similar to: Vectors out of lists?

Hello there, any ideas on how to save all the objects on my workspace inside a list object? For example, say my workspace is as follows ls() [1] "x" "y" "z" and suppose I want to put these objects inside a list object, say object.list <- list() without having to explicitly write down their names as in object.list$x = x object.list$y = y object.list$z = z Is

Hi! I was wondering if there are any other functions for numerical integration, besides 'integrate' from the stats package, but which wouldn't require the integrand to be vectorized. Oh, and must be capable of integrating over (-inf,+inf). Thanks in advance, Eduardo Horta [[alternative HTML version deleted]]

Hello, Are there any packages which allow for a good integration between R and LaTex / LyX? I'm interested mainly in automatic (automagic?) imports of plots/graphics. Thanks in advance and best regards, Eduardo de Oliveira Horta [[alternative HTML version deleted]]

Hello! I have a csv file of intra-day financial data (5-min closing prices) that looks like this: (obs - the dates are formated as day/month/year, as is usual here in Brazil) Date;Time;Close 01/09/2009;10:00;56567 01/09/2009;10:05;56463 01/09/2009;10:10;56370 ##(goes on all day) 01/09/2009;16:45;55771 01/09/2009;16:50;55823 01/09/2009;16:55;55814 ##(jumps to the subsequent day)

Hello, this is probably a recurrent question, but I couldn't find any answers that didn't involve the expression "data frame"... so perhaps I'm looking for something new here. I wanted to find a code equivalent to > x=sqrt(1:10) > y=log(1:10) > plot(1:10, x, type="lines", col="darkgreen") > lines(1:10, y, col="red") to use with

Hello there I wish to get the "coordinates" of the minimum element of an array. For example, if the array were > H = array(c(8:5,1:4),dim=c(2,2,2)) > H , , 1 [,1] [,2] [1,] 8 6 [2,] 7 5 , , 2 [,1] [,2] [1,] 1 3 [2,] 2 4 then > min(H) [1] 1 and > max(H) [1] 8 Say "idx" were the function I'm looking for. Then, what

Hello, say I have an object > x = list(first = 10, second = 20, third = "Yeah!") and want to remove the $second element of that list. My first try was, of course, > rm(x$second) which gave me the following error message > Error in rm(x$second) : ... must contain names or character strings Any ideas here, folks? Best regards, and Merry Christmas! Eduardo Horta

Hello, I'm willing to plot a sequence of densities on a 3d graph, something like ----------------------------------------------------------------- x <- sapply(1:10, function(i)rnorm(1000)) f <- sapply(1:10, function(i)density(x[,i], from=-5,to=5)$y) grid <- density(x[,1], from=-5,to=5)$x win.graph() persp(grid1, 1:10, f,theta=-50, phi=30, d=2)

Hello, I was trying to define a set of functions inside a loop, with the loop index working as a parameter for each function. Below I post a simpler example, as to illustrate what I was intending: f<-list() for (i in 1:10){ f[[i]]<-function(t){ f[[i]]<-t^2+i } } rm(i) With that, I was expecting that f[[1]] would be a function defined by t^2+1, f[[2]] by t^2+2 and so on.

Hi there, yet on the topic of greek letters and pdf plotting: when I run the following code pdf(file="temp.pdf") mu=seq(from=-pi, to=pi, length=100) plot(mu, sin(mu^2), type="l", xlab=expression(mu%in%(list(-pi,pi))), ylab=expression(sin(mu^2)), main=expression((list(mu,sin(mu^2))))) dev.off() I get a "proportional to" symbol in place of a

Hello, I was trying to build some functions which I would like to integrate over an interval using the function 'integrate' from the 'stats' package. As an example, please consider the function h(u)=sin(pi*u) + sqrt(2)*sin(pi*2*u) + sqrt(3)*sin(pi*3*u) + 2*sin(pi*4*u) Two alternative ways to 'build' this function are as in f and g below: coeff<-sqrt(1:4)

Is there a way to achieve lbl=c("a", "b", "c", "d") par(mfrow=c(2,2), ann=FALSE) for (t in 1:4){ plot(seq(from=1,to=2*pi,length=100), sin(t*seq(from=1,to=2*pi,length=100)), type="l") title(main=paste("(", lbl[t], ")", sep="")) } without having to use an object like 'lbl'? More generally: is it possible

Hello, I've googled for a while and couldn't find anything on this topic: say I have a matrix A and want to build matrices B1, B2,... using blocks from A (or equivalently an array B with B[,,i] being a block from A), and that I must sum the B[,,i]'s. I've come up with this rather non-elegant code: > n = 6 > p = 3 > > A <- matrix(1:(n^2), n, n, byrow=TRUE) >

Hello, I want to save a pdf plot using Cairo, but the canvas of the saved file seems too large when compared to the actual plotted area. Is there a way to control the relation between the canvas size and the size of actual plotting area? Thanks in advance, and best regards, Eduardo Horta [[alternative HTML version deleted]]

Please excuse me for asking such basic questions: Here is my code > y=c(1.21,0.51,0.14,1.62,-0.8,0.72,-1.71,0.84,0.02,-0.12) > ybar=mean(y) > ll=length(y); > ybarv=rep(ybar,ll) > x=1:ll > plot(x,ybarv,pch=1) > segments(x[1],ybar,x[ll],ybar) What I get is a collection of small circles, with a segment "on top" of the circles, which is almost what I want. But I

Dear forum, Does anyone know any functions for plotting vertical error bars, if the x-axis variable is a date? I can produce a pretty plot with dates on the x-axis, if I don't want to include error bars. I can also produce a nice plot with error bars, if I just use Julian days for the x-axis, but I'd much rather have the actual dates. I have tried writing my own function, plotCI

#See David Williams' book "Weighing the odds", p286 y <- c(1.21, 0.51, 0.14, 1.62, -0.8, 0.72, -1.71, 0.84, 0.02, -0.12) ybar <- mean(y) ylength <- length(y) ybarv <- rep(ybar, ylength) x <- 1:ylength plot(x,y,asp=1,xlab="position",ylab="ybar",type="n",ylim=c(-1,1)) segments(x[1], ybar, x[ylength], ybar) segments(x,ybarv,x,y)

Hi im trying to add error bars to my barplots, there very basic, i have a few grapghs where the y variable is different but on all the X variable is Age (Adult and Juvenile) however this is split into two levels so i have males and females, so my graph basically has four bars on it. I know how to add eror bars for instance when there is only one level eg lookng at the diffrence between male and

Hello: If anyone could guide me with this I would greatly appreciate it. Thanking you in advance for your assistance. Using a 3-level input factor alternative so that a function(below) can compute both a two-sided and one-sided p-values. Making the two-sided test the default. And produce output information about which alternative was tested. Where would I place the ifelse statement?

Dear friends. On p 95 in 3. ed. MASS a zero-truncated Poisson distribution is analyzed. I understand the probability distribution and expected mean. The Newton iteration scheme is Lam(m+1)=Lam(m)-[Lam(m)-Ybar(1-exp(-Lam(m))]/[1-Ybar*exp(-Lam(m)], and I suppose the latter part should be f(Lam(m))/f ' (Lam(m)) and f(Lam(m)) is Lam(m)/(1-exp(-Lam(m)), right ? But then f ' (Lam(m)) is