similar to: Predicted points in splines

Hi, I am looking at the change in N concentration in plant roots over 4 time points and I have fit a spline to the data using ns and lme: fit10 <- lme( N~ns(day, 3), data = rcn10G) I may want to adjust the model a little bit, but for now, let's assume it's good. I get output for the fixed effects: Fixed: N ~ ns(day, 3) (Intercept) ns(day, 3)1 ns(day, 3)2 ns(day, 3)3

Hello R-help I have the following data for a standard curve concentration(nM),fluorescence 0,48.34 2,58.69 5,70.83 10,94.73 20,190.8 50,436.0 100, 957.9   (1)Is there function in R to plot a spline. (2)How can I interpolation,say 1000 point from 0nM-100nM and store this as a data frame of concentration,fluorescence (3)How can I modify the code below so that instead of retrieving a concentration

Hello, I am trying to estimate a multivariate regression of Y on X with regression splines. Y is (nx1), and X is (nxd), with d>1. I assume the data is generated by some unknown regression function f(X), as in Y = f(X) + u, where u is some well-behaved regression error. I want to estimate f(X) via regression splines (tensor product splines). Then, I want to get the predicted values for some new

Hi, I am looking at the change in N concentration in plant roots over 4 time points and I have fit a spline to the data using ns and lme: fit10 <- lme( N~ns(day, 3), data = rcn10G) I may want to adjust the model a little bit, but for now, let's assume it's good. I get output for the fixed effects: Fixed: N ~ ns(day, 3) (Intercept) ns(day, 3)1 ns(day, 3)2 ns(day, 3)3 1.15676524

Hello, I am using R and the smooth.Pspline function in the pspline package to smooth some data by using natural cubic splines. After fitting a sufficiently smooth spline using the following call: (ps=smooth.Pspline(x,y,norder=2,spar=0.8,method=1) [the values of x are age in years from 1 to 100] I tried to check that R in fact had fitted a natural cubic spline by checking that the resulting

Hello, I have fit a simple spline model to the following data. Data x y 0 1.298 2 0.605 3 0.507 4 0.399 5 0.281 6 0.203 7 0.150 8 0.101 Model Sp.1=lm(y~bs(x,df=4)) Now I wish to inverse predict the x for y=.75, say. Optimize works fine for a polynomial but I can figure out how to get the spline model into the function argument. Can anyone help me out. Thanks!! Jeff Jeff Morris Sanofi

Hai everbody,  Is there anyone have simple application b-spline in r language? I need it for make me understanding about b-spline for polynomial spline. thank u Arif  New Email addresses available on Yahoo! Get the Email name you&#39;ve always wanted on the new @ymail and @rocketmail. Hurry before someone else does! [[alternative HTML version deleted]]

Hi, I am looking for an R package that would calculate multivarite (mostly 2d and 3d, tensor) cubic interpolating splines, so that I could evaluate these splines (and their derivatives) at many points (unkown at the time of calculating the spline polynomials) repeatedly. To make things concrete, I have an array V with dim(V) = k and gridpoint vectors grid=list(...), length(grid[[i]])==k[i],

Does anyone know if R has a function for splines under tension. I know there are numerous packages for spline interpolation within R i just can't find one that lets you determine the tension factor. Any help would be much appreciated! Sam -- View this message in context: http://r.789695.n4.nabble.com/Splines-under-tension-tp2173887p2173887.html Sent from the R help mailing list archive at

Hi - I am fitting various Cox PH models with spline predictors. After fitting the model, I would like to use termplot() to examine the functional form of the fitted model (e.g., to obtain a plot of the relative risk (or log r.r.) versus the predictors). When there is only 1 predictor in the model, termplot returns a "?". In this case, I have not been able to figure out how to create

Hi all, I try to interpolate a data set in the form: time Erg 0.000000 48.650000 1.500000 56.080000 3.000000 38.330000 4.500000 49.650000 6.000000 61.390000 7.500000 51.250000 9.000000 50.450000 10.500000 55.110000 12.000000 61.120000 18.000000 61.260000 24.000000 62.670000 36.000000 63.670000 48.000000 74.880000 I want to get smoothed splines by using the class gam The first way I tried , was

Hi, If I set df=2 in my smooth.spline function, is that equivalent to running a linear regression through my data? It appears that df=# of data points gives the interpolating spline and that df = 2 gives the linear regression, but I just want to confirm this. Thank you, Steven

I want to fit a cubic spline of x on y. where : x [1] 467 468 460 460 450 432 419 420 423 423 y [1] 1 2 3 4 5 6 7 8 9 10 using the syntax spline(y, x) I got following output : $x [1] 1.000000 1.310345 1.620690 1.931034 2.241379 2.551724 2.862069 [8] 3.172414 3.482759 3.793103 4.103448 4.413793 4.724138 5.034483 [15] 5.344828 5.655172

I've got a problem with the R function spline.des. I use the following arguments: x_seq(0,1,length=200) knot_quantile(x,(1:8)/9) ord_3 k_sort(c(rep(range(x),ord,knot))) derivs_rep(2,200) When I do spline.des(k,x,ord,derivs)$design on Splus and on R, I don't have the same results. However, if I take an order ord=4 (ie a spline of degree 3), then, I have the same

Hello, I'm running mixed models in GAMM4 with 2 (non-nested) random intercepts and I want to include a spline term for one of my exposure variables. However, when I include a spline term, I always get reported degrees of freedom of less than 1, even when I know that my spline is using more than 1 degree of freedom. For example, here is the code for my model: >

hola soy nuevo usuario de R y necesito crear un objeto p tal que, p=lambda*integral[f´´(x)^2 dx ] donde "lambda" es uno de los parametros que resultan de la funcion "smooth.spline()" y la integral es sobre la derivada 2 de esa misma funcion... dos cuestiones: 1) como extraigo lambda de los resultados de smooth.spline() para usarlo como objeto cuando lo requiera y 2) como

Hey all, I'm using the gam() function inside the mgcv package to fit a penalised spline to some data. However, I don't quite understand what exactly the intercept it includes by default is / how to interpret it. Ideally I'd like to understand what the intercept is in terms of the B-Spline and/or truncated power series basis representation. Thanks!

BUG IN SPLINE()? Version R-1.0.1, system i486,linux If the spline(x,y,method="natural") function is given values outside the range of the data, it does not give a warning. Moreover, the extrapolated value reported is not the ordinate of the natural spline defined by (x,y). Example. Let x <- c(2,5,8,10) and y <- c(1.2266,-1.7606,-0.5051,1.0390). Then interpolate/extrapolate with

Dear all, I am new to R, and I am a biotechnologist, I want to fit a smoothing spline with smoothing parameter selected by generalized maximum likelihood. I was wondering what function implement this, and, if possible how I can find the fitted results for a certain point (or predict from the fitted spline if this is the correct language) -- View this message in context:

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