similar to: Turn any vector

Dear R-users, I'd like to fit a sine function to my data. The result should have a format (and thus the formula, too) y ~ a + sin(x+b) where y and x are vectors, and a and b are (yet) unknown values. The data sets (vectors x and y) are OK, and I can do a simple lm fitting lm(y~x), or lm(y~I(sin(2*pi*x/360))), succesfully My issue is that I'm not able to do the optional linear shift in

Dear R-users, I'm tackling with a problem which causing me a head-ache for a long time. I would like to create a nice x-axis to my plots, but I do not know how to implement the method. Imagine a matrix where you have rownames real numbers -- these rownames should be written in the x-axis nicely. I could not find any way how. I'll describe what I'm doing now: I have a matrix where

Solved. I tried to create a minimal example, but my script is too complicated to just cut out lines (too many dependencies). Instead of it, I sketch my idea. I do not say it is the nicest solution, but it works as I wanted. The first five rows of my data-matrix: > data_svd$u[1:5,] [,1] [,2] [,3] [,4] [,5] [,6] [,7] 25.12333 -0.05845695

Hi, Given a simple example, test <- matrix(c(0.1, 0.2, 0.1, 0.2, 0.1, 0.1, 0.3, 0.1, 0.1), 3, 3) How to generate row indexes for which their corresponding row values are less than or equal to 0.2 ? For this example, row 2 and 3 are the correct ones. Thanks [[alternative HTML version deleted]]

Hello I am a relatively new user of R and am struggling to use the 'network' package. I have a correlation matrix (produced using 'cor'), and want to draw a network where each item showing correlation above a threshold (say 0.5) is joined by a green line, and each item showing correlation below a threshold (say -0.5) is joined by a red line. Does anyone have any hints of how to

useR's, Consider: y <- c(20, 25, 30) > m <- matrix(c(0.0,1,NA,0.5,1.25,0.75, 0.5, NA, > NA),byrow=TRUE,nrow=3,ncol=3) > m [,1] [,2] [,3] [1,] 0.0 1.00 NA [2,] 0.5 1.25 0.75 [3,] 0.5 NA NA For each numeric value, I want to replace them with their corresponding y-value. The result should look like (here, each row represents a variable rather than the columns):

Dear R users, I plot data with points(my_data[x]~x, col = x, type = "o", lwd="4") where x is an integer running from 1 to 10, I get points drawn at the plot. When want to do a legend to this I try legend(leg.txt[x], col = x, text.col = 1, pch = 1, bty = "n") where leg.txt contains the names of the variable and x behaves the same. The difference between the

Dear List, i know there are some solutions for this in the archive, but they're not very good for numeric matrices, since they usually convert rows/columns to character strings. Is there an easy way to do $subject for numeric matrices properly, or i need to do it by hand? Thanks, Gabor

Hi, I'm looking for a manual (we based or pdf) which would explain in detail with graphical examples what all the option can do in plot and par. Does anybody now anything like this? A couple of manuals to R I went through do have plot parameters mentioned, but sometimes it is hard to understand or imagine what the parameter can do (the same with the related help page) e.g. I met "From

Greetings, R-ians: I would like to know which version or R was used to create a given package. I think I remember seeing that topic discussed recently but cannot find it among my notes. Can anyone tell me how to determine which version of R created a package? Thanks. Charles Annis, P.E. <mailto:Charles.Annis@StatisticalEngineering.com> Charles.Annis@StatisticalEngineering.com

Dear Contributors: I have the next vector: "Z" 526 723 110 1110 34 778 614 249 14 I want to generate a vector containing the ratios of all the values versus all the values of the z vector. I mean a vector containing the values of 526/723, 526/110, and so on, 723/723, 723/110, and so on, and so on. Is this doable in a simple way?? Thanks in advance again, Juan Pablo Fededa

Hi Consider the following code > x <- rep(1:13, 13) > y <- 1:3 I want to select all elements in x which are equal to 1, 2 or 3. I know that I could use > sel <- x==y[1] | x==y[2] | x==y[3] > x[sel] to obtain the values, but in my analysis, the y-vector is thousands of elements long. Is there any way, that I can do that easily? Thanks Rainer -- Rainer M. Krug, Dipl.

Hallo, I am running R-2.6. on Windows. I have a code which uses library(convert). Can anyone tell me which package I need to install to run this code. Everytime I receive the error message library (convert) not found. Thanks, Corinna

Dear all, "Writing R Extensions" explicitly says that A method must have all the arguments of the generic, including ... if the generic does. A method must have arguments in exactly the same order as the generic. If the generic specifies defaults, all methods should use the same defaults. This is clear. R CMD check even checks for this. But then how is it possible that

Dear colleagues, I'm trying to have a look at the Assortative and Disassortative ( http://en.wikipedia.org/wiki/Assortative_mixing) of the network I have. But it seems that the igraph hasn't mentioned that yet. I have to get the in/out degree of the vertices of each edge and calculate the Pearson's Correlation coefficient which seems to be quite a huge task for me. :( So I wonder if

Dear everybody! I am making a graph in R and employ pstoedit to expot the .pdf-output to .svg. When I open the .svg with firefox I get the .svg-code shown wit the following header: "Mit dieser XML-Datei sind anscheinend keine Style-Informationen verkn?pft. Nachfolgend wird die Baum-Ansicht des Dokuments angezeigt." Which information should how be included? Thank you in advance. Yours,

Hi there! I use an gnome-terminal for using R. When I resize the termial to the maximum size, R uses only the left side of the window. Can I tell R to use the whole window somehow? Thanks, Martin -- Ihr Partner f?r Webdesign, Webapplikationen und Webspace. http://www.roomandspace.com/ Martin Kaffanke +43 650 4514224 -------------- next part -------------- A non-text attachment was

I have a folder full of pngs and jpgs, and would like to consolidate them into a pdf with appropriate title and labels. Can this be done via R ? _________________________________________________________________ Easily publish your photos to your Spaces with Photo Gallery. [[alternative HTML version deleted]]

Dear R-users, in a minimal example exists() gives FALSE on an object which obviously does exist. How can I check on that list object anyway else, please? > SmoothData <- list(exists=TRUE, span=0.001) > SmoothData $exists [1] TRUE $span [1] 0.001 > exists("SmoothData") TRUE > exists("SmoothData$span") FALSE > exists("SmoothData[[2]]") FALSE

Dear All, I would like to do run R without having write permissions to any directory on the system. It seems that I need to modify the R source code for this, to make R start without creating a temporary directory. So far, so good. But should I expect any more complications? Does R really need the temporary directory that much? Thanks, Gabor -- Gabor Csardi <Gabor.Csardi at unil.ch>