df rather than knots; bs() then chooses df-degree-1 knots at suitable quantiles of x (which will ignore missing values)." So if one were to specify a spline with 6 degrees of freedom (and no intercept) then a basis

>>>>> Anupam Tyagi >>>>> on Tue, 9 Jul 2024 16:16:43 +0530 writes: > How can I do automatic knot selection while fitting piecewise linear > splines to two variables x and y? Which package to use to do it simply? I > also want to visualize the splines (and the scatter plot) with a graph. > Anupam NB: linear splines, i.e. piecewise

Suppose I have two variables, x and y. For a fixed number of knots, I want to create a spline transformation of x such that a loss function is minimized. Presumably, this loss function would be least squares, i.e. sum (f(x)-y)^2. The spline transformations would be linear, quadratic or cubic. I know I can solve this problem using some optimization function in R, but I was wondering if anyone

Dear List, I'm using gam in a multiple imputation framework -- specifying the knot locations, and saving the results of multiple models, each of which is fit with slightly different data (because some of it is predicted when missing). In MI, coefficients from multiple models are averaged, as are variance-covariance matrices. VCV's get an additional correction to account for how

Hi All, I am trying to figure out how to get the position of the knots in a pspline used in a cox model. my.model = coxph(Surv(agein, ageout, status) ~ pspline(x), mydata) # x being continuous How do I find out where the knot of the spline are? I would like to know to figure out how many cases are there between each knot. Best, Federico -- Federico C. F. Calboli Department of Epidemiology

Apologies if this is this too obscure for R-help. In package splines, ns(x,,knots,intercept=TRUE) produces an n by K+2 matrix N, the values of K+2 basis functions for the natural splines with K (internal) knots, evaluated at x. It does this by first generating an n by K+4 matrix B of unconstrained splines, then postmultiplying B by H, a K+4 by K+2 representation of the nullspace of C (2 by K+4),

dear all, I apologize for my delay in replying you. Here my contribution, maybe just for completeness: Similar to "earth", "segmented" also fits piecewise linear relationships with the number of breakpoints being selected by the AIC or BIC (recommended). #code (example and code from Martin Maechler previous email) library(segmented) o<-selgmented(y, ~x, Kmax=20,