similar to: proper work-flow with 'formula' objects and lm()

I am relatively new to R and object oriented programming. I have relied on SAS for most of my data analysis. I teach an introductory undergraduate forecasting course using the Diebold text and I am considering using R in addition to SAS and Eviews in the course. I work primarily with univariate or multivariate time series data. I am having a great deal of difficulty understanding and working with

Hi all, I'm extracting the name of the term in a regression model that dropterm specifies as the least significant one, and I'm assigning this name to an object. However, when I use update(), it ignores this object. Is there a way I can make it not ignore it? A reproducible example is below: > lm(x1~1+y1*y2+y3+y4,data=anscombe)->my.lm >

Hi, probably just a quick question: can I somehow change the formula used with predict? E.g., the regression was run on "y ~ u + v + w" but for the prediction the term v should be removed from the formula contained in the regression object and only "y ~ u + w" be used. I could use model.matrix etc. to do the predictions but it would be very helpful to know a simpler way.

Dear R users I would like to do some spreadsheet style expansion of dates. For example, I would need to obtain a vector of months. I approached in an obviously wrong way: > paste(01:12) [1] "1" "2" "3" "4" "5" "6" "7" "8" "9" "10" "11" "12" > as.Date(paste(01:12),

Dear all, Given the following vector: > (z <- c('R project', 'hello world', 'something Else')) [1] "R project" "hello world" "something Else" I know how to obtain all capitals or all lower case letters: > tolower(z) [1] "r project" "hello world" "something else" > toupper(z) [1] "R

Dear all, I'm trying to gsub() "%" with "\%" with no obvious success. > temp1 <- c("mean", "sd", "0%", "25%", "50%", "75%", "100%") > temp1 [1] "mean" "sd" "0%" "25%" "50%" "75%" "100%" > gsub("%",

Dear all How does one convert a "non-symmetric" list to a vector? See below: > x <- list() > x[[1]] <- letters[1:5] > x[[2]] <- letters[6:10] > x[[3]] <- letters[11:12] > x [[1]] [1] "a" "b" "c" "d" "e" [[2]] [1] "f" "g" "h" "i" "j" [[3]] [1] "k"

Dear all I'd like to make a beeping sound in R, but alarm() doesn't beep? I checked ?alarm but I couldn't find any pointers to system configuration. Any ideas? Regards Liviu > sessionInfo() R version 2.14.2 (2012-02-29) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8 [5]

Dear all This may be obvious, but I cannot get it working. I'm trying to subset & sort a data frame in one go. x <- iris x$Species1 <- as.character(x$Species) ##subsetting alone works fine with(x, x[Sepal.Length==6.7,]) ##sorting alone works fine with(x, x[order(Sepal.Length, rev(sort(Species1))),]) ##gets subsetted, but not sorted as expected with(x, x[(Sepal.Length==6.7) &

Dear all Is there a simpler method to achieve the following: When I obtain an empty data.frame after subsetting, I need for it to contain one line of NAs. Here's a dummy example: > (.xb <- iris[ iris$Species=='zz', ]) [1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species <0 rows> (or 0-length row.names) > dim(.xb) [1] 0 5 > (.xa <-

Dear all I'm using Xubuntu Lucid and I keep getting the same random numbers whenever I start a new session of R. For example, I keep getting > sample(1:1000, 1) [1] 87 or > rnorm(1:10) [1] -1.3618103 0.4241701 1.0720076 0.2208145 -0.5375314 -0.4846588 [7] 0.7576768 0.6527407 -0.6868786 0.8718527 I expected that some set.seed() instruction woudl be present in a config file in

Dear all How does print.htest display the p-value in scientific notation? > (x <- cor.test(iris[[1]], iris[[3]])) Pearson's product-moment correlation data: iris[[1]] and iris[[3]] t = 21.65, df = 148, p-value < 2.2e-16 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.8270 0.9055 sample estimates: cor 0.8718 Above the p-value comes

Dear all Is there any R function that would perform currency conversion using up-to-date exchange rates? I would be looking for a function that allows to download recent exchange rates (say, from Yahoo!) and then use these in converting currencies (say, USD to EUR). I am not sure whether r-sig-finance would be more appropriate, but the (off-)topic feels general enough to me. Thank you Liviu --

Dear all, I'm looking for an alternative way to replicate the "2," string for an x number of times, and end up with one string containing "2," x times. I can partly achieve this using replicate(). > y <- rep("2,", times=3) > y [1] "2," "2," "2," The output that I am looking for is, however, "2,2,2,". I also tried

Hi, I would like to create a plot of y1,y2,y3,y4 against x for several subjects such that y1 and y2 are plotted against x in one panel and y3 and y4 against x in another panel. Thus if there are 3 subjects I should end up with 6 panels. Is there a simple way of doing so (i.e. without calling xyplot() several times, and then padding the results together)?? Regards S?ren

Dear all, The other day I stumbled on this article, "A critique of R and S-PLUS" [1], and got curious on whether the points outlined are (still) valid. The article is quite old, dating 2004, but was updated several times. Regards, Liviu [1] http://fluff.info/blog/arch/00000041.htm -- Do you know how to read? http://www.alienetworks.com/srtest.cfm Do you know how to write?

Dear all Is there an easy way to display only one half (top-right or bottom-left) of a correlation matrix? > require(Hmisc) > rcorr(as.matrix(mtcars[ , 1:4])) mpg cyl disp hp mpg 1.00 -0.85 -0.85 -0.78 cyl -0.85 1.00 0.90 0.83 disp -0.85 0.90 1.00 0.79 hp -0.78 0.83 0.79 1.00 n= 32 P mpg cyl disp hp mpg 0 0 0 cyl 0 0 0 disp 0 0

Dear all It is difficult to use round(..., digits=2) on a data frame since one has to first take care to remove non-numeric variables such as 'character' or 'factor': > head(round(iris, 2)) Error in Math.data.frame(list(Sepal.Length = c(5.1, 4.9, 4.7, 4.6, 5, : non-numeric variable in data frame: Species > head(round(iris[1:4], 2)) Sepal.Length Sepal.Width Petal.Length

Dear all How should one parse all.equal() output? I'm specifically referring to the 'mean relative difference' messages. For example, > all.equal(pi, 355/113) [1] "Mean relative difference: 8.491368e-08" But I'm not sure how to understand these messages. When they're close to 0 (or 1xe-16), then it's intuitive. But when they're big, > all.equal(1, 4)

Dear all I have a 'character' vector containing mixed formats (thanks Excel!) and I'd like to translate it into a default "%Y-%m-%d" Date vector. x <- c("1/3/2005", "13/04/2004", "2/5/2005", "2/5/2005", "7/5/2007", "22/04/2004", "21/04/2005", "20080430", "13/05/2003",