similar to: strptime format = "%H:%M:%OS6"

Is there a way for an apply-type function to return a data frame? the closest thing I think of is foo <- as.data.frame(sapply(...)) names(foo) <- c(....) is there a more "elegant" way? Thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://dhimmi.com http://honestreporting.com

Hi, I have a data frame df and a list of names of columns that I want to turn into factors: df.names <- attr(df,"names") sapply(factors, function (name) { pos <- match(name,df.names) if (is.na(pos)) stop(paste(name,": no such column\n")) df[[pos]] <- factor(df[[pos]]) cat(name,"(",pos,"):",is.factor(df[[pos]]),"\n")

what does the output for [[]] mean here: > all$X.Time[5] [1] "2011-02-15 09:32:26.37222" > all$X.Time[[5]] [1] 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 > all$X.Time[1] [1] "2011-02-15 09:31:29.18761" > all$X.Time[[1]] [1] 29.18761 34.30949 36.38144 12.28500 26.37222 47.00837 40.20271 32.83765 [9] 54.56998 28.56961 55.96641 28.91920 32.29962 10.94081 34.31731

> x <- rnorm(1000) > h <- hist(x,plot=FALSE) > sum(h$density) [1] 2 ----------------------------- shouldn't it be 1?! > h <- hist(x,plot=FALSE, breaks=(-4:4)) > sum(h$density) [1] 1 ----------------------------- now it's 1. why?! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://www.childpsy.net/ http://www.memritv.org

I have two datasets: A with columns Open and Name (and many others, irrelevant to the merge) B with columns Time and Name (and many others, irrelevant to the merge) I want the dataset AB with all these columns Open from A - a difftime (time of day) Time from B - a difftime (time of day) Name (same in A & B) - a factor, does NOT index rows, i.e., there are _many_ rows in both A & B with

When a sparse matrix is multiplied by a regular one, the result is usually not sparse. However, when matrix.csr is multiplied by a regular matrix in R, a matrix.csr is produced. Is there a way to avoid this? Thanks! -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org http://truepeace.org

Hi, I find this behavior unexpected: --8<---------------cut here---------------start------------->8--- > strsplit(c("a,b;c","d;e,f"),c(",",";")) [[1]] [1] "a" "b;c" [[2]] [1] "d" "e,f" --8<---------------cut here---------------end--------------->8--- I thought that it should be identical to this:

I see this: --8<---------------cut here---------------start------------->8--- > length(which(is.na(z$language))) [1] 0 > locals <- z[z$country == mycountry,] > length(which(is.na(locals$language))) [1] 229 --8<---------------cut here---------------end--------------->8--- where are those locals without the language coming from?! -- Sam Steingold (http://sds.podval.org/) on

How do I concatenate two vectors of factors? --8<---------------cut here---------------start------------->8--- > a <- factor(5:1,levels=1:9) > b <- factor(9:1,levels=1:9) > str(c(a,b)) int [1:14] 5 4 3 2 1 9 8 7 6 5 ... > str(unlist(list(a,b),use.names=FALSE)) Factor w/ 9 levels "1","2","3","4",..: 5 4 3 2 1 9 8 7 6 5 ...

How do I read/write liblinear models to files? E.g., if I train a model using the command line interface, I might want to load it into R to look the histogram of the weights. Or I might want to train a model in R and then apply it using a command line interface. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/

> write.matrix.csr(mx, y = y, file = file) > table(y) 0 1 5194394 23487 $ cut -d' ' -f1 f | sort | uniq -c 23487 2 5194394 1 i.e., 0 is written as 1 and 1 is written as 2. why? is there a way to disable this? -- Sam Steingold (http://sds.podval.org/) on Ubuntu 12.04 (precise) X 11.0.11103000 http://www.childpsy.net/ http://palestinefacts.org

I know it does not look very good - using the same column names to mean different things in different data frames, but here you go: --8<---------------cut here---------------start------------->8--- > x <- data.frame(a=c(1,2,3),b=c(4,5,6)) > y <- data.frame(b=c(1,2),a=c("a","b")) >

Hi, It appears that deal does not support missing values (NA), so I need to remove them (NAs) from my data frame. how do I do this? (I am very new to R, so a detailed step-by-step explanation with code samples would be nice). Some columns (variables) have quite a few NAs, so I would rather drop the whole column than sacrifice all the rows (observations) which have NA in that column. How do I

I find myself doing --8<---------------cut here---------------start------------->8--- tab <- table(...) tab <- tab[tab > 0] tab <- sort(tab,decreasing=TRUE) --8<---------------cut here---------------end--------------->8--- all the time. I am wondering if the "drop 0" (and maybe even sort?) can be effected by some magic argument to table() which I fail to discover

Hi, I need this plot: given: x,y - numerical vectors of length N plot xi vs mean(yj such that |xj - xi|<epsilon) (running mean?) alternatively, discretize X as if for histogram plotting and plot mean y over the center of the histogram group. is there a simple way? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.6 (Final) X 11.0.60900031 http://thereligionofpeace.com

I want to count attributes of IDs: --8<---------------cut here---------------start------------->8--- z <- data.frame(id=c(10,20,10,30,10,20), a1=c("a","b","a","c","b","b"), a2=c("x","y","x","z","z","y"),

How do I convert a list to a matrix? --8<---------------cut here---------------start------------->8--- list(c(50000, 101), c(1e+05, 46), c(150000, 31), c(2e+05, 17), c(250000, 19), c(3e+05, 11), c(350000, 12), c(4e+05, 25), c(450000, 19), c(5e+05, 16)) as.matrix(a) [,1] [1,] Numeric,2 [2,] Numeric,2 [3,] Numeric,2 [4,] Numeric,2 [5,] Numeric,2 [6,] Numeric,2 [7,]

summary() for a factor prints: ColName SNDK : 72 VXX : 36 MWW : 30 ACI : 28 FRO : 28 (Other):1801 it would have been much more useful if it additionally printed frequency stats as if by summary(aggregate(frame$ColName,by=list(frame$ColName),FUN=length)$x) -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://jihadwatch.org

When I do > all(all$X.Time == all$Y.Time); [1] TRUE as expected, but > all.equal(all$X.Time,all$Y.Time); Error in target[[i]] : subscript out of bounds why? thanks! -- Sam Steingold (http://sds.podval.org/) on CentOS release 5.3 (Final) http://mideasttruth.com http://honestreporting.com http://dhimmi.com http://jihadwatch.org http://pmw.org.il http://ffii.org The dark past once was the

(summary.default): show the vector length in addition to quantiles diff -u -i -p -F '^(def' -b -w -B /home/sds/src/R-3.0.1/src/library/base/R/summary.R.old /home/sds/src/R-3.0.1/src/library/base/R/summary.R --- /home/sds/src/R-3.0.1/src/library/base/R/summary.R.old 2013-03-05 18:02:33.000000000 -0500 +++ /home/sds/src/R-3.0.1/src/library/base/R/summary.R 2013-09-10 10:19:02.682946339