similar to: Improving on edit()?

How do I parse a date "yyyymmdd"? I tried asking chron(s, "ymd") but that didn't work. Would the date parsing routines of the Date class of 1.9 grok this? -- Ajay Shah Consultant ajayshah at mayin.org Department of Economic Affairs http://www.mayin.org/ajayshah Ministry of Finance, New Delhi

I am using R 2.1 on Apple OS X. When I get the ">" prompt, I find it works well with emacs commandline editing. Keys like M-f C-k etc. work fine. The one thing that I really yearn for, which is missing, is bracket matching When I am doing something which ends in )))) it is really useful to have emacs or vi-style bracket matching, so as to be able to visually keep track of whether I

Hi Ajay: A casual search of google using terms like "MCMC", "Markov Chains", "Monte Carlo", and "tutorial" pulled up over 4000 hits. A few links that may interest you, I thought, like: http://csep1.phy.ornl.gov/CSEP/MC/MC.html http://www.stat.fi/isi99/proceedings/arkisto/varasto/gree0167.pdf http://www.maths.soton.ac.uk/staff/Sahu/utrecht/ ... and so

When using get.hist.quote(), I find the dates are broken. This is with R 2.1.1 on Mac OS X `panther'. > library(tseries) Loading required package: quadprog 'tseries' version: 0.9-27 'tseries' is a package for time series analysis and computational finance. See 'library(help="tseries")' for details. > x <-

I am discovering sapply! :-) Could you please help me with a very elementary question? Here is what I know. The following two programs generate the same answer. --------------------------------+---------------------------------------- Loops version | sapply version --------------------------------+----------------------------------------

In order to get around the problems of my posting a few minutes ago, I thought: $ awk -F\| '(NR > 2) {print $2}' cmie_firm_data.text > col2 $ awk -F\| '(NR > 2) {print $4}' cmie_firm_data.text > col4 $ paste col2 col4 | head -2 -510.45 -510.27 60700 101900 $ paste col2 col4 | tail -2 28648.12 31617.02 491014.77 494308.52 $ wc -l col2 col4 89323 col2

I have a situation with a large dataset (3000+ observations), where I'm doing lags as regressors, where I get: Call: lm(formula = rj ~ rM + rM.1 + rM.2 + rM.3 + rM.4) Residuals: 1990-06-04 1994-11-14 1998-08-21 2002-03-13 2005-09-15 -5.64672 -0.59596 -0.04143 0.55412 8.18229 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) -0.003297 0.017603

I learned R & MLE in the last few days. It is great! I wrote up my explorations as http://www.mayin.org/ajayshah/KB/R/mle/mle.html I will be most happy if R gurus will look at this and comment on how it can be improved. I have a few specific questions: * Should one use optim() or should one use stats4::mle()? I felt that mle() wasn't adding much value compared with optim, and

Hello, folks! I'm trying to use R as a graphics program, to make some pretty graphs that will go into prosper slideshows. I wrote this fragment, from the R manual, into a file demo.R: x=seq(-3,3,0.1) postscript("cm_test.eps", width = 4.0, height = 3.0, horizontal = FALSE, onefile = FALSE, paper = "special", family =

I'm in this situation: factorlabels <- c("School", "College", "Beyond") with data for 8 families: education.man <- c(1,2,1,2,1,2,1,2) # Note : no "3" values education.wife <- c(1,2,3,1,2,3,1,2) # 1,2,3 are all present. My goal is to create this table: School College Beyond

Folks, I asked a question on this mailing list about the subset support of lm(). In a flash, I got three helpful responses from Rajarshi Guha <rxg218 at psu.edu> <http://jijo.cjb.net> Erin Hodgess <hodgess at gator.uhd.edu> and Peter Dalgaard <p.dalgaard at biostat.ku.dk> :-) and it was just great. The mistake I was making was in not understanding the notion of a

Suppose I make two discrete variables -- > D <- data.frame(f1=sample(1:5,100,replace=T), f2=sample(1:5,100,replace=T)) I know I can do: > table(D$f1, D$f2) 0 1 2 3 4 0 5 5 5 5 4 1 4 2 6 7 3 2 5 3 5 3 6 3 3 1 3 1 2 4 6 4 3 3 6 > table(D$f1) 0 1 2 3 4 24 22 22 10 22 > table(D$f2) 0 1 2 3 4 23 15 22 19 21 which is all great. But how do I produce the

I have : raw <- read.table("monthly.text", skip=3, sep="|", col.names=c("junk", "junk2", "wpi", "g.wpi", "wpi.primary", "g.wpi.primary", "wpi.fuel", "g.wpi.fuel", "wpi.manuf", "g.wpi.manuf",

Folks, I have placed an example of a self-contained R program later in this mail. It generates a file inflation.pdf. When I stare at the picture, I see the "X label string" and "Y label string" sitting lonely and far away from the axes. How can these distances be adjusted? I read ?par and didn't find this directly. I want to hang on to 2.8 x 2.8 inches as the overall size

Folks, I'm in a situation where I do a few thousand regressions, and some of them are bad data. How do I get back an error value (return code such as NULL) from lm(), instead of an error _message_? Here's an example: > x <- c(NA, 3, 4) > y <- c(2, NA, NA) > d <- lm(y ~ x) Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : 0 (non-NA) cases

I have a program which is doing a few thousand runs of lm(). Suppose it is a simple model y = a + bx1 + cx2 + e I have the R object "d" where d <- summary(lm(y ~ x1 + x2)) I would like to obtain Var(x2) out of "d". How might I do it? I can, of course, always do sd(x2). But it would be much more convenient if I could snoop around the contents of summary.lm and

I have a data frame with one column "x": > str(data) `data.frame': 20 obs. of 1 variable: $ x: num 0.0495 0.0986 0.9662 0.7501 0.8621 ... Normally, I know that the notation dataframe[indexes,] gives you a new data frame which is the specified set of rows. But I find: > str(data[1:10,]) num [1:10] 0.0495 0.0986 0.9662 0.7501 0.8621 ... Here, it looks like the operation

I know that if: x = seq(1,10) d = c(7,3,2) and if I say y = x[d] then I get the vector y as (7,3,2). Very clever! This idea is used intensively with the boot library. Now consider the following code (which works): --------------------------------------------------------------------------- library(boot) sdratio <- function(D, d) { return(sd(D$x[d])/sd(D$y[d])) } x =

I am trying to simulate the trajectory of the pension assets of one person. In C-like syntax, it looks like this: daily.wage.growth = 1.001 # deterministic contribution.rate = 0.08 # deterministic 8% Wage = 10 # initial Asset = 0 # initial for (10,000 days) { Asset += contribution.rate * Wage

I know that this is correct: library(chron) x = dates("01-03-04", format="d-m-y", out.format="day mon year") print(x) It gives me the string "01 Mar 2004" which is correct. I also know that I can say: print(day.of.week(3,1,2004)) in which case he says 1, for today is monday. My question is: How do I combine these two!? :-) I have a