similar to: How to use "subset" in lm function

I have a dynamic time series model like this: dyn$lm( y ~ lag(y,-1) + x + lag(x,-1)+lag(x,-2) ) I need to do an out of sample forecast with this model. Is there any way I can do this with R? It would be greatly appreciated if some one can give me an example. Thanks. Harry [[alternative HTML version deleted]]

Hi, R users, I'm wondering how I can aggregate data in R with different functions for different columns. For example: x<-rep(1:5,3) y<-cbind(x,a=1:15,b=21:35) y<-data.frame(y) I want to aggregate "a" and "b" in y by "x". With "a", I want to use function "mean"; with "b", I want to use function "sum". I tried:

Hi, R users, I'm using while() in R to set a condition. My condition is: i<523 or i>535. How should I write this in R? I try "while (i<523 or i>535)" and it does not work. Thanks. Garry On Tue, Nov 10, 2009 at 11:26 AM, Hongwei Dong <pdxdong@gmail.com> wrote: > Exactly! Thanks, Duncan. > > Let me re-phrase me question like this: > > 1) X_i values

Hi, R users, Can anyone tell me how I can change the size of points in my plot? For example: x <- c(1,3,6,9,12) y <- c(1.5,2,7,8,15) plot(x,y,pch=20) How do I reduce the size of those points? Thanks. Gary [[alternative HTML version deleted]]

Hi, R users, I'm wondering if I can identify an element in a column by an element in another column. For example: x<-1:10 y<-11:20 z<-cbind(x,y) z x y [1,] 1 11 [2,] 2 12 [3,] 3 13 [4,] 4 14 [5,] 5 15 [6,] 6 16 [7,] 7 17 [8,] 8 18 [9,] 9 19 [10,] 10 20 What I want to do is: when x=5, y=y-1 Anyone can tell me how to do this? Thanks. Gary [[alternative

Hi, all, I'm doing a discrete choice model in R and kept getting this error: Error: cannot allocate vector of size 198.6 Mb. Does this mean the memory limit in R has been reached? > memory.size() [1] 1326.89 > memory.size(TRUE) [1] 1336 > memory.limit() [1] 1535 My laptop has a 4G memory with Windows Vista (32 bit). I increased the memory limit to 2500 M. But still getting the same

Hi, R users, I'm using the function "vglm" to estimate a multinomial logit model. Every time I use "summary()" to ask for the coefficients and std error, I got this: Length Class Mode 1 vglm S4 Anyone know what's wrong here? Thanks. Harry [[alternative HTML version deleted]]

All: How do I obtain one step ahead out-of-sample forecasts from a model using "dyn" or "dynlm" ? Thanks! Best, John [[alternative HTML version deleted]]

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I want to set up a model with a formula and then run dynlm(formula) because I ultimately want to loop over a set of formulas (see end of post) R> form <- gas~price R> dynlm(form) Time series regression with "ts" data: Start = 1959(1), End = 1990(4) <snip> Works OK without a Lag term R> dynlm(gas ~ L(gas,1)) Time series regression with "ts" data: Start =

Hi, please use the following the matrix z as the example: x<-c(2,4,5,7,6,9,8,2,0) y<-matrix(x,3,3) z<-apply(y,2,function(x)x/sum(x)) z On Tue, Mar 23, 2010 at 6:59 PM, David Winsemius <dwinsemius@comcast.net>wrote: > > On Mar 23, 2010, at 9:05 PM, Hongwei Dong wrote: > > Hi, R-helpers, >> >> I'm trying to use R to do a Monte Carlo simulation and

Hi, R users, I'm trying to transform a matrix A into B (see below). Anyone knows how to do it in R? Thanks. Matrix A (zone to zone travel time) zone z1 z2 z3 z1 0 2.9 4.3 z2 2.9 0 2.5 z3 4.3 2.5 0 B: from to time z1 z1 0 z1 z2 2.9 z1 z3 4.3 z2 z1 2.9 z2 z2 0 z2 z3 2.5 z3 z1 4.3 z3 z2 2.5 z3 z3 0 The real matrix I have is much larger, with more than 2000 zones. But I think it should

Exactly! Thanks, Duncan. Let me re-phrase me question like this: 1) X_i values are independent Gammas, with the shape 0.067 and scale 0.008 2) Min(X)=1 and Max(X)=85 3) SUM(X)=2000 4) Do I also have to define the number of draws? if yes, it could be 250. Based on these restrictions, I want to generate random draw. I'm wondering how I can do this in R. Thanks. Garry On Tue, Nov 10, 2009

Hi, I was wondering why the results from lm and dynlm are not the same for what I think is the same model. I have just modified example 4.2 from the Pfaff book, please see below for the code and results. Can anyone tell my what I am doing wrongly? Many thanks, Werner set.seed(123456) e1 <- rnorm(100) e2 <- rnorm(100) y1 <- ts(cumsum(e1)) y2 <- ts(0.6*y1 + e2) lr.reg <- lm(y2

Hi, I'm trying to plot two variables in one graph. One ranges between 0 and 1, while the other ranges between 50 and 500. Can I plot them in one graph with similar scale? Thanks Harry [[alternative HTML version deleted]]

Hi, I used R to run a linear regression and keep getting the following error message. I do not understand it very well. Anyone can help out? Thanks. Error in storage.mode(y) <- "double" : invalid to change the storage mode of a factor In addition: Warning message: In model.response(mf, "numeric") : using type="numeric" with a factor response will be ignored

Hi, R users, I'm using the "lme" function in R to estimate a 2 level mixed effects model, in which the size of the subject groups are different. It turned out that It takes forever for R to converge. I also tried the same thing in SPSS and SPSS can give the results out within 20 minutes. Anyone can give me some advice on the lme function in R, especially why R does not converge?

Hi: I have two time series y(t) and x(t). I want to regress Y on X. Because Y is a time series and may have autocorrelation such as AR(p), so it is not efficient to use OLS directly. The model I am trying to fit is like Y(t)=beta0+beta1*X(t)+rho*Y(t-1)+e(t) e(t) is iid normal random error. Anybody know whether there is a function in R can fit such models? The function can also let me specify

Hi, R-helpers, I'm trying to use R to do a Monte Carlo simulation and need the help. What I have is a matrix that consists of the probabilities for the persons to choose zones. For example, in the matrix shown below, each column represents a person, and each row represents a zone. So, the probability that the first person will choose the 2nd zone is 30%. 25% 30% 10% 30% 20% 0% 20% 50% 60%

Hi, I would like to regress y (dependent variable) on x (independent variable) and y(-1). I have create the y(-1) variable in this way: ly<-lag(y, -1) Now if I do the following regression lm (y ~ x + ly) the results I obtain are not correct. Can someone tell me the code to use in R in order to perform a regression using as explanatory variable a lagged dependent variable? My best regards,