similar to: Puzzled at rpart prediction

I am doing > library(rpart) > m <- rpart("y ~ x", D[insample,]) > D[outsample,] y x 8 0.78391922 0.579025591 9 0.06629211 NA 10 NA 0.001593063 > p <- predict(m, newdata=D[9,]) Error in model.frame(formula, rownames, variables, varnames, extras, extranames, : invalid result from na.action How do I persuade him to give me NA

I have a situation where this is fine: > if (length(x)>15) { clever <- rr.ATM(x, maxtrim=7) } else { clever <- rr.ATM(x) } > clever $ATM [1] 1848.929 $sigma [1] 1.613415 $trim [1] 0 $lo [1] 1845.714 $hi [1] 1852.143 But this variant, using ifelse(), breaks: > clever <- ifelse(length(x)>15, rr.ATM(x, maxtrim=7), rr.ATM(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

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

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 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

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

I am using R 2.1 on Apple OS X. When I get the ">" prompt, I find it works well with emacs commandline editing. Keys like M-f C-k etc. work fine. The one thing that I really yearn for, which is missing, is bracket matching When I am doing something which ends in )))) it is really useful to have emacs or vi-style bracket matching, so as to be able to visually keep track of whether I

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 <-

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 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 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

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

Folks, The betareg package appears to be polished and works well. But I would like to look at the exact formulas for the underlying model being estimated, the likelihood function, etc. E.g. if one has to compute \frac{\partial E(y)}{\partial x_i}, this requires careful calculations through these formulas. I read "Regression analysis of variates observed on (0,1): percentages, proportions and

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

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

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

Folks, I am using R 2.1.1 on Apple OS X 10.3. Earlier, I used to say $ sudo R > update.packages() and all the packages used to get installed. For several weeks, I noticed that nothing has been coming through. I used the R-for-Mac graphics console and I find that there are many packages where new versions have come out which I don't have. Is something wrong with update.packages()? I

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