similar to: matrix indexing

Dear R users, I've been struggling with the following problem: I want to extract the Wald p-value from an lmer() fit, i.e., consider library(lme4) n <- 120 x1 <- runif(n, -4, 4) x2 <- sample(0:1, n, TRUE) z <- rnorm(n) id <- 1:n N <- sample(20:200, n, TRUE) y <- rbinom(n, N, plogis(0.1 + 0.2 * x1 - 0.5 * x2 + 1.5 * z)) m1 <- lmer(cbind(y, N - y) ~ x1 + x2 + (1 | id),

Consider the following little "benchmark" > require(Matrix) > tmp <- Matrix(c(rep(1,1000),rep(0,9000)),ncol=1) > ind <- sample(1:10000,10000) > system.time(tmp[ind,]) user system elapsed 0.004 0.001 0.005 > ind <- sample(1:1000,10000,replace=TRUE) > system.time(tmp[ind,]) user system elapsed 0.654 0.006 0.703 >

Hi, I'd like to divide each element of a matrix by a specific value per column. These specific values are stored in a list. For example: > x <- c(1,2,3,4,5) > y <- matrix(c(1:30), nrow = 6) Now I want to divide each element in y[,1] by x[1], y[,2] by x[2] etc. I have tried this > my_function <- function(data, ind) data/ind > apply(y, 2, my_function, x) [,1] [,2]

Dear R users, I've been struggling some days with the following problem: I'm interesting in producing the following plot x <- seq(0.01, 10, length = 20) plot(c(0, 10), c(-20, 20), type = "n", xlab = "x", ylab = expression(2 * alpha * log(x))) pch. <- rep(NA, length(x)) for(i in 1:4){ pch.[7] <- as.character(i) lines(x, 2 * i * log(x), type

Dear R users, I've been trying to build R from sources (in Windows) using Dr. Goto's BLAS, unsuccessfully! I've followed the instructions in Section 3.1.2-3.1.3 of "R Installation and Administration" manual (but maybe I did something wrong), but I keep receiving the following error: -- initially I get -- make: ./Rpwd.exe: Command not found make[1]: ./Rpwd.exe: Command not

I have a very large dataframe and wish to extract a subset of rows. I have a two column matrix listing the starting and ending indices of one subset on each row. My idea is to create a vector of indices that could be applied to the dataframe and I have a solution using a for loop (below). But surely there is some more elegant way to do this! I looked thorough the archives without

Dear list, As a minimal test of a more complex grid layout, I'm trying to find a clean and efficient way to arrange text grobs in a rectangular layout. The labels may be expressions, or text with a fontsize different of the default, which means that the cell sizes should probably be calculated using grobWidth() and grobHeight() as opposed to simpler stringWidth() and stringHeight().

Hi I have the following array: test <- array(c(1:16), dim = c(3,4,3)) test ## I call some enries using an index array test.ind <- array(rbind(c(1,2,1), c(3,3,2)), dim = c(2,3)) test[test.ind] ## suppose I want all values in the 2nd row and 4th col over ## all three 3rd dimensions test[2,4,] how to specify a test.ind array with the last index left with ',' i.e test.ind should be

I have a matrix that I want to turn into a transformed matrix that includes the indices from the original matrix and the value. The matrix is simply real-valued and is square (and large (8k x 8k)). I want something that looks like (for the 3x3 case): i j value 1 1 1.0 1 2 0.783432 1 3 -0.123482 2 1 0.783432 2 2 1.0 2 3 0.928374 and so on.... I can do this with for loops, but there is

Hello R-help - I have vec <- c("string1", "string2", "string3") ind <- list(c(1,2),c(1,2,3)) I want "vec" indexed by each vector in the list "ind". The first element of the list I want would be vec[c(1,2)], the second element would be vec[c(1,2,3)], like the following. [[1]] [1] "string1" "string2" [[2]] [1]

Hi, I cannot go through the archives with which() as key-word... so common. Though I am sure to have seen something about this subject in the past could somebody put me on the track. I have a matrix (actually a data.frame) in which I would replace the non-null values by 1. I tried the following: indices<-which(myforetbin > 0,arr.ind=T) myforetbin[indices[,1],indices[,2]]<-1 and get

I' ve seen that several people are looking for a function that creates a barplot with an error indicators (I was one of them myself). Maybe you will find the following code helpful (There are some examples how to use it at the end): # Creates a barplot. #bar.plot() needs a datavector for the height of bars and a error #indicator for the interval #many of the usual R parameters can be set:

Hi all, I wanted to mark the estimation sample: mark what rows (observations) are deleted by lm due to missingness. For eg, from the original example in help, I have changed one of the values in trt to be NA (missing). # code below # ---- # original example > ctl <- c(4.17,5.58,5.18,6.11,4.50,4.61,5.17,4.53,5.33,5.14) > trt <- c(4.81,4.17,4.41,3.59,5.87,3.83,6.03,4.89,4.32,4.69) #

Folks, I have a matrix: set.seed(123) a <- matrix(rnorm(100), 10) And a vector: b <- rnorm(10) Now, I want to switch the signs of those rows of a corresponding to indices in b whose values exceed the 75 %-ile of b which(b > quantile(b)[4]) [1] 2 6 10 so I want, in effect: a[2, ] <- -a[2, ] a[6, ] <- -a[6, ] a[10, ] <- -a[10, ] I thought I could do a[which(b >

Hi, I noticed that by() returns an object of class 'by', regardless of what its argument 'simplify' is. ?by says that it always returns a list if simplify=FALSE, yet by.data.frame shows: ---<--------------------cut here---------------start------------------->--- function (data, INDICES, FUN, ..., simplify = TRUE) { if (!is.list(INDICES)) { IND <-

Hi All, Just hoping some one can give me a hand with a problem... I have a dataframe (DF) with about 5 million entries that looks something like the following: >DF ID Cl Co Brd Ind A AB AB 1 S-3 IND A BR_F BR_F01 1 0 0 2 S-3 IND A BR_F BR_F01 1 0 0 3 S-3 IND A BR_F BR_F01 1 0 0 4 S-3 IND A BR_F BR_F01 1 0 0 5 S-3 IND A BR_F BR_F01 1 0 0 6 S-3 IND A BR_F

summary: The function 'by' inconsistently strips class from the data to which it is applied. quick reason: tapply strips class when simplify is set to TRUE (the default) due to the class stripping behaviour of unlist. quick answer: This can be fixed by invoking tapply with simplify=FALSE, or changing tapply to use do.call(c instead of unlist executable example:

Dear friends, I have a basic question with R. I'm generating a set of random variables and then combining them using the cbind statement. The code for that is given below. for (i in 1:100) { y <- rpois(i,lambda=10) X0 <- seq(1,1,length=i) X1 <- rnorm(i,mean=5,sd=10) X2 <- rnorm(i,mean=17,sd=12) X3 <- rnorm(i,mean=3, sd=24) ind <- rep(1:5,20) }

Hello, I’m trying to follow the syntax of a script from a journal website. In order to create a regression formula used later in the script, the regression matrix must have column names “X1”, “X2”, etc. I have tried to assign these column names to my matrix ScoutRSM.mat using a for loop, but I don’t know how to interpret the error message. Suggestions? Thanks, Paul

Hi all, Suppose that I've two data frames, a and b say, both containing a column 'id'. While data frame 'a' contains multiple rows sharing the same id, data frame 'b' contains just one entry per id (i.e. a 1 to n relationship). For the ease of modeling I now want to generate a new data frame c, which is basically a copy of data frame 'a' augmented by the values