search for: grads

Dear group, Here is my data.frame : df <- structure(list(DESCRIPTION = c("PRM HGH GD ALU", "PRM HGH GD ALU", "PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL", "PRIMARY NICKEL", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ", "STANDARD LEAD ",

...roblem?) The dataframe that I use to build the model is the same as the data I'm using to predict with. Here is a portion of what is happening.. This is the value it is predicting = > [1] 9.482975 Summary of the model Call: lm(formula = reservesub$paid ~ reservesub[, 3 + i] + reservesub$grads[, i] + reservesub$Sun + reservesub$Fri + reservesub$Sat) Residuals: Min 1Q Median 3Q Max -15.447 -4.993 -1.090 3.910 27.454 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 5.71370 1.46449 3.902 0.000149 ***...

Found this solution. It is maybe not the most elegant way, but it does the job. > a=as.data.frame(substr(lme$DESCRIPTION,1,14)) > colnames(a)=c("DESCRIPTION") > lme=as.data.frame(c(a,lme[,2:3])) > lme DESCRIPTION CLOSING.PRICE POSITION 1 PRIMARY NICKEL 25,755.7100 0 2 PRIMARY NICKEL 25,760.8600 0 3 PRM HGH GD ALU 2,415.9000 0

libswfdec/swfdec_bits.c | 52 ++++++++++++++---------------------------------- 1 file changed, 16 insertions(+), 36 deletions(-) New commits: diff-tree 66b4014d57faa03a37e3d11788110bad17c860e6 (from ca10e69db70a8396253e0fa2114bd6e3787d61d2) Author: Benjamin Otte <otte at gnome.org> Date: Mon Sep 17 11:31:24 2007 +0200 more bad duplication diff --git a/libswfdec/swfdec_bits.c

libswfdec/swfdec_bits.c | 24 ++++++++++++++++++------ libswfdec/swfdec_font.c | 43 +++++++++++++++++++++++++++---------------- libswfdec/swfdec_movie.c | 4 ++++ 3 files changed, 49 insertions(+), 22 deletions(-) New commits: diff-tree 26a33bf08b5d9feeb8047274b312cee2999824ff (from ea79f997727fcd34b23b206be84b95c7e2f6d152) Author: Benjamin Otte <otte@gnome.org> Date: Sun Apr 22

Full_Name: Joerg Polzehl Version: 2.3.1 OS: x86_64, linux-gnu Submission from: (NULL) (62.141.176.22) I observed an incorrect behavior of function deriv when evaluating arguments of dnorm deriv(~dnorm(z,0,s),"z") expression({ .value <- dnorm(z, 0, s) .grad <- array(0, c(length(.value), 1), list(NULL, c("z"))) .grad[, "z"] <- -(z * dnorm(z))

I'm trying to obtain numerical derivative of a probability computed with mvtnorm with respect to its parameters using grad() and jacobian() from NumDeriv. To simplify the matter, here is an example: PP1 <- function(p){ thetac <- p thetae <- 0.323340333 thetab <- -0.280970036 thetao <- 0.770768082 ssigma <- diag(4) ssigma[1,2] <- 0.229502120

Hi there, I have a function which has a variable called show as an input: richardson.grad <- function(func, x, d=0.01, eps=1e-4, r=6, show=F){ # do some things if(show) { cat("\n","first order approximations", "\n") print(a.mtr, 12) } #do more things and return } The show variable is being used as a flag to show intermediate

Hello everyone, I have programmed a function and, among other things, it has to compute the differential of a function. I have been using the function grad() and I have programmed some simulated studies. I have testes with 72,576 different combinations of the initial values and now I got an error in only 4 situations. The error comes from the function grad() and I don't understand why. This

Dear R-helpers I have a problem related to the use of NLME I think is simply a matter of getting the nlme coding correct, but i cannot get my brain around it I am analysing some 24 growth curves of some cells , and i wanted to say that there are significant differences between the curves in two parameters that describe the pattern of growth. these parameters are from a logistic (r & k)

Dear everyone, the following is obviously used to compute the nth derivative, which seems to work (deriv(sqrt(1 - x^2),x,n)) However, before using this, I wanted to make sure it does what I think it does but can't figure it out when reading the ?deriv info or any other documentation on deriv for that matter: deriv(expr, namevec, function.arg = NULL, tag = ".expr", hessian = FALSE,

Quick question: Which function do you use to calculate partial derivatives from a model equation? I've looked at deriv(), but think it gives derivatives, not partial derivatives. Of course my equation isn't this simple, but as an example, I'm looking for something that let's you control whether it's a partial or not, such as: somefunction(y~a+bx, with respect to x,

I am attempting to extract the derivative/ gradient from this expression df1p <- deriv(f1, "P") > df1p expression({ .value <- s - c - a * P .grad <- array(0, c(length(.value), 1L), list(NULL, c("P"))) .grad[, "P"] <- -a attr(.value, "gradient") <- .grad .value }) So in this case I want to extract the "-a".

I am trying to code a steepest ascent algorithm to optimize parameters used in a survivor function type problem. My unknown parameters (alpha, Beta0, and Beta1) for which I have been able to optimize using Newton's method. I keep getting an error because my alpha becomes negative and I can't calculate the likelihood. Here is my log likelihood I am optimizing (in LaTex): l=\sum _{ i=1 }^{

Hi, I'm fitting a standard nonlinear model to the luminances measured from the red, green and blue guns of a TV display, using nls. The call is: dd.nls <- nls(Lum ~ Blev + beta[Gun] * GL^gamm, data = dd, start = st) where st was initally estimated using optim() st $Blev [1] -0.06551802 $beta [1] 1.509686e-05 4.555250e-05 7.322720e-06 $gamm [1] 2.511870 This works fine but I

The deriv() function takes an 'expression' as its first argument). I was wondering if the this function can take an array or a vector of expressions as its first argument. Aside, I saw how to give a vector argument to the second argument. like to have something like: deriv(c(~x^2+y^3, ~x^5+y^6), c("x","y")) the documentation for this function talks about being able to

Also, this should have gone in R-bugs quite a while ago : ------- start of forwarded message ------- From: Martin Maechler <maechler@stat.math.ethz.ch> To: R-core@stat.math.ethz.ch Subject: PROTECT() bugs in deriv(*, *, function.arg = ) Date: Mon, 16 Apr 2001 21:02:10 +0200 In R versions 0.50 and 0.64.2 , the following worked > deriv(expression(sin(cos(x) * y)),

Full_Name: bernardo moises lagos alvarez Version: 2.4.0 OS: Windows XP professional Submission from: (NULL) (152.74.219.16) I need obtained the MLE of weibull parameters using the nlm with exact gradient an hessian. I am doing. bug report :Erro en log(b) : el argumento "b" est? ausente, sin default 1.Construction to objectiv functin with n=1 data

Hi , I am trying to get some estimator based on lognormal distribution when we have left,interval, and right censored data. Since, there is now avalible pakage in R can help me in this, I had to write my own code using Newton Raphson method which requires first and second derivative of log likelihood but my problem after runing the code is the estimators were too high. with this email ,I provide

Hi, 2 questions: Question 1: example of what I currently do: for(i in 1:6){sink("temp.txt",append=TRUE) dput(i+0) sink()} x=scan(file="temp.txt") print(prod(x)) file.remove("C:/R-2.5.0/temp.txt") But how to convert the output of the loop to a vector that I can manipulate (by prod or sum etc), without having to write and append to a file? Question 2: >