similar to: I think a simple question

A couple days ago, Mark Leeds asked about a solution that would basically stagger two lists, a and b, to return a list in the form of a[1], b[1], a[2], b[2], a[3].... In particular, the summary of his question was in reference to lists defined by x <- 5 tempin <- seq(1,1411, by=30) a <- tempin b <- tempin + x I offered the following function everyOther <-

I have the following set of indices, call it idx, that correspond to the indices of a vector say temp. [1] 31 36 41 61 66 71 91 96 101 121 126 131 151 156 161 181 186 191 211 216 221 241 246 251 271 276 281 301 306 311 331 336 341 361 366 [36] 371 391 396 401 421 426 431 451 456 461 481 486 491 511 516 521 541 546 551 571 576 581

I have a vector of character strings such as mainnames<-c("CAD","AUD") and another vector say checknames<-c("CAD.l1","AUD.l1","JPY.l1","EUR.l1","CAD.l2","AUD.l2","JPY .l2","EUR.l2") I want a new vector of character strings that is just

I'm having that problem where I am sometimes using an object that's from a previous workspace when I don't want to be doing that. I was thinking of putting rm(objects=ls()) in my first.R function But, the problem with doing this, is that it doesn't prompt you with "are you sure" and there could be very rare cases where I don't want to delete the workspace ? Is there a

Hello all, I looking at package dse or vars or mAr I know how to simulate a VAR(p) process, my problem is that most of those processes are unstable (not weakly stationary). Do anybody know how to generate a random VAR (or VARMA even better) process that is weakly stationary? Thanks -- View this message in context: http://r.789695.n4.nabble.com/simulating-stable-VAR-process-tp4261177p4261177.html

Viswanath Puttagunta wrote: > if OPUS_ARM_NEON_INTR > CELT_ARM_NEON_INTR_OBJ = $(CELT_SOURCES_ARM_NEON_INTR:.c=.lo) \ > - %test_unit_rotation.o %test_unit_mathops.o > -$(CELT_ARM_NEON_INTR_OBJ): CFLAGS += $(OPUS_ARM_NEON_INTR_CPPFLAGS) > + $(CELT_SOURCES_ARM_NE10:.c=.lo) \ > + %test_unit_rotation.o %test_unit_mathops.o \ > +

Hi, Having several daily wind speed time series I want to extract those consecutive days over and below certain values (i.e. 5 < x <8) Don't know which funtion to use (aggregate, lapply?) and how to do it. Thanks in advance Antonio

I have the code below which gives me what I want for temp based on logvec but I was wondering if there was a shorter way ( i.e : a one liner ) without having to initialize temp to zeros. This is purely for learning purposes. Thanks. logvec <- c(TRUE,FALSE,FALSE,TRUE,FALSE,FALSE,TRUE,FALSE) temp<-numeric(length(invec)) temp[invec]<-which(invec) temp [1] 1 0 0 4 0 0 7 0 obviously, the

I just noticed something by accident with R syntax that I'm sure is correct but I don't understand it. If I have a simple numeric vector x and I subscript it, it seems that I can then subscript a second time with TRUE or FALSE, sort of like a 2 dimensional array in C. Does someone know if this is documented somewhere Because it's neat but I never knew it existed. To me it seems like a

I am constructing plots ( regular not lattice ) and my initial command is par(mar=c(3,4,2,2), mfcol=c(5,2)) and then I create 10 plots on the page. It looks great but the plots on the page go in the order 1 6 2 7 3 8 4 9 5 10 Where the numbers denote decile breakdowns. Is there an easy way to make them go from left to right so 1 2 3 4 5 6 7 8 9 10 I could try

does anyone know how to put a column name on a zoo object. I think achim and gabor are off line or they have gotten totally tired of me an decided to ignore me ( which is totalyy understandable ). logbidask<-log((aggfxdata[,"bid"] + aggfxdata[,"ask"])/2.0) logbidask doesn't have a name and I can't figure out how to get one on it ? aggfxdata is a zoo object.

Dear all R users, I have a small doubt about panel data analysis. My basic understanding on Panel data is a type of data that is collected over time and subjects. Vector Autoregressive Model (VAR) model used on this type of data. Therefore can I say that, one of statistical tools used for analysis of panel data is VAR model? If you clarify my doubt I will be very grateful. Thanks and regards,

Hi! I've made great progress in my R programming, but I am again stuck on a beginner's problem. I would like to start R with a command line that loads a file, and if possible, executes a function. Can anyone give me an example of how to do this? For example, in lisp, I would say: $ lisp -load toto.lisp -eval '(do-something $PORT)' to load the file "toto.lisp", then

I think I asked a similar question 3 years ago to the Splus list and I think the answer was no or noone answered so noone should spend more than 5 minutes on this because it could definitely be a waste of time. My question is whether the function below can be rewritten without a for loop. apply is fine if it can be done that way but i doubt it. I call it a lot and would prefer to not loop.

Dear all R users, I am really struggling to determine the most appropriate lag order of ARIMA model. My understanding is that, as for MA [q] model the auto correlation coeff vanishes after q lag, it says the MA order of a ARIMA model, and for a AR[p] model partial autocorrelation vanishes after p lags it helps to determine the AR lag. And most appropriate model choosed by this argument gives

Greetings, I'm learning R and I'm stuck on a basic concept: how to specify a logical condition once and then perform multiple transformations under that condition. The program below is simplified to demonstrate the goal. Its results are exactly what I want, but I would like to check the logical state of gender only once and create both (or any number of) scores at once.

I wrote an ugly algorithm to set certain elements of a matrix to 1 without looping and below works and you can see what The output is below the code. K<-6 lagnum<-2 restrictmat<-matrix(0,nrow=K,ncol=K*3) restrictmat[((col(restrictmat) - row(restrictmat) >= 0 ) & (col(restrictmat)-row(restrictmat)) %% K == 0)]<-1 restrictmat[,(lagnum*K+1):ncol(restrictmat)]<-0 restrictmat

i'm doing a bar plot and there are 16 column variables. is there a way to make the variable names go down instead of across when you do the barplot ? because the names are so long, the barplot just shows 3 names and leaves the rest out. if i could rotate the names 90 degrees, it would probably fit a lot more. or maybe i can use space to make the horizontal width longer ? I looed up ?barlot but

I was going through the example below which is taken from the example section in the R documentation for accessing the function call stack. I am confused and I have 3 questions that I was hoping someone could answer. 1) why is y equal to zero even though the call was done with gg(3) 2) what does parents are 0,1,2,0,4,5,6,7 mean ? I understand what a parent frame is but how do the #'s relate

The code bekow works so this is why I didn't include the data to reproduce it. The loops about 500 times and each time, a zoo object with 1400 rows and 4 columns gets created. ( the rows represent minutes so each file is one day worth of data). Inside the loop, I keep rbinding the newly created zoo object to the current zoo object so that it gets bigger and bigger over time. Eventually,