similar to: How do I obtain the design matrix of an lm()?

How does one convert objects c("a","b","c") and "d" into "abcd"? > paste(c("a","b","c"), "d") of course yields [1] "a d" "b d" "c d" -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org

I have two data frames. Suppose the first has rows r1 r2 r3 and the second has rows R1 R2 R3 I'd like to generate the data frame: r1 R1 r1 R2 r1 R3 r2 R1 r2 R2 r2 R3 r3 R1 r3 R2 r3 R3 How would I go about doing this? I'm sure there's a clean way to do it but I find myself thinking in loops. -- Ajay Shah

Suppose one has x <- c(1, 2, 7, 9, 14) y <- c(71, 72, 77) How would one write an R function which alternates between elements of one vector and the next? In other words, one wants z <- c(x[1], y[1], x[2], y[2], x[3], y[3], x[4], y[4], x[5], y[5]) I couldn't think of a clever and general way to write this. I am aware of gdata::interleave() but it deals

1. I have used gdata::read.xls() with much happiness. But every now and then it breaks. I have not, as yet, been able to construct a mental model about the class of .xls files for which it works. Does someone have a simple rule for predicting the circumstances under which it will work? 2. Just like there is a read.xls(), it'd be great if we have a read.ods() which directly

Folks, I'm doing fine with using orthogonal polynomials in a regression context: # We will deal with noisy data from the d.g.p. y = sin(x) + e x <- seq(0, 3.141592654, length.out=20) y <- sin(x) + 0.1*rnorm(10) d <- lm(y ~ poly(x, 4)) plot(x, y, type="l"); lines(x, d$fitted.values, col="blue") # Fits great! all.equal(as.numeric(d$coefficients[1] + m

I wondered was people on this list felt about this article: http://www.voxeu.org/index.php?q=node/2363 which talks about the problems of obtaining sound answers in numerical optimisation in settings such as MLE or NLS. -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org http://ajayshahblog.blogspot.com <*(:-? -

How would I do something like this: f <- function(x, g) { s <- as.character(g) # THIS DOES NOT WORK sprintf("The %s of x is %.0f\n", s, g(x)) } f(c(2,3,4), "median") f(c(2,3,4), "mean") and get the results "The median of x is 3" "The mean of x is 3" -- Ajay Shah

I have a situation with a large dataset (3000+ observations), where I'm doing lags as regressors, where I get: Call: lm(formula = rj ~ rM + rM.1 + rM.2 + rM.3 + rM.4) Residuals: 1990-06-04 1994-11-14 1998-08-21 2002-03-13 2005-09-15 -5.64672 -0.59596 -0.04143 0.55412 8.18229 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.003297 0.017603

Folks, A nice new data resource has come up -- http://data.un.org/ I thought it would be wonderful to setup an R function like tseries::get.hist.quote() which would be able to pull in some or all of this data. I walked around a bit of it and I'm not able to map the resources to predictable URLs which can then be wget. There's some javascript going on that I'm not understanding.

Folks, I am holding a dataset where firms are observed for a fixed (and small) set of years. The data is in "long" format - one record for one firm for one point in time. A state variable is observed (a factor). I wish to make a markov transition matrix about the time-series evolution of that state variable. The code below does this. But it's hardcoded to the specific years that I

I went to the article on np in R news 7/2 (October 2007). What's the general technique to get the source code associated with the article as a .R file that I can play with? -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org http://ajayshahblog.blogspot.com <*(:-? - wizard who doesn't know the answer.

The pretty picture that I saw at: http://chartsgraphs.wordpress.com/2009/02/09/r-panel-chart-beats-excel-chart/#more-1096 inspired me to try something similar. The code that I wrote is: ------snipsnip--------------------------------------------------------------------- M <- structure(list(date = structure(c(13634, 13665, 13695, 13726, 13757, 13787, 13818, 13848, 13879, 13910, 13939, 13970,

Folks, I often build up R objects starting from NULL and then repeatedly using rbind() or cbind(). This yields code like: a <- NULL for () { onerow <- craft one more row a <- rbind(a, onerow) } This works because rbind() and cbind() are forgiving when presented with a NULL arg: they act like nothing happened, and you get all.equal(x,rbind(NULL,x)) or all.equal(x,cbind(NULL,x)).

I have a program which is doing a few thousand runs of lm(). Suppose it is a simple model y = a + bx1 + cx2 + e I have the R object "d" where d <- summary(lm(y ~ x1 + x2)) I would like to obtain Var(x2) out of "d". How might I do it? I can, of course, always do sd(x2). But it would be much more convenient if I could snoop around the contents of summary.lm and

Consider this code fragment: --------------------------------------------------------------------------- set.seed(42) x <- runif(20) y <- 2 + 3*x + rnorm(20) m1 <- lm(y ~ x) m2 <- lm(y ~ -1 + x) summary(m1) summary(m2) cor(y, fitted.values(m1))^2 cor(y, fitted.values(m2))^2 --------------------------------------------------------------------------- m1 is the

Here's a small R program: --------------------------------------------------------------------------- a <- rep(1,10000000) system.time(a <- a + 1) system.time(for (i in 1:10000000) {a[i] <- a[i] + 1}) --------------------------------------------------------------------------- and here's its matlab version:

Folks, Based on http://www.biostat.wustl.edu/archives/html/s-news/1999-06/msg00125.html I thought I should experiment with using survreg() to estimate tobit models. I start by simulating a data frame with 100 observations from a tobit model > x1 <- runif(100) > x2 <- runif(100)*3 > ystar <- 2 + 3*x1 - 4*x2 + rnorm(100)*2 > y <- ystar > censored <- ystar <= 0

I know that this is correct: library(chron) x = dates("01-03-04", format="d-m-y", out.format="day mon year") print(x) It gives me the string "01 Mar 2004" which is correct. I also know that I can say: print(day.of.week(3,1,2004)) in which case he says 1, for today is monday. My question is: How do I combine these two!? :-) I have a

How do I parse a date "yyyymmdd"? I tried asking chron(s, "ymd") but that didn't work. Would the date parsing routines of the Date class of 1.9 grok this? -- Ajay Shah Consultant ajayshah at mayin.org Department of Economic Affairs http://www.mayin.org/ajayshah Ministry of Finance, New Delhi

I think I have isolated a problem with integration between Sweave and beamer. Could you please see the file: http://www.mayin.org/ajayshah/tmp/bugdemo.Rnw Unfortunately, it uses some of my internal libraries, so you can't run it. When I put it through Sweave, I get: http://www.mayin.org/ajayshah/tmp/bugdemo.tex which is, of course, a generic latex file which you can read and