similar to: constrained OLS on coefficient

Hi everyone, I have been trying to compute numbers of days between two dates as follows: > x <- c("1jan1960", "2jan1960", "31mar1960", "30jul1960") > z <- format(x, "%d%b%Y") > ex <- c("1jan1961", "15jan1960", "21mar1975", "10jul1981") > ez <- format(ex, "%d%b%Y") > ez-z

Hi, I am trying to figure out how to lable the second y-axis after the following codes: plot(x, y, xlab="time", ylab="pay1" ) par(new=TRUE) plot(x,y2, ann=FALSE, xaxt="n", yaxt="n", pch=7 ) axis(4) Then, I want to label the second y axis "pay2". I tried "title(ylab="pay2")", but it put this lable on the first

Hi, I would like to know if anyone has any idea of how to run an OLS with constraints? thank you Soyoko ______________________________________ Ms. Soyoko Umeno Graduate Research Assitant for the Illinois-Missouri Biotechnology Alliance (IMBA) at http://www.imba.missouri.edu/ Ph.D. Student at the Department of Agricultural and Consumer Economics at the University of Illinois at Urbana-Champaign

< In Stata, I can just use "bs" function ...> in STATA the bs command runs a bootstrap If you want to use the bootstrap function (or you could linearize b*c/a :) look under ?boot (you may need to load the package first). Usage: boot(data, statistic, R, sim="ordinary", stype="i", strata=rep(1,n), L=NULL, m=0, weights=NULL,

Hi, I would like to know if anyone knows how to compute a variance of the non-linear combination of the coefficient estimates. Say, I get a model of y~c+ax+bz (1) where x and z are the independent variables, c is the constant estimate, and a and b are the coefficient estimates. Then, I want to know the variance of b*c/a (2). How am I going to get it? In Stata, I can just use

Hi everyone, I am having a problem reading my data file in R. It only reads part of it, and stops doing it in half way through the data. I looked at the observation where reading stops, but it is really not different from the observations read it. I expanded the matrix size, but it did not help. I am using a text format to read the data file. Could someone help me? Thank you Soyoko

Hi, I would like to know if anyone knows how to draw a plot with two Y-axises and one X-axis? When you have two sets of y values that do not have the same scale, but correspond to the same x value, I would like to plot them on one graph. Could you please help me? Thank you Soyoko

how do I find out what avahi services my server is advertising? I installed avahi-tools and tried avahi-browse -a (and -ac) and nothing was listed. I find it hard to believe that my system that has apache up and running is not advertising anything?

I might be sorry for asking this question :-) I have two equations and I tried to estimate them individually with "lm" and "gls", and then in a system (using systemfit) with "OLS", "WLS" and "SUR". Quite surprisingly (for myself at least) the results are identical to the last digit. Could someone (please!) give a hint as to what am I

I've almost got this thing working. I have it set up on a centos machine to authenticate logins and automounts to windows file servers. But it won't allow me to specify a domain as part of the userid. I can set a default domain in smb.conf and logging into that domain works like a champ. And I can list the other domains with "wbinfo --online-status" (not sure what

Dear R committee: I am Renzhi, Ph.D student in computer science in the University of Missouri. I have one question for you. I try to install R in the linux server, but I don't have the root permission, is there any way to install the R locally? Thank you very much for helping me. Renzhi Cao Graduate Research Assistant Department of Computer Science University of

How can I extract the adjusted R^2 value from an ols object (using rms package)? library(rms) x <- rnorm(10) y <- x + rnorm(10) ols1 <- ols(y ~ x) Typing "ols1" displays adjusted R^2 among other things, but how can I assign it to a variable? I tried str(ols1) but couldn't see where to go from there. Thanks, Mark Seeto

Hi to all, In other statistical software, such as Eviews, it is possible to regress a model with the Least Squares method, assuming that the residuals follow an AR(q) process. For example the resulting regression is something like y = 1.2154 + 0.2215 x + 0.251 AR(1) How is it possible to do the same in R? Thank you very much in advance, Constantine Tsardounis http://www.costis.name

Hi, I am trying to fit an AR model, maximum order =4, order selection criterion is aic. I wonder why these two give different results: m1<-ar.ols(x, aic=TRUE, method="ols", order.max=4) m1<-auto.arima(x,d=0, D=0, max.p=4, max.P=0, max.q=0, max.Q=0, ic="aic") Could they both use the function predict to do forecasting? Is there any function that works better?

Hello everybody! I am currently using the quantreg package on R 2.2.1. for my diploma thesis and want to create plots with plot.summary.rqs, which should work like this: Usage: plot.summary.rqs(x, nrow=3, ncol=2, alpha = 0.1, ols = TRUE, ...) Arguments: x: an object produced by rq() fitting nrow: rows in mfrow ncol: columns in mfrow alpha: alpha level of

Hi, I have 142 observations off different variables and I'm trying to do a OLS. And I get this error. Any Ideas ? > f <- ols(lnmigr3~popden78+income+modern+spareha+rain) Error in La.chol2inv(x, size) : size cannot exceed nrow(x) = 1 D. ----------------------------------------- Stay ahead of the information curve. Receive GIS news and jobs on your desktop daily. Subscribe today

Hi folks! Does anybody know how to use the OLS function in fMultivar to do dynamic regression? I've tried specifying lags in OLS using a data series created in fSeries and it doesn't seem to work. I've done dynamic regression using dyn$lm and I was wondering how to accomplish the same thing using the OLS function from fMultivar. Thanks! John [[alternative HTML version

Hi, I was hoping to clarify the exact behavior associated with this incantation: validate(fit.ols, method='cross', B=50) Output: index.orig training test optimism index.corrected n R-square 0.5612 0.5613 0.5171 0.0442 0.5170 50 MSE 1.3090 1.3086 1.3547 -0.0462 1.3552 50 Intercept 0.0000 0.0000 -0.0040 0.0040

dear R wizards: apologies for two queries in one day. I have a long form data set, which identifies about 5,000 regressions, each with about 1,000 observations. unit date y x 1 20060101 <two values> 1 20060102 <two values> ... 5000 20081230 <two values> 5000 20081231 <two values> I need to run such regressions many many times, because they are part of an

Dear R users, Please let me know how to plot the forecast in such a model: First I do it simple with ARIMA model that works ok with the codes provided to me at the lecture: arima<-arima(HCPIlong, order=c(1,1,0)) arima.predict<-predict(arima, n.ahead= 5 ) ts.plot(HCPIlong,arima.predict$pred,lty=1:2, main="Forecast of HCPI") But I need to include the additional variable in my