similar to: RE: summaries (was: SUMMARY: "elementary sapply question")

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 --------------------------------+----------------------------------------

I am grateful to Andy Liaw, Douglas Grove, Brian Ripley, Tony Plate, Dirk Eddelbuettel and Sundar Dorai-Raj all of whom got together and drilled sense into my skull. I would like to take some effort into explaining what the question was, that I was grappling with, and the (nice) R way of solving the question. My apologies: I am still a victim of too many years of writing C, so I'm a bit dense

How does one convert objects c("a","b","c") and "d" into "abcd"? > paste(c("a","b","c"), "d") of course yields [1] "a d" "b d" "c d" -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org

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

I have two data frames. Suppose the first has rows r1 r2 r3 and the second has rows R1 R2 R3 I'd like to generate the data frame: r1 R1 r1 R2 r1 R3 r2 R1 r2 R2 r2 R3 r3 R1 r3 R2 r3 R3 How would I go about doing this? I'm sure there's a clean way to do it but I find myself thinking in loops. -- Ajay Shah

Suppose one has x <- c(1, 2, 7, 9, 14) y <- c(71, 72, 77) How would one write an R function which alternates between elements of one vector and the next? In other words, one wants z <- c(x[1], y[1], x[2], y[2], x[3], y[3], x[4], y[4], x[5], y[5]) I couldn't think of a clever and general way to write this. I am aware of gdata::interleave() but it deals

Folks, I am holding a dataset where firms are observed for a fixed (and small) set of years. The data is in "long" format - one record for one firm for one point in time. A state variable is observed (a factor). I wish to make a markov transition matrix about the time-series evolution of that state variable. The code below does this. But it's hardcoded to the specific years that I

1. I have used gdata::read.xls() with much happiness. But every now and then it breaks. I have not, as yet, been able to construct a mental model about the class of .xls files for which it works. Does someone have a simple rule for predicting the circumstances under which it will work? 2. Just like there is a read.xls(), it'd be great if we have a read.ods() which directly

Hi, I did not find any web page about using R in economics and econometrics so far. However, this does not mean that there is none (searching with google for "R" and "economics" gives many pages about economics and a name like Firstname R. Lastname on it ;-)). Does anybody in the list does know such a web page? If not, I will be happy if you, Ajay, could build and

I have an elementary programming question. Could someone please point me in the right direction? I have a function which will run for thousands of companies. At each invocation, it returns 2 numbers. I plan to do something like: think_one_firm <- function(filename) { # Do stuff return(list(x=x,y=y)) } So for each of the firms in my dataset, I will call

Folks, I'm doing fine with using orthogonal polynomials in a regression context: # We will deal with noisy data from the d.g.p. y = sin(x) + e x <- seq(0, 3.141592654, length.out=20) y <- sin(x) + 0.1*rnorm(10) d <- lm(y ~ poly(x, 4)) plot(x, y, type="l"); lines(x, d$fitted.values, col="blue") # Fits great! all.equal(as.numeric(d$coefficients[1] + m

I wondered was people on this list felt about this article: http://www.voxeu.org/index.php?q=node/2363 which talks about the problems of obtaining sound answers in numerical optimisation in settings such as MLE or NLS. -- Ajay Shah http://www.mayin.org/ajayshah ajayshah at mayin.org http://ajayshahblog.blogspot.com <*(:-? -

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

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

How would I do something like this: f <- function(x, g) { s <- as.character(g) # THIS DOES NOT WORK sprintf("The %s of x is %.0f\n", s, g(x)) } f(c(2,3,4), "median") f(c(2,3,4), "mean") and get the results "The median of x is 3" "The mean of x is 3" -- Ajay Shah

I find it's just great to be able to say: library(tseries) x <- get.hist.quote(instrument="ongc.ns") and it gets a full time-series of the stock price of the symbol ongc.ns from Yahoo quote. However, once my hopes have been raised by such beauty :-) I get disappointed when I do > plot(x) and the annotation is horrible! The x axis is not labelled as dates. The default

Folks, A nice new data resource has come up -- http://data.un.org/ I thought it would be wonderful to setup an R function like tseries::get.hist.quote() which would be able to pull in some or all of this data. I walked around a bit of it and I'm not able to map the resources to predictable URLs which can then be wget. There's some javascript going on that I'm not understanding.

I know this: > library(date) > x="1979-04-04" > try=as.date(x, "ymd") > print(try) [1] 4Apr79 and that `x' here has to be a string, e.g.: > x=1979-04-04 > print(x) [1] 1971 I'm stuck in reading from a file. I say: > A <- read.table(file="try") > print(A) V1 V2 1 1979-04-04

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

The pretty picture that I saw at: http://chartsgraphs.wordpress.com/2009/02/09/r-panel-chart-beats-excel-chart/#more-1096 inspired me to try something similar. The code that I wrote is: ------snipsnip--------------------------------------------------------------------- M <- structure(list(date = structure(c(13634, 13665, 13695, 13726, 13757, 13787, 13818, 13848, 13879, 13910, 13939, 13970,