similar to: MA coefficients

Hi, i've been trying to apply the lme apprach to growth curves of children, but lme keeps running for ever and ever as soon as I use a reasonable basis. First Example: Data are 39 boys from the Berkeley growth study, each one measured 31 times at the ages of 1.00 1.25 1.50 1.75 2.00 3.00 4.00 5.00 6.00 7.00 8.00 8.50 9.00 9.50 10.00 10.50 11.00 11.50 12.00 12.50 13.00 13.50

Hello ! I'd like to know to which of the FANN package network corresponds the R nnet network ? In more details, what is the R nnet activation function, what is the training algorithm (rprop, quickprop, ...) ? Also, it seems that the R nnet "decay" parameter in nnet corresponds to the "learning_rate" parameter in FANN. Correct ? Many thanks in advance ! Luc Moulinier

Dear Sir, I am Pankaj Singhal from Jaipur, India. I am very much interested and strongly looking forward in getting involved in this project Learning-to-Rank. My previous experience in this field is good. Last semester I did a similar job of ranking the URLs of the given huge dataset based on their attribute values. The dataset consisted hundreds of thousands of URLs and each url

Dear all, I am trying to implement Simar & Wilson's (2007) second algorithm and have the following question: If I use a truncated regression on the m<n observations, how do I get fitted values for all n observations, instead of for m observations, which is what the command fitted returns; I would need these to construct the left-truncation needed to draw n random deviates.

On Fri, Sep 14, 2007 at 11:42:18PM -0700, Tanya Lattner wrote: > The 2.1 pre-release (version 1) is available for testing: > http://llvm.org/prereleases/2.1/version1/ > > [...] > > 2) Download llvm-2.1, llvm-test-2.1, and the llvm-gcc4.0 source. > Compile everything. Run "make check" and the full llvm-test suite > (make TEST=nightly report). > > Send

Hello, To indicate probability densities instead of counts on a histogram, I specify freq = FALSE. However, I expect that summing all top y coordinates over all the intervals of the histogram will provide 1. 1) v <- c(0.2885, 0.2988, 0.3139, 0.2615, 0.3179, 0.3163, 0.2583, 0.3052, 0.2527, 0.3147, 0.3235, 0.2408, 0.2480, 0.3108, 0.3577, 0.2829, 0.2694, 0.3275, 0.3314, 0.2639, 0.3076,

Dear all, I have finally managed to get a "Analysis of Variance Table": Response: LogHand Df Sum Sq Mean Sq F value Pr(>F) Prey 3 5.3125 1.7708 20.672 2.066e-11 *** Residuals 164 14.0488 0.0857 I want to compare the means of the four different groups, and see if there is a difference. I searched for help,

Hello I want to know if the summary of the logistic model with survey Pr (> | t |) to test if the coefficient of the model is significant, ie is the p_valor wald test for the model coefficients, for I am interested to know if the three levels of the variable educational level are significant to the model (significance of handling 0.2), I present below the results of my model

Hello all, This is in reference to "Learning to Rank" Project Idea. [I know, i made the entry a bit late, but hope you are still in interest to help out] I am looking for suggestions to help me narrowing down the choices of algorithms. I had been readily surveying on the referred algorithms for the purpose of choosing the right one. I am mentioning here some of my doubts to discuss and

Hi everyone, So I am trying to see which ecological parameter of different parks in nyc influence the most the diversity of the medium-sized mammals in those parks. I have a bunch of different parameters for each park I'm done studying and the presence (1) and absence (0) of each mammal. I wanted to do a principal component analysis of those data to know which ones of those parameters are the

Dear R readers: I have written a short lme.R function, which adds normalized coefficients and White heteroskedasticity-adjusted statistics to the standard output. Otherwise, it behaves like lm. This is of course trivial for experts, but for me and other amateur users perhaps helpful. y= rnorm(15); x= rnorm(15); z= rnorm(15); m= lme( y ~ x + z); print(summary(m)); produces something

Hi all,  I was trying to use glht() from multcomp package to construct a contrast on interaction term in a linear model to do some comparisons. I am little uncertain on how to construct contrasts on a 3-way interaction containing a continuous variable, and hope someone can confirm what I did is correct or wrong: The linear model has a continuous dependent variable “y”, with treatment factor

Hi all there I am enjoying R since 2 weeks and I come to my first deadlock, il am trying to use predict.Arima in the ts package. I get a "Error in cbind(...) : cannot create a matrix from these types" -- Start R session ----------------------------------------------------- > fitdiv <- arima(data, c(2, 0, 3), xreg = y ) ; print(fitdiv) Call: arima(x = data, order = c(2, 0, 3),

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Hi all there I am enjoying R since 2 weeks and I come to my first deadlock, il am trying to use predict.Arima in the ts package. I get a "Error in cbind(...) : cannot create a matrix from these types" -- Start R session ----------------------------------------------------- > fitdiv <- arima(data, c(2, 0, 3), xreg = y ) ; print(fitdiv) Call: arima(x = data, order = c(2, 0, 3),

I am trying to use the survfit() function with the newdata argument to produce predicted survivor curves for a particular covariate profile. The main purpose of the plot will be to visualise the effect of snp1, coded 0 and 1. In my Cox model I have stratified by one variable, edu, and so I know I will automatically get a separate curve for each strata. My problem is how to deal with the

Hi, I have a question about using "lme" and "aov" for the following dataset. If I understand correctly, using "aov" with Error term in the formula is equivalent to using "lme" with default settings, i.e. both assume compound symmetry correlation structure. And I have found that equivalency in the past. However, with the follwing dataset, I got different

Hi, I have a question about using "lme" and "aov" for the following dataset. If I understand correctly, using "aov" with an Error term in the formula is equivalent to using "lme" with default settings, i.e. both assume compound symmetry correlation structure. And I have found that equivalency in the past. However, with the follwing dataset, I got different

Hi all, Sent the following question yesterday, but haven't got any suggestions yet. So just trying again, can anyone comment on the problem that I have? Thank you! ------------- Hi, I have a question about using "lme" and "aov" for the following dataset. If I understand correctly, using "aov" with an Error term in the formula is equivalent to using

Hello, I am Mohiuddin Abdul Qader, final year student from dept of CSE in Bangladesh University of Engineering & Technology(BUET). My major was artificial intelligence & i finished my course on Machine Learning and Pattern Recognition this year. I am very keen to contribute in open source community. I have just completed my thesis on 'Location Based Structured Web Search'. For the