similar to: a question about function eval()

Dear R-help, In the rms package, I have fitted an ols model with a variable represented as a restricted cubic spline, with the knot locations specified as a previously defined vector. When I save the model object and open it in another workspace which does not contain the vector of knot locations, I get an error message if I try to predict with that model. This also happens if only one workspace

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

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 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

>>>>> 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

I've just encountered a segfault when using DierckxSpline::percur function. Below is the minimal example which triggers the error: --- library(DierckxSpline) x <- 1:10 y <- rep(0, 10) pspline <- percur(x, y) --- *** caught segfault *** address (nil), cause 'memory not mapped' Traceback: 1: .Fortran("percur", iopt = as.integer(iopt), m = as.integer(m), x =

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),

I would like to fit a logistic regression using a smothing spline, where the spline is a piecewise cubic polynomial. Is the knots option used to define the subintervals for each piece of the cubic spline? If yes and there are k knots, then why does the coefficients field in the returned object from gam only list k coefficients? Shouldn't there be 4k -4 coefficients? Sincerely, Bill

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

Hi, there, I have 5 datasets. I would like to choose a basis spline with same knots in GAM function in order to obtain same basis function for 5 datasets. Moreover, the basis spline is used to for an interaction of two covarites. I used "cr" in one covariate, but it can only smooth w.r.t 1 covariate. Can anyone give me some suggestion about how to choose a proper smoothing spline

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

Hola a todos: quiero consultarles para estar seguro de que estoy entendiendo bien el funcionamiento de la función smoothSpline() del paquete 'timeSeries'. Tengo una serie temporal con datos mensuales a la cual quiero suavizar usando splines para, por ejemplo, comparar con otras series temporales. Por lo que estuve viendo, me conviene usar la función smoothSpline() que se basa en

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,

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

I'm trying to use mars{mda} to model functions that look fairly close to a sequence of straight line segments. Unfortunately, 'mars' seems to totally miss the obvious places for the knots in the apparent first order spline model, and I wonder if someone can suggest a better way to do this. The following example consists of a slight downward trend followed by a jump up after

Hello everyone,          Dr. wood told me that I can adapting his example to force cubic spline to pass through certain point.          I still have no idea how to achieve this. Suppose we want to force the cubic spline to pass (1,1), how can I achieve this by adapting the following code? # Penalized example: monotonic penalized regression spline ..... # Generate data from a monotonic truth.

Hi, I would like to fit an (interpolating) spline to data where the derivatives at the endpoints of the interval are nonzero, thus the natural spline endpoint-specification does not make sense. Books (de Boor, etc) suggest that in this case I use not-a-knot splines. I know what not-a-knot splines are (so if I were solving for the coefficients directly I knew how to do this), but I don't

Hi all, I am using GAM to model time trends in a logistic regression. Yet I would like to extract the the fitted spline from it to add it to another model, that cannot be fitted in GAM or GAMM. Thus I have 2 questions: 1) How can I fit a smoother over time so that I force one knot to be at a particular location while letting the model to find the other knots? 2) how can I extract the matrix

Here's the latest batch of words to get shipped out to Allison Smith. Please submit reasonably small changes to me by tomorrow 10:00 AM Eastern time, and I'll add them. As usual, donations to what will be a ~$110 USD expense would be appreciated, as I am paying for this round out of my pocket. Please send to paypal address "jtodd@loligo.com". I did not include all