similar to: sequential for loop

Greetings, I am trying to add a boxplot to the bottom of a histogram, right between the histogram bars and the x axis. Here is the code I am using at the moment (the par line is probably not relevant for our discussion): hs <- hist(x, breaks = 20, plot = F) par(mar = c(3,3,2,1)) hist(x, breaks = 20, main = NULL, ylim = c(-2, max(hs$counts))) boxplot(x, horizontal = T, axes = T, add =

Hello, I am an advanced user of R. Recently I found out that apparently I do not fully understand vectors and lists fully Take this code snippet: T = c("02.03.2008 12:23", "03.03.2008 05:54") Times = strptime(T, "%d.%m.%Y %H:%M") Times # OK class(Times) # OK is.list(Times) # sort of understand and not understand that length(Times)

I just acquired a copy of "Statistical Models in S", I guess most commonly known as the "white book", and realized to my dismay that most of the code is not directly executable in R, and I was wondering if there was a source discussing the things that are different and what the new ways of calling things are. For instance, the first obstacle was the solder.balance data

I have the following relatively simple problem. Say we have three factors, and we want to create a cross-tabulation against each of the other two: x <- factor(rbinom(5, 1, 1/2)) y <- factor(rbinom(5, 1, 1/2)) z <- factor(rbinom(5, 1, 1/2)) table(x,y) table(x,z) This looks like: y x 0 1 0 2 0 1 1 2 z x 0 1 0 1 1 1 2 1 I would like to get (surely this will

Hi all, Is there a simple function already implemented for getting the ISO weeks of a Date object? I couldn't find one, and so wrote my own function to do it, but would appreciate a pointer to the "default" way. If a function is not yet implemented, could the code below be of interest to submit to CRAN? Best Regards, Gustaf --------------------

Greetings, I have a problem that I am sure is very straightforward, but I just can't wrap my head around it. I've read the help pages on text, plotmath, expression, substitute, but somehow I can't find the answer to this simple question. Basically consider the following example: plot( NULL, xlim = c(0,2), ylim = c(0,2) ) expressions <- expression( -infinity, infinity )

I am getting some unexpected results from some functions of igraph and it is possible that I am misinterpreting the vertex numbers. Eg., the max betweenness measure seems to be from a vertex that is not connected to a single other vertex. Below if my code snippet: require(igraph) my.graph <- graph.adjacency(adjmatrix = my.adj.matrix, mode=c("undirected")) most.between.vert <-

Dear R Users, I have data for more than 3 years. For each year I want to find the day corresponding to Jaunary 1 of that year. For example: > x <- c('5/5/2007','12/31/2007','1/2/2008') > #Convert to day of year (julian date) - > strptime(x,"%m/%d/%Y")$yday+1 [1] 125 365 2 I want to know how to do the same thing but with time added. But I still

Dear R-Help, I'm new to R and struggling with weighting data when I run regression. I've tried to use search to solve my problem but haven't found anything helpful so far. I (successfully) import data from SPSS (15) and try to run a linear regression on a subset of my data file where WEIGHT is the name of my weighting variable (numeric), e.g.: library(foreign)

Hi, I got following error in write.table() : > write.table(dataa, file="c:/data1.csv", row.names=F, col.names=T, sep=",") Error in file(file, ifelse(append, "a", "w")) : cannot open the connection In addition: Warning message: In file(file, ifelse(append, "a", "w")) : cannot open file 'c:/data1.csv': Permission denied

I have a data frame named "database" with panel data, a little piece of which looks like this: Symbol Name Trial Factor1 Factor2 External 1 548140 A 1 -3.87 -0.32 0.01 2 547400 B 1 12.11 -0.68 0.40 3 547173 C 1

Hello, new to the list, first message. This question perhaps might be more appropriate to R-sig-teaching, and I'd be happy to take it there if this is not the right place for it. I am teaching applied statistics at a small liberal arts college with limited resources, and we are currently using SPSS for our courses. Mainly the reason for this, as I understand it, is that this is what

Hi, Below I have a function mlogl_k, later it's called with "nlm" . __BEGIN__ vsamples<- c(14.7, 18.8, 14, 15.9, 9.7, 12.8) mlogl_k <- function( k_func, x_func, theta_func, samp) { tot_mll <- 0 for (comp in 1:k_func) { curr_mll <- (- sum(dgamma(samp, shape = x_func, scale=theta_func, log = TRUE))) tot_mll <- tot_mll + curr_mll }

Hello all, I ran into the following, to me unexpected, behavior. I have (for reasons that don't necessarily pertain to the question at hand, hence I won't go into them) the need/desire to use an empty string for the name of a vector entry. Perhaps I did not read ?"[" very carefully, but it seems to me that he following lines should return "1" at the end:

Hi there, I set up my company's back up server using rsync. And I've got a strange problem. I searched in the archives of this list, but none of them seems not giving me an idea to solve the problem. If anyone can help, it would be grateful. I'm using cron by a user (non wheel/admin) to rsync everyday during the night. The cron is set in the server to transfer the backing-up

Full_Name: John Brzustowski Version: R-devel-trunk OS: linux (problem under Windows too) Submission from: (NULL) (74.101.124.238) (This bug was discovered by Phil Taylor, Acadia University.) I'm not sure from reading the documentation whether strptime(x, "%j") is meant to be supported, but if so, there is a bug which prevents it from working on the first day of months after

I'm trying to do a linear regression between the columns of matrices. In example below I want to regress column 1 of matrix xdat with column1 of ydat and do a separate regression between the column 2s of each matrix. But the output I get seems to give correct slopes but incorrect intercepts and another set of slopes with value NA. How do I do this correctly? I'm after the slope and

Hi, I have a dataframe column of the form v<-c("Fri Feb 05 20:00:01.43000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:02.274000 2010","Fri Feb 05 20:00:06.34000 2010") I need to convert this to datetime form. I did the following.. lapply(v,function(x){strptime(x, "%a %b %d %H:%M:%OS %Y")}) This gives me a list that looks like

I ran into a weird, to me at least, problem, and hoping someone can shed some light into it. In a nutshell, there seems to be some problem when one calls plot with a formula, from within another function, using ... to pass arguments, and one of those arguments being xlim (and only xlim shows this problem). Here is an example: > plotw <- function(obj,...) { + plot(k~j,

I have data in the following format: > DATE [1] "01/13/2004" In order to find the difference between two data points, I presently use brute force to calculate the day of the year: > strptime(DATE, format="%m/%d/%Y")$yday [1] 12 Although this works, it may not be robust over different years. I assume that R is sufficiently clever that a much simpler approach