similar to: sorting & subsetting a data.frame

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 What is the easy way to drop a variable by using its name (and not its number)? Example: > data(iris) > head(iris) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 5.1 3.5 1.4 0.2 setosa 2 4.9 3.0 1.4 0.2 setosa 3 4.7 3.2 1.3 0.2 setosa 4 4.6 3.1

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",

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 can I obtain the union of a list of logical values? Consider the following: x <- head(iris) x[,c(2,4)] <- NA x[c(2,4),] <- NA # > x # Sepal.Length Sepal.Width Petal.Length Petal.Width Species # 1 5.1 NA 1.4 NA setosa # 2 NA NA NA NA <NA> # 3 4.7 NA 1.3

Dear all How do I access individual elements of a "summary.aov" object? > data(iris) > AnovaModel.1 <- aov(Sepal.Length ~ Species, data=iris) > tmp <- summary(AnovaModel.1) > tmp Df Sum Sq Mean Sq F value Pr(>F) Species 2 63.2 31.6 119 <2e-16 *** Residuals 147 39.0 0.3 --- Signif. codes: 0 '***' 0.001 '**' 0.01

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, I installed the amd64 cran2deb [1] JGR binaries on a fresh Debian testing, and JGR is missbehaving. When JGR starts up, one core of the CPU goes to 100% and stays there, even though I am keeping it idle and not performing any operation. I tried JGR with both sun-java6-jdk and openjdk-6-jdk, debian-liv:/home/liviu# update-alternatives --config java There are 4 choices for the alternative

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 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 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 unable to find an example of extracting the rotated scores of a principal components analysis. I can do this easily for the un-rotated version. data(mtcars) .PC <- princomp(~am+carb+cyl+disp+drat+gear+hp+mpg, cor=TRUE, data=mtcars) unclass(loadings(.PC)) # component loadings summary(.PC) # proportions of variance mtcars$PC1 <- .PC$scores[,1] # extract un-rotated scores of

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'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 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, How should I add several variables to a data frame without losing the attributes of the df? Consider the following: > require(Hmisc) > Xa <- iris > label(Xa, self=T) <- "Some df label" > str(Xa) 'data.frame': 150 obs. of 5 variables: $ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ... $ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9

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 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, 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