similar to: Ragged time series data

Hi All, I'm wanting to convert a ragged list of values into a structured matrix for further analysis later on, i have a solution to this problem (below) but i'm dealing with datasets upto 1GB in size, (i have 24GB of memory so can load it) but it takes a LONG time to run the code on a large dataset. I was wondering if anyone had any tips or tricks that may make this run faster? Below is

How can I add an extra column containing the rank to a ragged array indexed by more than one grouping factors? E.g. with the barley dataset: How can I to add an additional column ``rank'' containing the rank of the ``yield'' of the different varieties in relation to the indices ``year'' and ``site'' to the barley dataframe? I achieved to calculate the ranks with:

Hello, I have a big problem which I’m just not able to solve. I created the following mean value from the following dataset structure: Id |value 1 | 2 1 | 3 1 | 4 2 | 2 2 | 1 3 | 5 4 | 3 etc.|etc. with the command: mean_rating <- tapply(ratok$value, ratok$project_id , mean,simplify = FALSE) this gives me a ragged array: > mean_rating [1] $`14`

I have a dataset that looks like: set.seed(144) sam<-sample(1000,100) dat<-data.frame(id=letters[1:10],value=rnorm(1000),day=c(rep(1,100),rep(2,100),rep(3,100),rep(4,100),rep(5,100))) I want to "normalise" it using the following function (unless you have a better idea...): adj.values<-function(dframe){ value_mean<-mean(dframe$value) value_sd<-sd(dframe$value)

Can anyone please suggest how I can print: a <- matrix(c( "Heading 1", "This is some info\nabout heading 1", "Heading 2", "This is some info\nabout heading 2", ), byrow=T, nrow=2) to look like: Heading 1 This is some info about heading 1 Heading 2 This is some info about heading 2 (if you're not using a fixed width

I wonder what's the right way in R to do the following -- placing objects of the same kind together in subarrays of varying length. Here's what I mean: > word <- c("a","b","c","d","e","f","g","h","i","j") > kind <- c(1,1,1,2,3,4,5,5,7,7) > d <-

Dear R users, I am currently writing a MCMC algorithm for spatial Poisson model. One problem which I need to solve is as follows: I have a vector X[1:J] which describes characteristics of cells of a rectangular grid. Further, for each cell I have a list of adjascent cells adj[] and the vector listing numbed of adjascent cells num[]. Thus if I want to list cells adjascent to cell i I write

Hi, I have a list made up of character strings with each item a different length (each item is between 1and 6 strings long). Is there a way to convert a "ragged" list to a matrix such that each item is its own row? Here is a simple example: a=list(); a[[1]] = c("a", "b", "c"); a[[2]] = c("d", "e"); a[[3]] = c("f",

Hello, I'm running an algorithm for graph structural cohesion that requires a depth-first search of subgraphs of a rather large network. The algorithm will necessarily be redundant in the subgraphs it recurses to, so to speed up the process I implemented a check at each subgraph to see if it's been searched already. This algorithm is very slow, and takes days to complete on a

Hello, I've been learning R functions recently and I've come across a problem that perhaps someone could help me with. # I have a a chron() object of times > hours=chron(time=c("01:00:00","18:00:00","13:00:00","10:00:00")) # I would like to subtract 12 hours from each time element, so I created a vector of 12 hours (actually, more like a

i know it''s probably really simple, how do i work out someone''s age if i have their d.o.b. stored as a date in my db. cheers -- Posted via http://www.ruby-forum.com/.

Hi, I have a dataframe column of the form v<-c("Fri Feb 05 20:00:01.43000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:06.34000 2010") I need to convert this to datetime form. I did the following.. lapply(v,function(x){strptime(x, "%a %b %d %H:%M:%OS %Y")}) This gives me a list that looks like

Hi there, I set up my company's back up server using rsync. And I've got a strange problem. I searched in the archives of this list, but none of them seems not giving me an idea to solve the problem. If anyone can help, it would be grateful. I'm using cron by a user (non wheel/admin) to rsync everyday during the night. The cron is set in the server to transfer the backing-up

Full_Name: John Brzustowski Version: R-devel-trunk OS: linux (problem under Windows too) Submission from: (NULL) (74.101.124.238) (This bug was discovered by Phil Taylor, Acadia University.) I'm not sure from reading the documentation whether strptime(x, "%j") is meant to be supported, but if so, there is a bug which prevents it from working on the first day of months after

Dear R-users, I use unlist of POSIX dates to extract the year, hour etc. With that I can search for files in my database which are in the form 'yyyymmddhh_synops.txt' However, I get stucked during midnight where unlist just gives NA's. The script is given below, the problem accurs at acc.period[16] (midnight). However when I write out the character, unlist works well. But

Hi all, Is there a simple function already implemented for getting the ISO weeks of a Date object? I couldn't find one, and so wrote my own function to do it, but would appreciate a pointer to the "default" way. If a function is not yet implemented, could the code below be of interest to submit to CRAN? Best Regards, Gustaf --------------------

Hello, I've come across a problem in developing a set of custom functions to calculate the number of hours of daylight at a given latitude, and the number of days a date precedes or secedes the summer solstice. I discovered an inconsistency concerning leap years between my derived values and those from the US naval databases. It seems as far as I can figure that my inconsistency arises either

Full_Name: Petr Simecek Version: 2.5.1, 2.6.1 OS: Windows XP Submission from: (NULL) (195.113.231.2) Several times I have experienced that a length of a POSIXt vector has not been computed right. Example: tv<-structure(list(sec = c(50, 0, 55, 12, 2, 0, 37, NA, 17, 3, 31 ), min = c(1L, 10L, 11L, 15L, 16L, 18L, 18L, NA, 20L, 22L, 22L ), hour = c(12L, 12L, 12L, 12L, 12L, 12L, 12L, NA, 12L,

Can someone, please, explain the difference is results below (notice the isdst value) > unlist(as.POSIXlt('2005-7-1')) sec min hour mday mon year wday yday isdst 0 0 0 1 6 105 5 181 1 > unlist(as.POSIXlt(as.Date('2005-7-1'))) sec min hour mday mon year wday yday isdst 0 0 0 1 6 105 5 181 0

Hi all, I'm usually comfortable using the *apply functions for vectorizing loops in R. However, my particular problem now is using it in a sequential operation, which uses values evaluated in an offset of the loop vector. Here is my example using a for loop approach: dat <- data.frame(year=rep(1970:1980,each=365),yday=1:365) dat$value <-