search for: rle

...993,K.JUN1994,K.JUN1995,K.JUN1996,K.JUN1997,K.JUN1998,K.JUN1999,K.JUN2000,K.JUN2001,K.JUN2002) Our motivation is to count number of wet days (1's) in each weeks. But counting number of wet days for entire K.JUN will not do. Thus in r console, > k<-0;j<-0;i<-(1:7)+30*j+7*k;K.JUN[i];rle(K.JUN[i]) [1] 1 0 1 1 1 1 1 Run Length Encoding lengths: int [1:3] 1 1 5 values : num [1:3] 1 0 1 where k=0,1,2,3 for each j=0 to 18 (k indicating weeks of any June, and j indicates years 1984-2002 respectively). Now we need to sum the run 'lengths' corresponding to each 'values...

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...

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]) out$V2[i]<-sum(dat$V2[temp:temp2]) out$state[i]<-rle.dat$value[[i]] temp<-temp2 } to a very...

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 funct...

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 values. See the following example as an illus...

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)$length Now I would like to extract only the positive duration, e.g. signData <- c(1,1,1,1,-1,-1,-1,1,1,-1,-1) posduration <- c(4,2) I think I should use rle in a nested way, this is what I tried but it doesn't work: posduration<- rle(signData[=1])$length Could you plea...

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 perf...

I have a matrix of frequency counts from 0-160. x<-as.matrix(c(0,1,0,0,1,0,0,0,1,0,0,0,0,1)) I would like to apply a function creating a new column (x[,2])containing values equal to: a) log(x[m,1]) if x[m,1] > 0; and b) for all x[m,1]= 0, log(next x[m,1] > 0 / count of preceding zero values +1) for example, x[1,2] should equal log(x[2,1]/2) = log(1/2) = -0.6931472 whereas x[3,2] should

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") #Suppose the test file looks like with ofcourse more samples with values: Chr Start End Sample1 Sample2 Sample3 chr2 98966...

I have searched for info on how to apply the Western Electric rules for process control, to data and plots I have in R, but I have not been able to learn how. Any help would be greatly appreciated. Thank you, sjcrauhut at agere.com 05/20/02 -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read

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 not. This conceptual strictness isnt universall...

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?

...e among the common values and probes is number of similar values. Code: >m<-read.table("test.txt",sep='\t',header=TRUE,colClasses=c('character','integer','integer','numeric','numeric')) #reading the test file >s<-data.frame(c(rle(m$Sample1)[[2]],rle(m$Sample2)[[2]]),c(rle(m$Sample1)[[1]],rle(m$Sample2)[[1]])) # to get the last 2 columns > names(s)=c("Values","Probes") >G=1 > for(i in 1:length(s$Probes)){ + if(G==1){first<-unique(m$Chr[G:s$Probes[i]]) + second<-min(m$Start[G:s$Probes[...

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]]

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) return(list(len...

Dear List I hope you can help me: I?ve got a dataframe (df) within which I am looking for Peak Over Threshold values as well as the length of the events. An event starts when walevel equals 5.8 and it should end when walevel equals the lower threshold value (5.35). I tried ?clusters (?)? from ?evd package?, and varied r (see example) but it did not work for all events (again

Hello, I am searching for a function to calculate "partial" cumsums. For example it should calculate the cumulative sums until a NA appears, and restart the cumsum calculation after the NA. this: x <- c(1, 2, 3, NA, 5, 6, 7, 8, 9, 10) should become this: 1 3 6 NA 5 11 18 26 35 45 any ideas? thank you and best regards, stefan

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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 obscure, tools and functions in R do you find handy or 'cool' to use in your work...

...ind <- .Internal(split(seq_along(f), f)) + ind <- .Internal(split(seq_along(x), f)) for(k in lf) y[[k]] <- x[ind[[k]]] y } Maybe a little harder to argue the following, but in split.default, for a class that one might wish to develop factor-like behaviour, e.g., Rle = setClass("Rle", representation(values="integer", lengths="integer")) f = Rle(values=1:2, lengths=2:3) the code if (is.list(f)) f <- interaction(f, drop = drop, sep = sep) else if (drop || !is.factor(f)) f <- factor(f) requires...