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Hi, I guess that my problem has an obvious answer, but I have not been able to find it. Suppose I create a custom function, consisting of two beta-distributions: myfunction <- function(x) { dbeta(x,2,6) + dbeta(x,6,2) } How can I calculate the quantiles of myfunction? I have not seen any continous function treated in the docs, and applying the "quantile function" gives me an error (since it seems only to be defined on lists and atoms). Thank you in advance, Gerhard

...t find a good way to do it. Hopefully one of you will have something clever to propose. Here is a simplified example: I have a squared matrix: > nom.plac2 <- c("102", "103", "301", "303","304", "403") > poids2 <- matrix(NA, 6,6, dimnames=list(nom.plac2,nom.plac2)) > poids2 102 103 301 303 304 403 102 NA NA NA NA NA NA 103 NA NA NA NA NA NA 301 NA NA NA NA NA NA 303 NA NA NA NA NA NA 304 NA NA NA NA NA NA 403 NA NA NA NA NA NA I want to replace some of the NAs following specifi...

How to show panels for factor levels of conditioning variables which do have no values? E.g. there are panels for "Grand Rapids" when they have values: data( barley ) with( barley, dotplot(variety ~ yield | year * site, layout=c(6,2) ) ) There are no panels for "Grand Rapids" when there are no values for "Grand Rapids": my.barley <- subset( barley, ! ( site == "Grand Rapids" ) ) with( my.barley, dotplot(variety ~ yield | year * site, layout=c(6,2) ) ) But there is a level "Grand Rapi...

...t; argument in panel.arrows is in terms of grid units (which I assume are specific to the particular panel). Can anybody think of a way to set the length so that it appears uniform across all panels? Example code: summTable<-data.frame("X" = seq(1,12), "Y" = c(1,8,3,6,6,5,7,3,8,1,10,-2), "Location" = rep(c("Site1", "Site2"), times = c(6,6)), "Species" = rep(c("A", "B"), 6) ) #change scale for Site 1 summTable[1:6,2]<-100*summTable[1:6,2] #arbitrary confidence intervals summTabl...

...adding another DID and I need to trap it correctly. Thanks. Here are snippets from the relevant files: -- zaptel.conf -- span=1,0,0,esf,b8zs e&m=1-8 loadzone=us defaultzone=us -- extensions.conf -- ; Need an extension to pick up calls from the T1 that uses e&m wink ; This comes in as a 6 instead of 4 full digits ; then pass to the s extension exten => 6,1,Wait(1) exten => 6,2,Goto(incoming,s,1) -- zapata.conf -- [channels] context=incoming signalling=em_w ; rxwink=600 echocancel=yes echotraining=yes group=1 immediate=no channel => 1-8 -------------- next part -----------...

Hi, I have what I think is some kind of linear programming question. Basically, what I want to figure out is if I have a vector of numbers, > x <- rnorm(10) > x [1] -0.44305959 -0.26707077 0.07121266 0.44123714 -1.10323616 -0.19712807 0.20679494 -0.98629992 0.97191659 -0.77561593 > mean(x) [1] -0.2081249 Using each number only once, I want to find the set of five pairs where the magnitude of the differences between the mean(x) and each pairs sum is least. > y <-...

Dear all, I like to extract row names & column names from the named matrix...... like...... a<-matrix(1:6,2) ro<-c("aa","bb") co<-c("dd","ee","ff") dimnames(a)<-list(ro,co) a > dd ee ff aa 1 3 5 bb 2 4 6 from the above matrix "a" I like to extract rownames separately like rownames(a)= (aa,bb) column n...

Readers, I have a data set as follows: 1,1 2,2 3,3 4,4 5,3 6,2 7,-10 8,-9 9,-3 10,2 11,3 12,4 13,5 14,4 15,3 16,2 17,1 I entered this data set using the command 'read.csv'. I want to exclude values fewer than zero in column 2 so then I tried the following command: boxplot.stats(x[>0,2],do.conf=FALSE) Error: syntax error, unexpected GT, expecting...

Hello R gurus, I have to create 12 plots, I have been using the following script, which leaves a large white space between two plot. I would appreciate if someone can suggest an alternative to reduce the white space. par(mar=c(3,3,.5,.5)) split.screen(c(6,2)) # split display into two screens for (i in 1:12) { if (i<11) { screen(i) plot(1:10,xaxt='n', xlab='', ylab='') box() }else{ screen(i) plot(1:10, xlab='', ylab='', cex=0.75) box() } } Thanks Sharad -- View this message in context: http://...

There is a code from SPSS Syntax do if sub(ATVK,2,2)="01". comp strata=1. else if sub(ATVK,2,2)>="05" and sub(ATVK,2,2)<="27". comp strata=3. else if sub(ATVK,4,2)>"20" or sub(ATVK,6,2)>"20". comp strata=4. else if sub(ATVK,4,2)>"00". comp strata=2. end if. value labels strata 1 "R 2 "Li" 3 "M" 4 "La". How can i rewrite it in R code? -- View this message in context: http://r.789695.n4.nabble.com/From-SPSS-Syntax-to...

Hello everyone, I use Bob Muenchen's approach for reading in "in-stream" (to use SAS parlance) delimited data within a script. This works great: mystring <- "id,workshop,gender,q1,q2,q3,q4 1,1,f,1,1,5,1 2,2,f,2,1,4,1 3,1,f,2,2,4,3 4,2, ,3,1, ,3 5,1,m,4,5,2,4 6,2,m,5,4,5,5 7,1,m,5,3,4,4 8,2,m,4,5,5,5" mydata <- read.table( textConnection(mystring), header=TRUE, sep=",", row.names="id", na.strings=" ") closeAllConnections() mydata Can anyone suggest a similar approach for reading in tab-delimited or single sp...

...een event times (or "death times") , since computation only occurs at event times. For instance, removing observations when there is no event at that time in the whole dataset does not change the results: > set.seed(1) > data <- as.data.frame(cbind(start=c(1:5,1:5,1:4),stop=c(2:6,2:6,2:5),status=c(rep( 0,7),1,rep(0,5),1),id=c(rep(1,5),rep(2,5),rep(3,4)),x1=rnorm(14))) > data start stop status id x1 1 1 2 0 1 -0.6264538 2 2 3 0 1 0.1836433 3 3 4 0 1 -0.8356286 4 4 5 0 1 1.5952808 5 5 6 0 1 0.3295078 6 1 2 0 2 -0.8204684 7 2 3 0 2 0.4874291 8 3 4 1 2 0.7383247 9 4 5 0 2 0....

Hi I have just upgraded from R2.6.0 to R2.7.1 (running on Windows) and a part of my code that previously ran ok now gives an error. The following is a simple example to demonstrate my problem. > a <- array(c(1,2,3,4,5,6,rep(NA,6)),dim=c(6,2)) > apply(a,2,sd,na.rm=T) In R2.6.0 this gives (which is what I would like) [1]...

The "no factor combination" case is distinguishable by 'tapply' with simplify=FALSE. > D2 <- data.frame(n = gl(3,4), L = gl(6,2, labels=LETTERS[1:6]), N=3) > D2 <- D2[-c(1,5), ] > DN <- D2; DN[1,"N"] <- NA > with(DN, tapply(N, list(n,L), FUN=sum, simplify=FALSE)) A B C D E F 1 NA 6 NULL NULL NULL NULL 2 NULL NULL 3 6 NULL NULL 3 NULL NULL NULL NULL 6 6 There is...

...e to make a user defined function so that I can input the desired conditions and just get the results accordingly. Below is an arbitrary data set & sample statements with its outputs in accordance with specified conditions: x <- data.frame(ID=rep(letters[1:5],2), A1=rep(10:14,2), A2=rep(2:6,2), A3=c(101:105,95:99), A4=3*ceil(rt(10,2))) x1 <- subset(x, ID == "a" & A1 <= 10 & A2 > 1 & A3 > 100) #condition 1 x2 <- subset(x, ID == "a" & A1 >= 10 & A2 > 1 & A3 < 100) #condition 2 x3 <- subset(x, ID == "a" &...

...I have some data on parasites on apple leaves and want to do a goodness of fit test to a Poisson distribution. This seems to do it: mites <- c(rep(0,70), rep(1,38), rep(2,17), rep(3,10), rep(4,9), rep(5,3), rep(6,2), rep(7,1)) tab <- table(mites) NSU <- length(mites) N <- sum(mites) NSU.Mean <- N / NSU exp <- dpois(as.numeric(dimnames(tab)[[1]]), NSU.Mean) * NSU chi2 <- sum(((as.vector(tab) - exp)^2) / exp) tab exp chi2 I have some small expected value...

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...assume would be easy to fix, but I can't find anything about it... I wish to have people dialing my phone, and if it is busy, they are put into a queue. And then I am dialed back when the previous call is finished, and connected to the waiting caller. Easy enough? ----------exten exten => 6,1,Background(salesq-intro); exten => 6,2,Queue(salesq|tT|||300); ----------queue [salesq] music = default context=default Member => SIP/99559088@future-out ---------- So.. what is the problem... The first caller gets connected to my (mobile) phone. The second and all other caller gets tran...

Dear helpers I was working with kmeans from package mva and found some strange situations. When I run several times the kmeans algorithm with the same dataset I get the same partition. I simulated a little example with 6 observations and run kmeans giving the centers and making just one iteration. I expected that the algorithm just allocated the observations to the nearest center but think this is not the result that I get... Here are the simulated data > dados<-matrix(c(-1,0,2,2.5,7,9,0,3,0,6,1,4),6,2) &...

Dear helpers I'm sorry to insist but I still think there is something wrong with the function kmeans. For instance, let's try the same small example: > dados<-matrix(c(-1,0,2,2.5,7,9,0,3,0,6,1,4),6,2) I will choose observations 3 and 4 for initial centers and just one iteration. The results are > A<-kmeans(dados,dados[c(3,4),],1) > A $cluster [1] 1 1 1 1 2 2 $centers [,1] [,2] 1 0.875 2.75 2 8.000 2.50 $withinss [1] 38.9375 6.5000 $size [1] 4 2 If I do it by hand, aft...