similar to: re stricting points in a data frame

Consider a data frame (x) with 2 variables, x1 and x2, having equal values. It looks like: x1 x2 1 1 2 2 3 3 Now, consider a second data frame (xk): xk1 xk2 0.5 0.5 1.0 0.5 1.5 0.5 2.0 0.5 0.5 1 1.0 1 1.5 1 2.0 1 0.5 1.5 1.0 1.5 1.5 1.5 2.0 1.5 0.5 2 1.0 2 1.5 2 2.0 2 I have written code to calculate some differences between these

I keep getting the 'memory.size' error message when I run a program I have been writing. It always it cannot allocate a vector of a certain size. I believe the error comes in the code fragement below where I have multiple arrays that could be taking up space. Does anyone know a good way around this? w1 <- outer(xk$xk1, data[,x1], function(y,z) abs(z-y)) w2 <- outer(xk$xk2,

This is a continuation from a previous posting of mine: The following algorithm below is what I want to accomplish: Z(xk) = Average(Yi, i belongs to Ik), where Ik contains all i such that for each j, |Xi,j - xkj?? 2. Here, j = 1, 2 and i corresponds to the elements in each X and/or xk >data x1 x2 y 1 1 2 2 2 6 3 3 12 Now, consider a second data frame or matrix (xk):

useR's, Consider the variables defined below: yvals <- c(25,30,35) x1 <- c(1,2,3) x2 <- c(3,4,5) x3 <- c(6,7,8) x <- as.data.frame(cbind(x1,x2,x3)) delta <- c(2.5, 1.5, 0.5) h <- delta/2 vars <- 3 xk1 <- seq(min(x1)-0.5, max(x1)+0.5, 0.5) xk2 <- seq(min(x2)-0.5, max(x2)+0.5, 0.5) xk3 <- seq(min(x3)-0.5, max(x3)+0.5, 0.5) xks <- list(xk1,xk2,xk3) xk <-

Hi Mates, I have a very long R code that needs to go to production but my portfolio managers do not use R language and the software is not supported by my bank. Is there any way I can compile the code to an executable file and make it usable to my portfolio managers who have no knowledge at all of R? Thanks Mama ----- Mama Attiglah, PhD Quantitative Strategist Liability Driven Investment

Hi all, I am struggling to compile a massive R code that I have written through some years, aiming to provide automated investment strategies to my Portfolio Managers. You may ask me why do I not rewrite the code in C or C++; the answer is it is really massive and that will take me lots of time. The pb is I will be moving from one team to another therefore I want to make it an exec program that

I have a big matrix 'ret'. I want to multiply each row of it with a 2nd vector 'pos', resulting result, I want to save in a vector named 'port'. I wrote following code: > pos [1] 2593419 2130220 6198197 1673888 1980000 1784732 2052120 -7490228 -5275000 > dim(ret) [1] 500 9 > fu # user defined function function(x) { fu = x %*%

Hi, I have measured two response variables (y1, y2) at each treatment level (x = 0, 1.5 or 3). Now I would like to show the y1 and y2 against x in a bar plot. However, y1 and y2 differ in scale so I need two y-axises, one on the left side and one on the right side (and I dont want to standardize my responses). This is fairly easy if you want to show points,lines etc, but gets more complicated

Hello, I'm struggling with an elementary problem with R. I have a simple data frame such as this one giving the number of accidents subdivided by sex, age and region. sex age region no_of_accidents F young north 10 F young south 12 F old north 5 F old south 7 M young north 24 M

useR's, I want to apply this function to the columns of a data frame: u[u >= range(v)[1] & u <= range(v)[2]] where u is the n column data frame under consideration and v is a data frame of values with the same number of columns as u. For example, v1 <- c(1,2,3) v2 <- c(3,4,5) v3 <- c(2,3,4) v <- as.data.frame(cbind(v1,v2,v3)) uk1 <- seq(min(v1) - .5, max(v1) + .5,

Hello a <- c("Mama", "Papa", "Papa; Mama", "", "Sammy; Mama; Papa") a <- strsplit(a, "; ") mama <- rep(F, length(a)) mama[sapply(a, function(x) { sum(x=="Mama") }, simplify=T) > 0] <- T papa <- rep(F, length(a)) papa[sapply(a, function(x) { sum(x=="Papa") }, simplify=T) > 0] <- T # ... more

useRs, I am looking to find the minimum positive value of some data I have. Currently, I am able to find the minimum of data after I apply some other functions to it: > x [1] 1 0 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 > sort(x) [1] 0 1 1 2 3 3 4 5 5 5 6 7 8 8 9 9 10 10 > diff(sort(x)) [1] 1 0 1 1 0 1 1 0 0 1 1 1 0 1 0 1 0 > min(diff(sort(x))) [1] 0

useR's, I am trying trying to find out if there is a faster way to do a certain computation. I have successfully used FOR loops and the apply function to do this, but it can take some time to fully compute, but I was wondering if anyone may know of a different function or way to do this: > x [1] 1 2 3 4 5 > xk [1] 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 6.0 I want to do:

useR's, I want to randomly generate 500 numbers from the standard normal distribution, i.e. N(0,1), but I only want them to be generated in the range -1.5 to 1.5. Does anyone have a simple way to do this? Thanks, dxc13 -- View this message in context: http://www.nabble.com/simple-random-number-generation-tp18642611p18642611.html Sent from the R help mailing list archive at Nabble.com.

useR's I am trying to color the points on a scatter plot (code below) with two colors. Red for values 0.5 -1.0 and blue for 0.0 - .49. Does anyone know a easy way to do this? x <- runif(100, 0, 1) y <- runif(100, 0, 1) plot(y ~ x, pch=16) Thanks, dxc13 -- View this message in context: http://www.nabble.com/color-ranges-on-a-2D-plot-tp14893457p14893457.html Sent from the R help

http://bugzilla.mindrot.org/show_bug.cgi?id=257 ------- Additional Comments From markus at openbsd.org 2002-05-27 17:09 ------- openssh is not derived from ssh.com-2.x or 3.x. but adding support for 32bit int is not hard. you can attach patches to this bug. ------- You are receiving this mail because: ------- You are the assignee for the bug, or are watching the assignee.

useR's, I have a matrix given by the code: mat <- matrix(c(rep(NA,10),1,2,3,4,5,6,7,8,9,10,10,9,8,NA,6,5,4,NA,2,1,rep(NA,10),1,2,3,4,NA,6,7,8,9,10),10,5) This is a 10x5 matrix containing missing values. All columns except the second contain missing values. I want to delete all columns that contain ALL missing values, and in this case, it would be the first and fourth columns. Any column

Hi all, I cannot think of the technical name of this plot, but I want to create a plot with the data below that looks like two back-to-back horizontal bar plots. Ideally, there would be a vertical line in the center of the plot at zero, and on the right hand side would be 4 bars representing the values for 2007, and on the left side of the vertical line would be the corresponding values for 2005.

useR's, I have several matrices of size 4x4 that I want to calculate means of their respective positions with. For example, consider I have 3 matrices given by the code: mat1 <- matrix(sample(1:20,16,replace=T),4,4) mat2 <- matrix(sample(-5:15,16,replace=T),4,4) mat3 <- matrix(sample(5:25,16,replace=T),4,4) The result I want is one matrix of size 4x4 in which position [1,1] is the

useR's I am trying to compare two vectors that have the same length. More specifically, I am interested in comparing the corresponding positions of each element in the vector. Consider these two vectors of length 20: v1 <- 2 2 4 NA NA NA 10 NA NA NA NA NA NA NA NA NA NA NA NA NA v2 <- NA 4 NA NA NA NA 10 NA NA NA NA NA NA NA NA NA NA NA