similar to: NAs and rle

Dear R Community - I hope you might be able to provide some guidance regarding the use of the rle function. I have a set of time-series data where a measured value is recorded every 30 seconds after the start of an experiment. Many of the measured values repeat and I am interested only in the values when there is a change. If I turn the measured values into a vector, the rle function works

why? > rle Run Length Encoding lengths: int [1:1650061] 2 2 8 2 4 5 6 3 26 46 ... values : chr [1:1650061] "4bbf9e94cbceb70c BG bg" "4fbbf2c67e0fb867 SK sk" ... > as.data.frame(rle) Error in as.data.frame.default(vertices.rle) : cannot coerce class '"rle"' into a data.frame it seems that rle.df <-

I think need to do something like this: dat<-data.frame(state=sample(id=rep(1:5,each=200),1:3, 1000, replace=T,prob=c(0.7,0.05,0.25)),V1=runif(1,10,1000),V2=rnorm(1000)) rle.dat<-rle(dat$state) temp<-1 out<-data.frame(id=1:length(rle.dat$length)) for(i in 1:length(rle.dat$length)){ temp2<-temp+rle.dat$length[[i]] out$V1[i]<-mean(dat$V1[temp:temp2])

I want to make a histogram from the lengths vector which is part of the output of rle. But I don't know how to access that vector so that I use it as an argument of hist(). What argument must I use so that I use the lengths vector as an input to hist()? Example output is: Run Length Encoding lengths: int [1:4] 1 2 3 3 values : num [1:4] -1 1 -1 1 A printout of the function rle() may

Hallo, I have an other problem, I have this vector signData with an alternation of 1 and -1 that corrispond to the duration of two different percepts. I extracted the durations like this: signData<- scan("dataTR10.txt") dur<-rle(signData)$length Now I would like to extract only the positive duration, e.g. signData <- c(1,1,1,1,-1,-1,-1,1,1,-1,-1) posduration <- c(4,2) I

Hello there R-help, I'm not sure if this should be posted here - so apologies if this is the case. I've found a problem while using rle and am proposing a solution to the issue. Description: I ran into a niggle with rle today when working with vectors with NA values (using R 2.31.0 on Windows 7 x64). It transpires that a run of NA values is not encoded in the same way as a run of other

Hi group, This is how my test file looks like: Chr start end sample1 sample2 chr2 9896633 9896683 0 0 chr2 9896639 9896690 0 0 chr2 14314039 14314098 0 -0.35 chr2 14404467 14404502 0 -0.35 chr2 14421718 14421777 -0.43 -0.35 chr2 16031710 16031769 -0.43 -0.35 chr2 16036178 16036237 -0.43 -0.35 chr2 16048665 16048724 -0.43 -0.35 chr2 37491676 37491735 0 0 chr2 37702947 37703009 0 0

Full_Name: Jeff Hallman Version: 1.2.2 OS: Solaris Submission from: (NULL) (132.200.32.33) > rle(c(1, NA, 1) $lengths [1] 3 $values [1] 1 should be as in Splus: $lengths [1] 1 1 1 $values [1] 1 NA 1 The Splus implementation (which works fine in R) is: rle <- function(x){ if(!is.atomic(x)) stop("Argument must have an atomic mode") if(length(x) == 0)

Dear all, I would like to task you if you know a rle version that can work also in a non consecutive way too. B.R Alex [[alternative HTML version deleted]]

Dear R experts, code: >m<-read.table("test.txt",sep='\t',header=TRUE,colClasses=c('character','integer','integer','rep('numeric',150)) > s<-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]],rle(m$Sample3)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]],rle(m$Sample3)[[1]])) > names(s)=c("Values","Probes")

Dear List, Suppose we have a variable K.JUN defined as (with 1=wet, 0=dry): K.JUN1984 = c(1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1) K.JUN1985 = c(0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1) K.JUN1986 = c(0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1)

Hello R help, I have this data frame M2[160000,5] with NAs, a simple example would be: set.seed(1234) M2<-expand.grid(ID=182:183, year=2012, month=1:3, day=1:3, KEEP.OUT.ATTRS=FALSE) M2 <- M2[with(M2, order(ID, year, month, day)),] #sort the data M2$value <- sample(c(NA, rnorm(100)), nrow(M2), prob=c(0.5, rep(0.5/100, 100)), replace=TRUE) M2: ID year month day

Dear R-users, Say that I have a sequence of zeroes and ones: x <- c(1,1,1,0,0,0,0,1,1,1,0,0,0,0,1,1,1,0,0,0,0) The sequences of ones represent segments and I want to report the starting and endpoints of these segments. For example, in 'x', the first segment starts at location 1 and ends at 3, and the second segment starts at location 8 and ends at location 10. Is there an efficient

Hi All, I was looking for a function to find a small matrix inside a larger matrix in R similar to the one described in the following link: https://www.mathworks.com/matlabcentral/answers/194708-index-a-small-matrix-in-a-larger-matrix I couldn't find anything. The above function can be seen as a "generalisation" of the "which" function as well as the function described

On Fri, 11 Oct 2019 10:45 Duncan Murdoch, <murdoch.duncan at gmail.com> wrote: > On 11/10/2019 6:44 a.m., Morgan Morgan wrote: > > Hi All, > > > > I was looking for a function to find a small matrix inside a larger > matrix > > in R similar to the one described in the following link: > > > > >

zstd is accessible within R using the archive package [1]. I use it all the time when saving large objects, using code I adapted from [2]. Is your suggestion to import the libraries/source code into base? [1] https://CRAN.R-project.org/package=archive [2] https://coolbutuseless.github.io/2018/10/02/using-lz4-and-zstandard-to-compress-files-with-saverds/ On Fri, Jan 10, 2025 at 6:17?PM Jeroen

I need an analogue of "uniq -c" for a data frame. xtabs(), although dog slow, would have footed the bill nicely: --8<---------------cut here---------------start------------->8--- > x <- data.frame(a=1:32,b=1:32,c=1:32,d=1:32,e=1:32) > system.time(subset(as.data.frame(xtabs( ~. , x )), Freq != 0 )) user system elapsed 12.788 4.288 17.224 --8<---------------cut

Is the splash screen RLE is standard 640x480x4 or a modified one because I can neither open the file in Photoshop CS2 (Windows under ext2fsd) or Gimp 2.2 (Linux 2.6.12.16ubuntu) and I am unable to decipher Perl scripts. Will syslinux support standard RLE?

I suppose it could be possible to patent something and let the patent expire so that it is registered at the patent office but not enforcable. No one else could patent it then. I get curious about the RLE patent. I heard Someone has a patent on run length encoding and I wonder how long they have had it because I remember RLE code running on a sinclair spectrum in the 80's before the whole

Hi, I'm relatively new to R, so I don't know the full list of base (or popular add-on packages) functions and tools available. For example, I tripped across mention of rle() in a message about some other problem. rle() turned out to be a handy shortcut to splitting some of my data by magnitude (vaguely like a sequence-based histogram). So I thought I'd ask: what small, or