similar to: starting html help for a package?

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

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, 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 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 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 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, 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 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 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 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 I am writing an Sweave document and have encountered formatting issues with the "locale" part of toLatex(sessionInfo()). The fact that there is no spaces between the various "locale" variables means that LaTeX cannot easily find an appropriate place to break the lines, and some will get printed off screen. Below is the text output, and this .pdf document [1] shows the

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 I am getting this strange error when checking my package. Would you have an idea what causes it? Thank you Liviu * checking Rd cross-references ... WARNING Error in .find.package(package, lib.loc) : there is no package called 'KernSmooth' Calls: <Anonymous> -> lapply -> FUN -> .find.package Execution halted > sessionInfo () R version 2.10.0 (2009-10-26)

Dear all I would like to start R with Rcmdr from the cli, without tweaking Rprofile.site. This has been discussed in the past [1], but I don't see a solution that (1) could be used with any working directory and (2) would avoid starting Rcmdr on every R start-up. Personally I tried the following, which starts R but not Rcmdr liv at liv-laptop:~$ R --interactive -e 'require(Rcmdr)'

Dear all I have a work-flow issue with lm(). When I use > lm(y1~x1, anscombe) Call: lm(formula = y1 ~ x1, data = anscombe) Coefficients: (Intercept) x1 3.0001 0.5001 I get as expected the formula, "y1 ~ x1", in the print()ed results or summary(). However, if I pass through a formula object > (form <- formula(y1~x1)) y1 ~ x1 > lm(form, anscombe) Call:

Dear all Is there a clean plyr version of the following by() and do.call(rbind, ...) construct: > df<-data.frame(a=1:10,b=11:20,c=21:30,grp1=c("x","y"),grp2=c("x","y"),grp3=c("x","y")) > dfsum<-by(df[c("a","b","c")], df[c("grp1","grp2","grp3")], range) >

Dear all Is it possible to set globally the option .progress = "text" to all the apply functions in 'plyr'. For example, current default is daply(..., .progress = "none"). I would like to set it to daply(..., .progress = "text"), so as to avoid writing the argument every time I call such a function. I looked into ?daply and ?create_progress_bar without much

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, Does this make any sense: var() = cov() != acf(lag.max=0, type="covariance")? I have daily data of IBM for May 2005, and I'm using the logarithmic return: > ibm200505$LRAdj.Close [1] NA 0.0203152 0.0005508 -0.0148397 -0.0025182 0.0092025 -0.0013889 [8] 0.0098196 -0.0103757 -0.0274917 0.0005716 -0.0159842 -0.0074306 0.0091710 [15] 0.0002898 0.0226306