similar to: sweep and zoo objects

Given a long zoo matrix, the goal is to "sweep" out a statistic from the entire length of the sequences. longzoomatrix<-zoo(matrix(rnorm(720),ncol=6),as.yearmon(outer(1900,seq(0,length=120)/12,"+"))) cnames<-c(12345,23456,34567,45678,56789,67890) colnames(longzoomatrix)<-cnames longzoomatrix[1:24,] 12345 23456 34567 45678

Hello everybody, I have two datasets, observed and predicted. Since my observed dataset is not in regular intervals, I need to filter my predicted dataset based on the measurement date of my observed data. Here, is an example similar to what I have library(chron);library(zoo) DATE<- seq(as.Date("2009-01-01"), as.Date("2009-05-01"), by = 1) mydat<- rnorm(length(DATE),

I have date data as a numeric and hourly data in 0 to 2300 hours in a dataframe. d <- rep(20110101,24) h <- seq(from = 0, to = 2300, by = 100) df <- data.frame(LST_DATE = d, LST_TIME = h, data = rnorm(24, 0, 1)) S <- chron(dates. = as.character(df$LST_DATE), times. = paste(as.character(df$LST_TIME/100), ":0:0", sep = ""), format =

A friend of mine just got bitten by the fact that sweep() will happily sweep out a STATS vector of an arbitrary length -- even one whose length is not a divisor of any of the margins -- without complaining. I know the answer to this could be "well just don't do that", but it's easy to make a mistake in which margin you're sweeping ... What would R-core think of the

The zoo package as a merge function which merges a set of zoo objects result<-merge(zoo1,zoo2,...) Assume your zoo objects are already collected in a list # make a phony list to illustrate the situation. ( hat tip to david W for constructing a list in a loop) ddat <- as.list(rep("", 20)) ytd<-seq(3,14) for(i in 1:20) { + ddat[[i]] <- zoo(data,ytd ) + } ddat [[1]] 1 2

A quick question x <- as.yearmon(2000 + seq(0, 23)/12) x [1] "Jan 2000" "Feb 2000" "Mar 2000" "Apr 2000" "May 2000" "Jun 2000" "Jul 2000" "Aug 2000" "Sep 2000" "Oct 2000" "Nov 2000" "Dec 2000" "Jan 2001" [14] "Feb 2001" "Mar 2001" "Apr

I am having a hard time understanding just what 'sweep' does. The documentation states: Return an array obtained from an input array by sweeping out a summary statistic. So what does it mean "weeping out a summary statistic"? Thank you. Kevin

I have a matrix of zoo series. each series is in a column. x <- as.yearmon(2000 + seq(0, 23)/12) # 24 months of data, lets make 20 sets of random data testData <- matrix(rnorm(480),ncol=20) # make a zoo object and columns will hold the 20 series TestZoo <- zoo(testData,order.by=x) # now run lm for just one series. m <- lm(TestZoo[,1]~time(TestZoo))$coeff[2] m time(TestZoo)

Hi, I would like to use sweep to "sweep out" proportions and confidence intervals for an array, however when I supply a function which returns a string (containing something like "9% (3-18%)") I get back a list instead of an array, here is a simplified example: # example showing that sweep does not return an array with same dimensions as STATS as advertised

Hi Jean-Marc and others, I put Speex support in Sweep (a sound editor) yesterday, and had some fun with it :) I'd like to get some feedback before releasing it publically. I'm not really used to speech codecs (I work more with music) but I've been quite impressed at the file size and quality (especially when you hear what I did with some of the sample files ;) I didn't have any

Hi, I'm updating speex support in sweep to RC2 and I'd like some advice on encoder options, and would appreciate some testing of the user interface. a screenshot of the updated speex encoding dialog: http://www.vergenet.net/~conrad/tmp/sw_speexRC2_1.png you can grab a tarball of sweep for testing here: http://www.vergenet.net/~conrad/tmp/sweep-0.8.0-speexRC2.tar.gz general

I''ve got a Rails project that has been on the shelf for a few months, and I''m picking it back up. I''ve frozen Rails at 1.1.4 in vendor/, and did the "rails ." and "rake rails:update" to get it up to a current Rails level. The code initially seems to be working fine, but when I go to my login method, I get a 500 error thrown. Looking in the

Dear list, I have a matrix, spc, that should row-wise be interpolated: E.g. spc <- matrix (1:1e6, ncol = 250, nrow = 4000, byrow = TRUE) spc [1:10, 1:10] shifts <- seq_len (nrow (spc)) wl <- seq_len (ncol (spc)) interpolate <- function (spc.row, shift, wl) spline (wl + shift, spc.row, xout = wl, method = "natural")$y interpolate (spc [1,], shift = shifts [1], wl = wl)

I'm looking for an R equivalent of Beaton's (1964) Sweep algorithim for partial inversion of a matrix by pivoting. It implemented in SAS/IML as sweep(matrix, indices), described here http://support.sas.com/documentation/cdl/en/imlug/59656/HTML/default/langref_sect266.htm and here for python http://adorio-research.org/wordpress/?p=262 -- Michael Friendly Email: friendly AT yorku

Dear List: I am wondering if there is a more efficient way to compute the following. For the example I am using the star data frame in the mlmRev package. This has 80 schools and includes grades K, 1, 2, and 3. First I compute the grade level mean in each school using tapply as: tapply(star$math, list(star$sch,star$gr), mean, na.rm=T) This results in a table of means by school for each grade.

For example, subtracting 1:2 from the rows of a two-column matrix: > t(apply(matrix(1:6,ncol=2),MARGIN=1,function(y) y - 1:2)) [,1] [,2] [1,] 0 2 [2,] 1 3 [3,] 2 4 > sweep(matrix(1:6,ncol=2),MARGIN=2,1:2,FUN="-") [,1] [,2] [1,] 0 2 [2,] 1 3 [3,] 2 4 Is there a logic to this difference, or is it just a quirk of the history of these

Hi I had a hard-to-find bug in some of my code the other day, which I eventually traced to my misusing of sweep(). I would expect sweep() to give me a warning if the elements don't recycle nicely, but X <- matrix(1:36,6,6) sweep(X,1,1:5,"+") [,1] [,2] [,3] [,4] [,5] [,6] [1,] 2 9 16 23 30 32 [2,] 4 11 18 25 27 34 [3,] 6 13 20 22

Hi, I'd appreciate help with this. I have a data matrix with one column, called f in the example below, a factor. I'd like to subtract the means from each of other columns for each level of the factor. That is, in the example, to go from the first matrix below to the second. I know SWEEP will take out means, but I want to do this for each level of the factor. f x 1 2 1

I've used barplot(), including the anmes.arg parameter, on data frames successfully, but I'm even newer to using zoo than I am to R. :-} I am working on a functon that accepts a data frame ("df") as its primary argument, extracts information from it to create a zoo, then generates a plot based on that. The data frame has a column called "time" which is a standard

Dear R users, When dumping an R matrix object into a file -- typically via the 'write.table' function -- the 'eol' option can be used to specify the end-of-line character(s) which should appear at the end of each row. However the argument to 'eol' seems to be restricted to have length 1, whereas ideally I would like different rows to be written to file each with its own