similar to: R and screen (UNIX question)

Folks: Say I have a matrix: test=matrix(c(1,2,3),nrow=10,ncol=3) I would like to have an output character vector where each line is row's values delimited by commas, e.g.: "1,2,3" "2,3,1" "3,1,2" ... "1,2,3" What is the fastest way of doing this? I can paste() row-by-row but this seems an inefficient approach to doing this. Thanks! I'm not

R-helpers: It looks like http://finzi.psych.upenn.edu/search.html has stopped spidering the mailing lists -- this used to be my go-to site for searching for R solutions. Are there any good replacements for this? I want to be able to search both functions and mailing lists at the same time. Cheers! --j -- Jonathan A. Greenberg, PhD Assistant Professor Department of Geography University of

The code base is a bit too complicated to paste in here, but the gist of my question is this: given I have a function myfunction <- function(x) { # Do something A # Do something B # Do something C } Say "#Do something B" returns this error: Error in cat(list(...), file, sep, fill, labels, append) : argument 2 (type 'list') cannot be handled by 'cat' A standard

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) Please: Show me the tree. Mark -------- Original Message -------- Subject: Re: [R] help with rpart From: "Stephen Milborrow" <[1]milbo at sonic.net>

I thought some of you might be amused by this. In my non-work time, I'm an avid weaver and teacher of weaving. I'm working on a project involving creating many detailed weaving patterns, so I wrote R code to automate it. Details here: http://stringpage.com/blog/?p=822 If the overlap between R users and avid tablet weavers turns out to be >> 1, I'll polish it up and turn it

Hi Sarah, Thank you for your help. I tried using CR1<-as.matrix(CR1) but gives error Error in corrplot(CR1, method = "circle") : The matrix is not in [-1, 1]!. I am using a corrplot library. Please find the reproducible example: dput(head(CR1,10)) structure(c(26L, 46L, 39L, 38L, 47L, 59L, 56L, 61L, 43L, 60L, 78L, 63L, 2L, 58L, 8L, 1L, 1L, 9L, 11L, 2L, 1037500L, 46747L, 346300L,

Hi All, I generated 5000 samples using the following script test<- rnorm(5000,1000,100) test1 <- subset(test, subset=(test > 1100)) d <- density(test) plot(d, main="Density of production") abline(v=mean(test1) I wanted to do the following but faced difficulties 1. to shade or color (blue) the curve using the criterion that any

I am trying to replicate the SAS proc univariate in R. I got most of the stats I needed for a by grouping in a data frame using: all1 <- ddply(all,"ACT_NAME", summarise, mean=mean(COUNTS), sd=sd(COUNTS), q25=quantile(COUNTS,.25),median=quantile(COUNTS,.50), q75=quantile(COUNTS,.75), q90=quantile(COUNTS,.90), q95=quantile(COUNTS,.95), q99=quantile(COUNTS,.99) )

Hi, I have a set of x, y points where x represents dates and y actual values. I am trying to plot a line graph of the data with points on top, but R is connecting the wrong points with lines. Does anyone know how I can rectify this. Please see sample below: x= 24/09/2009 09:13 16/10/2009 11:17 24/10/2009 21:43 11/09/2009 18:34 22/08/2009 15:45 10/08/2009 00:30 14/08/2009 14:52 24/09/2009

Hi just a quick question in which I must use the "function and return" part of R. If I was asked to define a function called "median" where the function has a single argument x and returns the median value of a vector. How would I go about saying the median = element x in terms of n (say (n + 1)/2 or whatever it would be) if there is an even amount of elements, or x in terms

That does clarify for me that you're missing a step: I didn't clearly follow your description at first. corrplot expects a correlation matrix, not your original data. You need to use cor() first. That's pretty clear in the documentation. See for instance the examples: data(mtcars) M <- cor(mtcars) corrplot(M) Sarah On Sat, Mar 17, 2018 at 12:00 PM Shivi Bhatia <shivipmp82 at

Hi, Can R read in .xyz files? If so, what is the package, thanks -- Shane [[alternative HTML version deleted]]

I need the output to have groups and the probability any given record in that group then has of being in the response class. Just like my email in the beginning i need the output that looks like if A and if B and if C then %77 it will be D. The examples you provided are just simply not similar. They are different and would take interpretation to get what i need. On Apr 14, 2016 1:26 AM,

Hi. Can you please answer to my questions about R ? 1.how can I write command for vector ? for exaple in this sample : I have this : a1 <- c (1:10) now how can I put in the vector ? bye for now, Thanks a lot. Majid. [[alternative HTML version deleted]]

I have a column of 1000 datapoints from the normal distribution with mean 2 and variance 4. How can I get a histogram of these observations with 20 bins with each bin having 50 observations? -- Thanks, Jim. [[alternative HTML version deleted]]

Dear Users! I encountered with some problem in data reading while I challenged R (and me too) in a validation point of view. In this issue, I tried to utilize some reference datasets ( http://www.itl.nist.gov/div898/strd/index.html). And the result departed a bit from my expectations. This dataset dedicated to challenge cancellation and accumulation errors (case SmLs07), that's why this

Dear list, I was curious how to subtract a vector from matrix? Say, I have mat <- matrix(1:40, nrow=20, ncol=2) x <-c(1,2) I want, x-mat[1,] and x-mat[2,], and so on... Basically, subtract column elements of x against column elements in mat. But x-mat won't do it. Thanks, Mike [[alternative HTML version deleted]]

? stato filtrato un testo allegato il cui set di caratteri non era indicato... Nome: non disponibile URL: <https://stat.ethz.ch/pipermail/r-help/attachments/20130624/2836bcf8/attachment.pl>

Sir, I would like to create a combined line-bar plot, with the line up top and bar plot below, but with the x-axes suppressed on the topmost plot, the line and bar plot areas joined by a common line, and the x-axes (of dates) joined to the bottom part of the bar plot. i have been able to format the area using the layout(matrix( ... commands, but cannot figure out the final step. here's the

Hi, Can any one please explain why the following code doesn't work? Or can anyone suggest an alternative. Suppose       x<-c(23,67,2,87,9,63,8,2,35,6,91,41,22,3)        mat<-0;        for(j in 1:length(x))        {              for(i in 1:p)                      mat[i,j]<-x[j]^i;        }    Actually I want to have a matrix with p columns such that each column will have the