similar to: buglet in curve?

Hi R collective, I quite like the "curve" function because you can chuck a R function into it, and see the graph in one line of R. I had a google and found some threads for filling under the line; http://tolstoy.newcastle.edu.au/R/e2/help/07/09/25457.html However they seem to miss the point of the simplicity of going, "I wonder what that looks like, and can I have some colour

Hello, everybody I run the following program, and depending on the size of eps I get different results. With eps=1e-05, the program calculates wrong values for x=65:67 and others. The program runs fine with eps=1e-07. Why is this so? Also, I am using area() instead of integrate() because I cannot make integrate to work, especially with imaginary numbers. Maybe someone can show me how to use

Hi, I've got a matrix, Z, of values representing (as it happens) optical power at each pixel location. Since I know in advance I've got a single, convex peak, I would like to do a 2D parabolic fit of the form Z = poly((x+y),2) where x and y are the x,y coordinates of each pixel (or equivalently, the row, column numbers). Is there an R function that lets me easily implement that?

Hi, Suppose I have a multivariate response Y (n x k) obtained at a set of predictors X (n x p). I would like to perform a linear regression taking into consideration the covariance structure of Y within each unit - this would be represented by a specified matrix V (k x k), assumed to be the same across units. How do I use "lm" to do this? One approach that I was thinking of

Hello List! I asked this before (with no solution), but maybe this time... I'm trying to project a surface to the XY under a 3d cloud using lattice. I can project contour lines following the code for fig 13.7 in Deepayan Sarkar's "Lattice, Multivariate Data Visualization with R", but it fails when I try to "color them in" using panel.levelplot. ?utilities.3d says there

Hello R Users, How to make a variable equidistance with time i.e. how to interpolate a variable if it is not sampled at equal time interval. Many thanks, Regards, Yogesh [[alternative HTML version deleted]]

After a bunch of discussion in the core group, we have decided to deprecate the delay() function (which was introduced as "experimental" in R 0.50). This is the function that duplicates in R code the delayed evaluation mechanism (the promise) that's used in evaluating function arguments. The problem with delay() was that it was handled inconsistently (e.g. sometimes you would see

Hi, I am a biologist (relatively new to R) analyzing data which we predict to fit a power function. I was wondering if anyone knew a way to model piecewise functions in R, where across a range of values (0-x) the data is modeled as a power function, and across another range (x-inf) it is a linear function. This would be predicted by one of our hypotheses, and we would like to find the AICs

I need to make a fairly complex animated graphic and decided to use grid for it. A very simple example of what I need: ##============================================================================== library(grid) grid.newpage() pushViewport(plotViewport()) pushViewport(viewport(xscale = extendrange(c(0, 100)), yscale = extendrange(c(0, 100)))) grid.xaxis() grid.yaxis()

Hello dear developers, I find myself often having the result of "range" oder "extendrange", which I want to create a sequence with. But "seq" needs two seperate arguments "from" and "two". Could an argument "range" be added? Otherwise I will have to create an object with the range (may come from a longer calculation), index twice from

Hi all, Running R 0.90.1 on a RH 6.1 system. Installation of the integrate_2.1-2 package went smoothly. My code contains a loop in which integrate() is called several times in each pass. I get a segmentation fault after what seems to be a random number of calls to integrate(). Debug output shows: Program received signal SIGSEGV, Segmentation fault. promiseArgs (el=0x40276414,

Dear R experts, I am plotting the population of students who live in a city, and in successive circular bands made of the contiguous districts that surround the city. This is a stylized figure, where I specify the area of each successive circle based on the cumulative population of students. I want to compare two sets of concentric circles across different populations - such as 'All

Hello, ?delayedAssign presents substitute() as a way to look at the expression in the promise. However, msg <- "old" delayedAssign("x", msg) msg <- "new!" x #- new! substitute(x) #- x (was 'msg' ?) Here, we just got 'x'... shouldn't we got 'msg'? Same result when the promise is not evaluated yet: delayedAssign("x",

I'm not sure if this is a known peculiarity or a bug, but I stumbled across what I think is very odd behavior from delayedAssign. In the below example x switches values the first two times it is evaluated. > delayedAssign("x", {x <- 2; x+3}) > x==x [1] FALSE > delayedAssign("x", {x <- 2; x+3}) > x [1] 5 > x [1] 2 The ?delayedAssign documentation says

Duncan, Thank you for the clarification on how delayedAssign works. Should R-level interfaces to promise objects ever become available, I expect they would at time come in handy. On the subject of substitute and delayedAssign, I do have a follow-up question for the list. I'm trying to convert a named list of expression objects into an environment of promise objects. After conversion, each

The help files for delayedAssign and substitute both say that substitute() can be used to see the expression associated with a promise. However, I can't see how to do that. When I try the example in help file for delayedAssign I don't see substitute() extracting the promise, e.g.: > msg <- "old" > delayedAssign("x", msg) > msg <-

1. Is there some way to copy a promise so that the copy has the same expression in its promise as the original. In the following we y is a promise that we want to copy to z. We want z to be a promise based on the expression x since y is a promise based on the expression x. Thus the answer to the code below is desired to be z=2 but its 1, 1 and y in the next three examples so they are not the

>>>>> Henrik Bengtsson <hb at biostat.ucsf.edu> >>>>> on Mon, 26 Jan 2015 12:41:48 -0800 writes: > On Mon, Jan 26, 2015 at 12:24 PM, Hadley Wickham <h.wickham at gmail.com> wrote: >> If it was any other environment than the global, you could use substitute: >> >> e <- new.env() >>

Hi all, The help for delayedAssign suggests that you can use substitute to access the expression associated with a promise, and the help for substitute says: "If it is a promise object, i.e., a formal argument to a function or explicitly created using ?delayedAssign()?, the expression slot of the promise replaces the symbol. But this doesn't seem to work: > a <- 1 > b <- 2

Hi everyone, newbee query! I've installed R 2.8.0 and tried to run this simple glm - x is no of cars in a given year, y is the number voted in an election that year while n is the population 18+: votes <- data.frame(x = c(0.62,0.77,0.71,0.74,0.77,0.86,1.13,1.44), + y=c(502,542,711,653,771,806,934,1123), n=