similar to: graphics layout

Folks, I'm trying to get stats from a matrix for each transition from one state to another. I have a matrix x as below. structure(c(0, 2, 2, 2, 0, 0, 0, 1, 1, 1, 1, 2, 2, 1, 1, 1, 0, 0, 2, 2, 0.21, -0.57, -0.59, 0.16, -1.62, 0.18, -0.81, -0.19, -0.76, 0.74, -1.51, 2.79, 0.41, 1.63, -0.86, -0.81, 0.39, -1.38, 0.06, 0.84, 0.51, -1, -1.29, 2.15, 0.39, 0.78, 0.85, 1.18, 1.66, 0.9, -0.94,

# polygon ignores requests to have its border transparent, look at par(bg="transparent") plot(c(0, 3), 0:1) polygon(c(0, 1, 1, 0), c(0, 0, 1, 1), border=NA, col = 0) polygon(c(1, 2, 2, 1), c(0, 0, 1, 1), border="transparent", col = 0) polygon(c(2, 3, 3, 2), c(0, 0, 1, 1), border=0, col = 0) # a quick fix for erase.screen() is the following erase.screen <- function (n =

Hi folks, Not sure what this sort of estimation is called. I have a 2-column time-series x(i,t) [with (i=1,2; t=1,...T)], and I want to do the following 'simultaneous' regressions: x(1,t) = (d - 1)(x(1, t-1) - mu(1)) x(2,t) = (d - 1)(x(2, t-1) - mu(2)) And I want to determine the coefficients d, mu(1), mu(2). Note that the d should be the same for both estimations, whereas the

Folks, I have a 'matching' matrix between variables A, X, L, O: > a <- structure(c(1, 0.41, 0.58, 0.75, 0.41, 1, 0.6, 0.86, 0.58, 0.6, 1, 0.83, 0.75, 0.86, 0.83, 1), .Dim = c(4L, 4L), .Dimnames = list( c("A", "X", "L", "O"), c("A", "X", "L", "O"))) > a A X L O A 1.00 0.41

Hello. I am having trouble understanding the use of split.screen. I want to divide the device surface first into 4 equal screens: split.screen(figs=c(2,2)) This works. I next want to subdivide each of these 4 screens into 10 subscreens. I do, for the first of these 4 screens: screen(1,new=T) and then: split.screen(figs=c(10,2)) My understanding is that this should split screen 1 (i.e.

Hi folks, I want to sort a matrix row-by-row and create a new matrix that contains the corresponding colnames of the original matrix. E.g. > set.seed(123) > a <- matrix(rnorm(20), ncol=4); colnames(a) <- c("A","B","C","D") > a A B C D [1,] -0.56047565 1.7150650 1.2240818 1.7869131 [2,]

Folks, I have the following dataframe: > x <- structure(list(name = c("EU B", "EU B", "EU B", "EU B", "EU B", "EU B", "AU A", "AU A", "AU A", "AU A", "AU A", "AU A"), date = c("2010-10-11", "2010-10-12", "2010-10-13",

I want to populate the matrix prb through the function HWMProb <- function (a,j,dt) that encapsulates different functions (please see code below), using j= 0:2 for each j. It only populates prb if I specify each function independently in the global environment and then run the loop with the iF statement, as per below. for (j in 0:2) { if (j==0) { prb["0","1"] <-

Hi Petr, Many thanks for your response. Basically I want to create a probability matrix to be used in a trinomial tree going forward. This is the reason why I thought to build the matrix around 0 would be much more efficient. I need to loop through because the probabilities will depend on my node and is not always the same per row (e.g. if N> jmax, jmax being defined in another function) I

Hi Eric I did not see any answer and frankly speaking I cannot provide you with canned help. AFAIK if a function is defined within another function (which is your case) it cannot be called directly so it is necessary to define it in global environment. > fff <- function(x) { + myf <- function(a) a+2 + myf(x)^2} > > fff(5) [1] 49 > myf(5) Error in myf(5) : could not find

Hi Well, I am not an expert in this field so I cannot comment your approach. I wanted only to point out that building matrix your way is like scratching your left ear with right hand, especially in R. What if you want increase size of your matrix? E.g. you use function ProbUP once for row "0" and than for rows different from jmax (if I correctly understand your code). Use of any

This means that the margins for 10 screens would take up more room than you have - essentially the plot area is being squeezed to nothing. You can try reducing your margins using par. Also, it looks like you're trying to split into 20 screens there. Hope this helps, Matt -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch] On

Hi, I want to save a array (say, array[6,7,8]) write a cvs file. How can I do that??? can I write in one file? if I could not write in one file, i want to use a loop to save in different files (in the matrix[6,7,8], should be 8 csv files), such as the filename structure should be: file ="filename" +str(i) +"." +"csv" Many thanks. Dong [[alternative HTML version

Hi, I want to save an array(say, array[6,7,8]) write a cvs file. How can I do that??? can I write in one file? if I could not write in one file, i want to use a loop to save in different files (in the array[6,7,8], should be 8 csv files), such as the filename structure should be: file ="filename" +str(i) +"." +"csv" Many thanks. [[alternative HTML version

folks, is there a clever way to compute the sum of the product of two vectors such that the common indices are not multiplied together? i.e. if i have vectors X, Y, how can i compute Sum (X[i] * Y[j]) i != j where i != j also, what if i wanted Sum (X[i] * Y[j] * R[i, j]) i != j where R is a matrix? thanks, murali

Sorry to plague the list, but I think I got the answer. The following would do: > signalList <- list(tradingRules$Signal[tradingRules$Enabled]) [[1]] > length(signalList) [1] 2 Now my problem is shifted: I have the Signal column in the original data frame referring to actual matrices previously created in R. That is, bar_signal and cif_signal are extant matrices. What I need is the

I have tried and tried unsuccessfully to place a legend outside the plot frame of a graph. This would allow me to draw a legend for an image() plot. But any legend I add is cut off outside the axes. Is there a way of doing this? Thanks for any advice on how to make this work (I'm using R1.5.1 with Windows 2000). Richard Condit Smithsonian Tropical Research Institute Unit 0948 APO AA

Folks, I have a 3000 x 4 matrix (y), which I need to regress row-by-row against a 4-vector (x) to create a matrix lm.y of intercepts and slopes. To illustrate: y <- matrix(rnorm(12000), ncol = 4) x <- c(1/12, 3/12, 6/12, 1) system.time(lm.y <- t(apply(y, 1, function(z) lm(z ~ x)$coefficient))) [1] 44.72 18.00 69.52 NA NA Takes more than a minute to do (and I need to do many

Folks, I have three dataframes storing some information about two currency pairs, as follows: R> a EUR-USD NOK-SEK 1.23 1.33 1.22 1.43 1.26 1.42 1.24 1.50 1.21 1.36 1.26 1.60 1.29 1.44 1.25 1.36 1.27 1.39 1.23 1.48 1.22 1.26 1.24 1.29 1.27 1.57 1.21 1.55 1.23 1.35 1.25 1.41 1.25 1.30 1.23 1.11 1.28 1.37 1.27 1.23 R> b EUR-USD NOK-SEK 1.23 1.22 1.21 1.36 1.28 1.61 1.23 1.34 1.21 1.22

Folks, If I have a vector such as the following: x <- c(0, -1, -1, -1, 0, 0, 1, -1, 1, 0) and I want to replace the zeroes by the nearest non-zero number to the left, is there a more elegant way to do this than the following loop? y <- x for (i in 2 : length(x)) { if (y[i] == 0) { y[i] <- y[i - 1] } } > y [1] 0 -1 -1 -1 -1 -1 1 -1 1 1 You can see the