similar to: Feed rle() output to hist()

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])

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

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

Hi All, A twitter user, Mike fc (@coolbutuseless) mentioned today that he was surprised that repeated NAs weren't treated as a run by the rle function. Now I know why they are not. NAs represent values which could be the same or different from eachother if they were known, so from a purely conceptual standpoint there is no way to tell whether they are the same and thus constitute a run or

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")

There are probably many mistakes with this code. I am used to coding in C with a debugger, so I am very new to coding in R without one and different syntax. My code in the upper left panel of R Studio isz<-c(3,1,4,5,2,1,0,3,5,8)zlength(z)y<-c(0,0,0,0,0,0,0,0,0,0)ylength(y)zdiff = diff(z)zdiffn<-length(zdiff)nx<-zdifff<-function(x) { for (k in 1:n){

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

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

getnybble() needs to return four bits at a time from every byte. During rle decode, rows are rounded to an integer number of bytes. The rle length needs to be able to hold values > 255. Signed-off-by: Chas Williams <3chas3 at gmail.com> --- core/graphics.c | 35 +++++++++++++++++++++-------------- 1 file changed, 21 insertions(+), 14 deletions(-) diff --git a/core/graphics.c

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

Hello everybody, I have problem with a lapply command, which rather proves that I don't fully understand it. I want to extract from a list that consists of dataframes, the length of the first sequences from a given variable (its part of a simulation exercises). Below is code which does the job, but I think it should be possible to make it more compact. ### Example Data dat <-list()

I'm running rle() on a long vector, and get a result which looks like > uc Run Length Encoding lengths: int [1:16753] 1 1 1 1 1 1 1 1 1 1 ... values : int [1:16753] 29462748 22596107 18322820 14323315 12684505 9909036 7296916 6857692 5884755 5883697 ... I can print uc$names or uc$levels separately. Is there any way to print them together as tuples, looking like (29462748, 1)