similar to: Memory consumption, integer versus factor

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

#### I checked through every 3 factor * 3 loading case. #### While, 4 factor * 3 loading failed. #### the data is 6 factor * 3 loading require(sem); cor18<-read.moments(); 1 .68 1 .60 .58 1 .01 .10 .07 1 .12 .04 .06 .29 1 .06 .06 .01 .35 .24 1 .09 .13 .10 .05 .03 .07 1 .04 .08 .16 .10 .12 .06 .25 1 .06 .09 .02 .02 .09 .16 .29 .36 1 .23 .26 .19 .05 .04 .04 .08 .09 .09 1 .11 .13 .12 .03 .05 .03

Is there a way to apply paste to?list(form1 = EQ1, form2 = EQ2, form3 = EQ3, form4 = EQ4)?such that I don't have to write form1=EQ1 for all my models?(I might have a list of 20 or more)? I also need the EQs to read the formulas associated with them. For example, below, I was able to automate the name assignment but I could not figure out how to?to set up the list using?paste or other

Hi, Using S=1000 and simdata <- replicate(S, generate(3000)) #If you want both "m1" and "m0" #here the missing values are 0 res1<-sapply(seq_len(ncol(simdata.psm1)),function(i) {x1<-merge(simdata.psm0[,i],simdata.psm1[,i],all=TRUE); x1[is.na(x1)]<-0; x1}) res1[,997:1000] #????? [,1]???????? [,2]???????? [,3]???????? [,4]??????? #x1??? Numeric,3000 Numeric,3000

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 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 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 <-

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))

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 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

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

I am trying to use the reshape package for the first time. I have two waves of a survey, so the id variables include a subject identification number and a variable denoting the wave of the survey. I used the following arguments: library(reshape) svy.melt <- melt(svy, id=c("id", "WAVE")) svy.wide <- cast(svy.melt, id ~ WAVE + ...) and got the following error:

hello r-helpers: there is a .txt file: x1 x2 x3 x4 x5 x6 x7 x8 x9 x10 x11 y1 17 5 77 18 19 24 7 24 24 72 52 100 2 6 72 18 17 15 4 12 18 35 42 97.2 17 2 58 10 5 3 4 3 3 40 28 98 17 2 69 14 13 12 4 6 6 50 37 93 2 3 75 20 38 18 6 12 18 73 67 99 14 4 59 16 18 9 4 3 15 47 40 99.95 17 4 87 18 17 12 4 15 12 69 46 100 14 3 74 15 9 12 1 15 12 44 35 98 17 6 76 15 33 21 15 9 18 46 41 100 17 5 76 17 22 18 1

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

After getting one list done, I am now struggling to form a data frame in C. I tried to do a list of lists which gives me : $<NA> $<NA>[[1]] [1] "BID" $<NA>[[2]] [1] 0.6718 $<NA>[[3]] [1] 3e+06 $<NA> $<NA>[[1]] [1] "BID" $<NA>[[2]] [1] 0.6717 $<NA>[[3]] [1] 5e+06 $<NA> $<NA>[[1]] [1] "BID"

I'm in a situation where I say: > predict(m.rpart, newdata=D[N1+t,]) 0 1 173 0.8 0.2 which I interpret as meaning: an 80% chance of "0" and a 20% chance of "1". Okay. This is consistent with: > predict(m.rpart, newdata=D[N1+t,], type="class") [1] 0 Levels: 0 1 But I'm puzzled at the following. If I say: > predict(m.rpart,