similar to: Question about design matrix

I want to know how to get trace of product of matrices **faster** when the matrices are really big. Unfortunately the matrices are not symmetric. If anybody know how to get the trace of it, please help me. An example is as below. n <- 2500 a <- matrix(rnorm(n*n),n,n) b <- matrix(rnorm(n*n),n,n) tr1 <- sum(diag(a %*% b)) tr2 <- sum(diag(a %*% b %*% a %*% b)) Thanks, Yongwan Chun

I want to try to get a result of element wise addition between a vector and a list. It can be done with "for statement." In order to reducing computing time, I have tried to avoid "for state." If anybody give me an idea, I would apprecite it much. for example, with a & b as below lines, a<- list(c(1,3),c(1,2),c(2,3)) b<-c(10,20,30) I would like to have a list (like

Hello, I work on large matrices and found something interesting. For multiplication of matrices, the order has a huge influence on computing time when one of them is a sparse matrix. In the below example, M is a full matrix and A is a sparse matrix in a regular matrix class. A %*% M takes much more time than M %*% A; moreover, t(t(M) %*% t(A)) is much faster than A %*% M with same result. I

Hello, I was trying to build R from source on Windows XP. I installed software which are mentioned from the follow web page http://www.murdoch-sutherland.com/Rtools/ (Last accessed on Nov. 13th, 2006) . Unfortunately, I got error messages whenever I run 'make all recommended' without modifying 'MkRules' file. I have removed software and reinstalled them several times but I still

Hi all, I believe this is a bug in the model.matrix function. I'd like a second opinion before filing a bug report. If I have a nested covariate B with multiple values for just one level of A, I can not get a non-singular design matrix out of model.matrix > df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x",

Hi useRs, Perhaps I am having a senior moment? I have a nested variable situation to model, toy example: > df <- data.frame(A = factor(c("a", "a", "x", "x"), levels = c("x", "a")), + B = factor(c("b", "x", "x", "x"), levels = c("x", "b"))) > >

Dear professor Fox and R helpers, I have a quick question about the Anova function in the car package. When using the default "type II" SS I get results that I don't understand (see below). library(car) Data <- data.frame(y=rnorm(10), x1=factor(c(rep("a",4), rep("b",6))), x2 = factor(c(rep("j", 2), rep("k", 3), rep("j", 2),

> On Mar 5, 2018, at 8:52 AM, Ding, Yuan Chun <ycding at coh.org> wrote: > > Hi Bert, > > I am very sorry to bother you again. > > For the following question, as you suggested, I posted it in both Biostars website and stackexchange website, so far no reply. > > I really hope that you can do me a great favor to share your points about how to explain the

This list provides help on R programming (see the posting guide linked below for details on what is/is not considered on topic), and generally avoids discussion of purely statistical issues, which is what your query appears to be. The simple answer is yes, you can fit the model as described, but you clearly need the off topic discussion as to what it does or does not mean. For that, you might try

Hi Bert, I am very sorry to bother you again. For the following question, as you suggested, I posted it in both Biostars website and stackexchange website, so far no reply. I really hope that you can do me a great favor to share your points about how to explain the coefficients for drug A and drug B if run anova model (response variable = drug A + drug B). is it different from running three

Dear experts: Is it possible to create a new function based on stats:::model.matrix.default so that an alternative factor coding is used when the function is called instead of the default factor coding? Basically, I'd like to reproduce the results in 'mat' below, without having to explicitly specify my desired factor coding (identity matrices) in the 'contrasts.arg'. dd

Dear List, I am realtively inexperienced so i apologise in advance and ask for understanding in the simplicity of my question: I have data on the amount of grass per km in a cell ( of which i have lots) "grass" and for each cell i have x/y coordinates - required due to spatial autocorrelation Cells can be classfied in a hierarchical nature into AREAS and STATES i.e Cell 1, Cell 2,

Hi, I don't understand what the meaning of the following lines returned by model.matrix(). Can somebody help me understand it? What can they be used for? attr(,"assign") [1] 0 1 2 2 attr(,"contrasts") attr(,"contrasts")$A [1] "contr.treatment" attr(,"contrasts")$B [1] "contr.treatment" Regards, Peng > a=2 > b=3 > n=4

Hi Bert and David, Thank you so much for willingness to spend some time on my problem!!! I have some statistical knowledge (going to get a master in applied statisitics), but do not have a chance to purse a phD for statistics, so I am always be careful before starting to do analysis and hope to gather supportive information from real statisticians. Sorry that I did not tell more info about

Dear r-devel list members, I'm posting this message here because it concerns the nlme package, which is maintained by R-core. The problem I'm about to describe is somewhere between a bug and a feature request, and so I thought it a good idea to ask here rather posting a bug report to the R bugzilla. I was made aware (by Ben Bolker) that the car::Anova() method for "lme"

> On Mar 5, 2018, at 2:27 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > David: > > I believe your response on SO is incorrect. This is a standard OFAT (one factor at a time) design, so that assuming additivity (no interactions), the effects of drugA and drugB can be determined via the model you rejected: >> three groups, no drugA/no drugB, yes drugA/no drugB,

> On Mar 5, 2018, at 3:04 PM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > > But of course the whole point of additivity is to decompose the combined effect as the sum of individual effects. Agreed. Furthermore your encoding of the treatment assignments has the advantage that the default treatment contrast for A+B will have a statistical estimate associated with it. That was a

Dear R Team, # Summary of the problem: setting contrasts as > contrasts(g) <- contr.treatment or > contrasts(g) <- matrix(c(1,-1,0),ncol=1) (i.e. without quotes around `contr.treatment' or `contr.sum', etc.) and fitting an lm model without an intercept results in a model matrix that lacks one column. (I do ask for forgiveness if this is not a bug but is due to my

Dear R users, I need to analyze data generated from a partial two-by-two factorial design: two levels for drug A (yes, no), two levels for drug B (yes, no); however, data points are available only for three groups, no drugA/no drugB, yes drugA/no drugB, yes drugA/yes drug B, omitting the fourth group of no drugA/yes drugB. I think we can not investigate interaction between drug A and drug B,

Dear R-devel list members, I'd like to suggest a more flexible procedure for generating contrast names. I apologise for a relatively long message -- I want my proposal to be clear. I've never liked the current approach. For example, the names generated by contr.treatment paste factor to level names with no separation between the two; contr.sum simply numbers contrasts (I recall an