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Hi, I need to aggregate rows of a data.frame by computing the mean for rows with the same factor-level on one factor-variable; here is the sample code: x <- data.frame(rep(letters,2), rnorm(52), rnorm(52), rnorm(52)) aggregate(x, list(x[,1]), mean) Now my problem is, that the actual data-set is much bigger (120 rows and approximately 100.000 columns) ? and it takes very very long

Hi, I would like to cut a vector of values in parts. Each part should have an equal number of elements. for example: x <- (rnorm(500)^2) now I want 5 vectors each with 100 elements. The first vector should include the 100 lowest values of x and so on (so that the fifth vector contains the 100 highest values of x). thanks for any help!

Hi, When using the aggregate function to aggregate a data.frame by one or more grouping variables I often have the problem, that I want the mean for some numeric variables but the unique value for factor variables. So for example in this data-frame: data <- data.frame(x = rnorm(10,1,2), group = c(rep(1,5), rep(2,5)), gender =c(rep('m',5), rep('f',5))) aggregate(data,

Hi, is there a function or an easy way to convert a variable with continuous values into a categorial variable (with x levels)? here is what I mean: I want to transform x: x <- c(3.2, 1.5, 6.8, 6.9, 8.5, 9.6, 1.1, 0.6) into a 'categorial'-variable with four levels so that I get: [1] 2 2 3 3 4 4 1 1 so each element is converted into its rank- value / categorial-value (in

Hi, I have the following problem: I have two vectors: i <- c('a','c','g','h','b','d','f','k','l','e','i') j <- c('a', 'b', 'c') now I would like to generate a vector with the length of i that has zeros where i[x] != any element of j and 1 where i[x] == any element of j.

Hi, I would like to fit lm-models to a matrix with 'samples' of a dependent variable (each row represents one sample of the dependent variable). The independent variable is a vector that stays the same: y <- c(1:10) x <- matrix(rnorm(5*10,0,1), 5, 10) now I would like to avoid looping over the rows, since my original matrix is much larger; for(t in 1:dim(x)[1]) {

Hi, I get a lot of error messages with this command, but I don't understand why; plot(c(),c(), xlim=c(1,10), ylim=c(0,10000), log="y") thanks for any help! [[alternative HTML version deleted]]

Hi, is there a way to replicate a data.frame like you can replicate the entries of a vector (with the repeat-function)? I want to do this: x <- data.frame(x, x) (where x is a data.frame). but n times. And it should be as cpu / memory efficient as possible, since n is pretty big in my case. thanks for any suggestions!

Hi, I have multiple three dimensional arrays. Like this: x1 <- array(rnorm(1000, 1, 2), dim=c(10, 10, 10)) x2 <- array(rnorm(1000, 1, 2), dim=c(10, 10, 10)) x3 <- array(rnorm(1000, 1, 2), dim=c(10, 10, 10)) Now I would like to compute the mean for each corresponding cell. As a result I want to get one 3D array (10 x 10 x 10) in which at position x, y, z is the mean of the

hi, I have a vector full of strings like; xy_100_ab xy_101_ab xy_102_ab xy_103_ab I want to seperate each string in three pieces and the separator should be the "_" at the end I want a data.frame like: column1 column2 column3 xy 100 ab xy 101 ab xy 102 ab xy 103 ab I tried strsplit but I couldn't figure out how to convert the list I get into a data.frame. I just

hi, is it possible to convert latex code to pdf in R (like a latex-program would do it)? Is there a package that comes with this capabilities? My problem is that I want to generate tables automatically - and I can't use a latex editor at that computer ... Besides latex ... are there good ways to generate tables in R? thanks for any suggestions!

Dear R-group, is there a way to perform calculations that are defined in a string format? for example I have different variables: x1 <- 3 x2 <- 1 x4 <- 1 and a string-variable: do <- 'x1 + x2 + x3' Is there any way to perform what the variable 'do'-describes (just like the formula-element but more elemental)? Perhaps my idea to solve my problem is a little

Hi, is there a way to place text at the upper left corner (or another corner) of the plot? I want to place it really at the upper left corner of the whole plot (the file I get), not at the upper left corner of the plot-region. I tried text() and mtext(), and corner.label() of the plotrix package but it didn't work out. thanks!

Hi, I know that I can perform a Mann-Whitney U test with wilcox.test(x,y) But instead of an U-value I get a W-Value. Is it possible to change this?

hi, is there an easy way to extract numbers from a string? for example I have; "this Item costs 3.32 Dollars" is there an easy way to extract the 3.32 as a number? thanks!

Hi, Is there a way to open help files in the default web browser instead of a new R-window when I use the help-functions (like ?, help.search() etc.)? thanks!

Hi, I have a somewhat abstract and tricky problem (at least for me). Perhaps someone here can help me: I would like to convert: x <- c(0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0) into x <- c(0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4) every 1 in x marks kind of a 'border' which is separated in different 'regions' in the

Hi, I have two data-frames which I want to match by a date-variable. The problem now is, that the date-variable has a different format in this two data-frames. So simply matching by this variable wouldn't work. In the one data-frame the date-variable contains strings that look like this: 14.Aug.2012-16:32 3.Sep.2012-16:50 etc. in the other data-frame the date-variable looks like this:

Hi, I have a vector like: r <- runif(100) Now I would like to split r into 10 pieces (each with 10 elements) ? but the 'pieces' should be roughly similar with regard to mean and sd. what is an efficient way to do this in R? thanks!

Dear R-list, I currently have to convert a data.frame with several factor-variables to a numeric matrix. Now the problem is, that the order of the factor-labels don't match the order I would like to use. for example, let's assume I have this factor-variable in my data-frame: x <- factor(rep(1:4, 5), labels=c("slightly disagree", "disagree", "agree",