search for: sumsquared

I need to calculate a large number of t statistics, and would like to do so via matrix operations. So far I have figured out a way to calculate the mean of each row of the matrix: d <- matrix(runif(100000,1,10), 1000, 10) # some test data s <- rep(1,ncol(d)) # a sum vector to use for matrix multiplication means <- (d%*%s)/ncol(d) This is at least 1 order of magnitude faster than

Mark Leeds <mleeds at mlp.com> wrote (to S-News): > does anyone know the R equivalent of the SPlus rowVars function ? Andy Liaw <andy_liaw at merck.com> replied: > More seriously, I seem to recall David Brahms at one time had created an R > package with these dimensional summary statistics, using C code. (And I > pointed him to the `two-pass' algorithm for variance.)

Hi people, If there is a fn "colMeans" why isn't there a "colSDs" etc Thanks, Phil. -- Philip Rhoades Pricom Pty Limited (ACN 003 252 275) GPO Box 3411 Sydney NSW 2001 Australia Mobile: +61:0411-185-652 Fax: +61:2:8923-5363 E-mail: pri at chu.com.au -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

Hi all! I 'm beginner and i develop a bio-application with VB and i need some statistic functions! could i calculate StdError, CoeffOfVariance, SumSquared with R langage? if yes, what are functions to use? I need also to use ANOVA and t-test... Thanks for your help! Laurent Houdusse Analyste Programmeur

I am using the R 2.2.1 in a Windows XP environment. I have a dataframe with 12 columns and 1,000 rows. (Some of the rows have 1 or fewer values.) I am trying to use rowVars to calculate the variance of each row. I am getting the following message: ?Error in na.remove.default(x) : length of 'dimnames' [1] not equal to array extent? Is there a good work-around?

The cautions people have given about starting values are worth heeding. That nlxb() does well in many cases is useful, but not foolproof. And John Fox has shown that the problem can be tackled very simply too. Best, JN On 2023-08-19 18:42, Paul Bernal wrote: > Thank you so much Dr. Nash, I truly appreciate your kind and valuable contribution. > > Cheers, > Paul > > El El

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 =

Is it possible to compute the variance of every column in a matrix without a loop? -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info", "help", or "[un]subscribe" (in the "body", not the subject !) To: r-help-request at stat.math.ethz.ch

Thank you so much Dr. Nash, I truly appreciate your kind and valuable contribution. Cheers, Paul El El s?b, 19 de ago. de 2023 a la(s) 3:35 p. m., J C Nash < profjcnash at gmail.com> escribi?: > Why bother. nlsr can find a solution from very crude start. > > Mixture <- c(17, 14, 5, 1, 11, 2, 16, 7, 19, 23, 20, 6, 13, 21, 3, 18, 15, > 26, 8, 22) > x1 <- c(69.98, 72.5,

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 =

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

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

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

Dear friends, Hope you are all doing well and having a great weekend. I have data that was collected on specific gravity and spectrophotometer analysis for 26 mixtures of NG (nitroglycerine), TA (triacetin), and 2 NDPA (2 - nitrodiphenylamine). In the dataset, x1 = %NG, x2 = %TA, and x3 = %2 NDPA. The response variable is the specific gravity, and the rest of the variables are the predictors.

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

I've seen a number of problems like this over the years. The fact that the singular values of the Jacobian have a ration larger than the usual convergence tolerances can mean the codes stop well before the best fit. That is the "numerical analyst" view. David and Jeff have given geometric and statistical arguments. All views are useful, but it takes some time to sort them all out and

I am not as expert as John, but I thought it worth pointing out that the variable substitution technique gives up one set of constraints for another (b=0 in this case). I also find that plots help me see what is going on, so here is my reproducible example (note inclusion of library calls for completeness). Note that NONE of the optimizers mentioned so far appear to be finding the true best