similar to: How to add a double quote to a string

Dear all, I have a multiline plot with each line labeled with a different letter. But I'm not able to make the legend display the same kind of pattern '-a-', instead the letter is overwritten by the line. A simpler legend with only the letter is not very visible and the pt.bg does nothing with letters. Any idea? plot(1:10,10:1,lty=1,type='b', lwd=2,pch='a')

Hi all, I am trying to call the R script/ R Program via my Shell Script. I need to pass a file to the R script as an argument. When I am trying to do that I am getting error and only first line of the file is being read. Can you tell me how should I do that so that it reads the entire file in a R object so that later I am able to process the matrix operation on that file. It is a tab delimited

    I created a couple of timeSeries objects - when I was merging them , I got an error. Looking at the data , I see that one of the time series has   06/30/2007  0.0028       0.0183  0.0122      0.0042  0.0095    -          07/31/2007 -0.0111       0.0255  0.0096     -0.0069 -0.0024  0.0043       08/31/2007 -0.0108      -0.0237 -0.0062     -0.0138 -0.0173 -0.0065       09/30/2007 

I just upgraded my Win7 32bits installation to 2.14.0 after deinstalling 2.12.x First thing I moved the win-library from 2.12 to 2.14 and executed a update.packages(ask='graphics',checkBuilt=TRUE) (Swiss mirror). This aborts with the console message: Error in if (any(diff)) { : missing value where TRUE/FALSE needed And the antivirus (AVG) pops up complaining that colorspace.dll

Hello I have some strange output from R and I try to understand how R works. Could you please help me with that? temp <- rbind (c(10,1),c(99,98)) > temp [,1] [,2] [1,] 10 1 [2,] 99 98 > dist(temp) 1 2 131.6435 > sqrt(dist(temp)) 1 2 11.47360 so far so good. until the nex line: when I try to do what i did before but adding the 1/(what I did

During some experimentation in preparing R lessons I encountered this behaviour which I can not explain fully mat <- matrix(1:16, 4,4) df1 <- data.frame(mat) > mean(df1) X1 X2 X3 X4 2.5 6.5 10.5 14.5 Expected, documented > median(df1) [1] 6.5 10.5 Rather weird, AFAIK there shall not be an issue with data frame at least I did not find any in help page. I tracked it

[sorry, forgot some details...] I'm using optim(param, fun, method='L-BFGS-B', lower=lo, upper=up) to minimize a certain function. Often the minimization ends with the message: ERROR: ABNORMAL_TERMINATION_IN_LNSRCH What is optim() trying to say? What have I to change in my function to make the minimization succeed? Do you think using BBoptim() instead of optim() changes anything?

I'm preparing to install 2.13, so I need a clarification about package location on disk. On window the packages I have installed are under C:\Users\mvalle\R\win-library\2.12 and the ones that come with R are under "C:\Program Files\R\library" If I execute update.packages() from the R gui, it fails because it finds e.g. Matrix as a package that should be updated but the process

On Linux 2.13 by default links cairo, so png() uses it for rendering. One of my users has performance problem, so ask if it possible to temporary disable cairo during png() rendering so it reverts to grGraphics. My workaround for now is to configure R without cairo, but this affects all users that are forced to load and use CairoDevice if they wants better quality. I want to avoid this, is it

Any advice when to use denstity() and when the KernSmooth package bkde() to smooth a histogram? No specific problem to use either one, but I'm curious why there are two so similar implementations. Thanks! mario -- Ing. Mario Valle Data Analysis and Visualization Group | http://www.cscs.ch/~mvalle Swiss National Supercomputing Centre (CSCS) | Tel: +41 (91) 610.82.60 v.

Hi all! I have an ordered vector of values. The distribution of these values can be modeled by a sum of Gaussians. So I'm using the package 'mclust' to get the Gaussians's parameters for this 1D distribution. It works very well, but, for input sizes above 100.000 values it starts taking really forever. Unfortunately my dataset has around 4.6M values... My question: is it

Some time ago someone asked for things that make R difficult to master. Here I want to record one R behavior that took me off-guard yesterday. Moral of the story: vectorialize, but don't exaggerate. Hope it helps mario ### A very simple data frame tc <- textConnection( "prefix idx val A 1 11 A 2 22 B 1 33 B 2 44") s <- read.table(tc,

Dear all, I'm using R 2.12.0 on Windows 7 (32bits) I created a filled contour from the attached data using the following code: load('bug.RData') pdf('bug.pdf', width=14, height=7) filled.contour(o4$x, o4$y, o4$z, color=rainbow, xlim=c(4,18), ylim=c(0,35), zlim=c(14,36)) dev.off() If you look (I used acrobat reader 9.4) at the attached bug.pdf, you see an incomplete grid of

-------------------------------------------------------------------------- Hello all My computer is MacBook and I want to draw a plot in R, for example for x <- c(1,3,6,9,12) y <- c(1.5,2,7,8,15) I use this command plot(x,y). but it dosn't work. Could you please help me? thank you khazaei

Hi, i have a function: zz<- (constrOptim(c(.5,0), fr, grr, ui=rbind(c(-1,0),c(1,-1)), ci=c(-0.9,0.1))) i can get the result by using command (for example): zz$par now if i can use multicore: zz<-parallel(constrOptim(c(.5,0), fr, grr, ui=rbind(c(-1,0),c(1,-1)), ci=c(-0.9,0.1))) result < collect(zz) i cant get my the result: result$par because multicore add process id. for example:

Hello, I'm looking for a R-package to solve such an equation : http://n4.nabble.com/file/n1590462/iv.png which is one current-voltage equation of a solar cell. Thanks in advance, Ptit Bleu. -- View this message in context: http://n4.nabble.com/looking-for-a-package-to-solve-current-voltage-equation-I-f-I-tp1590462p1590462.html Sent from the R help mailing list archive at Nabble.com.

Hi I am having a problem with outputting text inside an if function. I have the following code: for (j in 20:length(bb.up)) { if (up[j]==1 && up[j-1]==0) { sprintf("Long on %s at a price of %f",dates[j],data[j]) } if (down[j]==1 && down[j-1]==0) { sprintf("Short on %s at a price of %f",dates[j],data[j]) } } The loop runs fine and the sprintf

> sprintf("%d",4) [1] "4" > for(i in 1:4) sprintf("%d",4) > for(i in 1:4) print(4) [1] 4 [1] 4 [1] 4 [1] 4 > Why doesn't sprintf like the for loop here

Dear all I have an sprintf inside a loop to track changes in variable's value. This sprintf statement works if I copy and paste it inside R sprintf('I did the the %d,%d \n',k,l) but when this is inside a loop I get no message. listcounter<-1 for (k in c(1:mmax)){ # for (l in c(1:nmax)){ lst[[listcounter]] <-fun(estimatedsr) listcounter<-listcounter+1

Hello, Is there any literature there that says that the EM is better/worse than a Baysian model when it comes to differentiating univariate mixture of normal distributions? -- Thanks, Jim. [[alternative HTML version deleted]]