similar to: reading lisp file in R

Here are the beginning of a R program. I hope it can help you writing the rest of the program. Regards Martin M. S. Pedersen ---- filename <- "university.data" lines <- readLines(filename) first <- T for (ALine in lines) { ALine <- sub("^ +","",ALine) ALine <- sub(")","",ALine, fixed = T) if

The file also has a bunch of email headers stuck in the middle of it: ..... (QUALITY-OF-LIFE SCALE:1-5 4) (ACADEMIC-EMPHASIS HEALTH-SCIENCE) ) ------- ------- >From LEBOWITZ at cs.columbia.edu Mon Feb 22 20:53:02 1988 Received: from zodiac by meridian (5.52/4.7) Received: from Jessica.Stanford.EDU by ads.com (5.58/1.9) id AA04539; Mon, 22 Feb 88 20:59:59 PST Received: from

> On Jan 17, 2018, at 8:22 PM, Ranjan Maitra <maitra at email.com> wrote: > > 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? It's just a text file. What difficulties

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

It seems the file contains records, with each record having 18 fields. I would use awk (standard unix tool), creating an awk script to process the file into a new file with one line for each record, each line with 18 fields, say comma-separated. The csv file can then be easily read into R via the function read.csv. HTH, Eric On Thu, Jan 18, 2018 at 6:22 AM, Ranjan Maitra <maitra at

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

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 ) {

Hi, I am trying to read the "credit.lisp" file of the Japanese credit database in UCI repository, but it is in lisp format which I do not know how to read. I have not found how to do that in the foreign library http://archive.ics.uci.edu/ml/datasets/Japanese+Credit+Screening <http://archive.ics.uci.edu/ml/datasets/Japanese+Credit+Screening> Could anyone help me? Best

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

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

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

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 <maitra at email.com> wrote: > > 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)

Dear friends, I have two questions regarding the use of lattice. First some code: ## begin code z <- cbind(rep(c("BIC", "ICL", "s_v", "Q_v", "sig-q", "s_lsk", "s_lML", "s_mlsk", "s_mlML", "s_la8", "s_haar"), each = 250), rep(c(5, 10, 20, 30, 50), each = 50)) z

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:

Would package "teigen" help? Ranjan On Thu, 29 Jun 2017 14:41:34 +0200 vare vare via R-help <r-help at r-project.org> wrote: > Hello! > > I am new to R (before used python exclusively and would actually call the R solution for this issue inside a python notebook, hope that doesn?t disqualify me right of the batch). > > Right now I am looking for a piece of

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

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

I guess the question that is being asked here is what is the scaling matrix that is being returned in the qda object. The help file on qda() says: ... scaling: for each group ?i?, ?scaling[,,i]? is an array which transforms observations so that within-groups covariance matrix is spherical. ... This is a bit ambiguous. I tried a few cases (spectral, QR decomposition, especially given that it is an