similar to: Automatic Knot selection in Piecewise linear splines

>>>>> 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

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 want to fit a piecewise spline of degree 1, i.e. a spline consisting of a straight line over each piece. I downloaded the R package pspline, then I have following questions: 1. in the program, the degree of the spline is specified by 2*norder-1. Why do they adopt such scheme that we can only fit a spline with odd degree? 2. norder cannot be set to 1. Is there any specific reason

How can I fit a piecewise continuous logistic regression with a single free knot (i.e. the knot is not specified; the model produce an estimate of the value of the knot). Thank you, John John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center

Hello, In the following I tried 3 versions of an example in R help. Only the two first predict command work. After : library(splines) require(stats) 1) fm1 <- lm(weight ~ bs(height, df = 5), data = women) ht1 <- seq(57, 73, len = 200) ph1 <- predict(fm1, data.frame(height=ht1)) # OK plot(women, xlab = "Height (in)", ylab = "Weight (lb)") lines(ht1, ph1) 2)

Dear all, I have two variables, y and x. It seems that the relationship between them is Piecewise Linear Functions. The cutpoint is 20. That is, when x<20, there is a linear relationship between y and x; while x>=20, there is another different linear relationship between them. How can i specify their relationships in R correctly? # glm(y~I(x<20)+I(x>=20),family = binomial, data =

Hi - I'd like to construct and plot the percents by year in a small data set (d) that has values between 1988 and 2007. I'd like to have a breakpoint (buy no discontinuity) at 1996. Is there a better way to do this than in the code below? > d year percent se 1 1988 30.6 0.32 2 1989 31.5 0.31 3 1990 30.9 0.30 4 1991 30.6 0.28 5 1992 29.3 0.25 6 1994 30.3

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 Rusers, I am now using R and SAS to fit the piecewise linear functions, and what surprised me is that they have a great differrent result. See below. #R code--Knots for distance are 16.13 and 24, respectively, and Knots for y are -0.4357 and -0.3202 m.glm<-glm(mark~x+poly(elevation,2)+bs(distance,degree=1,knots=c(16.13,24)) +bs(y,degree=1,knots=c(-0.4357,-0.3202

windows 7, R 2.12 I am trying to run a piecewise linear regression with a single knot, i.e. a regression composed of two straight lines where the two lines intersect at an x value given by the variable knot. I wish to estimate the slope of both lines, the value of knot, the x value where the two lines intersect, and an intercept. I am using the nls code below, and get the following error

Dear All, I am trying to simulate data for a spline/piecewise regression model. I am missing something fundamental in my simulation procedure because when I try to fit my simulated data using the Gauss-Newton method in SAS, I am getting some wacky parameter estimates. Can anyone please check my simulation code and tell me what mistake I am making in generating data for spline model? Thank you

Hi List, I have to compare a relationship between y and x for two locations. I found logistic regression fits both datasets well, but I am not sure how to test if relationships for both sites are significantly different. I searched the r site, however no answers exactly match the question. I used Tukey's HSD to compare two means, but the relationship in my study was not simply linear. So I

The subject says it all really. Question 1. Here is some code created to illustrate my problem, can anyone spot where I'm going wrong? Question 2. The reason I'm following a manual specification of models relates to the fact that in reality I am using mgcv::gam, and I'm not aware that dredge is able to separate individual smooth terms out of say s(a,b). Hence an additional request,

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 there, I have a data frame as listed below: > Ca.P.Biomass.A P Biomass 1 334.5567 0.2870000 2 737.5400 0.5713333 3 894.5300 0.6393333 4 782.3800 0.5836667 5 857.5900 0.6003333 6 829.2700 0.5883333 I have fit the data using logistic, Michaelis?Menten, and linear model, they all give significance. > fm1 <- nls(Biomass~SSlogis(P, phi1, phi2, phi3), data=Ca.P.Biomass.A)

Hi, I'm using natural cubic splines from splines::ns() in survival regression (regressing inter-arrival times of patients to a queue on queue size). The queue size fluctuates between 3600 and 3900. I would like to be able to run predict.survreg() for sizes <3600 and >3900 by assuming that the rate for <3600 is the same as for 3600 and that for >4000 it's the same as for

Hi R experts, I'm trying to use nls() for a piecewise linear regression with the first slope constrained to 0. There are 10 data points and when it does converge the second slope is almost always over estimated for some reason. I have many sets of these 10-point datasets that I need to do. The following segment of code is an example, and sorry for the overly precise numbers, they are just

I'm trying to use the FollowMe app with Asterisk 1.4.17. I've followed the WIKI page on setting it up but I can't seem to get it to work. Here is my Asterisk version: pbx1*CLI> core show version Asterisk 1.4.17 built by root @ pbx1 on a i686 running Linux on 2008-01-10 12:08:48 UTC Here is a log of when the FollowMe is being called: NOTE: I've tried to use the AstDB as

Hi R Users, When I use package lme4 for mixed model analysis, I can't distinguish the significant and insignificant variables from all random independent variables. Here is my data and result: Data: Rice<-data.frame(Yield=c(8,7,4,9,7,6,9,8,8,8,7,5,9,9,5,7,7,8,8,8,4,8,6,4,8,8,9), Variety=rep(rep(c("A1","A2","A3"),each=3),3),

Windows Vista R 2.10 - I know it is old, I will update later today. How might I perform a piece-wise linear regression where two linear segments are separated by a single knot? In addition to estimating the slopes of the two segments (or the slope in one segment and the difference between the slope of the first and second segment), I would like the analysis to select the optimum knot. My first