similar to: Puzzled at generating combinations

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

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 <*(:-? -

In order to get around the problems of my posting a few minutes ago, I thought: $ awk -F\| '(NR > 2) {print $2}' cmie_firm_data.text > col2 $ awk -F\| '(NR > 2) {print $4}' cmie_firm_data.text > col4 $ paste col2 col4 | head -2 -510.45 -510.27 60700 101900 $ paste col2 col4 | tail -2 28648.12 31617.02 491014.77 494308.52 $ wc -l col2 col4 89323 col2

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

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

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.

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,

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.

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)).

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

I am using the clever formula notation of R to first do an OLS. E.g. I say m <- lm(y ~ x + f) where f is a factor, and R automatically constructs the dummy variables. Very nice. I need to then go on to do some other ML estimation using the same design matrix that's used for the OLS. I could, of course, do this manually. But it seems that lm() has done all this hard work. I wonder if

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

Folks, The contour() function wants x and y to be in increasing order. I have a situation where I have a grid in x and y, and associated z values, which looks like this: x y z [1,] 0.00 20 1.000 [2,] 0.00 30 1.000 [3,] 0.00 40 1.000 [4,] 0.00 50 1.000 [5,] 0.00 60 1.000 [6,] 0.00 70 1.000 [7,] 0.00 80 0.000 [8,] 0.00 90

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:

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

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 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