similar to: Reading Data from mle into excel?

Displaying 20 results from an estimated 1000 matches similar to: "Reading Data from mle into excel?"

2011 May 23
0
Error in backSpline.npolySpline(sp) : spline must be monotone
I get the following error: Error in backSpline.npolySpline(sp) : spline must be monotone Has anyone had this error before? any ideas on a workaround? > > vols=read.csv(file="C:/Documents and Settings/Hugh/My > Documents/PhD/Swaption vols.csv" + , header=TRUE, sep=",") > X<-ts(vols[,2]) > #X > > > dcOU<-function(x,t,x0,theta,log=FALSE){ +
2023 Aug 20
2
Issues when trying to fit a nonlinear regression model
Dear Bert, Thank you so much for your kind and valuable feedback. I tried finding the starting values using the approach you mentioned, then did the following to fit the nonlinear regression model: nlregmod2 <- nls(y ~ theta1 - theta2*exp(-theta3*x), start = list(theta1 = 0.37, theta2 = exp(-1.8), theta3 =
2023 Aug 20
3
Issues when trying to fit a nonlinear regression model
Dear friends, This is the dataset I am currently working with: >dput(mod14data2_random) structure(list(index = c(14L, 27L, 37L, 33L, 34L, 16L, 7L, 1L, 39L, 36L, 40L, 19L, 28L, 38L, 32L), y = c(0.44, 0.4, 0.4, 0.4, 0.4, 0.43, 0.46, 0.49, 0.41, 0.41, 0.38, 0.42, 0.41, 0.4, 0.4 ), x = c(16, 24, 32, 30, 30, 16, 12, 8, 36, 32, 36, 20, 26, 34, 28)), row.names = c(NA, -15L), class =
2004 May 15
2
questions about optim
Hi, I am trying to do parameter estimation with optim, but I can't get it to work quite right-- I have an equation X = Y where X is a gaussian, Y is a multinomial distribution, and I am trying to estimate the probabilities of Y( the mean and sd of X are known ), Theta1, Theta2, Theta3, and Theta4; I do not know how I can specify the constraint that Theta1 + Theta2 + Theta3 + Theta4 = 1 in
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Oh, sorry; I changed signs in the model, fitting theta0 + theta1*exp(theta2*x) So for theta0 - theta1*exp(-theta2*x) use theta1= -.exp(-1.8) and theta2 = +.055 as starting values. -- Bert On Sun, Aug 20, 2023 at 11:50?AM Paul Bernal <paulbernal07 at gmail.com> wrote: > Dear Bert, > > Thank you so much for your kind and valuable feedback. I tried finding the > starting
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
I got starting values as follows: Noting that the minimum data value is .38, I fit the linear model log(y - .37) ~ x to get intercept = -1.8 and slope = -.055. So I used .37, exp(-1.8) and -.055 as the starting values for theta0, theta1, and theta2 in the nonlinear model. This converged without problems. Cheers, Bert On Sun, Aug 20, 2023 at 10:15?AM Paul Bernal <paulbernal07 at
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Dear Bert, Thank you for your extremely valuable feedback. Now, I just want to understand why the signs for those starting values, given the following: > #Fiting intermediate model to get starting values > intermediatemod <- lm(log(y - .37) ~ x, data=mod14data2_random) > summary(intermediatemod) Call: lm(formula = log(y - 0.37) ~ x, data = mod14data2_random) Residuals: Min
2023 Aug 20
1
Issues when trying to fit a nonlinear regression model
Basic algebra and exponentials/logs. I leave those details to you or another HelpeR. -- Bert On Sun, Aug 20, 2023 at 12:17?PM Paul Bernal <paulbernal07 at gmail.com> wrote: > Dear Bert, > > Thank you for your extremely valuable feedback. Now, I just want to > understand why the signs for those starting values, given the following: > > #Fiting intermediate model to get
2008 Apr 22
2
optimization setup
Hi, here comes my problem, say I have the following functions (example case) #------------------------------------------------------------ function1 <- function (x, theta) {a <- theta[1] ( 1 - exp(-theta[2]) ) * theta[3] ) b <- x * theta[1] / theta[3]^2 return( list( a = a, b = b )) } #----------------------------------------------------------- function2<-function (x, theta) {P
2011 Jul 09
3
Confusing piece of R code
m0<-epxression((4*theta1*theta2-theta3^2)/(2*x*theta3^2)-0.5*theta1*x) params<-all.vars(m0) this reads all the params from m0 so theta1,2 and 3 correct? params<-params[-which(params=="x")] checks which params are multiplied by x? np<-length(params) for(i in 1:6){ esp<-get(sprintf("m%d",i-1))
2018 Feb 13
3
Help with regular expressions
R 3.4.2 OS X Colleagues I would appreciate some help with regular expressions. I have string that looks like: " ITERATION ,THETA1 ,THETA2 ,THETA3 ,THETA4 ,THETA5 ,THETA6 ,THETA7 ,SIGMA(1,1) ,SIGMA(2,1) ,SIGMA(2,2)? In the entries that
2009 Oct 27
1
Poisson dpois value is too small for double precision thus corrupts loglikelihood
Hi - I have a likelihood function that involves sums of two possions: L = a*dpois(Xi,theta1)*dpois(Yi,theta2)+b*(1-c)*a*dpois(Xi,theta1+theta3)*dpois(Yi,theta2) where a,b,c,theta1,theta2,theta3 are parameters to be estimated. (Xi,Yi) are observations. However, Xi and Yi are usually big (> 20000). This causes dpois to returns 0 depending on values of theta1, theta2 and theta3. My first
2012 Mar 16
2
Elegant Code
Hi, Can anyone help to write a more elegant version of my code? I am sure this can be put into a loop but I am having trouble creating the objects b1,b2,b3,...,etc. b1 <- rigamma(50,1,1) theta1 <- rgamma(50,0.5,(1/b1)) sim1 <- rpois(50,theta1) b2 <- rigamma(50,1,1) theta2 <- rgamma(50,0.5,(1/b2)) sim2 <- rpois(50,theta2) b3 <- rigamma(50,1,1) theta3 <-
2003 Feb 22
2
4-parameter logistic model
Dear R users I'm a new user of R and I have a basic question about the 4-parameter logistic model. According to the information from Pinheiro & Bates the model is: y(x)=theta1+(theta2-theta1)/(1+exp((theta3-x)/theta4)) == y(x)=A+(B-A)/(1+exp((xmid-input)/scal)) from the graph in page 518 of the book of the same authors (mixed models in S) theta 1 corresponds to the horizontal asymptote
2023 Nov 07
1
non-linear regression and root finding
G'day Troels, On Mon, 6 Nov 2023 20:43:10 +0100 Troels Ring <tring at gvdnet.dk> wrote: > Thanks a lot! This was amazing. I'm not sure I see how the conditiion > pK1 < pK2 < pK3 is enforced? One way of enforcing such constraints (well, in finite computer arithemtic only "<=" can be enforced) is to rewrite the parameters as: pK1 = exp(theta1) ##
2004 Mar 18
1
profile error on an nls object
Hello all, This is the error message that I get. > hyp.res <- nls(log(y)~log(pdf.hyperb(theta,X)), data=dataModel, + start=list(theta=thetaE0), + trace=TRUE) 45.54325 : 0.1000000 1.3862944 -4.5577142 0.0005503 3.728302 : 0.0583857346 0.4757772859 -4.9156128701 0.0005563154 1.584317 : 0.0194149477 0.3444648833 -4.9365149150 0.0004105426 1.569333 :
2018 Feb 13
0
Help with regular expressions
Hi Dennis, How about: # define the two values to search for x<-2 y<-3 # create your search string and replacement string repstring<-paste(x,y,sep=",") newstring<-paste(x,y,sep=".") # this is the string that you want to change thetastring<-"SIGMA(2,3)" sub(repstring,newstring,thetastring) [1] "SIGMA(2.3)" Use gsub if you want to change
2023 Nov 07
1
non-linear regression and root finding
Thanks a lot, Berwin. Unfortunately, pK1 may well be negative and as I understand the literature it may be poorly defined as such, and also seems to be at a boundary, since when lower is set to say rep(-4,3) pK1 is returned as -4 while pK2 and pK3 are undisturbed. Perhaps the point is that pK1 is not carrying any information at the pH around 5. Fair enough, I guess. Only, I believe I need
2017 Oct 18
4
Error messages using nonlinear regression function (nls)
Hi all, I am trying to use nonlinear regression (nls) to analyze some seed germination data, but am having problems with error codes. The data that I have closely matches the germination dataset included in the drc package. Here is the head of the data temp species start end germinated TotSeeds TotGerminated Prop 1 10 wheat 0 1 0 20 0 0.0 2 10 wheat
2018 Feb 13
1
Help with regular expressions
You can either use positive lookahead/lookbehind - but support for that is a bit flaky. Or write a proper regex, and use backreferences to keep what you need. R > x <- "abc 1,1 ,1 1, x,y 2,3 " R > gsub("(\\d),(\\d)", "\\1.\\2", x, perl = TRUE) [1] "abc 1.1 ,1 1, x,y 2.3 " B. > On Feb 12, 2018, at 9:34 PM, Jim Lemon <drjimlemon at