similar to: Predict nls new data with se.fit snf intervals

The option se.fit in predict.nls is currently ignored. Is there any other function available to calculate the error in the predictions? Thanks, Manuel ______________________________________________ LLama Gratis a cualquier PC del Mundo. Llamadas a fijos y m??viles desde 1 c??ntimo por minuto.

Hello- It has been several years since anyone has asked, so i am asking again- has anyone created a routine to estimate confidence intervals for predictions from nls models (ala Bates and Watts 1988)? Thanks - Alice Shelly [[alternative HTML version deleted]]

Dear all I have tried to estimate the confidence intervals for predicted values of a nonlinear model fitted with nls. The function predict gives the predicted values and the lower and upper limits of the prediction, when the class of the object is lm or glm. When the object is derived from nls, the function predict (or predict.nls) gives only the predicted values. The se.fit and interval aguments

Following an example on p 111 in 'Nonlinear Regression with R' by Ritz & Streibig, I have been fitting nls models using square brackets with the grouping variable inside. In their book is this example, in which 'state' is a factor indicating whether a treatment has been used or not: > Puromycin.m1 <- nls(rate ~ Vm[state] * + conc/(K[state] + conc), data = Puromycin,

Hello all, I'm trying to establish some confidence intervals on predictions I am making using >predict(nls(...)) and predict.nls (unfortunately) does not utilize the se.fit option. A little more background is that I am trying to match the output with older SAS routines to maintain consistency. Because predict.nls does not provide se's for individual predictions, I have been using a

Hi all I have fitted a model usinf nls function to these data: > x [1] 1 0 0 4 3 5 12 10 12 100 100 100 > y [1] 1.281055090 1.563609934 0.001570796 2.291579783 0.841891853 [6] 6.553951324 14.243274230 14.519899320 15.066473610 21.728809880 [11] 18.553054450 23.722637370 The model fitted is: modellogis<-nls(y~SSlogis(x,a,b,c)) It runs OK. Then I calculate

Hi all I am using r version 1.5.0 on a windows 2000 OS I have been converting an old S+ script I had that uses the nls function, to r. In S+ I could obtain the standard errors using predict(...,se = T) In my documentation for R it says for predict.nls(), "At present `se.fit' and `interval' are ignored." my questions are 1. is this still the case or am I just out of date

Dear All I am quite new to R and would appreciate some help fitting 95% confidence intervals to a nls function. I have the data DOY CET 90 5.9 91 8 92 8.4 93 7.7 95 6.6 96 6.8 97 7.1 98 9.7 99 12.3 100 12.8 102 11 103 9.3 104 9.8 105 9.9 107 7.7 110 6.2 111 5.9 112 5.9 113 3.4 114 3.5 116 3.3 117 5.4 118 6.3 119 9.7 120 11.2 121 7.3 124 7.8 etc I am trying to use some code that has been

I've tried to predict the values from a new data.frame using the nls.predict function and keep getting the error message: Error in if (sum(wrong) == 1) stop(gettextf("variable '%s' was fitted with class \"%s\" but class \"%s\" was supplied", : missing value where TRUE/FALSE needed I first thought that it was becuase there may have been something

I have data that looks like O.lengthO.age 176 1 179 1 182 1 ... 493 5 494 5 514 5 606 5 462 6 491 6 537 6 553 6 432 7 522 7 625 8 661 8 687 10 704 10 615 12 (truncated) with a simple VonB growth model from within nls(): plot(O.length~O.age, data=OS) Oto = nls(O.length~Linf*(1-exp(-k*(O.age-t0))), data=OS, start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE) mod <- seq(0, 12)

Hi, Does anyone know how to compute the confidence prediction intervals for a nonlinear least squares models (nls)? I was trying to use the function 'predict' as I usually do for other models fitting (glm, lm, gams...), but it seems that se.fit, and interval computation is not implemented for the nls... Cheers Enrique ~~~~~~~~~~~~~~~~~~~~~~~~~~~ Fisheries Research Services, Marine

Dear list, I have encountered a problem with predict.nls (Windows XP, R.2.5.0), but I am not sure if it is a bug... On the nls man page, an example is: DNase1 <- subset(DNase, Run == 1) fm2DNase1 <- nls(density ~ 1/(1 + exp((xmid - log(conc))/scal)), data = DNase1, start = list(xmid = 0, scal = 1)) alg = "plinear", trace =

Greetings, We are performing a meta-analysis of mink pup survival data versus chemical concentration. We have modeled percent survival successfully using nls as shown below and the plot. What we need to do is construct a confidence interval on the concentration at which we get 50% survival (aka the EC50, although we may want other percent survivals in the future). My first question is, what seems

hello, can anyone help with this: ########################################################### ###data: measurments (response = trans) run several times at the same predictor value level (press) por<-data.frame(list(structure(list(run = structure(c(1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c("1", "3", "4"), class = "factor"), press

Dear R users, Can I fit nls.lm to a non-continuous date data. looked at previous examples but still not able to fit the model to my data. There are 25 rows of observations as below; df <- data.frame(Date=as.Date(rownames(df),'%m/%d/%Y'),Y=df$height) df$days <- as.numeric(df$Date - df[1,]$Date) head(df) Date Y days 1 2009-12-01 0.2631250 0 2 2010-01-08 0.4436012

Dear List, my exploration of R goes on... And I REALLY enjoy it ! (thanks to all guRus). Yesterday a colleague ask me for fitting some data presented as a data.frame, something like : >a x X158.7 X150.0 ... 1 -0.25 506 183.1 2 -0.75 633 210.7 3 -1.25 674 220.3 4 -1.50 NA 244.6 5 -1.75 742 261.2 6 -2.25 787 269.1 7 -2.50 NA 283.5 8

Dear R colleagues: Am trying to fit a simple NL model to determine Economical Optimum Nitrogen Rates. The segmented (quadratic + plateau) model only works with some y's, in some cases I get a "singular gradient" error. I'll appreciate any ideas in how to solve the singular gradient error. Thanks, Jose # The following code works using yield2 in the nls model but not using

Buenas tardes, Quisiera obtener el intervalo de confianza (y también intervalos de predicción) para los valores predichos en un modelo nls. ¿Hay alguna manera que no sea por ggplot2 (me interesaría obtener el valor listado -además de en el gráfico-) o por bootstrap? Os copio el código del ajuste del modelo y predicción para los 3 días siguientes: *#Ajuste del modelo* model = nls(formula =

hi, i noticed that se.fit from predict.lm is the same whether interval="conf" or interval="pred". it is not clear to me from ?predict.lm whether this is intended or not. i suggest that se.fit should match the type of interval requested, if interval is specified. suggested change in lm.R line 700 if(se.fit || interval != "none") se <- sqrt(ip) to if(se.fit

for unweighted fits using `nls' I compute confidence intervals for the fitted model function by using: #------------------- se.fit <- sqrt(apply(rr$m$gradient(), 1, function(x) sum(vcov(rr)*outer(x,x)))) luconf <- yfit + outer(se.fit, qnorm(c(probex, 1 - probex))) #------------------- where `rr' contains an `nls' object, `x' is the independent variable vector, `yfit'