similar to: Lattice plot

> Suppose you have four lifetime operations on the same address in memory, > with loads and stores all around them: > > start1--end1 .. start2--end2 > > If you remove start1 then you have a bare pointer, the memory came from > somewhere and you lose the optimization that loads before start1 become > undef, but you don't miscompile. This is assuming no looping after

On 12/27/2012 12:35 PM, Rafael Espíndola wrote: >>> Oh, I was reading "precedes/following" as having static (dominance) >>> meaning. That is, in the above example you could not delete the store >>> since it is not true that >>> llvm.lifetime.end dominates it. >>> >>> Nick, is this what you had in mind? If not, then we must delete a

Dear all, I am a biologist. I have two sets of distance P(start1, end1) and Q(start2, end2). The distance will be like this. P ------------------------ Q ---------------------------------------- I want to know whether P falls closely to the right end or left end of Q. P and Q are of different lengths for each data point. There are more than 10000 pairs of P and Q. Is there any test or

I've been using the following code to plot using Sweave in version 1.9.1 library(RODBC) library(lattice) channel <-odbcConnectExcel("h:/water.xls") data <- sqlQuery(channel,"select * from `Sheet1$` where Test = 'TOC' and (Valve='5010-05' or Valve='8030-V26' or Valve='1180-08' or Valve='5040-08')") odbcClose(channel) srt <-

Dear R People: I have the following code that I use to convert a monthly zoo object to a data.frame, and it works perfectly: library(tseries) z <- get.hist.quote(instrument=inst1, start=start1,end=end1, quote=quot1,comp = "m") y <- as.ts(aggregate(z, as.yearmon, tail, 1)) y.df <- data.frame(y=y,time=time(y)) y.df$x <- ts(y.df[,1]) tsp(y.df$x) <-

Hi, I wanted to make a graph with the following table (2 rows, 3 columns): a b c x 1 3 5 y 5 8 6 The first column represents the start cordinate, and the second column contains the end cordinate for the x-axis. The third column contains the y-axis co-ordinate. For example, the first row in the matrix above represents the points (1,5),(2,5), (3,5). How would I go about making a discontinuous graph

Dear R-ers, In my question there are no statistics involved - it's all about data manipulation in R. I am trying to write a code that should replace what's currently being done in SAS and SPSS. Or, at least, I am trying to show to my colleagues R is not much worse than SAS/SPSS for the task at hand. I've written a code that works but it's too slow. Probably because it's

On 12/28/2012 04:20 AM, Rafael Espíndola wrote: >> Suppose you have four lifetime operations on the same address in memory, >> with loads and stores all around them: >> >> start1--end1 .. start2--end2 >> >> If you remove start1 then you have a bare pointer, the memory came from >> somewhere and you lose the optimization that loads before start1 become

>> Oh, I was reading "precedes/following" as having static (dominance) >> meaning. That is, in the above example you could not delete the store >> since it is not true that >> llvm.lifetime.end dominates it. >> >> Nick, is this what you had in mind? If not, then we must delete a >> matching llvm.lifetime.end, but it is not clear how we define

Just as an exercise I am tying to add colours to a geom_segment command. I can get one colour but not a sequence of colours. Can anyone suggest how I can get the green lines in the plot below to be different colours? I thought I could use a palatte of colours but that did not seem to work. Thanks ========================================================================= library(ggplot2) xx

Dear useRs, Thanks for any advices # I do not know where are the examples how to mark groups # based on signal occurence in the additional variable: cf. variable c2, # How to calculate different calculations for groups defined by (split by occurence of c2 characteristic data) #First example of simple data #mexample 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Hello, I really wonder how to distinguish between secs and mins in the example below. In other terms, how can I access the time unit in variable d ? start1 <- Sys.time(); stop1 <- Sys.time(); d <- stop1-start1; print(d); v<-unlist(strsplit(as.character(d), split=" ")); print(v) Time difference of 3.024054 secs [1] "3.02405381202698" stop1 <- Sys.time(); d

Hi, My example code is this; x11<-data.frame(A=c(.6,.6,.6),B=c(.20,.20,.20),C=c(0.20,.20,.20)) ggtern(data=x11,aes(A,B,C,xend = c(0.7,.00,0.7),yend = c(.30,.50,.0),zend =c(.0,.50,0.3)))+ geom_point()+ theme_showarrows()+geom_segment(size=.5)+ geom_text_viewport(x=c(.45,.27,.37),y=c(.32,.29,.22),label=as.expression("P_a","P_b","P_c"))

I'm doing a path plot with ggplot2, the result is looking very nice, but I want to give some indication of which direction the lines are going. I thought of using colour gradients, but it doesn't look right. What would be ideal is a line type that indicated direction, something like ">>>>>>>". Is there any way to achieve this? Or can anyone suggest another

D'oh! Thanks for pointing this out. I blame caffeine depletion at the time... -pd > On 4 Dec 2017, at 15:48 , Eik Vettorazzi <E.Vettorazzi at uke.de> wrote: > > reading ?plotmath you might notice that "_" isn't the propper syntax for > subscripts. This will work: > > ggtern(data=x11,aes(A,B,C,xend = c(0.7,.00,0.7),yend = > c(.30,.50,.0),zend

I have just started using ggplot2 and I seem to be doing something stupid in writing ggplot2 commands on more than one line. In the example below the commands on one line are working fine, but as soon as I put them on two lines I get an error. Can any one point out what I am doing wrong? It must be something blindingly simple. Thanks Example

reading ?plotmath you might notice that "_" isn't the propper syntax for subscripts. This will work: ggtern(data=x11,aes(A,B,C,xend = c(0.7,.00,0.7),yend = c(.30,.50,.0),zend =c(.0,.50,0.3)))+ geom_point()+ theme_showarrows()+geom_segment(size=.5)+ geom_text_viewport(x=c(.45,.27,.37),y=c(.32,.29,.22),label=c("P[a]","P[b]","P[c]"), parse=TRUE)

Question 1.. Do the "say number" and "say digits" commands in AGI scritps work? If I use "EXEC SayNumber 123" it works but is I try "say number 123" it doesn't.. I think I have the syntax right becaasue thats how its shown when typing "show agi" on a console and also on the agi pages I have looked at.. Question 2.. Can an AGI script be

Dear list, I've been using plotmeans {gplots} a lot before, and found the "connect" argument to be quite useful. I've moved to ggplot for several reasons, but would still like to connect lines conditionally, somehow. Is it possible to do? Small example: df <- data.frame(cat=LETTERS[1:4],num=rnorm(4)) ggplot(df, aes(x=cat, y=num)) + geom_point() +

Hola, Sí, lo puedes hacer de esta forma... #----------------- set.seed(20) DATOS <- data.frame ( ID = c (1:10) , TIEMPO = sample(1:40, 10, replace=F) , DEF = as.factor(sample(c(0,1), 10, replace=T)) ) library(ggplot2) ggplot( data = DATOS ) + geom_point( aes(x = TIEMPO, y = ID , shape = DEF, color = DEF), size = 5 ) + geom_segment( aes( x = 0, y = ID, xend =