similar to: Basic Question about local

Dear R-helpers, I´m in the context of writing a general function for error propagation in R. There are somehow a few questions I would like to ask (discuss), as my statistical knowledge is somewhat restricted. Below is the function I wrote, the questions are marked. Many thanks in advance. propagate <- function(expr, varList, type = c("stat", "raw"), cov = TRUE) {

Hi R-listers, Q1) What is the difference between the scatterplot and plot function? Q2) I am able to make a graph with the scatterplot function: scatterplot(DevelopIndex ~ Veg, + data = Turtle, + xlab = "Vegetation border (m)", + ylab = "Embryonic development index") And have been successful. But I do not know if the lines are for:

Hello, I am using package MuMIn to calculate AIC for a full model with 10 explanatory variables. Thanks in advance in sharing your experience. Q1 In the AIC list of all models, each model is differentiated by model number. Please kindly advise if it is possible to find the corresponding explanatory variable(s) for the model number. Q2 error message I tried to display sub-model with only

Hi, I'm a bit confused about what the '--copy-unsafe-links' option considers the 'source tree'. In the man page it says that for links pointing outside the 'source tree' the file will be copied, but when I try this, for all links pointing outside of the local directory the files are copied even though the links point to files that are in the directory tree that is

InDATA <-read.table("C:/Data/May 2004/season.txt",header=T) X <- decompose(InDATA) print(X) Period Connections Q1 67519 Q2 69713 Q3 68920 Q4 69452 Q1 70015 Q2 59273 Q3 57063 Q4 65596 Q1 73527 Q2 58586 Q3 69522 Q4 60091 Q1 51686 Q2 63490 Q3 55702 Q4 53200 Q1 51033 Q2 48175 Q3 52709 Q4 50106 Q1 50855 Q2 43466 Q3 48190 Q4 41702 Q1 48747 Q2 51441 Q3 42537

Hello, I have a function that creates an expression object with some variables substituted e.g foo <- function(s){ expression({ v <- s print(v) }) } Thus foo returns an expression, however the expression has the symbol 's' contained within it and thus returns an error when eval'd e.g x <- foo(10) eval(x) Error in eval(expr, envir, enclos) : object 's'

The following dummy data frame has factor Q (with 2 levels) nesting factor P (with levels p1 and p2 nested under q1, and p3 and p4 nested under q2), but both crossing the random variate s, which has 8 levels. The dependent measure is dv. > # The data frame: > testnest dv s P Q 1 1 s1 p1 q1 2 2 s2 p1 q1 3 1 s3 p1 q1 4 2 s4 p1 q1 5 1 s5 p1 q1 6 3 s6 p1 q1 7 3 s7

Hi R-people, I have a question about lattice in general, and histogram specfically. How do you control the ordering of factors that controls the placement of the conditional panels. I have a dataset with factors that go 'Q1','Q2',"Q3','Q5' and of course I want the plot to place Question Q1 at the top and Question Q5 at the bottom of the graphical output.

Dear R-help, I wonder if you could give me some suggestions in how to do a union join of two data frames as follow: -> union join the common column, and insert a 0 if one is missing. I made a function to perform the following, and I know it may not that quite welly written, but it works. Any suggestions are welcome, many thanks. Anthony > q1 =

Hi Guys, I'm using zoo package now. I found lag is not doing what I assumed. > x <- zoo(11:21) > z <- zoo(1:10, yearqtr(seq(1959.25, 1961.5, by = 0.25)), frequency = 4) > x 1 2 3 4 5 6 7 8 9 10 11 11 12 13 14 15 16 17 18 19 20 21 > lag(x) 1 2 3 4 5 6 7 8 9 10 12 13 14 15 16 17 18 19 20 21 > z 1959 Q2 1959 Q3 1959 Q4 1960 Q1 1960 Q2 1960 Q3 1960 Q4

sample report data that i want to forecast quarter quarter_index Revenue 2007 Q1 1 $3,856,799 2007 Q2 2 $4,243,328 2007 Q3 3 $4,930,369 2007 Q4 4 $5,443,579 2008 Q1 5 $5,164,830 2008 Q2 6 $5,104,413 2008 Q3 7

q1 = Question.find(1) q2 = Question.find(1) Now, we find q1.__id__ != q2.__id__. It appears the two aliases are not kept in sync: q1.name => ''old name'' q2.name = ''new name'' q2.save! q1.name => ''old name'' q2.name => ''new name'' q1.reload q1.name =>

I have a data.frame named "df". The dput of df is at the bottom of this e-mail. What I'd like to do is replace the "n/a " values with NA. On Mac OSX, it works to do this: df[df == "n/a"] <- NA However, it does not work on Ubuntu. See below. Thanks in advance, Garrett > x <- df[27, 4] # complete data.frame dput is below > dput(x) "n/a?"

I have a xts object with logical data . > tail(q1) ..1 2010-02-19 TRUE 2010-02-22 FALSE 2010-02-23 FALSE 2010-02-24 FALSE 2010-02-25 FALSE 2010-02-26 FALSE > I want to build a xts that records the dates that there is a change. If TRUE -> FALSE it is "down" if FALSE -> TRUE it is "up" I can do it like this... q2 <- matrix(ncol=2, nrow =

sample report data that i want to forecast quarter quarter_index Revenue 2007 Q1 1 $3,856,799 2007 Q2 2 $4,243,328 2007 Q3 3 $4,930,369 2007 Q4 4 $5,443,579 2008 Q1 5 $5,164,830 2008 Q2 6 $5,104,413 2008 Q3 7

I've been trying to get a basic 5 question IVR survey working in an AGI script, and am having trouble with the SQL portion not updating the table. When I take out all the AGI references, and run just the perl script, the table updates with no problem(DBname,username,password have been substituted in this example for the actual values): #!/usr/bin/perl # # use DBI; $DATETIME =

Hi, Why the qr() produces a negative Q compared with Gram-Schmidt? (note example below, except Q[2,3]) Here is an example, I calculate the Q by Gram-Schmidt process and compare the output with qr.Q() a <- c(1,0,1) b <- c(1,0,0) c <- c(2,1,0) x <- matrix(c(a,b,c),3,3) ########################## # Gram-Schmidt ########################## A <- matrix(a,3,1) q1 <-

Dear R-help, This is my first R day. I want to ask some more beginner's questions. Q1. How can I obtain the covariance matrix for parameter estimates of a multiple regression? I checked ?lm but didn't get the information. Q2. How can I see the old graphs in the graph window? Q3. Can R plot animated graph? For example, I want to see the dynamic change of a 2D graph during a time

Hello, Suppose I have an expression, E, which accesses some variables present in an environment V. I do this via eval(E,envir=V) however all assignments end up in V. I would like the results of assignments in E to end up in the .GlobalEnv ? Or at least the calling environment. Is there a quick way to this instead of iterating over all objects E and assigning into .GlobalEnv? Thank you Saptarshi

Hello, So I am afraid I am having a recurring problem that I just can't figure out. I am using the plm package to conduct a panel analysis - although I am not sure if the problem is arising as a result of the plm package or something more general. I am trying to run a fixed effects model with effects over time and individual. The model has various lags, and the problem is that these lags do