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I will very appreciate if anyone can provide some materials to draw a simplex plot of a Dirichlet distribution in R as shown in the page at http://en.wikipedia.org/wiki/File:Dirichlet_distributions.png .

I have some trouble to deal the value of 'NaN'. For example, > exp(1e3) [1] Inf > exp(1e3)*0 [1] NaN The correct answer should be 0 rather than NaN. I will very appreciate if anyone can share some technique...

Oh... I see!!! I very appreciate for your clear explanation! Thank you, Uwe Ligges. -- View this message in context: http://r.789695.n4.nabble.com/General-question-on-the-message-with-non-zero-exit-status-tp4551438p4568839.html Sent from the...

Thank you, David. I found some good stuffs for which I have been looking. On Sun, 2012-08-26 at 09:01 -0700, David Winsemius wrote: > On Aug 25, 2012, at 9:00 PM, Chel Hee Lee wrote: > > > I will...

I have a general question how to understand the message with "non-zero exit status" when new package is installed. Based on my experience, this message implies a package dependence. Am I correct to understand this message? Are there anyone...

> x <- c("hola mundo mundo"); > table(unlist(strsplit(x, " "))) hola mundo 1 2 > Is this what you are looking for? I hope this helps. Chel Hee Lee On 1/22/2015 8:25 AM, bgnumis bgnum wrote: > Hi all, > > I...

Your question is not clear to me. > x$alpha[1:2] [1] "a" "b" > x$alpha[2] [1] "b" > Is this what you are looking for? I hope this helps. Chel Hee Lee On 12/5/2014 11:12 AM...

> y = matrix(cbind(c(0, 0.5, 1),c(0, 0.5, 1)),ncol=2) > z = matrix(c(12, -6),ncol=2) > t(apply(y, 1, function(x) x*z)) [,1] [,2] [1,] 0 0 [2,] 6 -3 [3,] 12...

> paste(rep(v1, each=3), 1:3, sep="_") [1] "a_1" "a_2" "a_3" "b_1" "b_2" "b_3" "c_1" "c_2" "c_3" > Is this what you are looking for? I hope this helps. Chel Hee...

Could you also kindly check the following links in the rw-FAQ manual at <http://cran.r-project.org/bin/windows/base/rw-FAQ.html>?? The links list in the below seem to be broken. I hope these links are...

Or, you can also do the same job using 'colSums()' as shown in the below: > colSums(status=="I") 2010 2011 2012 3 4 4 > I hope this helps. Chel Hee Lee On 1/28/2015 7:31 PM, JSHuang wrote...

> do.call(rbind, lapply(split(data, data$Name), function(x) x[order(x$CheckInDate),][nrow(x),])) Name CheckInDate Temp John John 2014-04-01 99.0 Mary Mary 2014-03-01 98.1 Sam Sam 2014-04-01 97.5...

> lapply(1:2, function(x) t(A[rev(1:3),,x])) [[1]] [,1] [,2] [,3] [1,] "g" "d" "a" [2,] "h" "e" "b" [3,] "i" "f" "c" [[2]] [,1] [,2] [,3] [1,] "p" "m" "j" [2,] "q" "n" "k" [3...

> ex <- strptime(c("2014-01-30 15:39:46", "2012-04-20 14:49:02"), format="%Y-%m-%d %H:%M:%S") > ex [1] "2014-01-30 15:39:46 CST" "2012-04-20 14:49:02 CST" > format(ex...

Could you kindly check if the following links are working fine in the R-FAQ page at <http://cran.r-project.org/doc/FAQ/R-FAQ.html>? The links listed in the below seem to be broken. I hope these...

This approach may not be fancy as what you are looking for. > xl <- unlist(x) > xl[grep("A", names(xl))] f1.x1.A f1.x2.A f2.x3.A f2.x4.A 11 12 13 14 > I hope this helps...

Could you kindly test the following codes? It is because I found strange answer when 'aggregate()' is used with a formula. I am trying to count how many missing data entries are in each group. For this exercise, I created...

Here is another implementation: > b [1] 1 2 3 4 5 > c [1] 1 2 1 3 5 4 > outer(c,b, "==")*1 [,1] [,2] [,3] [,4] [,5] [1,] 1 0 0 0 0 [2,] 0 1 0 0 0...

The following example may give you an idea regarding your question. Please see what happens by typing the codes > x <- seq(from=-5, to=5, by=1) > y1 <- 0 + 0.5*x > y2 <- 0 - 0.5*x > > plot(x,y1...

> LDL <- NA_real_ > is.LDL <- is.na > v <- c(0.2, 0.28, LDL, 0.9) > v [1] 0.20 0.28 NA 0.90 > is.LDL(v) [1] FALSE FALSE TRUE FALSE > Hope this helps. Chel Hee Lee On...