similar to: Rbind for appending zoo objects

Dear All, I have to use loop over an array so I am using following procedure count<-1 repeat{ count<-count + 1 c(g[count],1:i[count]) ->qw if(count>5)break } as a result qw is [1] 0.9643836 1.0000000 2.0000000 3.0000000 4.0000000 5.0000000 [7] 6.0000000 7.0000000 8.0000000 9.0000000 10.0000000 11.0000000 [13] 12.0000000 13.0000000 14.0000000 15.0000000 16.0000000

Hello all! I have a problem with ggplot2 library. I want to do an heat map and the y variables are the year months. If I use the following code, he y values are in alphabetical order, but I want it in month order. The code is: library(reshape) library(ggplot2) library(scales) p <- ggplot(data.m, aes(variable, Month)) + geom_tile(aes(fill = value),

Buenas Tengo un Data table de la siguiente manera: datos<-data.table(grupo=rep(c("a","b"),5),x=c(1:10),y=rnorm(10,2,1)) Lo que quiero es añadir una fila por cada grupo y en esa nueva fila, al valor de la x ponerle el valor anterior de la y Lo que hago es añadir una nueva fila por grupo, con: datos[,.SD[1:(.N+1)],by=grupo] Y para añadir el valor anterior uso la función

is there a way to do element-by-element multiplication as in Gauss and MATLAB, as shown below? Thanks. --- a 1.0000000 2.0000000 3.0000000 x 1.0000000 2.0000000 3.0000000 2.0000000 4.0000000 6.0000000 3.0000000 6.0000000 9.0000000 a.*x 1.0000000 2.0000000 3.0000000 4.0000000

folks, I have been struggling with the "R" documentation for too long now and I need a simple answer on two questions. The documentation does not have adequate examples. Please help. given two equal vector lists: A <- c(0,1,2,3) B <- c(5,6,7,8) [Question #1] how do I dump them to a text file with the following format:

Since format, i.e., format.default(.), is a pretty basic function I thought I'd ask before just changing it... aa <- cbind(1:7, rnorm(7)) format(aa) or format(aa, digits=7) looks like [,1] [,2] [1,] " 1.0000000" " 0.2406669" [2,] " 2.0000000" "-0.4973221" [3,] " 3.0000000" " 0.4672260" [4,] "

Hello, I have a matrix mat1 of dim [1,8] and mat2 of dim[30,8], I want to replace the first row of mat2 with mat1, this is what I do: mat2[1,]<-mat1 but it transforms mat2 in a list I don't understand, I want it to stay a matrix... ----- Anna Lippel -- View this message in context: http://n4.nabble.com/Simple-question-on-replace-a-matrix-row-tp1427857p1427857.html Sent from the R help

Hello, I have 2 functions (a and b) a = function(n) { matrix (runif(n*2,0.0,1), n) } > > > b = function (m, matrix) { > n=nrow (matrix) > p=ceiling (n/m) > lapply (1:p, function (l,n,m) { > inf = ((l-1)*m)+1 > if (l<p) sup=((l-1)*m)+m > else sup=n >

below is my data frame. I would like to compute summary statistics for mgl for each river mile (mean, median, mode). My apologies in advance- I would like to get something like the SAS print out of PROC Univariate. I have performed an ANOVA and a tukey LSD and I would just like the summary statistics. thanks stephen RM mgl 1 215 0.9285714 2 215 0.7352941 3 215 1.6455696 4 215

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Hi. I am currently trying to run some Spearman correlations, and have encountered two issues. 1) When using cor.test() with a variable that includes ties, I get the "Cannot compute exact p-values with ties" error. I have read that this function now uses an asymptotic formula that allows for ties, so do not understand why I am getting this error. (I am running version 2.4.0.) I

I am doing analysis on hourly precipitation on a file that is disorganized. However, I managed to clean it up and store it in a dataframe (called CA1) which takes the form as followed: Station_ID Guage_Type Lat Long Date Time_Zone Time_Frame H0 H1 H2 H3 H4 H5 H6 H7 H8 H9 H10 H11 H12 H13 H14 H15 H16 H17 H18 H19 H20 H21 H22 H23 1

Fellow R-helpers, Suppose we create a histogram as follows (although it could be any vector with zeroes in it): R> lenh <- hist(iris$Sepal.Length, br=seq(4, 8, 0.05)) R> lenh$counts [1] 0 0 0 0 0 1 0 3 0 1 0 4 0 2 0 5 0 6 0 10 0 9 0 4 0 [26] 1 0 6 0 7 0 6 0 8 0 7 0 3 0 6 0 6 0 4 0 9 0 7 0 5 [51] 0 2 0 8 0 3 0 4 0 1 0 1 0 3

There's something peculiar that I do not understand here. However, did you realize that the thing you are assigning into parts of `a' is NULL? Check you're my.test.boot.ci.1: It's NULL. Be that as it may, I get: > a <- data.frame(matrix(1:4, nrow=2), X3=NA, X4=NA) > a X1 X2 X3 X4 1 1 3 NA NA 2 2 4 NA NA > a[a$X1 == 1,]$X3 <- NULL > a X1 X2 X3 X4 1 1

This 'strange behaviour' manifest itself within some quite complex code. When I created a *very* simple example the behaviour dissapeared. Here is the simplest version I have found which still causes the strange behaviour (it could be quite unrelated to the boot library, however). library(boot) ## boot statistic function my.mean.s <- function(data,subset){ mean(data[subset]) }

Hi @ all, how do I get the largest or lowest k values of list? It must similar to the min() / max() function, but I just don't get it. Best wishes, Markus

Hi all, As a molecular biologist by training, I'm fairly new to R (and statistics!), and was hoping for some advice. First of all, I'd like to apologise if my question is more methodological rather than relating to a specific R function. I've done my best to search both in the forum and elsewhere but can't seem to find an answer which works in practice. I am carrying out

Hi all, I am a bit mystified by this error message that I get when I try to apply lmer() to a simple dataset with one between factor (age) and one within factor (item): "$ operator is invalid for atomic vectors" I'll just provide the code, because I don't see where the problem is: library(lme4) options(contrasts=c("contr.helmert","contr.poly")) data =

The following snippet suggests that there is either a bug in qr(,LAPACK=T), or some bug in my understanding. Note that the detected rank is correct (= 2) using the default LINPACK qr, but incorrect (=3) using LAPACK. This is running on Linux Redhat 9.0, using the lapack library that comes with the Redhat distribution. I'm running R 1.7.1 compiled from the source. If the bug is in my

Why? And how to solve it? The code and result are following, > data=rnorm(50) > > kde=density(data,n=20,from=-1,to=10) > > kde$x;kde$y [1] -1.0000000 -0.4210526 0.1578947 0.7368421 1.3157895 1.8947368 [7] 2.4736842 3.0526316 3.6315789 4.2105263 4.7894737 5.3684211 [13] 5.9473684 6.5263158 7.1052632 7.6842105 8.2631579 8.8421053 [19] 9.4210526 10.0000000 [1]