similar to: mtext bug

Hi All, I am trying to run a loop that will have varying numbers of rows with each output. Previously I have had the same number of rows so I would use (and I appreciate that this will no doubt achieve some gasps as being thoroughly inefficient!): xdfrow<-(0) xdfrow1<-(1:32) xdfrow2<-(33:64) xdfrow3<-(65:96) xdfrow4<-(97:128) xdfrow5<-(129:160) xdfrow6<-(161:192)

Dear Rui, Not working. I have entirely removed the column containing % but am still bootless: > head(coredata(INFYTX)) INFY Historical Data INFY Historical Data INFY Historical Data INFY Historical Data [1,] "47.26" "44.28" "47.56" "44.28" [2,] "46.30" "44.92"

(Sorry, sent the message before I finished it) Hello, everyone I have a long script that uses zoo objects. In this script I used simple moving averages and these I can very efficiently calculate with filter() functions. Now, I have to use special "exponential" moving averages, and the only way I could write the code was with a for-loop, which makes everything extremely slow. I don't

Dear list, I would like to expand a DF with all the missing levels of a variable. a <- c(2,2,3,4,5,6,7,8,9) a.cut <- cut(a, breaks=c(0,2,6,9,12), right=FALSE ) (x <- data.frame(a, a.cut)) # In 'x' the level "[0,2)" is "missing". AddMissingLevel <- function(xdf) { xfac <- factor( c("[0,2)", "[2,6)", "[6,9)",

Hello, everyone I have a long script that uses zoo objects. In this script I used simple moving averages and these I can very efficiently calculate with filter() functions. Now, I have to use special "exponential" moving averages, and the only way I could write the code was with a for-loop, which makes everything extremely slow. I don't know how to optimize the code, but I need to

Dear members, I have an xts object: > head(INFYTX) INFY Historical Data INFY Historical Data.1 INFY Historical Data.2 2003-04-16 "47.26" "44.28" "47.56" 2003-04-17 "46.30" "44.92" "46.53" 2003-04-21 "45.82"

?s 19:57 de 12/04/2023, akshay kulkarni escreveu: > Dear members, > I have an xts object: > >> head(INFYTX) > INFY Historical Data INFY Historical Data.1 INFY Historical Data.2 > 2003-04-16 "47.26" "44.28" "47.56" > 2003-04-17 "46.30" "44.92"

Isn't this like trying to tie up the horse after it has left the barn? Why not figure all this out _before_ converting to xts? On April 12, 2023 12:29:49 PM PDT, akshay kulkarni <akshay_e4 at hotmail.com> wrote: >Dear Rui, > Not working. I have entirely removed the column containing % but am still bootless: > >> head(coredata(INFYTX)) > INFY

Dear R-helpers, It seems to me that a character zoo cannot be coerced to a numeric zoo. Below is a minimal example. Can someone tell me what I have done wrong? > z<-zoo(1:4,order.by=1:4) > coredata(z)<-as.character(coredata(z)) > str(z) ‘zoo’ series from 1 to 4 Data: chr [1:4] "1" "2" "3" "4" Index: int [1:4] 1 2 3 4 >

I have NexentaCP2 (Opensolaris b104+). Parameters for the kernel: ''zfs-bootfs=syspool/rootfs-nmu-000,bootpath=/xpvd/xdf@0:a'' Pygrub returns: ''zfs-bootfs=syspool/30,bootpath=/xpvd/xdf@0:a'' Result: Boot with pygrub don''t work. What is "30" in syspool/30 ? _______________________________________________ Xen-users mailing list

Dear all, To explain my problem I am attaching a demonstration package "myclasspkg": I have the following two S4 classes with similar inheritance: SubSubClassA <- SubClassB <- BaseClass SubSubClassB <- SubClassB <- BaseClass In R I am calling the following functions: > library(myclasspkg) > subA <-

Hello. I am new to R, so my question/problem might be very basic, but I cannot figure out how to solve it. So, I would really appreciate your help. I would like to rename a variable in a very large file (181GB) called "2013.xdf" that is saved on my external hard drive. I tried running the following code to rename variable # 2 in the file and save this change, which produced some

Dear list, I would like to do some calculation using different grouping variables. My 'df' looks like this: # Some data set.seed(345) id <- seq(200,400, by=10) ids <- sample(substr(id,1,1)) group1 <- rep(1:3, each=7) group2 <- rep(1:2, c(10,11)) group3 <- rep(1:4, c(5,5,5,6)) df <- data.frame(id, ids, group1, group2, group3) df <- rbind(df, df, df) df$time <-

I want to do a series of contour plots, 4 in all. The data is coming from a data frame named "nd.frame", which has elements "xdf", "ydf", "zdf", and "pndt". I am treating "pndt" as a factor, and it has four levels. I make a call to the lattice graphics routine "contourplot" like so:

Hi everyone, ? I am trying to do a horizonplot using my own time series data. I know that there is a horizonplot function in latticeExtra, but on closer examination i think that the graph itself is slightly wrong (it displays some regions as triangles and i think they should be trapezoids, and the red regions (that are below the baseline) are displayed on top of the blue areas ? while i think

My problem is the following : I create 2 zoo objects and then I try to subset one of them using logic. indicesthatpass is a vector of trues and falses but when I send it into bckret, it returns an empty bckret. Obviously it has something to do with bckret being a zoo object and if I do the same subsctripting off of coredata(bckret), I'm confident it will work. But, I need to keep the minute

Dear all, I have an error message, when I try to convert a zoo object (called test) to ts (on R 2.4.0, Package zoo version 1.2-1, Windows XP) > test 1994-05-10 1994-06-09 1994-07-09 0.0024943889 0.0024881824 0.0006955831 > str(test) atomic [1:3] 0.002494 0.002488 0.000696 - attr(*, "index")=Class 'Date' num [1:3] 8895 8925 8955 > is.regular(test) [1] TRUE

Dear list-reader, by running the following script: library(zoo) sessionInfo() search() packageDescription("zoo") data(EuStockMarkets) dax <- as.zoo(EuStockMarkets[1:10, "DAX"]) daxr <- diff(log(dax)) identical(as.vector(qnorm(daxr)), qnorm(coredata(daxr))) qqnorm(coredata(daxr)) qqnorm(daxr) qqnorm() produces an error: > qqnorm(daxr) Fehler in if (xi == xj) 0L

Dear all, I'm trying to plot, in the same window, two different series, using barplot() for the first one and plot() for the second. What happens is that the second chart has a different axes origin, therefore the final plot is wrong. This piece of code shows the differences between the values of par()$usr: barplot(coredata(Z0), beside=TRUE, ylim=c(0,100)); par()$usr;

Dear all, Currently, I am trying to create a test package "testS4" using S4 classes, which I am attaching. Running R CMD check gives the following error: coeurebooks-computer:/Volumes/CoreData/CRAN cs$ R CMD check testS4_0.1.1.tar.gz * checking for working latex ...sh: line 1: latex: command not found NO * using log directory '/Volumes/CoreData/CRAN/testS4.Rcheck' * using R