similar to: example of geom_contour() with function argument

Hello Ulrik, I apologize, but I can not see how to provide a pdf in place of the density function which calculates a KDE (that is, something from the dataset in the example). Can you please point to the specific example that might help? Here is what I get: require(mvtnorm) require(ggplot2) set.seed(1234) xx <- data.frame(rmvt(100, df = c(13, 13))) v <- ggplot(faithfuld, aes(waiting,

Hi BFD, ?geom_contour() *does* have helpful examples. Your Google-foo is weak: Searching for geom_contour brought me: http://ggplot2.tidyverse.org/reference/geom_contour.html as the first result. HTH Ulrik On Mon, 9 Oct 2017 at 08:04 Big Floppy Dog <bigfloppydog at gmail.com> wrote: > Can someone please point me to an example with geom_contour() that uses a > function? The help

Hi, This is not a HW problem, sadly: I was last in a classroom 30 years ago, and can no longer run off to the instructor :-( I apologize but I cut and paste the wrong snippet earlier and made a typo in doing so, but the result is the same with the more appropriate snippet. require(mvtnorm) require(ggplot2) set.seed(1234) xx <- data.frame(rmvt(100, df = c(13, 13))) v <- ggplot(data = xx,

library(mvtnorm) # you were misusing "require"... only use require if you plan to library(ggplot2) # test the return value and fail gracefully when the package is missing set.seed( 1234 ) xx <- data.frame( rmvt( 100, df = c( 13, 13 ) ) ) xx2 <- expand.grid( X1 = seq( -5, 5, 0.1 ) # all combinations... could be used to fill a matrix , X2 = seq( -5, 5, 0.1 )

> On Oct 9, 2017, at 6:03 AM, Big Floppy Dog <bigfloppydog at gmail.com> wrote: > > Hello Ulrik, > > I apologize, but I can not see how to provide a pdf in place of the density > function which calculates a KDE (that is, something from the dataset in the > example). Can you please point to the specific example that might help? > > Here is what I get: > >

Note: I have posted this on SO also but while the question has been upvoted, there has been no answer yet. https://stackoverflow.com/questions/46622243/ggplot-plot-2d-probability-density-function-on-top-of-points-on-ggplot Apologies for those who have seen it there also but I thought that this list of experts may have someone who knows the answer. I have the following example code:

Hi, I am no expert on ggplot2 and I do not know the answer to your question. I looked around a bit but could not find an answer right away. But one possibility could be, if a direct approach is not possible, to draw ellipses corresponding to the confidence regions of the multivariate t density and use geom_polygon to draw this successively? I will wait for a couple of days to see if there is a

Dear all, I have a question with regard to the use of hclust. I would like to be able to specify my own distance matrix instead of asking R to compute the distance matrix for me. It is computationally easier for me this way. My question is: How can I get hclust to accept this? Thanks, Ranjan -- *************************************************************************** Ranjan

Thanks! Yes, however, this seems a bit wasteful. Just wondering if there are other, more efficient options possible. Best wishes, Ranjan On Thu, 29 Mar 2018 22:20:19 -0400 Boris Steipe <boris.steipe at utoronto.ca> wrote: > If one is equal to the reverse of another, keep only one of the pair. > > B. > > > > > On Mar 29, 2018, at 9:48 PM, Ranjan Maitra

Dear friends, Is there a way to read data files written in lisp into R? Here is the file: https://archive.ics.uci.edu/ml/machine-learning-databases/university/university.data I would like to read it into R. Any suggestions? Thanks very much in advance for pointers on this and best wishes, Ranjan -- Important Notice: This mailbox is ignored: e-mails are set to be deleted on receipt. Please

Dear friends, I would like to get all possible arrangements of n objects listed 1:n on a circle. Now this is easy to do in R. Keep the last spot fixed at n and fill in the rest using permuations(n-1, n-1) from the gtools package. However, what if clockwise or counterclockwise arrangements are the same? I know that half of the above (n - 1)! arrangements are redundant. Is there an easy way to

Dear friends, Sorry for this somewhat generically titled posting but I had a question with using contrasts in a manova context. So here is my question: Suppose I am interested in doing inference on \beta in the case of the model given by: Y = X %*% \beta + e where Y is a n x p matrix of observations, X is a n x m design matrix, \beta is m x p matrix of parameters, and e is a

Jeff, I wanted to let you know that your function is faster than generating the directional circular permutations and weeding. Here is the time for n = 10. I compared with just doing the permutations, there is no point in proceeding further with the weeding since it is slower at the start itself. system.time(directionless_circular_permutations(10)) user system elapsed 1.576 0.000

I don't know if this is more efficient than enumerating with distinct directions and weeding... it seems kind of heavyweight to me: ####### library(gtools) directionless_circular_permutations <- function( n ) { v <- seq.int( n-1 ) ix <- combinations( n-1, 2 ) jx <- permutations( n-3, n-3 ) x <- lapply( seq.int( nrow( ix ) ) , function( i ) {

Dear list, I have been using the Rmath library for quite a while: in the current instance, I am calling dnt (non-central t density function) repeatedly for several million. When the argument is small, I get the warning message: full precision was not achieved in 'pnt' which is nothing unexpected. (The density calls pnt, if you look at the function dnt.) However, to have this happen a

Dear colleagues, I have an image which I can display in the greyscale using image. On this image, for some pixels, which I know, I want to display their activity based on a third measure. One way to do that would be to color these differently, and use an opacity measure to display the third measure. An example of what I am trying to do is at:

Dear colleagues, I had a question wrt the car package. How do I evaluate whether a simpler multivariate regression model is adequate? For instance, I do the following: ami <- read.table(file = "http://www.public.iastate.edu/~maitra/stat501/datasets/amitriptyline.dat", col.names=c("TCAD", "drug", "gender", "antidepressant","PR",

Thanks! I am trying to use it in R. (Actually, I try to give my students experiences with different kinds of files and I was wondering if there were tools available for such kinds of files. I don't know Lisp so I do not actually know what the lines towards the bottom of the file mean.( Many thanks for your response! Best wishes, Ranjan On Wed, 17 Jan 2018 20:59:48 -0800 David Winsemius

New function below is a bit faster due to more efficent memory handling. for-loop FTW! directionless_circular_permutations2 <- function( n ) { n1 <- n - 1L v <- seq.int( n1 ) ix <- combinations( n1, 2L ) jx <- permutations( n-3L, n-3L ) jxrows <- nrow( jx ) jxoffsets <- seq.int( jxrows ) result <- matrix( n, nrow = factorial( n1 )/2L, ncol = n ) k

Hello, I have been wondering of an efficient way to do this: I have an n x m x p array Z and a p x n matrix Y. I want to multiply each of the n matrices with the corresponding column vector of Y. In other words, I am wanting to matrix multiply: Z[i, ,] %*% Y[, i] which will give me a (two-dimensional) array or matrix of dimension n x p with the i'th row storing the above. Any pointers