similar to: help interpreting the output of functions - any sources of information

Hi, Being a R-newbie I am wondering how to calculate a correlation coefficient (preferably with an associated p-value) for data like: > d[,1] [1] 25.5 25.3 25.1 NA 23.3 21.5 23.8 23.2 24.2 22.7 27.6 24.2 ... > d[,2] [1] 0.0 11.1 0.0 NA 0.0 10.1 10.6 9.5 0.0 57.9 0.0 0.0 ... Apparently corr(d) from the boot-library fails with NAs in the data, also cor.test cannot cope with a

Hi All, Another nls related problem (for background, I'm migrating a complicated modelling package from S-plus to R). Below I've reduced this to the minimum necessary to demonstrate my problem (I think); the real situation is more complicated. Two similar selfStart functions, ssA and ssB. The 'initial' function for ssB modifies its arguments a little and then calls getInital

Hello, from Verzani, simpleR (pdf), p. 80, I created the following function to test the coefficient of lm() against an arbitrary value. coeff.test <- function(lm.result, var, coeffname, value) { # null hypothesis: coeff = value # alternative hypothesis: coeff != value es <- resid(lm.result) coeff <- (coefficients(lm.result))[[coeffname]] # degrees of freedom = length(var) -

Hello, I was trying to build some functions which I would like to integrate over an interval using the function 'integrate' from the 'stats' package. As an example, please consider the function h(u)=sin(pi*u) + sqrt(2)*sin(pi*2*u) + sqrt(3)*sin(pi*3*u) + 2*sin(pi*4*u) Two alternative ways to 'build' this function are as in f and g below: coeff<-sqrt(1:4)

Hello, I can`t figure out how can increase the velocity of the fitting data by nls. I have a long data .csv I want to read evry time the first colunm to the other colunm and analisy with thata tools setwd("C:/dati") a<-read.table("Normalizzazione.csv", sep=",", dec=".", header=F) for (i in 1:dim(a[[2]]]) { #preparazione dati da analizzare

Hi all, This is probably going to come off as unnecessary (and show my ignorance) but I am trying to understand the parameter estimates I am getting from R when doing an ANCOVA. Basically, I am accustomed to the estimate for the categorical variable being equivalent to the respective cell means minus the grand mean. I know is the case in JMP - all other estimates from these data match the

I am interested in testing two similar nls models to determine if the lines are statistically different when fitted with two different data sets; one corn, another soybean. I know I can do this in linear models by testing for interactions. See Introductory Statistics with R by Dallgaard p212-218 for an example. I have two different data sets I am comparing to lai. ci.re should have very

Hi, I would like to asign titles into strip of a panel of curves designed thanks to a xyplot function (lattice package) I ve tried the "strip" function, ... here is my code I would like to write in the strip of each panel conditionning varialbe (here, it is data[,1]) xyplot (data[,3] ~ data[,2] | data[,1] , strip.default (which.given=1, which.panel = c(1:15), var.name= c

Hi R users I am trying to run a non linear parameter optimization using the function optim() and I have problems regarding memory allocation. My data are in a dataframe with 9 columns. There are 656100 rows. >head(org_results) comb.id p H1 H2 Range Rep no.steps dist aver.hab.amount 1 1 0.1 0 0 1 100 0 0.2528321

Dear all, I am trying to calculate the fitted values using a ridge model (lm.ridge(), MASS library). Since the predict() does not work for lm.ridge object, I want to get the fitted_value from the coefficients information. The following are the codes I use: fit = lm.ridge(myY~myX,lambda=lamb,scales=F,coef=T) coeff = fit$coef However, it seems that "coeff" (or "fit$coef") is

Good day! I'm creating HDL IP CORE (for using in FPGA) for theora encoder (now only I-frames). I don't undestand one moment. Now i develop such stages: 1. From RBG(byer) to YCbCr converter 2. DCT processing (8x8 pixels blocks) 3. Quantizator of DCT coeff. 4. Zig-Zag of quantized DCT coeff. and now i have uresolved last stage of compression - how i must send 8x8 blocks to huffman

Hello, I have different results from these two softwares for a simple binomial GLM problem. >From Genmod in SAS: LogLikelihood=-4.75, coeff(intercept)=-3.59, coeff(x)=0.95 >From glm in R: LogLikelihood=-0.94, coeff(intercept)=-3.99, coeff(x)=1.36 Is there anyone tell me what I did wrong? Here are the code and results, 1) SAS Genmod: % r: # of failure % k: size of a risk set data

Hi R folks, In my previous post I forgot to mention that I was new to R. I was really grateful for your quick help. I have two further questions: 1) In the graph of a regression line I would like to show one specific residual yi obs - yi pred (let's take the person whose residual is 76). How do I add a bracket to this vertical distance and name it? I'am getting stuck after the

Dear all, running the example by D. Eddebuettel (http://dirk.eddelbuettel.com/blog/2011/04/23/) I get an error message. Specifically, the R code I was taking from the above example is ### BEGIN EXAMPLE ### suppressMessages(require(RcppArmadillo)) suppressMessages(require(Rcpp)) suppressMessages(require(inline)) code <- ' arma::mat coeff = Rcpp::as<arma::mat>(a); arma::mat

Hello R users, I am new to R and am wondering if anyone can help me out with the following issue: I wrote a function to build ts models using different inputs, but when R displays the call for a model, I cannot tell which variables it is using because it shows the arguments instead of the real variables passed to the function. (e.g Call: lm(formula = dyn(dep ~ lag(dep, -1) + indep)) --->

Dear Mac users, Hi, as you might have probably read the thread of "[R] Rsquared in summary(lm)" on May 10, it seems that Mac version of lm() seem to be working incorrectly. I enclose the script to produce the result both for lm() and manual calculation for a simple regression. Could you run the script and report with the version of R, so I don't have to go through every builds

I have a model (run with glm) that has a factor, type. Type has two levels, "general" and "regional". I am trying to get estimates (and SEs) for the model with type="general" and type ="regional" using fit.contrast but I can't get the syntax of the coefficients to use in fit.contrast correct. I hope someone can show me how to use fit.contrast, or some

Dear R users, I?m a graduate students and in my master thesis I must obtain the values of the parameters x_i which maximize this Multinomial log?likelihood function log(n!)-sum_{i=1]^4 log(n_i!)+sum_ {i=1}^4 n_i log(x_i) under the following constraints: a) sum_i x_i=1, x_i>=0, b) x_1<=x_2+x_3+x_4 c)x_2<=x_3+x_4 I have been using the ?ConstrOptim? R-function with the instructions

Hi, I have two data frames (u and v). > u   coe      nam 1   0     Time 2   0    Poten 3   0   AdvExp 4   0    Share 5   0   Change 6   0 Accounts 7   0     Work 8   0   Rating > v       coeff    enter 1 0.7272727 Accounts 2 0.3211112     Time 3 0.0500123    Poten I want to update the values of coe in u by using the values ofcoeff in v. That is, I want to get the following result >

Hi, I've a list containing parameters (intercepts & coefficients) of 12 regressions fitted > coeff [[1]] (Intercept) anno -427017.1740 217.0588 [[2]] (Intercept) anno -39625.82146 21.78025 ..... [[12]] (Intercept) anno 257605.0343 -129.7646 I want create a data frame with two columns (intercept and anno)using data in these list. Any help