similar to: Regarding anova result

Displaying 20 results from an estimated 10000 matches similar to: "Regarding anova result"

2008 Apr 10
1
(no subject)
Subject: nls, step factor 0.000488281 reduced below 'minFactor' of 0.000976563 Hi there, I'm trying to conduct nls regression using roughly the below code: nls1 <- nls(y ~ a*(1-exp(-b*x^c)), start=list(a=a1,b=b1,c=c1)) I checked my start values by plotting the relationship etc. but I kept getting an error message saying maximum iterations exceeded. I have tried changing these
2009 Jun 11
1
Error in 1:p : NA/NaN argument when running model comparisons
Hi there, I am trying to compare nonlinear least squares regression with AIC and anova. The simplest model is one nonlinear curve, and in the more complex model I have a categorical variable (producing parameter estimates for four curves). Both models run fine, but when I try to produce an AIC value for the second model I get the error: > AIC(pow.nls1) [1] 114408.3 > AIC(pow.nls2) Error in
2008 May 09
2
A problem with anova()
  Hi, I fitted tree growth data with Chapman-Richards growth function using nls. summary(fit.nls) Formula: HEIGHT ~ A * (1 - exp(-B * AGE))^C Parameters: Estimate Std. Error t value Pr(>|t|) A 29.007627 0.270485 107.24 <2e-16 *** B 0.030813 0.001095 28.13 <2e-16 *** C 1.849405 0.068659 26.94 <2e-16 *** --- Signif. codes: 0 ''***'' 0.001
2006 Oct 11
2
nls function does not use subset argument (PR#9290)
Full_Name: Tadashi Kadowaki Version: 2.4.0 OS: Redhat Linux 9 Submission from: (NULL) (58.12.166.67) Doesn't nls function support subset? It seems not to work. And, there are no information in the online help. Has it sunk into oblivion?
2005 Sep 08
1
Tip: I() can designate constants in a regression
Just thought I would share a tip I learned: The function I() is useful for specifying constants to formulas and regressions. It will prevent nls (for example) from trying to treat the variable inside I() as something it needs to estimate. An example is below. -David P.S. This may be obvious to some, but it is not made clear to be by the documentation or common books that I reviewed.
2008 Aug 10
2
ANOVA help
Hi, I'm doing anova on a matrix of multivariate data where I want to assess the effect of each column (element). My matrix is 86 rows x 31 columns. I've created a grouping factor of length 86 containing group assignments of 6 types. Then I run: x<- aov(matrix~grouping.factor) summary(aov.fit.raw, test="Wilks") This is working fine enough, but I'm getting different
2007 Sep 23
4
nls fits by groups
Dear Colleagues, I am trying to estimate several non-linear models simultaneously. I don't want to use non-linear mixed model, but non-linear model with same form, but it should be estimated separately according to variable group (I have lots of groups that have lots of observations....). I would like to have unique parameters for each group. e.g. something like this mod <- nls(y ~
2009 Oct 09
2
weigths in nls (PR#13991)
Potential bug: I mistyped weights in the call ('weigths') and it did not produce any error= message. The coefs were exactly the same like without weights, so I was su= spicious and when weights(nls1) gave NULL, I saw my typo. Usually the function will say "Unused arguments", which shows you the error= , but not nls. Regards Stephen [[alternative HTML version deleted]]
2010 Apr 30
2
Curve Fitting
I am having troubles in fitting functions of the form y~a*x^b+c to data, for example x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23) y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86) I tried for example with nls, which did only work with really good initial guessed values. Any suggestion, what I should use? Thanks a lot Thomas [[alternative HTML version deleted]]
2010 Sep 02
1
How using the weights argument in nls2?
Good morning gentlemen! How using a weighted model in nls2? Values with the nls are logical since values with nls2 are not. I believe that this discrepancy is due to I did not include the weights argument in nls2. Here's an example: MOISTURE <- c(28.41640, 28.47340, 29.05821, 28.52201, 30.92055, 31.07901, 31.35840, 31.69617, 32.07168, 31.87296, 31.35525, 32.66118, 33.23385,
2010 Mar 30
6
Error "singular gradient matrix at initial parameter estimates" in nls
I am using nls to fit a non linear function to some data. The non linear function is: y= 1- exp(-(k0+k1*p1+ .... + kn*pn)) I have chosen algorithm "port", with lower boundary is 0 for all of the ki parameters, and I have tried many start values for the parameters ki (including generating them at random). If I fit the non linear function to the same data using an external
2013 Jan 02
1
Need help with self-defined function to perform nonlinear regression and get prediction interval
Dear All, I was trying to call a self-defined function that performs nonlinear regression and gets the corresponding prediction upper limit using nls2 package. However, weird thing happened. When I called the function in the main program, an error message "fitted(nlsmodel): object 'nlsmodel' not found" came up. But when I directly ran the codes inside the function, no error came
2008 May 23
3
nls diagnostics?
Hi, All: What tools exist for diagnosing singular gradient problems with 'nls'? Consider the following toy example: DF1 <- data.frame(y=1:9, one=rep(1,9)) nlsToyProblem <- nls(y~(a+2*b)*one, DF1, start=list(a=1, b=1), control=nls.control(warnOnly=TRUE)) Error in nlsModel(formula, mf, start, wts) : singular gradient matrix at initial
2010 May 11
1
nls() and nls2() behavior?
first, apologies for so many posts yesterday and today. I am wrestling with nls() and nls2(). I have tried to whittle it down to a simple example that still has my problem, yet can be cut-and-pasted into R. here it is: library(nls2) options(digits=12); y= c(0.4334,0.3200,0.5848,0.6214,0.3890,0.5233,0.4753,0.2104,0.3240,0.2827,0.3847,0.5571,0.5432,0.1326,0.3481) x=
2012 Nov 03
6
Parámetros iniciales para ajustes no lineales
Hola a todos estoy aplicando la función polinómica de Hossfeld [1], y algunos otros modelos no lineales para tratar de ajustarlos a un grupo de datos forestales,   [1] Y= b*t*exp(c)/(t*exp(c)+a) Al colocar la función en R con parámetros estimados, me devuelve los siguiente: ## model1 <- nls(ho ~ (b*edad*exp(c)/(edad*exp(c)+a)), data=nigra,     start=list(a=0.005,b=0.08,c=-0.00006),
2011 Dec 11
1
nls start values
I'm using nls to fit periodic gene-expression data to sine waves. I need to set the upper and lower boundaries, because I do not want any negative phase and amplitude solutions. This means that I have to use the "port" algorithm. The problem is, that depending on what start value I choose for phase, the fit works for some cases, but not for others. In the example below, the fit works
2010 Jan 13
1
Problem fitting a non-linear regression model with nls
Hi, I'm trying to make a regression of the form : formula <- y ~ Asym_inf + Asym_sup * ( (1 / (1 + (n1 * (exp( (tmid1-x) / scal1) )^(1/n1) ) ) ) - (1 / (1 + (n2 * (exp( (tmid2-x) / scal2) )^(1/n2) ) ) ) ) which is a sum of the generalized logistic model proposed by richards. with data such as these: x <- c(88,113,128,143,157,172,184,198,210,226,240,249,263,284,302,340) y <-
2009 Aug 25
3
Covariates in NLS (Multiple nonlinear regression)
Dear R-users, I am trying to create a model using the NLS function, such that: Y = f(X) + q + e Where f is a nonlinear (Weibull: a*(1-exp(-b*X^c)) function of X and q is a covariate (continous variable) and e is an error term. I know that you can create multiple nonlinear regressions where x is polynomial for example, but is it possible to do this kind of thing when x is a function with unknown
2005 Mar 08
4
Non-linear minimization
hello, I have got some trouble with R functions nlm(), nls() or optim() : I would like to fit 3 parameters which must stay in a precise interval. For exemple with nlm() : fn<-function(p) sum((dN-estdata(p[1],p[2],p[3]))^2) out<-nlm(fn, p=c(4, 17, 5), hessian=TRUE,print.level=2) with estdata() a function which returns value to fit with dN (observed data vactor) My problem is that only
2003 Nov 28
1
problem with nls()
I wanted to use the nls() module to solve a Problem from Sen & Srivastava (1990, p.209). Here is the (basic) code used to perform the estimation: library(SenSrivastava) library(nls) data(E9.8) # Use Linear Least Square for estimating start values lm.obj <- lm(R.1 ~ I.1 + S.1, data = E9.8) nls1.obj <- nls(R.1 ~ b.0 + b.1*(I.1^a.1-1)/a.1 + b.2*(S.1^a.2-1)/a.2,