similar to: R commandline editor question

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

I'm in this situation: factorlabels <- c("School", "College", "Beyond") with data for 8 families: education.man <- c(1,2,1,2,1,2,1,2) # Note : no "3" values education.wife <- c(1,2,3,1,2,3,1,2) # 1,2,3 are all present. My goal is to create this table: School College Beyond

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

When using get.hist.quote(), I find the dates are broken. This is with R 2.1.1 on Mac OS X `panther'. > library(tseries) Loading required package: quadprog 'tseries' version: 0.9-27 'tseries' is a package for time series analysis and computational finance. See 'library(help="tseries")' for details. > x <-

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, 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, I have placed an example of a self-contained R program later in this mail. It generates a file inflation.pdf. When I stare at the picture, I see the "X label string" and "Y label string" sitting lonely and far away from the axes. How can these distances be adjusted? I read ?par and didn't find this directly. I want to hang on to 2.8 x 2.8 inches as the overall size

I learned R & MLE in the last few days. It is great! I wrote up my explorations as http://www.mayin.org/ajayshah/KB/R/mle/mle.html I will be most happy if R gurus will look at this and comment on how it can be improved. I have a few specific questions: * Should one use optim() or should one use stats4::mle()? I felt that mle() wasn't adding much value compared with optim, and

Folks, I'm in a situation where I do a few thousand regressions, and some of them are bad data. How do I get back an error value (return code such as NULL) from lm(), instead of an error _message_? Here's an example: > x <- c(NA, 3, 4) > y <- c(2, NA, NA) > d <- lm(y ~ x) Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : 0 (non-NA) cases

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

I have a data frame with one column "x": > str(data) `data.frame': 20 obs. of 1 variable: $ x: num 0.0495 0.0986 0.9662 0.7501 0.8621 ... Normally, I know that the notation dataframe[indexes,] gives you a new data frame which is the specified set of rows. But I find: > str(data[1:10,]) num [1:10] 0.0495 0.0986 0.9662 0.7501 0.8621 ... Here, it looks like the operation

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 am discovering sapply! :-) Could you please help me with a very elementary question? Here is what I know. The following two programs generate the same answer. --------------------------------+---------------------------------------- Loops version | sapply version --------------------------------+----------------------------------------

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

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 wrote a simple log likelihood (for the ordinary least squares (OLS) model), in two ways. The first works out the likelihood. The second merely calls the first, but after transforming the variance parameter, so as to allow an unconstrained maximisation. So the second suffers a slight cost for one exp() and then it pays the cost of calling the first. I did performance measurement. One would

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

The PDF file on the web, which is an appendix on nonlinear regression associated with the CAR book, is very nice. When I ran through the code presented there, I found something odd. The code does a certain model in 3 ways: Vanilla NLS (using numerical differentation), Analytical derivatives (where the user supplies the derivatives) and analytical derivatives (using automatic differentiation). The

Consider this situation: > x1 <- runif(100); x2 <- runif(100); y <- 2 + 3*x1 - 4*x2 + rnorm(100) > m1 <- summary(lm(y ~ x1)) > m2 <- summary(lm(y ~ x2)) > m3 <- summary(lm(y ~ x1 + x2)) Now you have estimated 3 different "competing" models, and suppose you want to present the set of models in one table. xtable(m1) is cool, but doing that thrice would give