similar to: Basic Linear Algebra

I was looking for an R-package that would reduce matrices to row-echelon form, but Google was not my friend; any leads? If not, I wonder if the problem could be expressed in terms of constraint satisfaction...

how to produce a Row Reduced Echelon Form for a matrix in R? Aimin Yan

Hi, all. I want to write some functions like glm() so i studied it. In glm.fit(), it calls a fortran subroutine named "dqrfit" to compute least squares solutions to the system x * b = y To learn how "dqrfit" works, I just follow how glm() calls "dqrfit" by my own example, my codes are given below: > qr <- >

Yes and no. Given your response it appears that "Proper Case" is not a term that everyone uses. In Excel there is a function "Proper" which in essence changes "this line into something like this" into "This Line Into Something Like This." My look at casefold seesm to be that is is a wrapper of two functions to change text into either Lower or Upper case.So

Hi All, I am attempting to work with some data from loggers. I have read in a .csv exported from MS Access that already has my dates and times (in 24 clock format), (with StringsAsFactors=FALSE). > head(tdata) LogData date time 1 77.16 2008/04/24 02:00 2 61.78 2008/04/24 04:00 3 75.44 2008/04/24 06:00 4 89.43 2008/04/24

Dear Group, I have the following data matrix which is a timeseries. > dput(tData) structure(list(A = c(0.2, 0.13, 0.05, 0.1, 0.02, 0.18, 0.09, 0.06, 0.13), B = c(0.15, 0.06, 0.09, 0.02, 0.03, 0.12, 0.01, 0.15, 0.06), C = c(-0.1, 0, -0.07, -0.06, -0.05, -0.05, -0.06, -0.08, -0.07), D = c(-0.15, -0.05, -0.1, -0.03, -0.13, -0.04, -0.1, -0.04, -0.15), E = c(-0.17, -0.16, -0.08, -0.07, -0.09,

I would like to incorporate multiple summary plots from the rms package into a single device and to control the titles, and also to open a new device when I reach a specified number of plots. Currently I am only getting a single "plot(summary(" graph in the upper left- hand corner of each successive device. However, in the rms documention I see instances of a loop being used with

Hello R users, I have been trying to output all my results (text, plots, etc) into the same postscript file as one document, but have been unable to...Can anyone help me improve my code below so that I can accomplish this? Currently I have to output them separately then piece them back together into one document.. Thanks in Advance for any help! options (scipen=999, digits=7)

Hi all, I have been previously been using scatterplot3d package to create some graphs but unfortunately it does not allow me to rotate the plot on all three axis. The cloud() function in the lattice package does allow me to do so. When I was using scatterplot3d I was using a script (Shown Below) to calculate the mean, quartiles and range limits for all three axis and I was representing that on the

Dear, When i multiply the y of a regression by 10, I would expect that the coefficient would be multiply by 10 and the z value to stay constant. Here some reproducible code to support the case. *Ex 1* library(mvtnorm) library(VGAM) set.seed(1) x=rmvnorm(1000,sigma=matrix(c(1,0.75,0.75,1),2,2))

I am attempting to add a calculated column to a data frame. Basically, adding a column called "newcol2" which are the stock closing prices from 1 day to the next. The one little hang up is the name of the column. There seems to be an additional data column name included in the attributes (dimnames?). So when i run HEAD(DATAFRAMENAME) i get the column name = "Open". but

Dear all, ?The code below is used to generate interval censored data but unfortunately there is an error with the ifelse which i am not able to rectify. ?Can somebody help correct it for me. Thank you t<-rexp(20,0.2)? v<-c(0,m,999)? y<-function(t,v){ ? z<-numeric(length(t (( ? ? s<-numeric(length(t (( ? ? ? for(i in 1:length(t)){ ? ? ? ? for(j in 1:length(v-1))? ? ? ? ? { ifelse

Dear all, After a thorough research, I still find myself unable to find a function that does linear regression of 2 vectors of data using the "total least squares", also called "orthogonal regression" (see : http://en.wikipedia.org/wiki/Total_least_squares) instead of the "ordinary least squares" method. Indeed, the "lm" function has a

man for `qr` says that the function uses LINPACK's DQRDC, while it in fact uses DQRDC2. ``` The QR decomposition of the matrix as computed by LINPACK or LAPACK. The components in the returned value correspond directly to the values returned by DQRDC/DGEQP3/ZGEQP3 ```

I assume that I have missed something fundamental and that it is there in front of me in "An Introduction to R", but I need someone to point me in the right direction. > x1 <- "1159 1129 1124 -5 -0.44 -1.52" > x2 <- c("1159","1129","1124","-5","-0.44","-1.52") > x3 <- unlist(strsplit(x1,"

A user sent me an example where coxph fails, and the root of the failure is a case where names(mf) is not equal to the term.labels attribute of the formula -- the latter has an extraneous newline. Here is an example that does not use the survival library. # first create a data set with many long names n <- 30? # number of rows for the dummy data set vname <- vector("character",

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)) --->

I am probably just looking in the wrong place. I am sure there are a number of ways to do this. If anyone could point me in the right direction it would be very much appreciated. Thanks _________________________________________________ Tom Mulholland Senior Policy Officer WA Country Health Service 189 Royal St, East Perth, WA, 6004 Tel: (08) 9222 4062 e-mail:

Hola a todos Estoy intentando hacer un análisis rápido de una serie temporal de datos diarios pero me encuentro con algunos problemas. Me gustaría en primera instancia hacer una regresión lineal pero no encuentro la forma. Tras leer los datos diarios creo un objeto de la clase zoo y sobre éste no puedo utilizar lm(). He leído algo sobre dynlm pero no encuentro la forma. Se agradece

G'day all, I had a look at the GLM code of R (1.4.1) and I believe that there are problems with the function "glm.fit" that may bite in rare circumstances. Note, I have no data set with which I ran into trouble. This report is solely based on having a look at the code. Below I append a listing of the glm.fit function as produced by my system. I have added line numbers so that I