similar to: how does one apply Western Electric / AT&T rules to R plots?

Displaying 20 results from an estimated 1000 matches similar to: "how does one apply Western Electric / AT&T rules to R plots?"

2008 May 14
1
custom iptables chain jumping
Hi all, When we create a custom chain in iptables, should we specifically create a rule to 'jump back' to the previous chain? For example: iptables -A INPUT -j CUSTOMCHAIN iptables -A CUSTOMCHAIN rule1 iptables -A CUSTOMCHAIN rule2 Should we add: iptables -A CUSTOMCHAIN -j INPUT ? Or, it will automatically go back to CHAIN when there's no more rule? Thank you very much, -- Fajar
2011 Mar 25
0
applying random functions to multisets
Hi, I am solving following problem: Suppose I have some multiset: > multiset <- c("a","a","c","d","d") and rules, which operate with it (for simplicity not writen in R functions) > rule1: "a" -> c("a","b") > rule2: "a" -> c("a","c") > rule3: "c" ->
2011 Nov 26
3
Time series merge?
I have two time series a <- ts(1:10, start=c(1,6), end=c(2,5), frequency=10) b <- ts(1:5, start=c(2,1), end=c(2,5), frequency=10) Obviously 'b' is a subset of 'a'. I want a single index value indicating where that start of 'b' lines up with the start of 'a'. So in this simple example I would expect an index of 5. I was playing with 'merge'.
2005 Jun 24
1
r programming help II
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)
2012 Oct 16
2
cannot coerce class '"rle"' into a data.frame
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 <-
2013 Mar 26
2
Feed rle() output to hist()
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
2011 Jun 17
3
rle on large data . . . without a for loop!
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])
2011 Jun 23
3
problem (and solution) to rle on vector with NA values
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
2008 May 27
4
help with simple function
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
2012 Jun 08
2
help with rle function on paired data
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
2009 Jul 07
2
rle
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
2005 Oct 26
3
splash screen
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?
2010 May 26
3
Peak Over Threshold values
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
2009 Nov 11
2
partial cumsum
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
2007 Oct 09
3
identify number of sequences from a vector
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2008 Jun 19
1
replacing segments of vector by their averages
Given a vector of numeric of length n, I need to find segments that are >= 0.2, compute the average of individual segments, and replace the original values in each segment by their corresponding averages. For example, there are three segments that are >= 0.2, the average of 1st segment is 0.3, 2nd is 0.5, and the 3rd is 0.5333333 >
2020 Aug 26
1
NAs and rle
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
2009 Aug 03
1
selectively altering variable value
Hello, I have run an eye-tracking experiment for which I now like to analyse the saccades. Participants looked from a fixation cross (ia = 5) to the target area (ia = 4) in following example of a data frame. ia = 9 stands for everything else. A saccade is indicated by saccade = 1. Sometimes the saccade just ends before the target area (see below). This is due to the parameters that determine a
2010 Nov 18
2
RowSums Question
I have a question on RowSums. Lets say i have a timeSeries table A B C 1/1/90 NA 1 1 1/2/90 NA 1 1 1/3/90 NA 1 1 1/4/90 NA 1 1 1/5/90 1 1 1 1/6/90 1 1 1 if i use RowSums, i will get 1/5/90 3 1/6/90 3 but i want 1/1/90 2 1/2/90 2 1/3/90 2 1/4/90 2 1/5/90 3 1/6/90 3 I cant
2007 Feb 22
2
Is there better alternative to this loop?
Dear List, Thanks to those who helped with my enquiry a few days ago. I have a another question on loops, in this case I am trying to print out the row of a data frame if the previous 3 values (daily values) in col5 are in descending order. I have this loop which works, but ask whether this can be done differently (without conventional loop) in R: flag="T" d= 3 # d represents